electric circuit & electron devices

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VEL TECH VEL MULTI TECH VEL HIGH TECH

STUDY MATERIAL

ELECTRIC CIRCUIT AND ELECTRON DEVICES

DEPARTMENT OF ECE, CSE, IT & Bio-Medical

Feb – 2010

Vel Tech


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Vel Tech Multi Tech Dr.Rangarajan Dr.Sakunthala Engineering

College

Vel Tech High Tech Dr. Rangarajan Dr.Sakunthala Engineering

College

SEM - II

INDEX

UNITS PAGE NO.

I. CIRCUIT ANALYSIS TECHNIQUES 06

II. TRANSIENT RESONANCE IN RLC CIRCUITS 55

III. SEMICONDUCTOR DIODES 129

IV. TRANSISTORS 180

V. SPECIAL SEMICONDUCTOR DEVICES 204


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Preface to the First Edition

This edition is a sincere and co-ordinated effort which we hope has

made a great difference in the quality of the material. “Giving the best to

the students, making optimum use of available technical facilities &

intellectual strength” has always been the motto of our institutions. In

this edition the best staff across the group of colleges has been chosen to

develop specific units. Hence the material, as a whole is the merge of the

intellectual capacities of our faculties across the group of Institutions.

45 to 60, two mark questions and 15 to 20, sixteen mark questions for

each unit are available in this material.

Prepared By : Ms. S. Santhana Lakshmi.


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EC2151 ELECTRIC CIRCUITS AND ELECTRON DEVICES (For ECE, CSE, IT and Biomedical Engg. Branches)

UNIT I CIRCUIT ANALYSIS TECHNIQUES 12

Kirchoff’s current and voltage laws – series and parallel connection of independent sources – R, L and C – Network Theorems – Thevenin, Superposition, Norton, Maximum power transfer and duality – Star-delta conversion. UNIT II TRANSIENT RESONANCE IN RLC CIRCUITS 12

Basic RL, RC and RLC circuits and their responses to pulse and sinusoidal inputs – frequency response – Parallel and series resonances – Q factor – single tuned and double tuned circuits. UNIT III SEMICONDUCTOR DIODES 12 Review of intrinsic & extrinsic semiconductors – Theory of PN junction diode – Energy band structure – current equation – space charge and diffusion capacitances – effect of temperature and breakdown mechanism – Zener diode and its characteristics. UNIT IV TRANSISTORS 12

Principle of operation of PNP and NPN transistors – study of CE, CB and CC configurations and comparison of their characteristics – Breakdown in transistors – operation and comparison of N-Channel and P-Channel JFET – drain current equation – MOSFET – Enhancement and depletion types – structure and operation – comparison of BJT with MOSFET – thermal effect on MOSFET. UNIT V SPECIAL SEMICONDUCTOR DEVICES

(Qualitative Treatment only) 12


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Tunnel diodes – PIN diode, varactor diode – SCR characteristics and two transistor equivalent model – UJT – Diac and Triac – Laser, CCD, Photodiode, Phototransistor, Photoconductive and Photovoltaic cells –

LED, LCD.

TEXT BOOKS:

1. Joseph A. Edminister, Mahmood, Nahri, “Electric Circuits” – Shaum series,Tata McGraw Hill, (2001)

2. S. Salivahanan, N. Suresh kumar and A. Vallavanraj, “Electronic Devices and Circuits”,Tata McGraw Hill, 2nd Edition, (2008).

3. David A. Bell, “Electronic Devices and Circuits”, Oxford University Press, 5th Edition, (2008). REFERENCES:

1. Robert T. Paynter, “Introducing Electronics Devices and Circuits”, Pearson Education, 7th Education, (2006).

2. William H. Hayt, J.V. Jack, E. Kemmebly and steven M. Durbin, “Engineering Circuit Analysis”,Tata McGraw Hill, 6th Edition, 2002.

3. J. Millman & Halkins, Satyebranta Jit, “Electronic Devices & Circuits”,Tata McGraw Hill, 2nd Edition, 2008.


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UNIT – I

PART – A

1. State superposition theorem.

The superposition theorem states that the response in a linear circuit with multiple sources is given by algebraic sum of responses due to individual sources acting alone.

2. What is property of additivity and homogeneity?

The property of additivity says that the response in a circuit due to a number of sources is given by sum of the response due to individual sources acting alone.

The property of homogeneity says that if all the sources are multiplied by a constant, then the response is also multiplied by the same constant.

3. Find the current through the ammeter shown in figure (a), by using superposition theorem.

Figure (a)

Since the resistance of ammeter is not specified it can be represented by short circuit. The condition of the given circuit when each source is acting separately are shown in figure (b) & figure (c).


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Figure: (b)

Figure: (c)

With reference to figure (b) & figure (c), we can write,

Response due to 10 V source,

Response due to 5 V source,

Total response, I = I’ + I” = 2.5 + (-2) = 0.5 A

4. Find the voltage VL in the circuit shown in figure (a) by principle of superposition.

Figure (a)

The condition of the circuit when each source is acting separately are shown in figure (b) & figure (c).

With reference to figure (b) & figure (c), we can write,


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5. In the circuit shown in figure (a), the power in resistance R is 9 W when V 1 is acting alone and 4 W when V2 is acting alone. What is the power in R when V1 & V2 are acting together?

6. State Thevenin’s theorem.

The Thevenin’s theorem states that a circuit with two terminals can be replaced by an equivalent circuit consisting of a voltage source in series with a resistance (or impedance).

7. State Norton’s theorem.

The Norton’s theorem states that a circuit with two terminals can be replaced by an equivalent circuit consisting of a current source in parallel with a resistance (or impedance).

8. Find Thevenin’s voltage across terminals A & B in the circuit shown in figure (a).

Figure (a)

Thevenin’s voltage, Vth = 5 + 10 = 15 V

(Note : Voltage across 5 is 5 V).


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9. The VI characteristics of a network is shown in figure (a). Determine the maximum power that can be supplied by the network to a resistance connected across A-B.

Figure (a)When V = 0, I = - 5A

The condition V = 0 is equivalent to short circuiting terminals A – B and the current flowing through the short circuit is the Norton’s current.

Norton’s current, In = - I = - (-5) = 5A

When I = 0, V = 20 V

The condition I = 0 is equivalent to open terminals A-B and the voltage across the open terminals is the Thevenin’s voltage.

Thevenin’s voltage, Vth = 20 V

Thevenin’s resistance,

The resistance, R to be connected for maximum power transfer across terminals A – B is Rth.

10. Determine the value of R in the circuit shown in figure (a) for maximum power transfer.

Figure (a)


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The value of R for maximum power transfer is given by the looking back resistance (or Thevenin’s resistance) from the terminals for R which is determined as shown below.

11. Determine the Thevenin’s equivalent of the circuit shown in figure (a).

The Thevenin’s voltage is the voltage across 20 resistance. By voltage division rule,

Thevenin’s voltage,

To find Thevenin’s resistance the 200 V source is replaced by short circuit as shown in figure (b). With reference to figure (c), we can write,

Thevenin’s resistance, .

12. In the circuit shown in figure (a) using Thevenin’s theorem determine the voltage across 70 resistance after the switch is closed.


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Figure (a)

Since the load is balanced when the switch is open, the voltage across 90 is 100V. This is also Thevenin’s voltage at terminals A – B.

To find Thevenin’s resistance the voltage sources are replaced by short circuit. When the voltage sources are shorted the three 90 resistances will be in parallel.

Thevenin’s resistance,

The Thevenin’s equivalent at A-B is shown in figure (b). With reference to figure (b) by voltage division rule,

Figure (b)

Voltage across 70 resistance,

13. Find Thevenin’s equivalent of the circuit shown in figure (a).

To find Thevenin’s voltage the current source is converted to voltage source as shown in figure (b).


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By voltage division rule,

By KVL, Vth = V1 + 6 = 4 + 6 = 10 V.

To find Thevenin’s resistance the voltage source is replaced by short circuit and current source is opened as shown in figure (c).

The Thevenin’s equivalent is shown in figure (d).

Figure (c) Figure (d)

Thevenin’s equivalent

14. Find at terminals A-B in the circuit shown in figure (a).

Figure (a)

In the given circuit the voltage across series combination of 4 & j3 elements is 60 0 V. hence by voltage division rule,


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15. State maximum power transfer theorem.

In purely resistive circuits maximum power transfer theorem states that “Maximum power is transferred from source to load when the load resistance is equal to source resistance”.

In general the maximum power transfer theorem states that “Maximum power is transferred to a load impedance if the absolute value of the load impedance is equal to the absolute value of the looking back impedance of the circuit from the terminals of the load”.

16. Determine the value of R in the circuit shown in figure (a) for maximum power transfer.

Figure (a)

The value of R for maximum power transfer is given by the looking back resistance (or Thevenin’s resistance) from the terminals of R which is determined as shown below.

Figure (b) Figure (c)

17. Find the value of R for maximum power transfer in the circuit shown in figure (a).


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Figure (a)

For maximum power transfer the value of R should be equal to absolute value of the looking back impedance from the terminals of R.

18. Find the equivalent Y circuit for the circuit shown in figure (a).

Figure (a)Solution:-

From the circuit of figure (a), we see that we have the following resistor values:

RA = 90 RB = 60 RC = 30

We have the following equivalent “Y” resistor values:


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The resulting circuit is shown in figure (b)

Figure (b)

19. Find the network equivalent of the Y network shown in figure (a).


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Figure (a)

Solution:- The equivalent network is shown in figure (b).

Figure (b)

The values of the resistors are determined as follows:-

20. Given the circuit of Figure (a), find the total resistance, RT, and the total current, I.


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Figure (a)

Solution:-

As is often the case, the given circuit may be solved in one of two ways. We may convert the “” into its equivalent “Y”, and solve the circuit by placing the resultant branches in parallel, or we may convert the “Y” into its equivalent “”. We choose to use the latter conversion since the resistors in the “Y” have the same value. The equivalent “” will have all resistors given as

The resulting circuit is shown in figure (b).

Figure (b) Figure (c)We see that the sides of the resulting “” are in parallel, which allows us to simplify the

circuit even further as shown in figure (c). The total resistance of the circuit is now easily determined as

This results in a circuit current of


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21. State Kirchoff’s Current Law.

For any lumped electric circuit, at any time the (algebraic) sum of the branch currents leaving any of its nodes is zero.

22. State Kirchoff’s Voltage Law.

For any lumped electric circuit, for any of its loops the (algebraic) sum of the branch voltage around the loop is zero at any instant.

PART – B

CIRCUIT ANALYSIS TECHNIQUES

1. The following three impedances are connected in series across a 40 V, 20 kHz supply: (i) a resistance of 8 , (ii) a coil of inductance 130 H and 5 resistance, and (iii) a 10 resistor in series with a 0.25 F capacitor. Calculate (a) the circuit current, (b) the circuit phase angle and (c) the voltage drop across each impedance.


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The circuit diagram is shown in figure (a). Since the total circuit resistance is 8 + 5 + 10, i.e. 23 , an equivalent circuit diagram may be drawn as shown in figure (b).

Figure (a)

Figure (b)

Inductive reactance,

= 16.34

Capacitive reactance,

Since XC > XL, the circuit is capacitive

(a) Circuit impedance, Z = [R2 + (XC – XL)2] = [232 + 15.492]

= 27.73

Circuit current, I =

Circuit phase angle =

i.e. = .

(b) V1 = 1R1 = (1.442) (8) = 11.54 V V2 = IZ2 = I (52 + 16.342) = (1.442) (17.09) = 24.64 V


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V3 = IZ3 = I (102 + 31.832) = (1.442) (33.36) = 48.11 V

The 40 V supply voltage is the phasor sum of V1, V2 and V3.

2. A 20 resistor is connected in parallel with an inductance of 2.387 mH across a 60 V, 1 kHz supply. Calculate (a) the current in each branch, (b) the supply current, (c) the circuit phase angle, (d) the circuit impedance and (e) the power consumed.

(a) Current flowing in the resistor

Current flowing in the inductance

(b) From the phase, diagram supply current, I = (IR2 + IL

2) = (32 + 42) = 5 A

(c) Circuit phase angle, = arctan

= 538’ lagging

(d) Circuit impedance,

(e) Power consumed P = VI cos = (60) (5) (cos 538’) = 180 W

(Alternatively, power consumed P = IR2R = (3)2(20) = 180 W)

3. A 30 F capacitor is connected in parallel with an 80 resistor across a 240 V, 50 Hz supply. Calculate (a) the current in each branch, (b) the supply current (c) the circuit phase angle, (d) the circuit impedance, (e) the power dissipated and (f) the apparent power.

(a) Current in resistor, IR = 9

Current in capacitor,


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= 2fCV

(b) Supply current, I = (IR2+IC

2) = (32 + 2.2622) = 3.757 A

(c) Circuit phase angle, = arctan

= 371’ leading

(d) Circuit impedance,

(e) True or active power dissipated, P = VI cos = 240(3.757)cos 371’

= 720 W (Alternatively, true power P = IR

2 R = (3)2(80) = 720 W)

(f) Apparent power, S = VI = (240) (3.757) = 901.7 VA.

4. A capacitor C is connected in parallel with a resistor R across a 120 V, 200 Hz supply. The supply current is 2 A at a power factor of 0.6 leading. Determine the values of C and R.

The circuit diagram is shown in figure (a)

Figure (a)Power factor = cos = 0.6 leading, hence = arcos 0.6 = 53.13 leading.

From the phasor diagram shown in figure (b),


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Figure (b)

IR = I cos 53.13 = (2) (0.6) = 1.2 A

and IC = I sin 53.13 = (2) (0.8) = 1.6 A

(Alternatively, IR and IC can be measured from the scaled phasor diagram). From the circuit diagram,

5. Determine the values of the resistance and the series-connected inductance or capacitance for each of the following impedances: (a) (12 + j5) (b) – j40 (c) 30 60 (d) 2.20 106 - 30 . Assume for each a frequency of 50 Hz.

(a) For an R – L series circuit, impedance Z = R + jXL. Thus Z = (12 + j5) represents a resistance of 12 and an inductive reactance of 5 in series. Since inductive reactance XL = 2fL,

Inductance

i.e., the inductance is 15.9 mH. Thus an impedance (12 + j5) represents a resistance of 12 in series with an inductance of 15.9 mH.


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(b) For a purely capacitive circuit, impedance Z = -jXc. Thus Z = - j40 represents zero resistance and a capacitive reactance of 40 . Since capacitive reactance XC = 1/(2fC).

Capacitance C =

= 79.6 F

Thus an impedance –j40 represents a pure capacitor of capacitance 79.6 F.

(c) 3060 = 30(cos 60 + j sin 60) = 15 + j25.98 Thus Z = 3060 = (15 + j25.98) represents a resistance of 15 and an inductive reactance of 25.98 in series. Since XL = 2fL.

Inductance

Thus an impedance 3060 represents a resistance of 15 in series with an inductance of 82.7 mH.

(d) 2.20 106 -30 = 2.20 106 [cos(-30) + j sin (-30)] = 1.905 106 – j1.10 106

Thus Z = 2.20 106 - 30 = (1.905 106 – j1.10 106) represents a resistance of 1.905 106 (i.e. 1.905 M) and a capacitive reactance of 1.10 106 in series. Since capacitive reactance XC = 1/(2fC).

Capacitance C =

= 2.894 10-9 F or 2.894 nF

Thus an impedance 2.2 106 - 30 represents a resistance of 1.905 M in series with a 2.894 nF capacitor.

6. Determine the current through the 8-V battery for the circuit shown in figure (a).


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Solution:- Convert the current source into an equivalent voltage source. The equivalent circuit may now be analyzed by using the loop currents shown in figure (b).

Figure (b)

Loop 1:

Loop 2:

Rewriting the linear equations, you get the following:

Loop 1:

Loop 2:

Solving the equations using determinants, we have the following:


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If the assumed direction of current in the 8-V battery is taken to be I2, then

I = I2 – I1 = -4.00 A – (-6.00 A) = 2.00 A

The direction of the resultant current is the same as I2 (Upward).

7. Solve for the currents through R2 and R3 in the circuit of figure (a)

Figure (a)

Solution:-

Step 1: Although we see that the circuit has a current source, it may not be immediately evident how the source can be converted into an equivalent voltage source. Redrawing the circuit into a more recognizable form, as shown in figure (a), we see that the 2-mA current source is in parallel with a 6-k resistor. The source conversion is also illustrated in figure (b).


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Figure (b)

Step 2: Redrawing the circuit is further simplified by labeling some of the nodes, in this case a and b. After performing a source conversion, we have the two-loop circuit shown in figure (c). The current directions for I1 and I2 are also illustrated.

Step 3: The loop equations are

Loop 1:

Loop 2:

In loop 1, both voltages are negative since they appear as voltage drops when following the direction of the loop current.

These equations are rewritten as


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Step 4: In order to simplify the solution of the previous linear equations, we may eliminate the units (k and V) from our calculations. By inspection, we see that the units for current must be in milliamps. Using determinants, we solve for the currents I1 and I2 as follow:

The current through resistor R2 is easily determined to be

I2 –I1 = 0.644 mA – (-0.894 mA) = 1.54 mA

The current through R3 is not found as easily. A common mistake is to say that the current in R3 is the same as the current through the 6-k resistor of the circuit in figure. This is not the case. Since this resistor was part of the source conversion it is no longer in the same location as in the original circuit.

Although there are several ways of finding the required current, the method used here is the application of Ohm’s law. If we examine figure, we see that the voltage across R3 is equal to Vab. From figure, we see that we determine Vab by using the calculated value of I1.

The above calculation indicates that the current through R3 is upward (since point a is negative with respect to point b). The current has a value of


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8. Differentiate between Loop Analysis and Nodal Analysis.

Mesh (Loop) Analysis

A better approach and one which is used extensively in analyzing linear bilateral networks is called mesh (or loop) analysis. While the technique is similar to branch-current analysis, the number of simultaneous linear equations tends to be less. The principal difference between mesh analysis and branch-current analysis is that we simply need to apply Kirchoff’s voltage law around closed loops without the need for applying Kirchoff’s current law.

The steps used in solving a circuit using mesh analysis are as follows:-

1. Arbitrarily assign a clockwise current to each interior closed loop in the network. Although the assigned current may be in any direction, a clockwise direction is used to make later work simpler.

2. Using the assigned loop currents, indicate the voltage polarities across all resistors in the circuit. For a resistor which is common to two loops, the polarities of the voltage drop due to each loop current should be indicated on the appropriate side of the component.

3. Applying Kirchoff’s voltage law, write the loop equations for each loop in the network. Do not forget that resistors which are common to two loops will have two voltage drops, one due to each loop.

4. Solve the resultant simultaneous linear equations.5. Branch currents are determined by algebraically combining the loop currents which are

common to the branch.

Nodal Analysis

In this section we will apply Kirchoff’s current law to determine the potential difference (voltage) at any node with respect to some arbitrary reference point in a network. Once the potentials of all nodes are known, it is a simple matter to determine other quantities such as current and power within the network.

The steps used in solving a circuit using nodal analysis are as follows:-

1. Arbitrarily assign a reference node within the circuit and indicate this node as ground. The reference node is usually located at the bottom of the circuit, although it may be located anywhere.

2. Convert each voltage source in the network to its equivalent current source. This step, although not absolutely necessary, makes further calculations easier to understand.

3. Arbitrarily assign voltages (V1, V2, ….Vn) to the remaining nodes in the circuit. (Remember that you have already assigned a reference node, so these voltages will all be with respect to the chosen reference).


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4. Arbitrarily assign a current direction to each branch in which there is no current source. Using the assigned current directions, indicate the corresponding polarities of the voltage drops on all resistors.

5. With the exception of the reference node (ground), apply Kirchoff’s current law at each of the nodes. If a circuit has a total of n + 1 nodes (including the reference node), there will be n simultaneous linear equations.

6. Rewrite each of the arbitrarily assigned currents in terms of the potential difference across a known resistance.

7. Solve the resulting simultaneous linear equations for the voltages (V1, V2, ….Vn).

9. Given the circuit of figure (a), use nodal analysis to solve for the voltage Vab.

Solution:-

Step 1: Select a convenient reference node.

Step 2: Convert the voltage sources into equivalent current sources. The equivalent circuit is shown in figure (b).


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Step 3 and 4 : Arbitrarily assign node voltages and branch currents. Indicate the voltage polarities across all resistors according to the assumed current directions.

Step 5: We now apply Kirchhoff’s current law at the nodes labeled as V1 and V2:

Node V1 :

Node V2 :

Step 6: The currents are rewritten in terms of the voltages across the resistors as follows:-

The nodal equations become


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Substituting the voltage expressions into the original nodal equations, we have the following simultaneous linear equations:

These may be further simplified as

Step 7: Use determinants to solve for the nodal voltages as

and


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If we go back to the original circuit of figure (a), we see that the voltage V 2 is the same as the voltage Va, namely

Va = 4.67 V = 6.0 V + Vab

Therefore, the voltage Vab is simply found as

10. Determine the nodal voltages for the circuit shown in figure (a).

Figure (a)

Solution:- By following the steps outlined, the circuit may be redrawn as shown in figure (b)

Figure (b)

Applying Kirchhoff’s current law to the nodes corresponding to V1 and V2 the following nodal equations are obtained:


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Node V1 : I1 + I2 = 2ANode V2 : I3 + I4 = I2 + 3A

The currents may once again be written in terms of the voltages across the resistors:

The nodal equations become

Node V1 :

Node V2 :

These equations may now be simplified as

Node V1 :

Node V2 :

The solutions for V1 and V2 are found using determinants:


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11. Determine the nodal voltages for the circuit shown in figure (a).

Figure (a)

Solution:- The circuit has a total of three nodes; the reference node (at a potential of zero volts) and two other nodes, V1 and V2.By applying the format approach for writing the nodal equations, we get two equations:

Node V1 :

Node V2 :

On the right-hand sides of the above, those currents that are leaving the nodes are given a negative sign.

These equations may be rewritten as

Node V1 : (0.533 S)V1 – (0.200 S) V2 = - 5 ANode V2 : (0.200 S)V1 + (0.450 S) V2 = - 3A

Using determinants to solve these equations, we have


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12. Use nodal analysis to find the nodal voltages for the circuit of Figure (a). Use the answers to solve for the current through R1.

Solution:- In order to apply nodal analysis, we must first convert the voltage source into its equivalent current source. The resulting circuit is shown in figure (b).

Labelling the nodes and writing the nodal equations, we obtain the following:

Node V1 :

Node V2 :

Because it is inconvenient to use kilo ohms and milliamps throughout our calculations, we may eliminate these units in our calculations. You have already seen that any voltage obtained by using these quantities will result in the units being “volts”. Therefore the nodal equations may be simplified as


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Node V1 : (0.7833)V1 – (0.2500)V2 = -1 Node V2 : -(0.2500)V1 + (0.750)V2 = 2

Using the values derived for the nodal voltages, it is now possible to solve for any other quantities in the circuit. To determine the current through resistor RL = 5k, we first reassemble the circuit as it appeared originally. Since the node voltage V1 is the same in both circuits, we use it in determining the desired current. The resistor may be isolated as shown in figure (c).

Figure (c)

The current is easily found as

I = 10 V -

13. Solve for the currents through R1 and R4 in the circuit of figure (a)


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Figure (a)

Solution:- We see that the bridge of the above circuit is balanced (since R1/R3 = R2/R4). Because the circuit is balanced, we may remove R5 and replace it with either a short circuit (since the voltage across a short circuit is zero) or an open circuit (since the current through an open circuit is zero). The remaining circuit is then solved by one of the methods developed in previous chapters. Both methods will be illustrated to show that the results are exactly the same.

Method 1: If R5 is replaced by an open, the result is the circuit shown in figure (b).

Figure (b)

The total circuit resistance is found as

The circuit current is


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The current in each branch is then found by using the current divider rule:

The determinant for the denominator will is

Notice that, as expected, the elements in the principal diagonal are positive and that the determinant is symmetrical around the principal diagonal.

The loop currents are now evaluated as

The current through R1 is found as

IR1 = I1 – I2 = 2.586 A – 0.948 A = 1.638 A

The current through R5 is found as

IR5 = I3 – I2 = 1.034 A – 0.948 A


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= 0.086 A to the right.

The previous example illustrates that if the bridge is not balanced, there will always be some current through resistor R5. The unbalanced circuit may also be easily analyzed using nodal analysis, as in the following example.

14. Determine the node voltages and the voltage VR5 for the circuit of figure (a).

Figure (a)

Method: If R5 is replaced with a short circuit, the result is the circuit shown in figure (b).

Figure (b)

The total circuit resistance is found as

The above result is precisely the same as that found using Method 1. Therefore the circuit current will remain as IT = 3.0 A.

The currents through R1 and R4 may be found by the current divider rule as


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and

Clearly, these results are precisely those obtained in Method 1, illustrating that the methods are equivalent. Remember, though, R5 can be replaced with a short circuit or an open circuit only when the bridge is balanced.

14. Find the current through R5 for the circuit shown in Figure (a)

Figure (a)Solution By inspection we see that this circuit is not balanced, since

Therefore, the current through R5 cannot be zero. Notice, also, that the circuit contains two possible configurations. If we choose to convert the top to its equivalent Y, we get the circuit shown in figure (b).


41


By combining resistors, it is possible to reduce the complicated circuit to the simple series circuit shown in figure (c).

2 + (3 + 3) || (6 + 3) = 5.6

Figure (c)

The circuit of figure (c) is easily analyzed to give a total circuit current of

Using the calculated current, it is possible to work back to the original circuit. The currents in the resistors R3 and R4 are found by using the current divider rule for the corresponding resistor branches, as shown in figure.


42


These results are exactly the same as those found. Using these currents, it is now possible to determine the voltage Vbc as

The current through R5 is determined to be

15. Consider the circuit of figure (a)

Figure (a)

a. Determine the current in the load resistor, RL.b. Verify that the superposition theorem does not apply to power.

Solution:

a. We first determine the current through RL due to the voltage source by removing the current source and replacing in with an open circuit (zero amps) as shown in figure (b).The resulting current through RL is determined from Ohm’s law as


43


Next, we determine the current through RL due to the current source by removing the voltage source and replacing it with a short circuit (zero volts) as shown in figure.

Figure (b)

The resulting current through RL is determined from Ohm’s law as

Next, we determine the current through RL due to the current source by removing the voltage source and replacing it with a short circuit (zero volts) as shown in figure (c).


44


Figure(c)

The resulting current through RL is found with the current divider rule as

The resultant current through RL is found by applying the superposition theorem:

The negative sign indicates that the current through RL is opposite to the assumed reference direction. Consequently, the current through RL will, in fact, be upward with a magnitude of 0.7 A.

b. If we assume (incorrectly) that the superposition theorem applies for power, we would have the power due the first source given as

and the power due the second source as

The total power, if superposition applies, would be

Clearly, this result is wrong, since the actual power dissipated by the load resistor is correctly given as

16. Determine the voltage drop across the resistor R2 of the circuit shown in figure (a).


45


Figure (a)Solution:-

Since this circuit has three separate sources, it is necessary to determine the voltage across R2

due to each individual source.

First, we consider the voltage across R2 due to the 16-V source as shown in figure (b).

Figure (b)

The voltage across R2 will be the same as the voltage across the parallel combination of R2||R3 = 0.8 k. Therefore,

The negative sign in the above calculation simply indicates that the voltage across the resistor due to the first source is opposite to the assumed reference polarity.

Next, we consider the current source. The resulting circuit is shown in figure (c).

Figure (c)


46


From this circuit, you can observe that the total resistance “seen” by the current source is

The resulting voltage across R2 is

Finally, the voltage due to the 32-V source is found by analyzing the circuit of figure (d).

The voltage across R2 is

By superposition, the resulting voltage is

17. State and Explain Thevenin’s Theorem.

Thevenin’s Theorem

Thevenin’s theorem allows even the most complicated circuit to be reduced to a single voltage source and a single resistance. The importance of such a theorem becomes evident when we try to analyze a circuit as shown in figure (a).

If we wanted to find the current through the variable load resistor when RL = 0, RL = 2 k and RL = 5 k using existing methods, we would need to analyze the entire circuit three separate


47


times. However, if we could reduce the entire circuit external to the load resistor to a single voltage source in series with a resistor, the solution becomes very easy.

Thevenin’s theorem is a circuit analysis technique which reduces any linear bilateral network to an equivalent circuit having only one voltage source and one series resistor. The resulting two-terminal circuit is equivalent to the original circuit when connected to any external branch or component. In summary, Thevenin’s theorem is simplified as follows:-

Any linear bilateral network may be reduced to a simplified two terminal circuit consisting of a single voltage source in series with a single resistor as shown in figure (b).

The following steps provide a technique which converts any circuit into its Thevenin equivalent:

1. Remove the load from the circuit.

2. Label the resulting two terminals. We will label them as a and b, although any notation may be used.

3. Set all sources in the circuit to zero.

Voltage sources are set to zero by replacing them with short circuits (zero volts).

Current sources are set to zero by replacing them with open circuits (zero amps).

4. Determine the Thevenin equivalent resistance, RTh, by calculating the resistance “seen” between terminals a and b. It may be necessary to redraw the circuit to simplify this step.

5. Replace the sources removed in Step 3, and determine the open-circuit voltage between the terminals. If the circuit has more than one source, it may be necessary to use the superposition theorem. In that case, it will be necessary to determine the combined effect. The resulting open-circuit voltage will be the value of the Thevenin voltage, ETh.

6. Draw the Thevenin equivalent circuit using the resistance determined in Step 4 and the voltage calculated in Step 5. As part of the resulting circuit, include that portion of the network removed in Step 1.


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18. Determine the Thevenin equivalent circuit external to the resistor RL for the circuit of figure (a). Use the Thevenin equivalent circuit to calculate the current through RL

Figure (a)

Solution:-

Step 1 and 2:- Removing the load resistor from the circuit and labeling the remaining terminals, we obtain the circuit shown in figure (b).

Figure (b)

Step 3: Setting the sources to zero, we have the circuit shown in figure (c).

Step 4: The Thevenin resistance between the terminals is RTh = 24 .

Step 5: From figure (b), the open-circuit voltage between terminals a and b is found as

Step 6: The resulting Thevenin equivalent circuit is shown in figure (d).


49


Figure (c)

Figure (d)

Using this Thevenin equivalent circuit, we easily find the current through RL as

19. State and Explain Norton’s Theorem.

Norton’s Theorem


50


Norton’s theorem is a circuit analysis technique which is similar to Thevenin’s theorem. By using this theorem the circuit is reduced to a single current source and one parallel resistor. As with the Thevenin equivalent circuit, the resulting two-terminal circuit is equivalent to the original circuit when connected to any external branch or component. In summary, Norton’s theorem may be simplified as follows:-

Any linear bilateral network may be reduced to a simplified two-terminal circuit consisting of a single current source and a single shunt resistor as shown in figure (a).

Figure (a) Norton equivalent circuit

The following steps provide a technique which allows the conversion of any circuit into its Norton equivalent:

1. Remove the load from the circuit.

2. Label the resulting two terminals. We will label them as a and b, although any notation may be used.

3. Set all sources to zero. As before, voltage sources are set to zero by replacing them with short circuits and current sources are set to zero by replacing them with open circuits.

4. Determine the Norton equivalent resistance, RN, by calculating the resistance seen between terminals a and b. It may be necessary to redraw the circuit to simplify this step.

5. Replace the sources removed in Step 3, and determine the current which would occur in a short if the short were connected between terminals a and b. If the original circuit has more than one source, it may be necessary to use the superposition theorem. In this case, it will be necessary to determine the short-circuit current due to each source separately and then determine the combined effect. The resulting short-circuit current will be the value of the Norton current IN.

6. Sketch the Norton equivalent circuit using the resistance determined in Step 4 and the current calculated in Step 5. As part of the resulting circuit, include that portion of the network removed in Step 1.

The Norton equivalent circuit may also be determined directly from the Thevenin equivalent circuit by using the source conversion technique developed in Chapter. As a result, the Thevenin and Norton circuits shown in figure (b) are equivalent.


51


Thevenin equivalent circuit Norton equivalent circuit

Figure (b)

From figure (b) we see that the relationship between the circuits is as follows:-

ETh= INRN

IN =

20. Find the Norton equivalent of the circuit external to resistor RL in the circuit in figure (a). Use the equivalent circuit to determine the load current IL when RL

= 0.2 k, and 5 k.

Figure (a)

Solution:-

Step 1, 2 and 3:- After removing the load resistor, labeling the remaining two terminals a and b, and setting the sources to zero, we have the circuit of figure (b).


52


Figure (b)

Step 4: The Norton resistance of the circuit is found as

Step 5: The value of the Norton constant-current source is found by determining the current effects due to each independent source acting on a short circuit between terminals a and b.

Voltage Source, E: Referring to Figure (c), a short circuit between terminals a and b eliminates resistor R2 from the circuit. The short-circuit current due to the voltage source is

Figure (c) Figure (d)

Current Source, I: Referring to Figure (d), the short circuit between terminals a and b eliminates both resistors R1 and R2. The short-circuit current due to the current source is therefore


53


The resultant Norton current is found from superposition as

Step 6: The Norton equivalent circuit is shown in figure (e)

Figure (e)

Let RL =0: the current IL must equal the source current, and soIL =7.50 mA

Let RL =2 k: The current IL is found from the current divider rule as

Let RL =5 k: using the current divider rule again, the current IL is found as

2. Consider the circuit of figure (a)

Figure (a)

a. Find the Norton equivalent circuit external to terminals a and bb. Determine the current through RL.


54


Solution:

a. Step 1 and 2: After removing the load which consists of a current source in parallel with a resistor), we have the circuit to Figure (b)

Step 3: After zeroing the sources, we have the network shown in Figure (c)

Step 4: The Norton equivalent resistance is found as

Step 5: In order to determine the Norton current we must again determine the short-circuit current due to each source separately and then combine the results using the superposition theorem.

Voltage Source, E: Referring to Figure (d), notice that the resistor R2 is shorted by the short circuit between terminals a and b and so the current in the short circuit is


55


Current Source, I: Referring to figure (e) , the short circuit between terminals a and b will now eliminate both resistors. The current through the short will simply be the source current. However, since the current will not be from a to b but rather in the opposite direction, we write

Now the Norton current is found as the summation of the short-circuit currents due to each source:

The negative sign in the above calculation for current indicates that if a short circuit were placed between terminals a and b, current would actually be in the direction from b to a. the Norton equivalent circuit is shown in Figure (F)

Figure (e) figure (f)


56


b. The current through the load resistor is found by applying the current divider rule:

22. State and Prove maximum power transfer theorem.

MAXIMUM POWER TRANSFER THEOREM

In amplifiers and in most communication circuits such as radio receivers and transmitters, it is often desired that the load receive the maximum amount of power from a source.

The maximum power transfer theorem states the following:

The load resistance will receive maximum power from a circuit when the resistance of the load is exactly the same as the Thevenin (Norton) resistance looking back at the circuit.

The proof for the maximum power transfer theorem is determined from the Thevenin equivalent circuit and involves the use of calculus.

(a) (b)Figure 1

From figure (1) we see that once the network has been simplified using either Thevenin’s or Norton’s theorem, maximum power will occur when

Examining the equivalent circuits of Figure 9-45, shows that the following equations determine the power delivered to the load:


57


Under maximum power conditions (RL =RTh =RN), the above equations may be used to determine the maximum power delivered to the load and may therefore be written as

UNIT – II

PART – A

1. When a sinusoidal voltage v = 200 sin (377t + 30o) V is applied to a load, it draw a current of 10 sin (377t + 60o) A. Determine the active and reactive power of the load.

The rms current and voltage phasors in the polar form are,

Complex power,

Since,


58


Active power, P = 866 W

Reactive power, Q = - 500 VAR (Capacitive)

2. A load consisting of 3 resistance and 4 inductive reactance draw a current of 10 A when connected to a sinusoidal source. Determine the voltage and power in the load.

Magnitude of Impedance,

Voltage,

Power,

3. When a sinusoidal voltage of 120 V is applied across a load, it draw a current of 8 A with a phase lead of 30o. Determine the resistance, reactance and impedance of the load.

Let, be reference phasor.

Impedance,

Since, Resistance, R = 12.99

Reactance, X = 7.5 ’- Capacitive

4. When a sinusoidal voltage of 100 V is applied across a load, it draw a current of 10 A with 60o phase lag. Determine the conductance, susceptance and admittance of the load.

Let, be reference phasor.

Admittance,

Since,

Susceptance, B = 0.05 - Inductive


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5. An inductive load consumes 1000 W power and draw 10 A current when connected to a 250 V, 25 Hz supply. Determine the resistance and inductance of the load.

We know that, P = I2R ;

Resistance,

Impedance,

We know that,

Inductive reactance,

Inductance,

6. In a RC series circuit excited by sinusoidal source the voltage across resistance and capacitance are 60 V and 80 V respectively. What will be the supply voltage?

Let current, through RC series circuit be the reference phasor. With reference to phasor diagram shown in fig. 1 we can write,

Figure:1

Supply voltage,


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7. In a RL series circuit with R = 20 and XL = 30 , for what value of R, the impedance of RL series combination will be doubled?

When R = 20 &

Let R2 be the value of resistance when impedance is doubled.

Now,

8. ARC series circuit with R = 1.2 k and C = 0.1 F is excited by a sinusoidal source of 45 V and frequency 1 kHz. Find the apparent power.

Magnitude of impedance,

Magnitude of current, ]

Apparent power,

9. Determine the impedance of the RLC parallel circuit shown in fig 1.

Figure:

Another method


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Let total current be .

10. Determine the power factor of a RLC series circuit with R = 5 , XL = 8 and XC= 12 .

Impedance,

Power factor angle,

Power factor = cos

11. What is resonance?

The resonance is a circuit condition at which a RLC circuit behave as purely resistive circuit.

12. Write the expressions for resonant frequency and current at resonance of a RLC series circuit.

Angular resonant frequency,

Resonant frequency,

Current at resonance,

13. Draw the frequency response of RLC series circuit.

The variation of current with frequency is called frequency response, which is shown in fig

Figure:


62


14. Define Quality factor.

The Quality factor is defined as the ratio of maximum energy stored to the energy dissipated in one period.

Quality factor,

15. Write the expressions for quality factor of series RLC circuit.

Quality factor at resonance,

When r ,

When r ,

16. Define Bandwidth of RLC series circuit.

The Bandwidth, of RLC series circuit is defined as the range of frequencies over which

the current is greater than or equal to times the maximum current.

Figure:

17. What are half power frequencies?

In RLC circuits, the frequencies at which the power is half the maximum/minimum power are called half power frequencies.

18. Write the expression for half power frequencies of RLC series circuit.


63


Lower cut-off frequency,

High cut-off frequency,

19. Write the expression for impedance of RLC series circuit at half power frequencies.

At half power frequencies in RLC series circuit the total reactance is equal to resistance.

20. Write the expressions for bandwidth of RLC series circuit.

Bandwidth in rad/s

Bandwidth in

21. How the resonant frequency is related to half power frequencies in RLC series/parallel circuit?

The resonant frequency is given by the geometric mean of the two half power frequencies.

22. Define selectivity.

The selectivity is defined as the ratio of bandwidth and resonant frequency.

Selectivity


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23. Write the characteristics of series resonance.

At resonance, impedance is minimum and equal to resistance, therefore current is maximum.

Below resonant frequency the circuit behave as capacitive circuit and above resonant frequency the circuit behave as inductive circuit.

At resonance the magnitude of voltage across inductance & capacitance will be Q times the supply voltage, but they are in phase opposition.

24. A RLC series circuit has R = 10 & XC = 62.833 . Find the value of L for resonance at 50 Hz.

At resonance,

Since XL = 2fL, Inductance,

25. Determine the quality factor of a RLC series with R = 10 , L = 0.01 h and C = 100 F.


26. The impedance and quality factor of a RLC series circuit at = 107 rad/s are 100 + j0 and 100 respectively. Find the value of R, L and C.

Since the impedance is resistive the circuit will be in resonance. Therefore, At = r, Z = R, Resistance, R = 100

We know that,

We know that,

Capacitance,


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27. A RLC series circuit with R = 10 , Xl = 20 & XC = 20 is excited by a sinusoidal source of voltage 200 V. What will be the voltage across inductance.

Since XL = XC, the circuit will be in resonance. At resonance voltage across inductance is Qr times supply voltage.


Voltage across inductance =

28. A RLC series circuit excited by a 10 V sinusoidal source resonate at a frequency of 50 Hz. If the bandwidth is 5 Hz, what will be the voltage across capacitor?


Voltage across capacitor =

29. What is anti-resonance?

In RLC parallel circuit, the current is minimum at resonance whereas in series resonance the current is maximum. Therefore the parallel resonance is called anti-resonance.

30. Write the expression for resonant frequency for the RLC network shown in fig 1. What happens when R1 = R2 = R & L = CR2?

Figure:

Resonant frequency,

When L = CR2 the circuit will resonate at all the frequencies.


66


31. Draw the frequency response of RLC parallel circuit.

The variation of current with frequency is called frequency response, which is shown in fig.

Fig:

32. Write the expressions for quality factor of parallel RLC circuit.


33. Write the expressions for bandwidth of RLC parallel circuit.

34. Write the expression for half power frequencies of RLC parallel circuit.

Lower cut-off frequency,


67


Higher cut-off frequency,

35. Write the expression for admittance of RLC parallel circuit at half power frequencies.

At half power frequencies in RLC parallel circuit the total susceptance is equal to conductance.

i.e.,

At

or

36. Write the characteristics of parallel resonance.

At resonance, admittance is minimum and equal to conductance, therefore current is minimum.

Below resonant frequency the circuit behave as inductive circuit and above resonant frequency the circuit behave as capacitive circuit.

At resonance the magnitude of current through inductance and capacitance will be Q times the current supplied by the source, but they are in phase opposition.

37. What is dynamic resistance? Write the expression for dynamic resistance of RL circuit parallel with C.

The resistance of the RLC parallel circuit at resonance is called dynamic resistance. For RL circuit parallel with C, the dynamic resistance is given by,

38. In fig (a), & are in phase. Find the value of C.

Since are in-phase, the circuit will be in resonance.


68


fig.

For RLC circuit shown in fig,

39. For the RLC circuit shown in fig (b). Find the resonant frequency.

Impedance,

Figure:

At resonance the imaginary part of impedance should be zero. Therefore the numerator of the imaginary part should be zero.


69


40. A RLC parallel circuit with G=10 draw a current of 5 A when excited by a sinusoidal source. Determine the current through inductance.

Since BL = BC the circuit will be in resonance. At resonance the current through inductance will be Qr times the current draw from the source.


Current through inductance =

PART – B

1. A series combination of 10 resistance and 50 m H inductance is connected to a 220 V, 50 Hz supply. Estimate the current, active power, reactive power and apparent power. Also estimate the voltage across R & L and draw the phasor diagram.

Solution:


70


Figure: 1

Given that, V = 220 V, f = 50 Hz R = 10 , L = 50 mH

The RL series circuit excited by a sinusoidal source is shown in figure.

Inductive reactance = j XL = j L = j 2 f L = j 2 50 50 10-3 = j 15. 708

Impedance, Z = R + j XL = 10 + j 15.708 = 18.621 57.50

Let the supply voltage be reference phasor.

V = V 00 = 200 00 V

Let, I be the current through RL circuit. Now by Ohm’s law,

Current,

Power factor angle,

Power factor = cos = cos 57.50 = 0.5373 lag

The power factor is lag, because the current lags the voltage.Apparent power, S = VI = 220 11.8146 = 2599.2 VA

Active power, P = VI cos = 220 11.8146 0.5373


71


(or power)

Reactive power, Q = VI sin = 220 11.8146 sin 57.50

Voltage across resistance,

Voltage across inductance,

The phasor diagram of RL circuit with V as reference phasor is shown in figure.

Figure:

2. A current of 50 - 300 A if flowing through a circuit consist of series connected elements, when excited by a source of 230 450 V, 50 Hz. Determine the elements of the circuit and power. Also draw the phasor diagram.

Solution:

Given that,

Impedance,


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Since the reactance is positive, the circuit is RL series circuit. (Also the current is lagging and so the circuit is inductive).

We know that, Z = R + jXL

R = 1.1906 XL = 4.4433

We know that, XL = 2 f L.

The complex power,

Also, S= P + j Q

Active power,

Reactive power,

Apparent power,

The RL series circuit is shown in figure. Let VR & VL be the voltage across R & L. Now by Ohm’s law,

The phasor diagram of the RL series circuit is shown in figure.


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3. Consider a RL series circuit with an impedance angle of 500 at a frequency of 60 Hz. At What frequency the magnitude of the impedance at 60 Hz.

Solution:

Let f1 = 60 Hz &1 = 2 f1 = 2 60 = 376.9911 rad / s

Let Z1, be the impedance of RL series circuit at 1 and R & L be the resistance & inductance of the circuit.

Now,

Where, Z1 = Magnitude of Z1 =

1 = Impedance angle of Z1 = tan-1

Given that, 1 = 500

or

Let f2 be the frequency at which the magnitude of the impedance is doubled and 2 be the corresponding angular frequency. Let Z2 be the impedance corresponding to 2.

Now,


74


Where, Z2 = Magnitude of Z2 =

1 = Impedance angle of Z2 = tan-1

Given that, Z2 = 2 Z1

On squaring we get,

On dividing by R2 we get,

Put in the above equation,

Now,

Now,

Result

The frequency at which the magnitude of the impedance doubles = 148.332 Hz.


75


4. A RLC series circuit consists of R = 75 , L = 125 mH & C = 200 F. The circuit is excited by a sinusoidal source of value 115 V, 60 Hz. Determine the voltage across various elements. Calculate the current and power. Draw the phasor diagram.

Solution:

Given that, V = 115 V, f = 60 Hz, R = 75 , L = 125 mH & C = 200 F

Inductive reactance = j XL = j L = j 2 fL = j 2 60 125 10-3 = j 47.1239

Capacitive reactance =

= - j 13.2629

Total reactance = j X = j XL – j XC = j 47.1239 – j 13.2629 = j 33.861

Impedance,

The RLC series circuit is shown in figure. Let I be the current through the circuit and VR , VL & vC be the voltage across R, L & C respectively. Let V be the reference phasor.

Now by Ohm’s law,

I

I = I = 1.3975 A


76


Figure:

Apparent power, S = VI = 115 1.3975 = 160.7125 VA

Power factor angle, = V - I = 00 – (-24.30)= 24.30

Active power, P = VI cos = 115 1.3975 cos 24.30

Reactive power, Q = VI sin = 115 1.3975 sin 24.30 = 66.1355 VAR (inductive)The phasor diagram of RLC series circuit is shown in figure.

Here, V x = VL + VC

= I jXL + I(-jXC)

= I j (XL – XC)

= I j X = 1.3975

= 1.3975

= 47.3207

Figure: Phasor diagram of RLC series circuit


77


5. A RLC parallel circuit consists of R=50, L=150m H&C=100F. The circuit is excited by a current source of 5 0o A, 100 Hz. Calculate the voltage & current in the various elements. Determine the apparent, active and reactive power delivered by the source. Draw the phasor diagram.

Solution:

Given that,

Let us analyze the parallel circuit in terms of admittance.

Conductance,

Inductive susceptance =-jBL=-

Capacitive susceptance = jBC=j2fC=j210010010-6 =j0.0628

Total susceptance = jB=JBC-jBL=j0.0628-j0.0106=j0.0522

Admittance, =G+jB=0.02+j0.0522 =0.0559 69o

The RLC parallel circuit excited by a current source is shown in fig. Let be the voltage across the source and parallel connected elements. Let R, L & C be the current through R,L&C respectively.

Fig.

Now by ohm’s law,


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=89.4454 -69o j0.0522=89.4454 -69o 0.0522 90o

=4.669 21o A

Apparent power, S= =447.227 VA

Also, =P+jQActive power, P=160.2718WReactive power, Q=-417.5224 VAR

Or Q=417.5224 VAR – capacitive

Fig. Phasor diagram


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The phasor diagram of RLC parallel circuit with as reference phasor is shown in fig. 6. A load absorbs 2.5 kW at a power factor of 0.707 lagging from a 230 V, 50 Hz source. A capacitor is connected in parallel to the load in order to improve the power factor to 0.9 lag. Determine the value of the capacitor.

Solution:

Method – 1

Case I:

Given that V=230V, P1 = 2.5 kW, cos 1=0.707 lag, f= 50 Hz

Power factor angle, 1=cos-1 0.707=-45o

Reactive power, Q1=S1 sin 1=3.5361 sin 45o=2.5 kVAR =2.5 kVAR-inductive

Case II:

Given that V=230 V, f=50 Hz, cos 2=0.9 lag

The addition of capacitor to the load does not alter the active power but decrease the reactive power supplied by the source. Hence the active power remains same as that of 2.5 kW.

P2=2.5 kW

Apparent power,

Power factor angle, 2=cos-1 0.9=25.8o

Reactive power, Q2=S2 sin 2 =2.7778sin25.8o=1.209kVAR =1.209 kVAR – inductive

Hero, Q2 is inductive because the power factor is still lagging. Now, the reactive power supplied by the capacitor QC is given by,

QC=Q2-Q1=1.209-2.5=-1.291 kVAR =-1291 VAR

We know that,

Where, IC is the magnitude of current through capacitor.


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The capacitive reactance,

Also, XC=

Method II:

Case I:

Given that V= 230V, f=50Hz, P1=2.5 kW, cos 1=0.707 lag Let be the current through inductive load as shown in fig.

Now,

Magnitude of load current,

Let the supply voltage be reference phasor.

Fig.

Since the power factor is lagging the current will tag the supply voltage by an angle 1, where 1 = cos-1 0.707.


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Case II: When capacitor is added to inductive load

Given that, cos 2=0.9 lag

The active power remains same after addition of capacitor.

P2=2.5kW

The inductive load with capacitance in parallel is shown in fig. here the current through the load remains same as that of . Let be the current supplied by the source and be the current through the capacitor.

Now =15.3742 -45o A

By KCL we can write,


82


=10.8735-j5.2565-(10.8712-j10.8712)=0.0023+j5.6147j5.6147

Note: The small value of real part 0.0023 is due to approximation in calculations. Magnitude of capacitor current, Ic=5.6147A

Ca

Note: The slight difference in the capacitance value is due to approximation in calculations.

7. An inductive coil of power factor 0.8 lagging is connected in series with a 120F capacitor. When the series circuit is connected to a source of frequency 50Hz, it was observed that the magnitude of voltage across the coil and capacitor are equal. Determined parameters of the coil.

Fig.

Solution:

The given circuit is RLC series circuit as shown in fig.

Let, = Impedance of the coil

Let, be the current through RLC series circuit.


83


Now,

Magnitude of voltage across coil=IZ1

Magnitude of voltage across capacitor = IXC

Given that, IZ1=IXC

Z1=Xc

Magnitude of impedance, Z1=Xc=

=26.5258

Given that, power factor of the coil = 0.8 lag Power factor angle of the coil, =cos-1 0.8=36.9o

Let us construct an impedance triangle for Z1by using R & XL as two sides as shown in fig. here the impedance angle is same as power factor angle.

Impedance triangleWith reference to fig. we can write,

Resistance, R=Z cos =26.5258 cos 36.9o=21.2123

Inductive reactance, XL=Z sin = 26.5258 sin 36.9o=15.9266

We know that, XL=L

Inductance,


84


Result:

The parameters of the coil are R&L.Resistance of the coil, R=21.2123Inductance of the coil, L=50.7 mH

8. Three impedances 12-j0, 5+j8 & 0-j7 are connected in parallel. This parallel combination is connected in series with an impedance of 4+j6 across a 230 V source. Determine the current through each impedance and the power.

Fig (a)

Solution

The series-parallel connections of the impedances are shown in fig. Let us name the impedances as as shown in fig. Let the current through the impedances be as shown in fig. let supply voltage be reference phasor.

Supply voltage = 230 0o V

Let, be the equivalent impedance of the parallel combination of and the circuit can

be modified as shown in fig.(b)

Fig. (b)


85


Let be the voltage across as shown in fig (b) now by voltage division rule we can

write,

By KVL we can write,

Here the voltage across and the voltage across are (because are in parallel). Now the current through the impedances can be evaluated by using ohm’s law as shown below.

We know that the complex power is given by the product of voltage and conjugate of current. Hence the complex power in each impedance can be obtained from the product of voltage and conjugate of current in the impedance.

Let be the complex power of the impedances respectively. Let P1, P2,

P3 & P4 be active power and Q1, Q2, Q3 & Q4 be the reactive power of the impedances.


86


For impedance ,

S4=3219.7VA=3.2197kVA

P4=17863.4W=1.7864kW

Q4=2678.6VAR=2.6786kVAR

For impedance

=141.5949 -40.2o 11.7996 40.2o=1670.8 0o VA

=1670.8+j0=P1+jQ1

S1=1670.8VA=1.6708kVA P1=1670.8W=1.6708kW Q1=0

For impedance

S2=2125.2VA=2.1252kVA

P2=1126.2W=1.1262kW

Q2=1802.3VAR=1.8023kVAR

For impedance


87


S3=2864VA=2.864kVA

P3=0

Q3=-2864VAR=-2.864kVAR

9. Two reactive circuits have an impedance of 20 each. One of them has a power factor of 0.75 lagging and the other 0.65 leading. Find the voltage necessary to send a current of 12A through the two circuits in series. Also determine the current drawn from 200V supply if they are connected in parallel to the supply.

Solution:

Case I: Impedances in series

Let be the impedances of the two circuits. The series combination of two circuits excited by a voltage source can be represented by the circuit shown in fig.

Fig.

When the current is lagging the impedance will be inductive. Hence the impedance angle is positive and it is given by the power factor angle. Now the impedance can be expressed as shown below.

When the current is leading the impedance will be capacitive. Hence the impedance angle is negative and it is given by the negative of power factor angle. Now the impedance


88


Let, be the total impedance of series combination.

Now, = + =15.0022+j13.2262+12.989-j15.2081 =27.9912-j1.9819=28.0613 -4.1o

Magnitude of impedance, Z=| |=28.0613Given that the magnitude of current, I= =12A

Magnitude of supply voltage, V= IZ

=1228.0613=336.7356V

If is the reference phasor then,

=336.7356 0oV & =12 4.1oA

If is the reference phasor then,

=336.7356 -4.1oV & =12 0oA

Case ii: When the impedances are in parallel

The two circuits in parallel and excited by a 200 V source can be represented by the circuit shown

in fig. Let & be the current through the impedances respectively. Let be the total

current supplied by the source.

Fig.

Let the supply voltage be the reference phasor.


89


=V 0o=200 0o V

By KCL we can write,

10. In the RLC circuit shown in fig 1, determine the values of R & L if and the

voltage are in phase quadrature.

Figure:

Solution:

Let series combination of R1 & L be impedance and the series combination of R2 & C be

impedance


90


Given that, R2 = 3 , C = 200 & f = 50 Hz

Let the supply voltage be the reference phasor.

By voltage division rule we can write,

where,

and,

Again by voltage division rule we can write,


91


where,

and,

Given that,

On squaring we get,

Also it is given that,


92


On substituting XL = 0.189 R1 in equation (4.13.1) we get,

Result

11. The parameters of a RLC parallel circuit excited by a current source are R=40 , L=2mH & C=3 F. Determine the resonant frequency, quality factor, bandwidth and cut-off frequencies.


93


Solution:

Given that, R=40, L=2mH & C=3F

Angular resonant frequency,

Resonant frequency,


Bandwidth,

Bandwidth in

12. A coil of inductance 31.8mH and resistance 10 is connected in parallel with a capacitor across a 250V, 50Hz supply. Determine the value of capacitance if no reactive current is taken from the supply.

Solution:

The parallel combines of the coil and capacitor excited by the voltage source is shown in fig.


94


Fig.

The circuit of fig. has two parallel branches. Let be the admittance of the coil and BC be the

capacitive substance. Here =1/ where is the impedance of the coil.

Let be the total admittance of the two parallel branches.

Given that the current supplied by the source does not have any reactive components. This happens only at resonance. At resonance the circuit behave as purely resistive, which means that the imaginary part of admittance is zero. Therefore the value of C can be determined by equating the imaginary part of admittance to zero.

On letting imaginary part of admittance to zero we get,

BC-0.05=0

BC=0.05


95


13. In the RLC network shown in fig. determine the value of RC for resonance. Also calculate the dynamic resistance.

Fig.

Solution:

The given network has two parallel branches. Let be the admittances of the parallel branches as shown in fig.

Fig.

Here,

Where are the impedances of the parallel branches.

Let, = Total admittance of the RLC parallel network.

Now,

Let us separate the real and imaginary part by multiplying the numerator and denominator of each term by the complex conjugate of denominator.


96


For resonance the imaginary part of should be zero.

The dynamic resistance is given by the inverse of the real part of the admittance at resonance.

Result:

For resonance, RC=1.7857The dynamic resistance, Rdynamic=32.4676.

14. Determine the value of RL for resonance in the network shown in fig.


97


Fig.

Solution:

The given network has two parallel branches. Let be the admittances of the parallel branches as shown in fig.

Fig.

Here,

Where are the impedances of the parallel branches.

Let, = Total admittance of the RLC Parallel network.



98


At resonance the imaginary part of admittance will be zero.

Result:

For resonance, RL=24.4949

15. In the RLC network shown in fig. determined the two possible values of C for the network to resonance at 2000 rad/s. Also determined the value of C for resonance at all frequencies.

Fig.

Solution:


99


Let be the admittances of the parallel branches.

Here,

Where, are the impedances of the parallel branches.



At resonance the imaginary part of admittance will be zero.

11.6009XC=16+XC2

The roots of the quadratic are,


100


The condition for resonance at all frequencies is R=

Result:

16. In the RLC network shown in fig. determined the two possible values of L for the network to resonate at 4000 rad/s.


101


Fig.Solution:

Let be the admittances of the parallel branches.

Here,

Where, are the impedances of the parallel branches.




102


At resonance the imaginary part of the admittance will be zero.

The roots of the quadratic are,


103


Result:

17. Write short Notes on Tuned circuits.

At radio frequencies, iron core transformers are note used as eddy current losses and hysteresis losses increase with frequency. Thus at radio frequency air core transformers are frequently used. As there is air path between windings, the leakage flux increase and coefficient of coupling decreases.

In RF circuit design, tuned circuits are generally employed either for obtaining maximum power transfer to the load connected to secondary or for obtaining maximum possible value of secondary voltage.

There are two types of tuned circuits namely

i. Singly tuned circuit andii. Doubly tuned circuit

1. Singly Tuned Transformer

Consider a singly tuned transformer as shown in the Fig. a


104


Figure:

Basically above type of circuit is used for coupling a amplifier and radio receiver circuits. In such applications maximum power transfer is not expected. Instead of that maximum possible secondary voltage V2 is desired. In this application, R1 may be considered as the output resistance of amplifier stage. As this resistance is very high, the resistance of coil L1 is added to R1 or many times neglected even. R2 is the resistance of secondary winding.

The secondary voltage is given by,

The equivalent circuit of the circuit considered is as shown in the Fig.(b)

Figure:Applying KVD to loop 1, we get,

Applying KVL to loop 2, we get,


105


For equation (3.81),

Substituting value of I1 in equation (3.80), we get,

To have maximum secondary voltage, I2 should be large. From equation (1) it is clear that we can get maximum I2 if the reactive term in denominator is zero. So by using variable capacitor we can make the reactive term zero as follows.


106


or

Since R1>>R2, we can neglect term Thus we can assume the condition for

maximum current I2 as,

Under the condition given in equation (2), we can rewrite the expression for maximum I2 as,

But from equation (3), we can write,

Substituting this value in equation (4)

Now the maximum secondary voltage is given by,


107


Differentiating above expression with respect to we can get condition for maximum voltage as

This is called condition for critical coupling. In general, coefficient of coupling is given by,

Thus, putting value of M for critical coupling, we can write,

Critical coupling

For critical coupling, the secondary voltage is given by,

From equation (5) it is clear that the maximum value of secondary voltage can be achieved

by selecting either M or the ratio .

The variation of secondary voltage with the coefficient of coupling is as shown in the Fig.C


108


Figure: Variation of output voltage with coefficient of coupling

2. Doubly Tuned Transformer

In doubly tuned transformer circuits, both primary and secondary circuits contain adjustable capacitance. With the help of adjustable capacitive reactance, impedance matching is possible if the coupling is critical, sufficient or above. It is also possible to adjust phase angle such that impedance at generator side becomes resistive. The magnitude matching can be achieved by adjusting mutual inductance to the critical value, which effectively fulfills maximum power transfer condition. Consider doubly tuned transfer circuit as shown in the Fig. R1 is the resistance of primary winding L1 while R2 is that of secondary winding L2. R2 may include the resistance connected as a load.

Figure (a):Finding Thevenin’s equivalent circuit across terminals (A) and (B).


109


Figure: (1)

Open circuit voltage V1 is given by,

Assume that

Consider Fig 1 (b),

Again according to standard assumption,

From above equation it is clear that a capacitor C1 appears in series with no change in effective capacitance. Replacing the original circuit across terminals (A) and (B) by its


110


Fig 2:

Thevenin’s equivalent circuit as shown in the Fig. 2

Figure 3:

Applying KVL to loop 1, we get

Applying KVL to loop 2, we get,


111


If primary and secondary circuits are adjusted by using variable capacitor to resonate, then current I2 will be maximum. The condition of resonance is given by

At resonance reactive term becomes zero, hence rewriting equations (1) and (2) as,

From equation (4) we can write,

Substituting value of I2 in terms of I1 in equation (3),

Similarly substituting from equation (5) value of I1 in equation (3) again, we get,


112


In doubly tuned transformer circuit also the condition for critical coupling is obtained as,

Then under critical coupling the current I1 and I2 are given as,

Equation (7) represents the value of I2 giving maximum power transfer. But in such applications maximum output voltage is important. The maximum voltage V2 under condition of resonance is given by,

As r we can rewrite V2 as

Let Q1 be the quality factor of primary circuit. This is also same as quality factor of primary coil neglecting very high value resistance Rg.


113


Similarly let Q2 be the quality factor of secondary circuit which is same as quality factor of secondary coil if R2 is the coil resistance,

The coefficient of coupling is given by,

Substituting values of L1 and L2 from equation (8) and (9),

Hence expression for maximum voltage is given by,

If above expression is maximized with respect to k, we get the condition for critical coupling.

Thus the maximum voltage under critical coupling is given by

Hence to get maximum secondary voltage with critical coupling high values Q1, Q2 and L2

are selected.


114


Under critical coupling secondary current is maximum, hence secondary voltage is also maximum. Frequency response of secondary voltage for different coupling coefficients is as shown in below figure.

Figure:

18.A 20 resistor is connected in series with an inductor, a capacitor and an ammeter across 25 volts variable frequency supply as shown in Fig. When the frequency is 400 Hz, the current is at its maximum value of 0.5 A and the potential difference across capacitor is 150 volts. Calculate,

(i) The capacitance of capacitor

(ii) The resistance and inductance of inductor.

Figure: (a)Solution:

When the current is maximum at resonating frequency f0 = 400 Hz, the total reactance is zero. That means,


115


But

Where VC is voltage across capacitance at resonance and I0 is maximum current in circuit at resonance.

At resonance, XL = 300

L = 0.119 H

Also at resonance,

Circuit resistance = circuit impedance

i.e.

Resistance of inductor is 30 .

19. A resistor and capacitor are in series with a variable inductor. When the circuit is connected to 200 V, 50Hz supply, the maximum current obtained by varying the inductance is 0.314 A. The voltage across capacitor, when current in the circuit is maximum, is 800 V. Find values of series circuit elements.


116


Solutions:

V = 200 volts, VC = 800 volts, I0 = 0.314 AConsider magnitude of capacitor voltage, it is given by,

800 = Q0 (200)

Q0 = 4

The current in series resonant circuit is maximum only at resonance and it is given by,

R = 636.95

To calculate L and C, we write,

Now circuit is operating at 50 Hz which is resonant frequency,

g

Now,

L = 8.1 H

Also


117


20. A series circuit is in resonance at 8x106 Hz and has a coil of 35 H and 10 resistor.

(i) Find current at resonance.(ii) What capacitance will require for resonance?(iii) Find impedance at frequency of 8.1 MHz.(iv) Find current at this frequency.

Applied voltage is sinusoidal 100 V.

Solution:

At resonance, current in the circuit is maximum and it is given by

Resonant frequency f0 is given by,

Substituting values of f0 and L,

To find impedance at 8.1 MHz, we have to use formula as,

= 175.93 176


118


The deviation of actual frequency from resonant frequency is given by ,

The impedance at 8.1 MHz is,

The current in the circuit at 8.1 MHz is,

The negative sign of angle shown that the current is lagging voltage which is indication of inductive load (This fact is also clear from angle of Z at 8.1 mHz which is positive).

21. In a series RLC circuit, if C is variable then show that the value of C for maximum voltage across it is given by,

Solution: The circuit is shown in the Fig. below

Figure:

The impedance of the circuit is,


119


The voltage across the capacitor is,

For maximum VC,

differentiating by the rule of u/v,

For maximum C, numerator must be zero and splitting the index – 3/2 as – ½ and -1, we get the equation as,


120


22. Calculate the maximum voltage across the inductor in a series resonant circuit with constant voltage and variable frequency and R =50 , L = 0.05 H, C = 20 F and V = 100 V.

Solution: R = 50 , L = 0.05 H, C = 20 F, V = 100 V

Maximum voltage across inductor in series resonant circuit occurs at frequency given by,


121


The impedance of series resonant circuit at fL is given by,

Total current in series circuit is,

Voltage across inductor = VL = I (j XL)

23. An inductive coil having a resistance of 20 and an inductance of 0.02 H is connected in series with 0.02 F capacitor. Calculate:

(i) Q factor of coil(ii) Resonant frequency(iii) Half power frequencies

Solution:

R = 20 , L = 0.02H, C = 0.02

Resonant frequency is given by;


122


Q factor of coil is given by,

Bandwidth

Lower Half power frequency

Upper Half power frequency

24. A RLC series circuit of 8 resistance should be designed to have bandwidth of 50Hz. Determine the values of L and C so that the circuit resonates at 250 Hz.

Solution:

R = 8 , B.W = 50 Hz, f0 = 250 Hz

Bandwidth

L = 25.46 mH

To find value of C,


123


25. A parallel circuit has a fixed capacitor and variable inductor having constant quality factor of 4. Find value of inductance and capacitance for circuit impedance of (1000) at resonating frequency 2.4 MHz. What is bandwidth of circuit?

Solution: Q0 = 4, Z0 = 1000 , far = 2.4 MHz

The impedance of the parallel circuit at resonance is given as,

But

The resonating frequency is given by,


124


From equation (1) and (2), we get values of L as,

Similarly,

Bandwidth =

= 0.6 x 106

Bandwidth = 0.6 MHz

26. A parallel circuit resonates at 1 MHz having inductance of 150 H with Q0 of 60. Find value of capacitance and resistance of inductor.

Solution:

L


125


Also,

27. A coil resonates at 2 MHz when a 18 pF capacitor is shunted across it. When shunting capacitor is 81 pF, the resonating frequency becomes 1 MHz. Find the distributed capacitor of coil and what is self resonating frequency.

Solution:

Let distributed capacitance of coil be Cd. Assume that quality factor of coil is very much large than 10.

Then,

Now, when

also, when

Dividing equation (1) by (2),


126


Distributed capacitance of coil is 3 x 10-12 F ie. 3 pF. Now, to find self resonating frequency we must know value of L. So substituting value of Cd in equation (1) we get,

L = 0.3 mH

Self resonating frequency with L = 0.3 mH and Cd = 3 x 10-12 F is given by,

28. In the circuit shown in Fig. below the inductance of 0.1H having Q factor of 5 is in parallel with capacitor. Determine the value of capacitance and coil resistance at resonant frequency of 500 rad/sec.

Figure:

Solution: The quality factor is given by,


127


For above circuit frequency of resonance is given by,

29. If R is positive, show that resonance is impossible in the circuit show in Fig. below

Figure:

Total admittance is,


128


For resonance, the susceptance part is zero.

Thus, for resonance, value of R is negative i.e. R is imaginary. This clearly shows that in the circuit, resonance is impossible for positive values of R.

30. Two impedance Z1 = 20 + j 10 and Z2 = 10 – j 30 are connected in parallel and this combination is connected in series with Z3 = 30 + j X. Find the value of X which will produce resonance.

Solution: From given information, given circuit is as shown in Fig. (1)

Total impedance is given by,

Figure:


129


The circuit shown in Fig. will resonate, if imaginary part is zero,

31. A parallel resonant circuit has a coil of 150 H with Q factor of 100 and is resonated at 1 MHz.

(i) Specify the required value of capacitance,(ii) What is resistance of coil?(iii) What is resistance of circuit at parallel resonance?(iv) What is absolute bandwidth of resonant circuit?(v) What is bandwidth of the circuit when it is matched with the generator impedance?

Solution:

L = 150 H, Q0 = 100, far = 1 x 106 Hz

(i) Value of C required for resonance,


130


(ii) Resistance of coil RL is given by

(iii) The resistance of circuit at resonance Zar or Z0 is given by

(iv) The absolute bandwidth is given by,

For the circuit, with generator resistance equal to the circuit resistance at resonance, bandwidth is given by,


131


But Zar = Rg

32. A coil of 10H and resistance of 10 is in shunt with 100 pF capacitor. The combination is connected across a generator of 100 V, having internal resistance of 100 k. Determine,

(i) Voltage across parallel circuit at resonance and(ii) Bandwidth.

Solution:

L = 10 H, RL = 10 , C = 100 pF, Vin = 100 V, Rg = 100 k

The resonating frequency is given by,

At resonance, parallel circuit offers resistance,

The circuit can be shown as below,


132


Figure:

By voltage divider formula voltage across parallel resonant circuit is given by,

Bandwidth of circuit with generator resistance is given by,

33. Find value of L at which circuit shown in the Fig. resonates at a frequency of = 500 rad/sec.

Figure:

Solution: The total admittance is given by,


133


At resonance, susceptance becomes zero. Hence at resonance,

Solving above quadratic equation for XL we have

But

34. For the series R-L-C combination shown in the Fig, find impedance Z (j) as a function of ‘’. At what frequency is the magnitude of the impedance at a minimum?

Figure:


134


Solution:

Referring above, the impedance of the series RLC combination is given by

From above expression it is clear that impedance Z depends on frequency , hence we can write,

By the property of resonance, at a frequency of series resonance, impedance is minimum canceling out reactive part. Hence we can write at resonating frequency =0,


135


UNIT – III

SEMICONDUCTOR DIODE

PART – B

1. Define Intrinsic Semiconductor.

A semiconductor in an extremely pure form is defined as intrinsic semiconductor or pure semiconductor.

2. Define Extrinsic Semiconductor.

The electrical conductivity of pure semiconductor can be increased by adding some impurity into it. The resulting semiconductor is called extrinsic semiconductor or impure semiconductor.

3. What are the types of semiconductor?

There are two types of semiconductor. They are:

Intrinsic semiconductorExtrinsic semiconductor

a. N-type semiconductorb. P-type semiconductor

4. Define doping.

The process of adding impurities to an intrinsic semiconductor is known as doping.

5. What are N-type semiconductors.

When a small amount of penta valent impurity(e. g. Antimony, Arsenic) is added to a pure semiconductor, it is known as N-type semiconductor.

6. What are P-type semiconductors.

When a small amount of tri-valent impurity(e. g. Indium, Galluim) is added to a pure semiconductor, it is known as P-type semiconductor.

7. Define the term drift current.


136


When a potential difference is applied across the semiconductor, an electric field is developed in the material. It causes free electron to move in one direction and holes to drift in other direction. Because the electrons move in opposite direction, from the holes these two components of current add rather than cancel. The total current due to the electric field is known as the drift current. 8. Define the term diffusion current.

Whenever there is a concentration of carriers in one region of a semiconductor and shortage in other region, then the carrier in the high density will move to the low density region until their distribution becomes uniform. During this time that carriers are moving from the higher concentration to lower concentration, there is a transfer of charge is taking place. This current is called diffusion current.

9. What is called P-N junction diode biasing?

When a PN junction is connected to an external voltage source, it is called PN junction diode biasing. The two types of bias are : a. Forward Bias and b. Reverse Bias.

10. What are the differences between drift current and diffusion current?

Sl No

Drift current Diffusion current

1 Developed due to the potential gradient Developed due to the change in concentration gradient.

2Phenomenon found both in semiconductor & metals

Only in semiconductors.

11. What is forward Bias?

If P type terminal is connected to anode(positive electrode), and N type terminal is connected to cathode(negative electrode), it is known as forward bias. At forward bias, large current will flow in the range of mille amperes. Forward bias is equivalent to short circuit.

12. What is reverse Bias?

If P type terminal is connected to cathode(negative electrode), and N type terminal is connected to anode(positive electrode), it is known as reverse bias. At reverse bias, small current will flow in the range of mille amperes. Reverse bias is equivalent to open circuit.

13. What is depletion region or transition region or space charge region in a PN junction diode?

The region around the junction from which the charge carriers are depleted is called as depletion region. Since this region has immobile ions, which are electrically charges, the depletion region is also known as space charge region.


137


14. What is junction or potential barrier?

Barrier potential is the voltage developed by the junction due to the movement of carriers from original to the opposite side. That is electrons are moving to P side and holes are moving to N side. So the resultant field is weakened and the barrier height is reduced at the junction. 15. What is the potential barrier for silicon & Germanium?

Potential barrier for Silicon - 0. 7 V. Potential barrier for Germanium - 0. 3 V.

16. Define knee voltage or threshold voltage.

The forward voltage at which the current through the PN junction starts increasing rapidly is known as knee voltage. It is also called as cut-in voltage or threshold voltage.

17. Define Break down voltage.

The reverse voltage at which the PN junction breaks down, with sudden rise in reverse current is called as breakdown voltage.

18. What is mobility?

Mobility of a charge carrier is defined as the average drift velocity PG unit electric field. Its unit square meters per volt-seconds.

µ=V/E where µ - Mobility

V - Drift velocity E - Applied electric field.

19. What is conductivity?

Conductivity is defined as the current density per unit applied electric field. It s unit is mho per meters.

where Ó – Conductivity J – Current Density E – Applied Electric field.

20. In an unbiased junction, what is the thickness of the depletion region?

In the order of 0. 5µ.


138


21. What is dynamic resistance of a diode and static resistance of a diode?

The dynamic of a. c. Forward resistance of a diode is the resistance offered by the diode to an ac signal. It is defined as the ration of change in voltage across the diode to the resulting change in current through it.

Dynamic Resistance , ac=Δvf/ΔIf22. What is the static resistance of a diode?

Static or do forward resistance of a diode is the resistance offered by the diode to the direct current. It is defined as the ratio of the dc voltage across the diode to the dc current following through it.

Static Resistance, Rf=

23. What is reverse resistance of a diode?

Reverse resistance is defined as the opposition offered by the diode to the reverse current, under reverse biased condition. Its value is very large as compared to forward resistance.

Rn=VR/IN=Reverse voltage/Reverse current.

24. Write down and explain the junction diode equation.

where I = forward( or reverse )diode current

I0 = Reverse saturation current V = Diode voltage. = a const. 1 for Ge diodes

2 for Si diodesV

T = volt equivalent of temperature

Its value is T/11,600T = absolute temperature

25. What is maximum power rating of a diode?

The maximum power that a diode can dissipate without damaging it, it is called its maximum power rating.

26. What is peak inverse voltage.

Peak inverse voltage(Piv) is the maximum voltage across the diode when it is not conducting.

27. What causes the diffusion of holes and electrons across PN junction ?


139


Concentration gradient causes the diffusion of holes & electrons across a PN junction.

28. Define the term diffusion capacitance or Storage capacitance.

The capacitance, which exists in a forward-biased junction, is called a diffusion or storage capacitance. It is different from the depletion layer capacitance, which exists in a reverse-biased junction. The diffusion capacitance arises due to the arrangement of minority carrier density. And its value is much larger than the depletion layer capacitance. Its value for an abrupt junction is given by the relation :

Where is = mean life time of the carriers I = value of forward current = A const. (1 for Ge and 2 for Si)

VT = Volt equivalent of temperature

29. Give the expression for depletion capacitance of a diode.

GT= єA/Wwhere A = cross sectional area of the junction W = Width of the depleted region. 30. State the different types of semiconductor diode

i. PN junction diodeii. Zener diodeiii. Varactor diodeiv. Tuned diodev. PIN diodevi. LED

31. List the applications of PN junction diode.

i. Used as the rectifier diodes in DC power supplier. ii. Used as the signal diodes in communication circuits for modulation and demodulation. Iii. They are used in clipper and clamper circuits. iv. They are used as a switch in logic circuits used in computers.

32. What is Bulk resistance.

The resistance of the P and N semiconductor materials of which the diode is made of is known as bulk resistance or body resistance. It also includes the resistance introduced by the


140


connection between the semiconductor material and the external metallic conductor also called contact resistance. It is generally designated by r

B . Mathematically, the bulk resistance,

rB

= rp

+ rn

Where rp

= ohmic resistance of P type semiconductor

rn

= ohmic resistance of N type semiconductor

33. What is junction resistance?

Its value for a forward biased Pn junction depends upon the value of forward dc current. It is given by the relation,

rj = 26/If

where If= forward current, in milli amperes.

34. The PN junction contact potential cannot be measured by placing a voltmeter across the diode terminals(True/False).

True.

35. The potential barrier at a PN junction is due to the charges on either side of the junction. These charges are immobile positive ions and immobile negative ions.

36. i. Across an open circuited PN junction diode there exists a barrier potential.

i. When a PN junction is forward biased, the depletion region width decreases. ii. In an unbiased junction the thickness of the depletion region is of the order of 0. 5µ .

37. The potential barrier is heavily doped and breakdown voltage will be small.

38. What is the effect of temperature on diode characteristics?

As temperature increases, the exponent will reduce hence the diode current also decreases. It is found that the variation of saturation current I0 is much greater than the exponential term.

For Ge or Si diode the following equation will express all mathematical relations.

I02

= I01

2(T

2-T

1)/10

where I02

- Saturation current at temperature T2

I01

- Saturation current at temperature T1

39. Give the expression for Power dissipation in diode.


141


The value of the power dissipated is given by the product of voltage and the current through it. For a forward biased diode, it is given by,

PD =

VF

* IF

where VF

– Voltage drop up forward bias

IF

– Forward bias current

Similarly, for a reverse biased diode the power dissipation is given by,P

D = V

R * I

Rwhere V

R – Voltage drop at reverse bias

IR

– Reverse bias current

40. State the relationship between diode capacitance and the reverse bias voltage.

The value of the depletion capacitance of a PN junction diode is given by

where K–Constant, depends on the nature of the semiconductor material

VB - Barrier voltage, 0. 7 V for Si and 0. 3 for Ge.

V– Reverse bias voltage. n-Constant depends on the nature of the junction.

41. Define recovery time of the diode.

The recovery time is the time difference between the 10% point of the diode voltage and the time when this voltage reaches and remains within 10 % of its final value.

42. Define peak inverse voltage.

It is the maximum safe rating voltage of a rectified diode when it is reverse biased.

43. The potential barrier at a PN junction is due to the charges on either side of the junction. These charges are electrons and holes.

44. A pure Ge crystal is an intrinsic semiconductor and a doped crystal is an extrinsic semiconductor


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PART – B

1. What is intrinsic semiconductor? Explain how the electrons and holes constitute current in an intrinsic semiconductor.

Intrinsic Semiconductors

A sample of semiconductor in its purest form is called an intrinsic semiconductor. The impurity content in intrinsic semiconductor is very very small, of the order of one part in 100 million parts of semiconductor.

Crystal Structure of Intrinsic Semiconductor

Consider an atomic structure of an intrinsic semiconductor material like silicon. An outermost shell of an atom is capable of holding eight electrons. It is said to be completely filled and stable, if it contains eight electrons. But the outermost shell of an intrinsic semiconductor like silicon has only four electrons. Each of these four electrons form a bond with another valence electron of the neighbouring atoms. This is nothing but sharing of electrons. Such bonds are called covalent bonds. The atoms align themselves to form a three dimensional uniform pattern called a crystal.

The crystal structure of germanium and silicon materials consists of repetitive occurrence in three dimensions of a unit cell. This unit cell is in the form of a tetrahedron with an atom at each vertex. But such a three dimensional structure is very difficult to represent pictorially. Hence a symbolic two dimensional structure is used to represent a three dimensional crystal form, as shown in the figure.

The figure (a) shows two dimensional representation of a germanium crystal structure. Germanium has a total of 32 electrons. So its first orbit consists of 2 electrons, second consists of 8, third consists of 18 and the valence shell consists of 4 electrons. As there are 4 valence electrons, it is called tetravalent atom.


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Fig. Two dimensional representation of silicon crystal The covalent bonds are represented by a pair of dotted lines encircling the two electrons

forming the covalent bond. The more clear understanding of the covalent bonds can be obtained from the figure (b) which shows the sharing of valence electrons. Both the electrons are shared by the two atoms. Hence the outermost shell of all the atoms is completely filled, and the valence electrons are tightly bound to the parent atoms. No free electrons are available at absolute zero temperature.

Electrons and Holes in Intrinsic Semiconductors

Let us see what happens at room temperature. At room temperature, the number of valence electrons absorb the thermal energy, due to which they break the covalent bond and drift to the conduction band. Such electrons become free to move in the crystal as shown in the figure (a).

(a) Breaking of covalent bond (b) Electron-hole pair in a (c) Energy band diagram silicon crystal

Fig. (a)


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Such an intrinsic semiconductor behaves as a perfect insulator at absolute zero temperature


Once the electrons are dislodged from the covalent bonds, then they become free. Such free electrons wander in a random fashion in a crystal. The energy required to break a covalent bond is 0.72 eV for germanium and 1.1 eV for silicon, at room temperature.

When a valence electron drift from valence to conduction band breaking a covalent bond, a vacancy is created in the broken covalent bond. Such a vacancy is called a hole. Whenever an electron becomes free, the corresponding hole gets generated. So free electrons and holes get generated in pairs. The formation of electron-hole pair is shown in the figure (b) while the corresponding energy band diagram is shown in the figure (c). Such a generation of electron hole pairs due to thermal energy is called thermal generation.

The concentration of free electrons and holes is always equal in an intrinsic semiconductor. The hole also serves as a carrier of electricity similar to that of free electron.

Thus in an intrinsic semiconductors both holes as well as free electrons are the charge carriers. Conductivity of Intrinsic Semiconductor

The property called conductivity indicates the ease with which a material can carry the current. Thus more conductivity means that material can carry high current, very easily. The conductivity of a good conductor is high while that of an insulator is low.

In intrinsic semiconductor, very few electron-hole pairs get generated at room temperature. Hence very small current can be constituted, due to the application of voltage to an intrinsic semiconductor. Thus the conductivity of an intrinsic semiconductor at room temperature is very low. Such a low conductivity has very little practical significance.

Due to low conductivity, the intrinsic semiconductors are not used in practice for manufacturing of electronic devices.

2. What is extrinsic semiconductor? Explain the formation of P-type semiconductor & n-type semiconductor.

Extrinsic Semiconductors

In order to change the properties of intrinsic semiconductors a small amount of some other material is added to it. The process of adding other material to the crystal of intrinsic semiconductors to improve its conductivity is called doping. The impurity added is called dopant. Doped semiconductor material is called extrinsic semiconductor. The doping increases the conductivity of the basic intrinsic semiconductors hence the extrinsic semiconductors are used in practice for manufacturing of various electronic devices such as diodes, transistors etc.

Depending upon the type of impurities, the two types of extrinsic semiconductors are, 1. n-type and 2. p-type


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An electron is negatively charged particle. Thus a hole getting created due to electron drift is said to be positively charged.


Types of Impurities

The impurity material having five valence electrons is called pentavalent atom. When this is added to an intrinsic semiconductor, it is called donor doping as each impurity atom donates one free electron to an intrinsic material. Such an impurity is called donor impurity. The examples of such impurity are arsenic, bismuth, phosphorous etc. This creates an extrinsic semiconductor with large number of free electrons, called n-type semiconductor.

Another type of impurity used is trivalent atom which has only three valence electrons. Such an impurity is called acceptor impurity. When this is added to an intrinsic semiconductor, it creates more holes and ready to accept an electron hence the doping is called acceptor doping. The examples of such impurity are gallium, indium and boron. The resulting extrinsic semiconductor with large number of holes is called p-type semiconductor.

The resulting extrinsic semiconductor with large number of holes is called p-type semiconductor. Types Semiconductor

When a small amount of pentavaent impurity is added to a pure semiconductor, it is called n-type semiconductor. The pentavalent impurity has five valence electrons. These elements are such as arsenic, bismuth, phosphorous and antimony. Such an impurity is called donor impurity.

Consider the formation of n-type material by adding arsenic (As) into silicon (Si). The arsenic atom has five valence electrons. An arsenic atom fits in the silicon crystal in such a way that its four valence electrons form covalent bonds with four adjacent silicon atoms. The fifth electron has no chance of forming a covalent bond. This spare electron enters the conduction band as a free electron. Such n-type material formation is represented in the figure (a). This means that each arsenic atom added into silicon atom gives one free electron. The number of such free electrons can be controlled by the amount of impurity added to the silicon. Since the free electrons have negative charges, the material is known as n-type material and an impurity donates a free electron hence called donor impurity.

Fig. (a) n-type material formation


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p-type Semiconductor

When a small amount of trivalent impurity is added to a pure semiconductor, it is called p-type semiconductor. The trivalent impurity has three valence electrons. These elements are such as gallium, boron or indium. Such an impurity is called acceptor impurity.

Fig. (a) Consider the formation of p-type material by adding gallium (Ga) into silicon (Si). The

gallium atom has three valence electrons. So gallium atom fits in the silicon crystal in such a way that its three valence electrons form covalent bonds with the three adjacent silicon atoms. Being short of one electron, the fourth covalent bond in the valence shell is incomplete. The resulting vacancy is called a hole. Such p-type material formation is represented in the figure (a). This means that each gallium atom added into silicon atom gives one hole. The number of such holes can be controlled by the amount of impurity added to the silicon. As the holes are treated as positively charged, the material is known as p-type material.

At room temperature, the thermal energy is sufficient to extract an electron from the neighbouring atom which fills the vacancy in the incomplete bond around impurity atom. But this creates a vacancy in the adjacent bond from where the electron had jumped, which is nothing but a hole. This indicates that a hole created due to added impurity is ready to accept an electron and hence is called acceptor impurity. Thus even for a small amount of impurity added, large number of holes get created in the p-type material.

3. What is meant by Barrier potential and derive the expression for the potential barrier in a step graded p-n junction?

Barrier Potential

Near the junction, on one side there are many positive charges and on other side there are many negative charges. According to Coulomb’s law, there exists a force between these opposite charges. And this force produces an electric field between the charges. The direction of an electric field is from positive charge towards negative charge.


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The opposite charges existing near the junction creates a potential difference (voltage) across the junction. The electric field between the charges is responsible to produce potential difference across the junction. This potential difference has a fixed polarity and it acts as a barrier to the flow of electrons and holes, across the junction. Hence this potential is called barrier potential, junction potential or built-in potential barrier of a p-n junction.

The barrier potential is expressed in volts. Its value is called height of the barrier. It is denoted as V1 or Vbi.

Barrier potential also indicates the amount of voltage with proper polarity, to be applied across the p-n junction, to restart the flow of electrons and holes across the junction.

The barrier potential is approximately 0.7 V for silicon and 0.3 V for germanium, at 25C.

The barrier potential of p-n junction mainly depends on the following factors:1. The type of semiconductor used. 2. The concentration of donor impurity on n side. 3. The concentration of acceptor impurity on p side.4. The intrinsic concentration of basic semiconductor. 5. The temperature.

The entire behaviour of unbiased p-n junction i.e. open circuited p-n junction is shown in the figure (a).

Fig. (a) Open circuited p-n junction

Important observation about depletion region:


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The width of the depletion region depends on the amount of doping on n side and p side. If the two sides are equally doped, the width of the depletion region is equal on both sides as shown in figure (a). But if n-side is heavily doped as compared to p-side, then depletion region is observed more on p-side as shown in the figure (b). If p-side is heavily doped as compared to n-side, then depletion region is observed more on n-side as shown in the figure (c).

(a) Both sides equally doped (b) n-side heavily doped (c) p-side heavily doped

Figure

The depletion region penetrates more on the lightly doped side.

Expression for the Barrier Potential

Let us derive the expression for the potential barrier.

It is known that the change in concentration, induces the voltage and is given by,

At the junction, there is a abrupt change in the concentration of holes from pp to pn.

So p1 = pp NA and p2 = pn

Substituting above we get,

……(1)

where VJ is called potential barrier, junction potential or contact potential.

But according to law of mass action,


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……(2)

But nn ND

….(3)

This equation gives the value of junction potential. It can be seen from this equation that the junction potential depends on voltage equivalent of temperature i.e. VT and the amount of doping of n-side and p-side i.e. ND and NA.

4. Explain the energy band structure of open circuited P-N junction.

Energy Band Structure of Open Circuited P-N Junction

It is known that the Fermi level in n type material lies just below the conduction band while in p type material, it lies just above the valence band. When p-n junction is formed, the diffusion starts. The charges get adjusted so as to equalize the Fermi level in the two parts of p-n junction. This is similar to the adjustment of water levels in two tanks of unequal level, when connected to each other. The charges flow from p to n and n to p side till, the Fermi level on the two sides get lined up.

The transfer of charges does not disturb the relative positions of conduction band, valence band and Fermi level in any region either p or n.

The transfer of charges and energy band structure showing equalization of Fermi levels in p and n regions is shown in the figure (a).


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Fig. (a) Energy band structure of p-n junction

In p region, the Fermi level EFp is near EVp just above edge of valence band. In n region, the Fermi level EFn is near ECn just below edge of conduction band. And there is difference between levels of EFn and EFp. The transport of charges, the edge of conduction band ECp in the p type material becomes higher than ECn in the n type material. Similarly the edge of valence band EVp in p type is higher than EVn in n type material. Thus there is a shift of E1 in the Fermi level on p side while there is a shift of E2 in the Fermi level on n side from their intrinsic levels. (i.e. centre of EC

and EV). This adjusts the Fermi level on n and p side to get equivalent Fermi level EF for the p-n junction.

The total shift in energy E0 is E1 + E2 which is responsible to produce contact difference of potential across the junction. This is nothing but barrier potential or junction potential or contact potential given by,


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It is known, that NA =PP and ND nn

While from law of mass action, Pn = and nP=

Using in the equation of VJ,

This is alternative expression for VJ where ‘o’ indicates the thermal equilibrium condition.

Contact difference potential is also denoted by Vo.

5. Explain the operation of forward biased diode.

Forward Biasing of p-n Junction Diode

If an external d.c. voltage is connected in such a way that the p-region terminal is connected to the positive of the d.c. voltage and the n-region is connected to the negative of the d.c. voltage, the biasing condition is called forward biasing. The p-n junction is said to be forward biased.

Forward biasing means connecting p-region to positive and n region to negative of the battery.

The figure (a) shows the connection of forward biasing of p-n junction. To limit the current, practically a current limiting resistor is connected in series with the p-n junction diode. The figure (b) shows the symbolic representation of a forward biased diode.

(a) Forward biasing (b) Symbolic representation


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Operation of Forward Biased Diode

When the p-n junction is forward biased as long as the applied voltage is less than the barrier potential, there cannot be any conduction.

When the applied voltage becomes more than the barrier potential, the negative terminal of battery pushes the free electrons against barrier potential from n to p region. Similarly positive terminal pushes the holes from p to n region. Thus holes get repelled by positive terminal and cross the junction against barrier potential. Thus the applied voltage overcomes the barrier potential. This reduces the width of depletion region.

As forward voltage is increased, at a particular value the depletion region becomes very much narrow such that large number of majority charge carriers can cross the junction.

The large number of majority carriers constitute a current called forward current. Once the conduction electrons enter the p-region, they become valence electrons. Then they move from hole to hole towards the positive terminal of the battery. The movement of valence electrons is nothing but movement of holes in opposite direction to that of electrons, in the p-region. So current in the p-region is the movement of holes which are majority carriers. This is the hole current. While the current in the n-region is the movement of free electrons which are majority carriers. This is the electron current. Hence the overall forward current is due to the majority charge carriers. The action is shown in the figure (c). These majority carriers can then travel around the closed circuit and a relatively large current flows. The direction of flow of electrons is from negative to positive of the battery. While direction of the conventional current is from positive to negative of the battery as shown in the figure (c).

Fig. (c) Forward current in a diode

The direction of flow of electrons and conventional current is opposite to each other.

Effect on the Depletion Region


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Due to the forward bias voltage, more electrons flow into the depletion region, which reduces the number of positive ions. Similarly flow of holes reduces the number of negative ions. This reduces the width of the depletion region. This is shown in the figure (d).

(i) Unbiased diode (ii) Forward biased diode

Depletion region narrows due to forward bias voltage.

Effect of the Barrier Potential

Under the influence of applied forward bias voltage, the free electrons get the energy equivalent to the barrier potential so that they can easily overcome the barrier, which is a sort of a hill, and cross the junction. While crossing the junction, the electrons give up the amount of energy equivalent to the barrier potential. This loss of energy produces a voltage drop across the p-n junction which is almost equal to the barrier potential.

The polarities of voltage drop across the p-n junction in forward biased condition are opposite to that of barrier potential but the value is almost equal to the barrier potential.

Due to the internal resistance, there is additional small voltage drop across the diode.

Thus the total voltage drop across a p-n junction diode in a forward biased condition is V f

and it is made up of

1. Drop due to barrier potential2. Drop due to internal resistance.

The total Vf is of the order of 0.7 V for silicon and 0.3 V for the germanium.

6. Explain the Operation of reverse biased diode.

Reverse Biasing of p-n Junction Diode


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If an external d.c. voltage is connected in such a way that the p-region terminal of a p-n junction is connected to the negative of the battery and the n-region terminal of a p-n junction is connected to the positive terminal of the battery, the biasing condition is called reverse biasing of a p-n junction.

Reverse biasing means connecting p-region to negative and n-region to positive of the battery.

The figure (a) shows the connection of a reverse biasing of a p-n junction while the figure (b) shows the symbolic representation of a reverse biased diode.

(a) Reverse biasing (b) Symbolic representation

Operation of Reverse Biased Diode

When the p-n junction is reverse biased the negative terminal attracts the holes in the p-region, away from the junction. The positive terminal attracts the free electrons in the n-region away from the junction. No charge carrier is able to cross the junction. As electrons and holes both move away from the junction, the depletion region widens. This creates more positive ions and hence more negative charge in the n-region. This is because the applied voltage helps the barrier potential. This is shown in the figure.

Fig. Depletion region widens in reverse bias

Reverse biasing increases the width of the depletion region.


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As depletion region widens, barrier potential across the junction also increases. However, this process cannot continue for long time. In the steady state, majority current ceases as holes and electrons stop moving away from the junction.

The polarities of barrier potential are same as that of the applied voltage. Due to increased barrier potential, the positive side drags the electrons from p-region towards the positive of battery. Similarly negative side of barrier potential drags the holes from n-region towards the negative of battery. The electrons on p-side and holes on n-side are minority charge carriers, which constitute the current in reverse biased condition. Thus reverse conduction takes place.

The reverse current flows due to minority charge carriers which are small in number. Hence reverse current is always very small.

The generation of minority charge carriers depends on the temperature and not on the applied reverse bias voltage. Thus the reverse current depends on the temperature i.e. thermal generation and not on the reverse voltage applied.

For a constant temperature, the reverse current is almost constant though reverse voltage is increased upto a certain limit. Hence it is called reverse saturation current and denoted as Io.

Reverse saturation current is very small of the order of few microamperes for germanium and few nanoamperes for silicon p-n junction diodes.

The reverse current and its direction is shown in the figure (a).

(i) Flow of minority charge carriers (ii) Direction of reverse current

Fig. Reverse biased diode

The reverse biasing produces a voltage drop across the diode denoted as VR which is almost equal to applied reverse voltage.

7. Explain the Current components in a p-n junction diode.

The Current Components in a p-n Junction Diode


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It is indicated earlier that when a p-n junction diode is forward biased a large forward current flows, which is mainly due to majority carriers. The depletion region near the junction is very very small, under forward biased condition. In forward biased condition holes get diffused into n side from p-side while electrons get diffused into p-side from n side. So on p-side, the current carried by electrons which is diffusion current due to minority carriers, decreases exponentially with respect to distance measured from the junction. This current due to electrons, on p-side which are minority carriers is denoted as Inp. Similarly holes from p side diffuse into n-side carry current which decreases exponentially with respect to distance measured from the junction. This current due to holes on n side, which are minority carriers is denoted as Ipn. If distance is denoted by x then,

Inp (x) = Current due to electrons in p side as a function of x

Ipn (x) = Current due to holes in n side as a function of x

At the junction i.e. at x – 0, electrons crossing from n side to p side constitute a current, I np

(0) in the same direction as holes crossing the junction from p side to n side constitute a current, Ipn (0).

Hence the current at the junction is the total conventional current I flowing through the circuit.

Now Ipn (x) decreases on n side as we move away from junction on n side. Similarly Inp(x) decreases on p side as we move away from junction on p side.

But as the entire circuit is a series circuit, the total current must be maintained at I, independent of x. This indicates that on p side there exists one more current component which is due to holes on p side which are the majority carriers. It is denoted by Ipp (x) and the addition of the two currents on p side is total current I.

Ipp(x) = Current due to holes in p side.

Similarly on n side, there exists one more current component which is due to electrons on n side, which are the majority carriers. It is denoted as Inn(x) and the addition of the two currents on n side is total current I.

Inn(x) = Current due to electrons in n side.


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Fig. (a) Current components

These current components are plotted as a function of distance in the figure (a).

The current Ipp decreases towards the junction, at the junction enters the n side and becomes Ipn which further decreases exponentially. Similarly the current Inn decreases towards the junction, at the junction enters the p side and becomes Inp which also further decreases exponentially.

In forward bias condition, the current enters the p side as a hole current and leaves the n side as an electron current, of the same magnitude.

So sum of the currents carried by electrons and holes at any point inside the diode is always constant equal to total forward current I. But the proportion due to holes and electrons in constituting the current varies with the distance, from the junction.

8. Explain the V-I characteristics of a diode.

The Volt-Ampere (V-I) Characteristics of a Diode

The response of a diode when connected in an electrical circuit, can be judged from its characteristics known as Volt-Ampere commonly called V-I characteristics. The V-I characteristics in the forward biased and reverse biased condition is the graph of voltage across the diode against the diode current.

Forward Characteristics of p-n Junction Diode

The response of p-n junction can be easily indicated with the help of characteristics called V-I characteristics of p-n junction. It is the graph of voltage applied across the p-n junction and the current flowing through the p-n junction.


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Fig. (a) Forward biased diode

The figure (a) shows the forward biased diode. The applied voltage is V while the voltage across the diode is Vf. The current flowing in the circuit is the forward current If. The graph of forward current If against the forward voltage Vf across the diode is called forward characteristics of a diode.

The forward characteristics of a diode is shown in the figure (b).

Fig. (b) Forward characteristics of a diode Basically forward characteristics can be divided into two regions:

1. Region O to P: As long as Vf is less than cut-in voltage (Vf), the current flowing is very small. Practically this current is assumed to be zero.

2. Region P to Q and onwards: As Vf increases towards V the width of depletion region goes on reducing. When Vf exceeds V i.e., cut-in voltage, the depletion region becomes very thin and current If increases suddenly. This increase in the current is exponential as shown in the figure by the region P to Q.

The point P, after which the forward current starts increasing exponentially is called knee of the curve.

The normal forward biased operation of the diode is above the knee point of the curve, i.e. in the region P-Q.

The forward current is the conventional current, hence it is treated as positive and the forward voltage Vf is also treated positive. Hence the forward characteristics is plotted in the first quadrant.


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Forward Resistance of a Diode

The resistance offered by the p-n junction diode in forward biased condition is called forward resistance. The forward resistance is defined in two ways:

1. Static Forward Resistance:

This is the forward resistance of p-n junction diode when p-n junction is used in d.c. circuit and the applied forward voltage is d.c. This resistance is denoted as RF and is calculated at a particular point on the forward characteristics.

Thus at a point E shown in the forward characteristics, the static resistance RF is defined as the ratio of the d.c. voltage applied across the p-n junction to the d.c. current flowing through the p-n junction.

2. Dynamic forward resistance:

The resistance offered by the p-n junction under a.c. conditions is called dynamic resistance denoted as rf.

The dynamic resistance is reciprocal of the slope of the forward characteristics.

Consider the change in applied voltage from point A to B shown in the figure. This is denoted as V. The corresponding change in the forward current is from point C to D. It is denoted as L. Thus the slope of the characteristics is . The reciprocal of the slope is dynamic resistance rf.

Generally the value of rf is very small of the order of few ohms, in the operating region i.e. above the knee.

Reverse Characteristics of p-n Junction Diode

The figure (c) shows the reverse biased diode. The reverse voltage across the diode is ZVR

while the current flowing is reverse current IR flowing due to minority charge carriers. The graph of IR against VR is called reverse characteristics of a diode.


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Fig. (c) Reverse biased diode

The polarity of reverse voltage applied is opposite to that of forward voltage. Hence in practice reverse voltage is taken as negative. Similarly the reverse saturation current is due to minority carriers and is opposite to the forward current. Hence in practice reverse saturation current is also taken as negative. Hence the reverse characteristics is plotted in the third quadrant as shown in the figure (d).

Figure (d)

Typically the reverse breakdown voltage is greater than 50 V for normal p-n junctions.

As reverse voltage is increased, reverse current increases initially but after a certain voltage, the current remains constant equal to reverse saturation current I0 though reverse voltage is increased. The point A where breakdown occurs and reverse current increases rapidly is called knee of the reverse characteristics. Reverse Resistance of a Diode

The p-n junction offers large resistance in the reverse biased condition called reverse resistance. This is also defined in two ways.

1. Reverse static resistance: This is reverse resistance under d.c. conditions, denoted as Rf. It is the ratio of applied reverse voltage to the reverse saturation current I0.


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2. Reverse dynamic resistance: This is the reverse resistance under the a.c. conditions, denoted as rr. It is the ratio of incremental change in the reverse voltage applied to the corresponding change in the reverse current.

The dynamic resistance is most important in practice whether the junction is forward or reverse biased.

Complete V-I Characteristics of a Diode

The complete V-I characteristics of a diode is the combination of its forward as well as reverse characteristics. This is shown in the figure (e).

Fig. (e) Complete V-I characteristics of a diode In forward characteristics, it is seen that initially forward current is small as long as the bias

voltage is less than the barrier potential. At a certain voltage close to barrier potential, current increases rapidly. The voltage at which diode current starts increasing rapidly is called as cut in voltage. It is denoted by V. Below this voltage, current is less than 1% of maximum rated value of diode current. The cut-in voltage for germanium is about 0.2V while for silicon it is 0.6 V.

It is important to note that the breakdown voltage is much higher and practically diodes are not operated in the breakdown condition. The voltage at which breakdown occurs is called reverse breakdown voltage denoted as VBR.

Reverse current before the breakdown is very very small and can be practically neglected.


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V-I Characteristics of Typical Ge and Si Diodes

The combined forward and reverse characteristics is called V-I characteristics of a diode. As mentioned earlier, the barrier potential for germanium (Ge) diode is about 0.3 V while for Silicon (Si) diode is as about 0.7 V. The potential at which current starts increasing exponentially is also called offset potential, threshold potential or firing potential of a diode. The figure shows the V-I characteristics of typical Ge and Si diodes.

Fig. V-I characteristics of typical Ge and Si diodes

The reverse saturation current in a germanium diode is about 1000 times more than the reverse saturation current in a silicon diode of a comparable rating. The reverse saturation current I0 is of the order of nA for silicon diode while it is of the order of A for germanium diode. Reverse breakdown voltage for Si diode is higher than that of the Ge diode of a comparable rating.

9. Write in detail about the two types of capacitances associated with a diode.

Transition Capacitance (CT)

Consider a reverse biased p-n junction diode as shown in the figure (a)


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Fig. (a) Transition capacitance in reverse biased condition

As seen earlier, when a diode is reverse biased, reverse current flows due to minority carriers. Majority charged particles i.e. electrons in n-region and holes in p-region move away from the junction. This increases the width of the depletion region. The width of the depletion region increases as reverse bias voltage increases. As the charged particles move away from the junction there exists a change in charge with respect to the applied reverse voltage. So change in charge dQ with respect to the change in voltage dV is nothing but a capacitive effect. Such a capacitance which comes into the picture under reverse biased condition is called transition capacitance, space-charge capacitance, barrier capacitance or depletion layer capacitance and denoted as CT. The magnitude of CT is given by the equation,

This capacitance is very important as it is not constant but depends on the magnitude of the reverse voltage.

Diffusion Capacitance

During forward biased condition, an another capacitance comes into existence called diffusion capacitance or storage capacitance, denoted as CD.

In forward biased condition, the width of the depletion region decreases and holes from p side get diffused in n side while electrons from n side move into the p-side. As the applied voltage increases, concentration of injected charged particles increases. This rate of change of the injected charge with applied voltage is defined as a capacitance called diffusion capacitance.


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The diffusion capacitance can be determined by the expression

where = mean life time for holes.

So diffusion capacitance is proportional to the current. For forward biased condition, the value of diffusion capacitance is of the order of nano farads to micro farads while transition capacitance is of the order of pico farads. So CD is much larger than CT.

However in forward biased condition, CD appears in parallel with the forward resistance which is very very small. Hence the time constant which is function of product of the forward resistance and CD is also very small for ordinary signals.

Hence for normal signals CD has no practical significance but for fast signals CD must be considered.

The graph of CD against the applied forward voltage is shown in the figure (a).

Fig. (a) Diffusion capacitance versus applied forward biased voltage

10. What is transition capacitance and derive its necessary expressions?

Transition Capacitance (CT)

Consider a reverse biased p-n junction diode as shown in the figure (a)


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Fig. (a) Transition capacitance in reverse biased condition

As seen earlier, when a diode is reverse biased, reverse current flows due to minority carriers. Majority charged particles i.e. electrons in n-region and holes in p-region move away from the junction. This increases the width of the depletion region. The width of the depletion region increases as reverse bias voltage increases. As the charged particles move away from the junction there exists a change in charge with respect to the applied reverse voltage. So change in charge dQ with respect to the change in voltage dV is nothing but a capacitive effect. Such a capacitance which comes into the picture under reverse biased condition is called transition capacitance, space-charge capacitance, barrier capacitance or depletion layer capacitance and denoted as CT. The magnitude of CT is given by the equation,

This capacitance is very important as it is not constant but depends on the magnitude of the reverse voltage.

Derivation of Expression for Transition Capacitance

Consider a p-n junction diode, the two sides of which are not equally doped. Impurity added on one side is more than the other. Assume that p-side is lightly doped and n-side is heavily doped. As depletion region penetrates lightly doped side, the most of depletion region is on p-side as it is lightly doped as shown in the figure (b).


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Fig. (b) Unequally doped p-n junction diode

It can be further assumed that concentration of acceptor impurity on p-side (NA) is much less than the concentration of donor impurity on n-side (ND). Hence the width of depletion region on n-side is negligibly small compared to width of depletion region on p-side. Hence the entire depletion region can be assumed to be on the p-side only.

The relationship between potential and charge density is given by Poisson’s equation as,

…(1)

where x = the distance measured from the junction

and = the permittivity of the semiconductor

= 0 r …(2)

where 0 = permittivity of free space

=

and r = relative permittivity of the semiconductor

= 16 for germanium

= 12 for silicon

Note: In Poisson’s equation, the concentration of lightly doped side is used. If we assume that n-type is lightly doped compared to p-type then as ND less than NA, Poisson’s equation modifies to,


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Integrating equation (1) w.r.t. x we get,

….(3)

Assume constant of integration as zero.

Now is the electric field intensity over the region 0 to W over which depletion region is

spreaded.

…. (4)

where E is electric field intensity.

To get the potential, integrating equation (3) we get,

….(5)

At x = W, V = VB which is barrier potential

Now barrier potential is the difference between internally developed junction potential and externally applied bias voltage.

….(6)

where VB is barrier potential and V must be taken as negative for reverse bias.

Substituting in equation (5) we get,

….(7)

From the above equation it can be observed that,


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…(8)

The width of barrier i.e. depletion layer increases with applied reverse bias.

If A is the area of cross-section of the junction, then net charge Q in the distance W is

Q = Number of charged particle charge on each particle

Q = [NA Volume] q

Q = NA A Wq …(9)

Now differentiating equation (7) with respect to V,

…(10)

….(11)

Now differentiating equation (9),

But is the transition capacitance CT hence

….(13)

Now from equation (6) we know that VB = VJ – V and for reverse bias V is negative. Hence for reverse biased condition we get VB = VJ + V where V is applied reverse biased voltage. So as reverse biased voltage increases, VB increases. From equation (8), we can conclude that the width of depletion layer increases as reverse bias increases. Increasing width W, deceases the transition capacitance CT. Hence transition capacitance CT decreases as the reverse bias voltage increases.


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……(14)

As the reverse biased applied to the diode increase, the width of the depletion region (W) increases. Thus the transition capacitance CT decreases. In short, the capacitance can be controlled by the applied voltage. The variation of CT with respect to the applied reverse bias voltage is shown in the figure (c).

Fig. (c) CT against reverse voltage

As reverse voltage is negative, graph is shown in the second quadrant. For a particular diode shown, CT varies from 80 pF to less than 5 pF as VR changes from 2 V to 15 V.

It can be observed that VB = VJ – V where,

VJ = Internally developed junction potential V = Applied reverse bias voltage

For reverse bias, V is negative hence VB = VJ + V for reverse bias. Thus barrier potential increases and hence width of depletion layer increases as reverse bias increases.

11. What is Diffusion capacitance and derive the expression for diffusion capacitance?

Diffusion Capacitance

During forward biased condition, an another capacitance comes into existence called diffusion capacitance or storage capacitance, denoted as CD.

In forward biased condition, the width of the depletion region decreases and holes from p side get diffused in n side while electrons from n side move into the p-side. As the applied voltage increases, concentration of injected charged particles increases. This rate of change of the injected charge with applied voltage is defined as a capacitance called diffusion capacitance.


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The diffusion capacitance can be determined by the expression

where = mean life time for holes.

So diffusion capacitance is proportional to the current. For forward biased condition, the value of diffusion capacitance is of the order of nano farads to micro farads while transition capacitance is of the order of pico farads. So CD is much larger than CT.

However in forward biased condition, CD appears in parallel with the forward resistance which is very very small. Hence the time constant which is function of product of the forward resistance and CD is also very small for ordinary signals.

Hence for normal signals CD has no practical significance but for fast signals CD must be considered.

The graph of CD against the applied forward voltage is shown in the figure (a).

Fig. (a) Diffusion capacitance versus applied forward biased voltage

Derivation of Expression for Diffusion Capacitance

In a p-n junction, the total current at the junction (x = 0) is given by,

I = Ipn(0) + Inp (0) A ….(1)


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where Ipn(0) = Current due to holes diffusing from p to n side Inp(0) = Current due to holes diffusing from n to p side

For simplicity assume that the one side say p side is heavily doped with respect to other. Hence Inp(0) is negligible compared to Ipn(0).

I = Ipn (0) …(2)For the diffusion current, the current density is given by,

….(3)

where Dp = Diffusion constant

= Concentration gradient (Change in concentration with respect to x)

Now J = Current density =

…(4)

where A = Area of cross-section

Now for a p-n junction,

….(5)

where LP = Diffusion length for holes

This LP is related to the diffusion constant DP such that LP = where = mean life

time of charge carrier.

Differentiating (5)

….(6)

Using in (4),


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….(7)

At x = 0, IP(x) = Ipn(0) = I hence using x = 0 in equation (7),

….(8)

….(9)

Now the excess minority charge Q exists only on n side and given by,

…(10)

Using (9) in (10),

But

Q = I …(11)

Now ….(12)

From (11), ….(13)

From diode current equation,


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…(14)

Using (13) and (14) in (12),

This is the required expression for diffusion capacitance.

12. Write a short note on temperature dependence of diode characteristics.

Effect of Temperature on p-n Junction Diode

It has been mentioned earlier that the reverse saturation current I0 depends on temperature while VT is voltage equivalent of temperature is also temperature dependent. The diode current involving I0 and VT is hence temperature dependent.

The overall diode characteristics depends on the temperature.

The dependence of I0 on temperature T is given by,

…..(1)

where K = constant independent of temperature (Not the Boltzmann’s constant)

m = 2 for Ge and 1.5 for Si

and VG0 = Forbidden energy gap = 0.785 V for Ge and 1.21 V for Si

Now as temperature increases, the value of I0 increases and hence the diode current increases. To keep diode current constant it is necessary to reduce the applied voltage V of the diode.

Let us calculate, with what rate the applied voltage must be changed in order to keep the

diode current constant. For a constant diode current, So we have to calculate such a

change in voltage for which

A diode current is given by the equation,

I = I0


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For a forward current, neglecting 1 we get,

….(2)

Substituting equation (1) in (2) we get,

1 = K Tm

1 = K Tm ….(3)

as VT = kT where k is Boltzmann’s constant,

1 = K Tm ….(4)

Now for constant diode current, hence differentiating equation (4) with respect to T,

Note that, VG0 is forbidden energy gap at 0K and hence constant from differentiation point of view.

Taking Tm outside and ( k T2) as L.C.M. we get,

Replacing kT = VT we get,


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…..(5)

Now for constant diode current hence equating the equation (5) to zero we get,

…(6)

…(7)

….(8)

This is the required change in voltage necessary to keep diode current constant.

Hence for germanium, at cut-in voltage V = V = 0.2 V and with m = 2, = 1, T = 300K and VG0 = 0.785 V in equation (8) we get,

…(9)

The negative sign indicates that the voltage must be reduced at a rate of 2.12 mV per degree change in temperature to keep diode current constant.

Similarly for Si we get,

….(10)

Practically the value of is assumed to be – 2.5 mV/C for either Ge or Si at room

temperature.

Thus,

The negative sign indicates that decreases with increase in temperature.


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The equation (8) represents dependence of forward voltage on temperature. It contains two

terms. The term is due to dependence of VT on temperature while the other negative term is due

to temperature dependence of I0 and does not depend on voltage V across the diode. The

equation states that for increasing V, becomes less negative and reaches zero at V = VG0 + m

VT. The practical diodes show such behaviour in the operating region. Effect of Temperature on Reverse Saturation Current

Let us now study by what rate I0 changes with respect to temperature. Consider equation (1) again,

I

Taking logarithm of both sides we get,

Using VT = kT,

….(11)

Differentiating this equation with respect to T we get,

Replacing kT = VT,

…(12)

For germanium, substituting the values of various terms at room temperature we get,

This indicates that I0 increases by 11% per degree rise in temperature.


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For silicon, we get,

This indicates that I0 increases by 8% per degree rise in temperature.

But experimentally it is found that the reverse saturation current I0 increases by 7% per C change in temperature for both silicon and germanium diodes. If at TC I0 is 1 A then at (T + 1) C it becomes 1.07 A and so on. From this it can be concluded that reverse saturation current approximately doubles i.e. 1.0710 for every 10C rise in temperature.

The equations (8) and (12) explain the overall temperature dependence of diode characteristics.

Mathematical Interpretation: The above result can be mathematically represented as,

…(13)

where I02 = Reverse saturation current at T2

I01 = Reverse saturation current at T1

Thus the equation (13) is easy to use, than using equation (1), to obtain the effect of temperature on I0.

The temperature dependence of I0 for Si and Ge diodes is shown in the figure (a) and (b).

The temperature dependence is approximately same for both the types of diodes. It can be seen that at high temperatures, Ge diode produces excessively large reverse current while for Si diode the I0 is much smaller. So for rise in temperature from 25C to 90C the I0 increases to 100 A for Ge diode while it increases to only some tenths of A for Si diode.


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The various other parameters like cut in voltage, reverse breakdown voltage and forward current are also temperature dependent. The effect of temperature on these parameters is shown in the figure (c).

Fig. (c) Effect of temperature on diode

13. Explain the Breakdown mechanisms in a diode.


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Breakdown Mechanisms in Diode

Fig. (a) Diode characteristics

Consider the diode characteristics as shown in the figure (a). In reverse biased condition as long as the reverse voltage is less than breakdown voltage, the diode current is small and almost constant at I0. But when reverse voltage increases beyond certain value, large diode current flows. This is called breakdown of diode and corresponding voltage is called reverse breakdown voltage VBR of diode.

There are two distinct mechanisms due to which the breakdown may occur in the diode. These are, 1. Avalanche breakdown 2. Zener breakdown

Avalanche Breakdown

As seen earlier, the applied reverse bias causes a small reverse current I0 to flow in the device. This is due to movement of minority charge particles, viz., electrons from the p-material and holes from the n-material. The polarity of reverse bias voltage is such that only the minority charge particles are able to cross the p-n junction, while the majority charge particles move away from the junction. As the applied reverse bias voltage becomes larger, the minority charge carriers increasingly accelerate. There are collisions between these particles and electrons involved in the covalent bonds of the crystal structure.

If the applied voltage is such that the traveling electrons do not have high velocity, then the collisions take some energy away from them, altering their velocity. If the applied voltage is

increased, the velocity and hence the kinetic energy of electron increases. If such

an electron dashes against an electron involved in covalent bond, then the collision gives bond-valence electron enough energy to enable it to break its covalent bond. Thus, one electron by collision creates an electron-hole pair. These secondary particles are also accelerated and participate in collisions that generate new electron-hole pairs. This phenomenon is known as carrier multiplication. Electron-hole pairs are generated so quickly and in such large number that there is an apparent avalanche or self-sustained multiplication process. At this stage junction is said to be in breakdown and current starts increasing rapidly.


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The figure (b) shows the process of carrier multiplication which is a geometric progression 1,2, 4, ……..

Fig. (b) Carrier multiplication in avalanche breakdown

The series resistance must be used to limit the large reverse current flowing during breakdown.

The diodes having reverse breakdown voltage greater than 6V, show the avalanche mechanism of breakdown. The avalanche breakdown occurs for lightly doped diodes.

Zener Breakdown

The breakdown of a p-n junction may occur because of one more effect called zener effect. When a p-n junction is heavily doped the depletion region is very narrow. So under reverse bias conditions, the electric field across the depletion layer is very intense. Electric field is voltage per distance and due to narrow depletion region and high reverse voltage, it is intense. Such an intense field is enough to pull the electrons out of the valence bands of the stable atoms. So this is not due to the collision of carriers with atoms. Such a creation of free electrons is called zener effect which is different than the avalanche effect. These minority carriers constitute very large current and mechanism is called zener breakdown.

The zener effect is dominant for heavily doped diodes. For heavily doped diodes, the depletion region width is small. The field intensity for voltages less than 5 V becomes intense, of the order of 0.3 MV/cm. Such an intense field causes zener effect to pull the charge carriers from parent atoms and make them available as free carriers.

The diodes having reverse breakdown voltage less than 5 V show the zener mechanism of breakdown. This occurs for heavily doped diodes.

For the diodes having reverse breakdown rating between 5 V to 6 V, both zener and avalanche mechanisms occur simultaneously. The breakdown is due to combination of the two. The breakdown voltage rating of a diode can be adjusted by changing the doping levels in the junction, at the time of manufacturing. This decides the practical safe operating voltage rating of a diode caused Peak Inverse Voltage (PIV) rating of a diode. Practically in reverse biased condition, to avoid reverse breakdown the voltage appearing across the diode must be less than its PIV rating.


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14. Compare Zener breakdown and Avalanche breakdown.

Comparison of Breakdown Mechanisms

Zener Breakdown Avalanche Breakdown

1.

Breaking of covalent bonds is due to intense electric field across the narrow depletion region. This generates large number of free electrons to cause breakdown.

Breaking of covalent bonds is due to collision of accelerated charge carriers having large velocities and kinetic energy with adjacent atoms. The process is called carrier multiplication.

2.This occurs for zener diodes with VBR less than 6 V.

This occurs for zener diodes with VBR greater than 6 V.

3. The temperature coefficient is negative. The temperature coefficient is positive.

4.The breakdown voltage decreases as junction temperature increases.

The breakdown voltage increases as the junction temperature increases.

5.The V-I characteristics is very sharp in breakdown region.

The V-I characteristics is not as sharp as zener breakdown in breakdown region.

6. Occurs for heavily doped diodes. Occurs for lightly doped diodes.7.

15. Compare Zener diode and p-n junction diode.

Comparison of Zener Diode and p-n Junction Diode

No. Zener diode P-N junction diode

1. Operated in reverse breakdown condition.Operated in forward biased condition and never operated in reverse breakdown condition.

2. The characteristics likes in third quadrant. The characteristics lies in first quadrant.


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3.Dynamic zener resistance is very small in reverse breakdown condition.

The diode resistance in reverse biased condition is very high.

4.

Zener diode symbol is, The p-n junction diode symbol is,

5.The conduction in zener is opposite to that of arrow in the symbol, as operated in breakdown region.

The conduction when forward biased is in same direction as that of arrow in the symbol, when forward biased.

6.The power dissipation capability is very high.

The power dissipation capability is very low compared to zener diodes.

7.Applications of zener diode are voltage regulator, protection circuits, voltage limiters etc.

Applications of p-n junction diode are rectifiers, voltage multipliers, clippers, clampers and many electronic devices.

16. Explain the Characteristics and equivalent circuit of a Zener diode.

The zener diode is a silicon p-n junction semiconductor device, which is generally operated in its reverse breakdown region. The zener diodes are fabricated with precise breakdown voltages, by controlling the doping level during manufacturing. The zener diodes have breakdown voltage range from 3V to 200V. In 1934, a physicist Carl Zener investigated the breakdown phenomenon in the p-n junction diode.

The figure (a) shows the symbol of zener diode. The d.c. voltage can be applied to the zener diode so as to make it forward biased or reverse biased. This is shown in the figure (b) and (c). Practically zener diodes are operated in reverse biased mode.


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(a) Symbol (b) Forward biasing (c) Reverse biasing

Fig. Zener diode

Characteristics of Zener Diode

In the forward biased condition, the normal rectifier diode and the zener diode operate in similar fashion. But the zener diode is designed to be operated in the reverse biased condition. In reverse biased condition, the diode carries reverse saturation current till the reverse voltage applied is less than the reverse breakdown voltage. When the reverse voltage exceeds reverse breakdown voltage, the current through it changes drastically but the voltage across it remains almost constant. Such a breakdown region is a normal operating region for a zener diode. The normal operating regions for a rectifier diode and a zener diode are shown in the figure (a) and (b).

Fig. (a) Operating regions shown Fig. (b) Operating regions shown shaded for normal diode shaded for zener diode

The breakdown characteristic for a zener diode is significantly important, as it is an operating region for the diode. When the reverse voltage applied to a zener diode is increased, initially the current through it is very small, of the order of few A or less. This is the reverse leakage current of the diode, denoted by Io. At a certain reverse voltage current through zener diode increases rapidly. The change from a low value to large value of current is very sharp and well defined. Such a sharp change in the reverse characteristics is called knee or zener knee of the curve. At this knee, a breakdown is said to occur in the device. The reverse bias voltage at which the breakdown occurs is called zener breakdown voltage, denoted as VZ.


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The voltage VZ is set by carefully controlling the doping level during manufacturing process.

The current corresponding to a knee point is called zener knee current and it is a minimum current zener must carry to operate in reverse breakdown region. It is denoted as IZK or IZmin.

From the bottom of the knee, the zener breakdown voltage remains almost constant, though it increases slightly as the zener current IZ, increases. The current at which the nominal zener breakdown voltage is specified is called zener test current, denoted as IZT. This value and corresponding zener voltage VZ are specified on a datasheet of a zener diode. Every zener diode has a capacity to carry current. As current increases, the power dissipation PZ = VZ IZ increases. If this dissipation increases beyond certain value, the diode may get damaged.

Fig. (c) V-I characteristics of zener diode

The maximum current a zener diode can carry safely is called zener maximum current and is denoted as IZM or IZmax.

In practical circuits to limit the zener current between IZmin and IZmax, a current limiting resistor is used in series with the zener diode.

The complete V-I characteristics of the zener diode is shown in the figure (c).

Equivalent Circuit of Zener Diode

When the breakdown occurs then IZ may increase from IZmin to IZmax but voltage across zener remains constant. Hence actually the internal zener impedance decreases as current increases in the zener region. But this impedance is very small. Hence ideally the zener diode is indicated by a battery of voltage VZ, which remains fairly constant in the zener region. This is shown in the figure (d).


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Practically though very small, zener has its internal resistance. In the zener region, this resistance is called dynamic resistance of the zener denoted as rZ. Practically zener region is not exactly vertical. The small change in zener current IZ produces a small change in zener voltage VZ. The ratio of VZ to IZ is called zener resistance rZ. This is shown in the figure (e). Hence practically zener equivalent circuit is shown with a battery of VZ alongwith a series resistance rZ as indicated in the figure (f).

(e) Dynamic resistance (f) Practical equivalent circuit

From the graph the dynamic resistance is defined as,

This value is specified generally at zener test current IZT. In most of the cases this value is almost constant over the full range of zener region i.e. from IZmin to IZmax. It is of the order of few tens of ohms.

17. With the help of a neat diagram explain use of Zener diode as a voltage regulator.

The various applications of zener diode are, 1. As a voltage regulating element in voltage regulators.2. In various protection circuits. 3. In zener limiters i.e. clipping circuits which are used to clip off the unwanted portion of the

voltage waveform.


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Let us see how zener diode can be used as a voltage regulator. The figure (a) shows use of a zener diode to regulate a varying d.c. input voltage. This is called input regulation or line regulation, using zener diode.

Fig. (a) Line regulation using zener

As the input voltage varies, the zener current IZ changes. But the zener diode maintains constant voltage across the output terminals, over the certain range. The limitations on the input variations are set by the minimum (IZK) and maximum (IZM) zener current values, with which zener can operate in its breakdown region. The resistance R is used as a current limiting resistor.

For example for a particular zener if IZK = 5 mA and IZM = 50 mA, VZ = 6.8 V and R = 1k then for the minimum current the voltage across R is,

VR = IZK R = 5 10-3 1 103 = 5VNow VR = Vin - VZ

Vin = VR + VZ = 5 + 6.8 = 11.8 V

For the maximum current the voltage across R is, VR = IZM R = 50 10-3 1 103 = 50V

Vin = VR + VZ = 53 + 6.8 = 56.8 V

This shows that the zener diode can regulate an input voltage approximately from 11.8 V to 56.8 V and maintains it at 6.8 V at the output. The output voltage varies slightly due to the changes in the zener impedance due to the changes in the zener current. But as these changes are very small, can be neglected.


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UNIT – IV

PART - A

1. What is bipolar junction transistor?

A bipolar junction transistor is a three terminal semiconductor device in which the operation depends on the interaction of both majority and minority carriers and hence the name bipolar. It is used in amplifiers and oscillator circuits and as a switch in digital circuits.

2. What is a PNP and NPN transistors?

A transistor in which two blocks of P-type semiconductor separated by a thin layer of n – type semiconductor is know as PNP transistor.

A transistor in which two blocks of n-type semiconductor’s are separated by a thin layer of p-type semiconductor is known as NPN transistor.

3. Mention the different configurations of a transistor.

The transistor can be connected in three different ways

(i) Common Base configuration (CB)(ii) Common Emitter configuration (CE)(iii) Common Collector configuration (CC)

4. Give the relationship between ,.


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5. Draw the Ebers – Moll model for a PNP transistor and give the equations for emitter current and collector current.

Fig.

N Current gain in normal operation ICO Collector junction reverse saturation current

I inverted current gain IEOEmitter junction reverse saturated current.

6. What a reverse gate voltage of 12V is applied to a JFET, the gate current is 1 MA? Determine the resistance between gate and source.


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Solution:

VGS=12V

7. Draw the transfer characteristics of both enhancement and depletion type MOSFET on the same graph.

8. Calculate the values of IC and IE for a transistor dc=0.97 and ICBO=10A and IB is measured as 50A.

Solution:

IC=1.95mAIE=IC+IB

IE=1.95 10-3+5010-6

IE=210-3AIE=2mA

9. Depletion MOSFET is commonly known as “Normally – ON – MOSFET” why?

The depletion MOSFET can conduct even if the gate to source voltage (Vgs) is zero. Because of the reason depletion MOSFET is commonly known as “Normally – ON – MOSFET”

10. What are all internal capacitance in MOSFET?

The internal capacitance in MOSFET are

(a) Gate capacitance


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(b) Source – drain depletion capacitance

11. When a transistor is used as a switch, in which regions of output characteristic it is operated?

The transistors is used as a switch in cut off region and saturation region, transistor carry heavy current hence considered as ON State. In cut off it carry no current and it is equivalent to open switch.

12. Draw the low frequency hybrid model of BJT in CB configuration.

13. Mention the three regions that are present in the drain – source characteristic of JFET.

The three regions that are present in the drain – source characteristic of JFET are atomic region, constant current region (or) pinch off region and breakdown region.

14. Mention the characteristics of CC configuration.

The voltage gain is less than unity The current gain is high for low values of RL

The input resistance is the highest of all the configurations The output resistance is the lowest of all the configurations.

15. What is zener break down?

Zener break down takes place when both sides of the junction are very heavily doped and consequently the depletion layer is thin. When a smaller reverse bias voltage is applied a very strong electric field is set up across the tin depletion layer. This electric field is enough to break the covalent bonds. Now extremely large number of the charge carriers are produced which constitute the zener current. This process is known as zener break down.

16. Name the special features of a FET.

(i) High input resistance (ii) Low noise (iii) Better thermal stability (iv) High power gain


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(v) High frequency response.

17. How is a FET used as VVR?

At low voltages, the depletion regions are thin and the drain current increases with voltages. So, in the region where voltage is less than pinch – off voltage (VP), FET is behaving as a voltage Resistor (VVR). That is the drain to source resistance is controlled by VGS.

18. What is early effect?

As the collector voltage VCC is made to increase the reverse bias, the space change width between collector and base lends to increase, with the result that the effective width of the base decreases. This dependency of base-width on collector – to – emitter voltage is known as the early effect.

19. Define the current amplification factor of transistor in CB, CE and CC configuration.

For CB configuration the current amplification factor is given by “”

For common emitter the current amplification factor

For common –collector the current amplification factor

20. “CC” configuration is called an emitter follower circuit?

In a common collector circuit, the o/p voltage is in phase with the input voltage and also the same in magnitude. Thus emitter voltage follows the input voltage in step and hence the name emitter – follower.

21. Why hybrid equivalent circuit is widely used for small signal low frequency applications?

The hybrid parameters are widely used because They can be measured easily


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They are more independent of each other They h- parameters are real numbers at audio frequencies They are practically suitable for circuit analysis and design and are specified by the

transistor manufactures.

22. Give the shockly’s equation for FET?

The shockly’s equation give the relation between drain current [IO] in the pinch off region and the gate to source voltage VGS.

where,

IDSS Maximum value of drain current when VGS=0VP pinch – of voltage.

23. What is the difference between a FET and conventional transistor?

FET BJT(i) It is a chipolar device It is a bipolar device(ii) It is a voltage controlled devices It is a current controlled device(iii) It’s input resistance is very high It’ s input resistance is very low(iv) It is less noisy It is comparatively move noisy(v) No thermal run away There is thermal run away(vi) High switching speed Lower switching speed

24. Define pinch off voltages in FET.

Pinch off voltage is the minimum drain to source voltage where the drain current approach.

Constant value. Beyond the pinch off voltage the channel width can not be reduced.

25. List out some typical applications of FET.

(i) FET can be used in phase shift oscillation to minimize the loading effect. (ii) FET can be employed as a buffer amplifier for isolation between input and output.

26. State the two types of MOSFET. State also the modes in which they can operate.

The two types of MOSFET are


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(i) N-Channel MOSFET(ii) P-Channel MOSFET

The modes in which they can operate are

(i) Depletion mode:- In this mode the gate is maintained at negative potential with respect to source

(ii) Enhancement mode:- In this mode both the gate and drain are maintained at positive potential with respect to source.

27. Mention the points of superiority of FET’s over BJT’s.

The FET has the following advantage’s over BJT

1. The noise level is very low in FET since there are no junctions2. FET has very high power gain3. Offers perfect isolation between input and output since it has very high input impedance. 4. FET is a negative temperature co-efficient device hence avoids thermal run away.

28. Name the hybrid parameters of a bipolar junction transistor.

The hybrid parameters of a bipolar junction translators are

(i) hi Input impedance(ii) hoOutput admittance(iii) hfForward current gain (iv) hr Reverse voltage gain

29. Define drain resistance.

Drain resistance (rd) is defined as the ratio of small change in drain to source voltage (VDs) to the corresponding change in drain current ID at constant gate to source voltage VGS

30. Define “Transconductance” of JFET.

Transconductance (gm) is defined as the ratio of small change in drain current (Io) to the corresponding change in gate to source voltage ( VGS) at constant drain to source voltage (VDS)


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31. When the reverse gate voltage of JFET changes from 4.0 to 3.9V, the drain current changes from 1.3 to 1.6 M.A. Find the value of transconductance?

Solution:

VGS=4.0-3.9=0.1VID=1.6-1.3=0.3MA

32. Define “Amplification Factor” of FET.

Amplification factor () is defined as the ratio of small change in drain to source voltage (VdS) to the corresponding change in gate to source voltage (VGS) at a constant drain current

(ID)

33. What ratings limit the operation of a transistor?

The ratings limit the operation of a transistors

(i) Maximum voltage(ii) Maximum current(iii) Power handling capacity(iv) Minimum saturation voltage

34. Why is a coupling capacitor is used to connect a signal source to an amplifier?

A capacitor blocks d.c. signals and passes a.c. signals The primary function of the coupling capacitor connected between a signal source and an amplifier is to block unwanted d.c. signals and allow a.c. signals which are to be amplified.

35. Draw the output characteristics for a CE configuration and mark the cut off and saturation region.


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36. Differentiate enhancement and Depletion MOSFET.

Enhancement Depletion MOSFET(i) positive voltage at the gate Negative voltage at the gate(ii) Inversion layer is mode Depletion of majority carriers happens(iii) Negative charges are formed Positive charges are formed

37. Why is the input impedance of a MOSFET higher than that of a FET?

A thin layer of metal oxide (normally silicon dioxide) is deposited over the left side of the channel. A metallic gate is mounted on the oxide layer. As the oxide layer is an insulator, therefore, gate is insulated from the channel. The layer acts as a capacitor and increase the input impedence of a MOSFET. This arrangement is not present in the FET.

38. Define Threshold voltage.

The gate voltage at which the channel is formed to let through, the flow of current I D of predefined small value is called the gate source threshold voltage VGST(or) VT.

39. Compare JFET and MOSFET.

JFET and MOSFET differs in two aspects

1. JFET can be operated only in the depletion mode where as MOSFET can be operated in both modes i.e. enhancement and depletion.

2. The gate current of JFET is larger even if it is operated with a reserve bias on the junction

3. The MOSFET’s are easier to manufacture than JFET’s are easier to manufacture than JFET’s and hence MOSFET’s are more widely used than JFET’s

40. Draw the diagram for small signal low frequency FET model.


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41. Define delay time and rise time in the switching characteristics of transistor.

The time that elapses during this delay added with the time required for the current to rise to 10% of its maximum value is called the delay time (tr).

42. Define the hie and hfe for a common emitter transistor configuration.

Input impedance [hie]:-

It is defined as the ratio of the change in (input) base voltage to the change in (input) base current with the (output) collector voltage VCE kept constant.

Forward Current gain [hfe]:-

It is defined as a ratio of the change in collector current to the corresponding change in the base current keeping the collector voltage VCE constant hence,

43. Define the hoe and hrc for a CE transistor configuration.

Output admittance [hoe]:-

It is defined as the ratio of change in the collector current to the corresponding change in the collector voltage with the base current ID kept constant.

44. Define fall time and storage time.


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The fall time is specified as the time required for IC to go from 90% to 10% of its maximum level.

The storage time is the result of charge carriers being trapped in the depletion region when a junction polarity is reversed.

45. What is knee point?

When VDS = 0, there is no attracting potential at the drain and hence IB = 0, although the channel between the gate is fully open as VGS = 0 as VOD is increased, the drain current ID

increases linearly up to a knee point, this shows that FET behaves like an ordinary resistor till the point is reached.

PART - B

1. Draw the circuit diagram of an NPN junctions transistor in CE configuration and justify for the shape of the static input and output characteristics. Mark the active, saturation and cut off regions.

CE Configuration:- This is also called grounded emitter configuration. In this configuration base is the i/P terminal, collector is the o/p terminal and emitter is the common terminal.


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Input characteristics:-

To determine the input characteristics, the collector to emitter voltage is kept constant at zero volt and base current is increased from zero in equal steps by increasing VBE in the circuit shown in fig. (a)

When VCE =0, the emitter – base junction is forward biased and the junction behaves as a forward biased diode. Hence the input characteristic for VCE=0 is similar to that of a forward – biased diode. When vCE is increased, the width of the depletion region at the reverse biased collector – base junction will increase. Hence the effective width of the base will decrease. This effect causes a decrease in the base current IB. Hence to get the same value of IB as that for VCE=0, VBE should be increased therefore the curve shifts to the right as VCE increases.

Output characteristics:-

To determine the output characteristics, the base current IB is kept constant at a suitable value by adjusting base – emitter voltage, VBe the magnitude of collector emitter voltage VCE is increased in suitable equal steps from zero and the collector current IC is noted.


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Fig.

For each setting VCE. Now the circles of IC VS VCE are plotted for different constant values of IB. The outpur characteristics thus obtained are shown in fig. © from equation

for longer values of VCE, due to early effect, a very small

change in is reflected in a very large change in . For e.g. =0.98, If

increases to 0.985, then =66. Here is a slight increase in by about 0.5% results in increases in by about 34% Hence the o/p characteristics of CE configuration shows a larger slope taken compared with CB configuration.

The o/p characteristics have three regions namely. Saturation region, cut off region and cut of region.

2. Explain the operation of NPN transistor.

As shown in fig. the forward bias applied to the emitter base junction of an NPN transistor causes a lot of electrons from the emitter region to cross over to the base region. As the base is lightly doped with P-type impurity, the number of holes in the base region is very small and hence the number of electrons that combine with holes in the P – type base region is also very small. Hence a few electrons combine with holes to constitute a base current IB. The remaining electrons (more than 95%) crossover into the collector region to constitute a collector current IC. Thus the base and collector current summed up give the emitter current i.e. IE=-(IC+IB).


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Fig. Current in NPN transistor

In the external circuit of the NPN bipolar junction transistor, the magnitudes of the emitter current IE, the base current IB and the collector current IC are related by IE=IC+IB.

3. Explain the operation of PNP transistor.

As shown in fig. the forward bias applied to the emitter – base junction of a PNP transistor causes a lot of hoses from the emitter regions to cross over to the base region as the base is lightly doped with N-type impurity. The number of electrons in the base regions is very small and hence the number of holes combined with electrons in the N – type base region is also very small. Hence a few holes combined with electrons to constitute a base current IB.

Fig. Current in PNP transistor

The remaining holes ( more than 95%) cross over in to the collector region to constitute a collector current IC. Thus the collector and base current when summed up gives the emitter current. i.e. IE=- (IC+IB).

In the external circuit of the PNP bipolar junction transistor, the magnitudes of the emitter current IE, the base current IB and the collector current IC are related by


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IE=IC+IB

The equation gives the fundamental relationship between the currents in a bipolar transistor circuit. Also, this fundamental equation shows that there are current amplification factors and in common base transistor configuration and common emitter transistor configuration respectively for the static (d.c) currents, and for small changes in the currents.

Large – signal current gain (). The large signal current gain of a common base transistor is defined as the ratio of the negative of the collector – current increment to the emitter – current change from cut off (IE=0) to IE,i.e.

where ICBO (or ICO) is the reverse saturation current flowing through the reverse biased collector – base junction. i.e. the collector to base leakage current with emitter open. As the magnitude of ICBO is negligible when compared to IE, the above expression can be written as

Since IC and IE are flowing in opposite directions, is always positive. Typical value of ranges from 0.90 to 0.995. Also, is not a constant but varies with emitter current IE, collector voltage VCB and the temperature.

4. Compare the three types of transistor configuration.

Table A comparison of CB CE and CC configurations

Property CB CE CCInput resistance Low (about 100) Moderate (about 750 ) High (about 750 k)Output resistance High (about 450 ) Moderate (about 45 ) Low (about 25)Current gain 1 High HighVoltage gain About 150 About 500 Less than 1Phase shift 0 or 360o 180o 0 or 360o

Between input & output voltages Applications

For high frequency circuits

For audio frequency circuits

For impedance matching

5. Describe the two types of breakdown in transistors.

There is a possibility of voltage breakdown in the transistor at high voltages even through the rated dissipation of the transistor is not exceeded. Therefore, These is an upper limit to the maximum allowable collector function voltage. There are two type of breakdown, namely.

(1) Avalanche multiplication (or) avalanche breakdown


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(2) reach- through (or) punch – through

Avalanche breakdown and multiplication:

When a diode is reverse biased, there is a limit on the voltage that can be applied which is the avalanche voltage. Similarly, in the transistor, the maximum reverse biasing voltage which may be applied before breakdown between the collector and base terminals with the emitter open is called breakdown voltage BVCBO. Therefore, an upper limit is set on the collector voltage VCB by avalanche breakdown in the reverse biased collector – base junction.

Breakdown may occur because of avalanche multiplication of the current ICo that crosses the collector junction. As a result of this multiplication, the current becomes MICO where M is the avalanche multiplication, factor. At the breakdown voltage BVCBO, multiplication factor M becomes infinite and the current rises abruptly in the breakdown region as shown in fig. (a), there will be large changes in current with small changes in applied voltage.

Fig.

The avalanche multiplication factor depends on the voltage VCB between collector and base, which has been found to be given empirically by

As a result, in the presence of avalanche multiplication the current gain of CB transistor has become M. For the CE configuration, the collector to emitter breakdown voltage BVCEO with base open is


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In general, BVCEO is 40% to 50% of BVCBO. This is the upper limit of VCE that can be placed across the transistor without damaging it.

Reach – Through (or) punch – Through:

According to Early effect, the width of the collector junction transistor region increases with increased collector – junction voltage as the voltage applied across the junction increases the transition region penetrates deeper into the base and will have spread. Completely across the base to reach the emitter junction as the base is very thin. Thus the collector voltage has reached through the base region this effect. Known as reach through.

It is possible to raise the punch- through voltage by increasing the doping concentration in the base, but this automatically reduces the emitter efficiency punch through taken place at a fixed voltage between collector and base and is not dependent on circuit configuration, where as avalanche multiplication takes place at different voltages depending upon the circuit configuration therefore the voltage limit of a particular transistor is determined by either of the two types of breakdown which ever occurs at lower voltage. .

6. Derive the drain current equation.

EXPRESSION FOR SATURATION DRAIN CURRENT

For the transfer characteristics, VDS is maintained constant at a suitable value greater than the pinch – off voltage VP. The gate voltage VGS is decreased from zero till IP is reduced to zero. The transfer characteristics ID versus VGs is shown in fig. The shape of the transfer characteristic is very nearly a parabola. It is found that the characteristic is approximately represented by the parabola.


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where IDS is the saturation drain current, IDSS is the value of IDS when VGS when VGS=0, and VP is the pinch – off voltage. Differentiating Eq. with respect to VGs we can obtain an expression for gm.

We know that VDS is constant

Therefore,

Now from Eqn. we have

Substituting this value in Eqn. we get,

Suppose gm=gmo, when VGS=0, then from Eqn.

Equation shows that gm varies as the square root of the saturation drain current IDS, and Eqn. shows the gm decreases linearly with increase of VGS.

7. With the help of suitable diagram explain the working of an n- channel enhancement MOSFET?

MOSFET is the common term for the insulated Gate field effect Transistor (IGFET) there are two basic forms of MOSFET.

(I) Enhancement MOSFET(II) Depletion MOSFET


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Principle:

By applying transverse electric field across an insulator, deposited on the semi conducting material the thickness and hence the resistance of a conducting channel of a semi conducting material can be controlled. Enhancement MOSFET Construction:

The construction of N- channel enhancement MOSFET is shown in fig (a) and the circuit symbols for an N-channel and a p-channel enhancement MOSFET are shown in fig. (b) & (c) respectively. As there is no continuous channel in an enhancement MOSFET, this condition is represented by the broken line in the symbols.

Two highly doped N* regions are diffused in a lightly doped substrate P-type silicon substrate. One N* region is called the source S and the other one is called the drain D. They are separated by 1 mil (10-3 inch). A thin insulating layer of Sio2 is grown over the surface of the structure and hole are get into the oxide layer allowing contact with source and drain.

Fig.

A thin layer of metal aluminum is formed over the layer of Sio2. This metal layer covers the entire channel region and it forms the gate G.

The metal area of the gate, in conjunction with the insulating oxide layer of Sio2 and the semiconductor channel forms a parallel plate capacitor. This device is called the insulated gate FET because of the insulating layer of Sio2. This layer give extremely high input impedance for the MOSFE.

Operation:-

If the substrate is grounded and a positive voltage is applied at the gate, the positive charge on G induces an equal negative charge on the substrate side between the sources and drain regions thus an electric field is produced between the source and drain regions.


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Fig.Drain Characteristic:

The direction of the electric field is perpendicular to the plates of the capacitor through the oxide. The negative charge of electrons which are minority carriers in the p-type substrate forms a inversion layer. As the positive voltage on the gate increases, the induced negative charge in the semiconductor increases. Hence the conductivity increases and current flows from source to drain through the induced channel. Thus the drain current in enhanced by the positive gate voltage as shown in fig. (d).

Fig.

8. Explain the construction, working principle and characteristics of n-channel depletion MOSFET.

Principle: By applying a transverse electric field across an insulator, deposited on the semi conducting material the thickness and hence the resistance of a conducting channel of a semi conducting material can be controlled.

In a depletion MOSFET, the controlling electric field reduces the number of majority carriers available for conduction. Whereas in the enhancement MOSFET. Application of electric field causes an increase in the majority carrier density in the conducting regions of the transistor.


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Construction and Working:

The construction of an N-channel depletion MOSFET is shown in fig.(a) where an N channel is diffused between the source and drain to the basic structure of MOSFET.

With VGS =0 and the drain D at a positive potential with respect to the source, the electrons (majority carriers) flow through the N – channel from S to D.

Fig. N-channel depletion MOSFET

Fig. Drain characteristics

Therefore the conventional current ID flows through the channel D to S. If the gate voltage is made negative positive charge consisting of holes is induced in the channel through Sio2 of the gate channel capacitor. The introduction of the positive charge causes depletion of mobile electrons in the channel. Thus a depletion region is produced in the channel. The shape of the depletion region depends on VGS. Hence the channel will be wedge shaped as shown in fig. (a)


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when VDS is increased ID increases and it becomes practically constant at a certain value of VDS

called the pinch off voltage. The drain current ID almost gets saturated beyond the pinch off voltage. Since the current in a FET is due to majority carriers, the induced positive charges make the channel less conductive, and IDdrops as VGS is made negative.

The depletion MOSFET may also be operated in an enhancement mode. It is only necessary to apply a positive gate voltage so that negative.

The depletion MOSFET may also be operated in an enhancement mode. It is only necessary to apply a positive gate voltage so that negative charges are induced in to the N- type channel. Hence the conductivity of the channel increases and ID increases. As the depletion MOSFET can be operated with bipolar input signals irrespective of doping of the channel. It is also called as duel mode MOSFET. The volt – ampere characteristics are indicated in fig. (b) and, The curve of ID versus VGS for constant VDS is call Transfer characteristics of MOSFET and id shown in fig. (c).

9. Compare JFET & MOSFET.

1. In enhancement and depletion types of MOSFET, the transverse electric field induced across an insulating layer deposited on the semiconductor material control the conductivity of the channel. In the JFET the transverse electric field across the reverse biased PN junction controls the conductivity of the channel.

2. The gate leakage current in a MOSFET is of the order of 10 -12A. Hence the input resistance of a MOSFET is very high in order of 1010 to 1015. The gate leakage current of a JFET is of the order of 10-9A and its input resistance is of the order of 108.

3. The output characteristics of the JFET are flatter than those of the MOSFET and hence, the drain resistance of a JFET (0.1 to 1M) is much higher than that of a MOSFET (1 to 50 k).

4. JFETs are operated only in the depletion mode. The depletion type mode. The depletion type MOSFET may be operated in both depletion and enhancement mode.

5. Comparing to JFET, MOSFETs are easier to fabricate. 6. MOSFET is very susceptible to over load voltage and needs special handling during

installation. It gets damaged easily if it is not properly handled. 7. MOSFET has zero offset voltage. As it is a symmetrical deice, the source and drain can be

interchanged. These two properties are very useful in analog signal switching. 8. Special digital CMOS circuits are available which involve near – zero power dissipation and

very low voltage and current requirements. This makes them most suitable for portable systems.

MOSFETs are widely used in digital VLSI circuits then JFETs because of their advantages

10. Compare JFET and BJT.


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1. FET operation depends only on the flow of majority carriers – holes for P – channel FETs and electrons for N-channel FETs. Therefore, they are called Unipolar devices. Bipolar transistor (BJT) operation depends on both minority and majority current carriers.

2. As FET has no junctions and the conduction is through an N-type or P-type semiconductor materials, FET is less noisy than BJT.

3. As the input circuit of FET is received biased, FET exhibits a much higher input impedance (in the order of 100M) and lower output impedance and there will be a high degree of isolation between input and output. So, FET can act as an excellent buffer amplifier but the BJT has low input impedance because its input circuit is forward biased.

4. FET is a voltage controlled device, i.e voltage at the input terminal controls current, whereas BJT is a current controlled device, i.e. the input current controls the output current.

5. FETs are much easier to fabricate and are particularly suitable for ICS because they occupy less space than BJTs.

6. The performance of BJT is degraded by neutron radiation because of the reduction in minority – carrier lifetime, whereas FET can tolerate a much higher level of radiation since they do not rely on minority carriers for their operation.

7. The performance of FET is relatively unaffected by ambient temperature changes. As it has a negative temperature coefficient at high current levels, it prevents the FET from thermal breakdown. The BJT has a positive temperature coefficient at high current levels which leads to thermal breakdown. The BJT has a positive temperature coefficient at high current levels which leads to thermal breakdown.

8. Since FET does not suffer from minority carrier storage effects, it has higher switching speeds and cut-off frequencies. BJT suffers from minority carrier storage effects and therefore has lower switching speed and cut – off frequencies.

9. FET amplifiers have low gain bandwidth product due to the junction capacitive effects and produce more signal distortion except for small signal operation.

10.BJTs are cheaper to product than FETs.

11. Explain the construction and operation of n-channel JFET.

A junction field effect consist of a p-type or n-type silicon bar. The bar is the conducting channel for the charge carriers. If the bar is made up of N-type material it is known as N-channel FET and if the bar is made up of p-type material it is known as p-channel FET. To form a JFET, two junction diodes are connected internally. The three terminals are namely source, gate and drain.

Operation of N-channel JFET:


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When VGS=0 and VDS=0 when no voltage is applied between drain and source, and gate source, and gate source, the thickness of the depletion regions round the PN junction is uniform as shown in fig. (a)

Fig.

When VDS=0 and VGS is decreased from zero.

In this case the PN junctions are reverse biased and hence he thickness of the depletion region increases. As VGS is decreased from zero, the reverse bias voltage across the PN junction is increased and hence, the thickness of the depletion region in the channel until the two depletion regions make contact with each other. In this condition, the channel is said to be cut-off. The value of VGS which is required to cut – off the channel is called the cut-off voltage VC.

When VGS =0 and VDS is increased from zero.

Drain is positive with respect to the source with VGS=0. Now the majority carriers flow through the N-channel from source to drain. Therefore the conventional current ID flows from drain to source As VDS is increased the cross sectional area of the channel will be reduced. At a certain value of VP of VDS, the cross sectional area becomes minimum. At this voltage, the channel is said to be pinched off and the drain voltage VP is called the pinch – off voltage.

(i) As VDS is increased from zero. ID increases along VP, and the rate of increases of ID with VDS decrease as shown in fig. (c) The region from VDS=0 to VDS=VP is called the holmic region.

(ii) When VDS=VP ID becomes maximum when VDS is increased beyond VP, the length of the pinch – off or saturation region increases. Hence, there is no further increase of ID.

(iii) At a certain voltage corresponding to the point B, ID suddenly increases

When VGS is negative and VDS is increased


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Fig.

When the gate is maintained at a negative voltage less than the negative cut off voltage. The reverse voltage cross the junction is further increased. Hence for a negative value of VGS, the curve of ID versus VDS is similar to that for VGS=0 but the values of VP and BVDGO are lower as shown in fig. (c).

Drain characteristics

From the curves, it is seen that above the pinch off voltage, at a constant value of VDS, ID

increases with an increases of VGS. Hence a JFET is suitable for use as a voltage amplifier, similar to a transistor amplifier.


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UNIT - V

Tunnel Diode, Pin Diode, SCR characteristic and two transistor equivalent model –UJT , DIAC & Triac Laser, CCD, Photodiode, Phototransistor, Photo conductive Cell & Photo voltaic cell –LED & LCD

PART – A

1. What is Tunnel Diode or Esaki Diode?

A tunnel diode is a two terminal negative resistance device that can be employed as an amplifier, an oscillator or switch. Some times called as Esaki diode after its inventor.

2. What is tunneling effect?

A tunnel diode is a heavily doped semiconductor material so the depletion region is very narrow. It does not constitute a large barrier; even a small forward or reverse bias the charge carrier sufficient energy to cross the depletion region. This effect is known as tunnelloy effect.

3. What is negative resistance ?

In normal resistance, the relation between voltage and current are linear. In negative resistance devices a rise in current while there is decrease in device voltage or vice versa. This kind of devices is employed in oscillator(eg) tunnel diode , UJT

4. State some application of tunnel diode.

1. Ultra high speed switch2. Logic memory device.3. Microwave oscillator4. Relaxation oscillator circuit5. Amplifier

5. What is Pin Diode?

The pindiode has heavily doped p-type and n-type region separated by Intrinsic region. When it is reverse biased acts like an capacitor, and when forward biased it behaves as a variable resistor.

6. Draw PIN diode diagram.


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P I N

N typeP typeSemiconductort

IntrinsicPure semicondor


7. State some application of PIN DIODE.

It is used as a modoizhing device for AC signal. It is used in microwave switching applications.

8. What is Unijunction Transistor?

Lightly doped N-type siliconbar with a heavily doped p type material alloyed to its one side. Three terminal semiconductor switching device has only one PN Junction, hence it is commonly called as unijunction transistor, (UJT),It is used as a control switch and Relaxation oscillator.

9. Define Intrinsic standoff ration .

one typical value of n ranges from 0.56 to 0.75

10. Explain the difference between UJT and Bipolar Transistor.

BIOPOLAR JUNCTION UJT

Two junction Base emitter and Collector Base Only one junction emitter and baseConduction due to both majority and minority carrier

Conduction due to only majority carrier

Bipolar device Unipolar deviceUsed as an amplifier and switch Used as an oscillator and switchNo negative resistance region Exhibit negative resistance

11. Name some application of UJT.

UJT can be employed in variety of application U1z saw tooth wave generator, pulse generator, switching , timing and phase control circuits

12. Find Intrinsic stand of ration of a UJT when RBB = 10 KRB2 =3.5 K

RBB =RB1 +RB2

RB1 = 10-35 =6.5 k


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13. What do you mean by Thyristor family?

Thyristor is a semiconductor device having three or more junction. It uses internal feedback for switching action. This device has high voltage and current restrings. Thyristor family consist of PNPN Diode, SCR, TRIAC DIAC, UJT.

14. Draw two transistor model of a SCR.

15. Explain Breakover Voltage, Latching current, holding current.

Breakover Voltage VB0 is the voltage at which the junction J2 breakdown and the lathc (switch) is ON from OFF state latching current. It is the current at which the device is hermed on I L holding current In is the minimum current so that the device is on. Otherwise when the current falls below holding current, the device switch to OFF condition.

16. Draw the diagram of SCR characteristic

17. Why SCR termed as controlled rectifier? How it is different from ordinary diode rectifier circuit?


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In normal rectifier the output of the circuit is either single half or both. No way to control the power.

In SCR we can control the output by controlling the gate trigger circuit and Power output controlled over a wide range, hence it is known as controlled rectifier.

18. Explain the operation of SLR in shortnotes.

The output voltage can be controlled by triggering the Gen voltage at suitable hence -Firing angle at which the SCR conducts. By controlling - we can control the power output to the load.

19. Draw a SCR Halfwave rectifier and its output.

20. Name few Application of SCR.

SCR Rectifier SCR Crowbar SCS – act as a switch

21. What is DIAC?

DIAC is a three layer two terminal device. It acts as bidirectional evalanche diode. Short form of Diode AC (DIAC)

22. What is bidirectional Thysistor?

Thysistor in which conduction takes place in both direction.

23. Draw DIAC Symbol a Basic Structure


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24. What is Triac?

Triac Inode AC Switch. A three terminal semiconductor switching device which can control alternating current in a load. Two SCRs are connected back to back.

25. Draw Basic Structure and Circuit symbol of Triac.

26. Write the application of Triac.

Triac is used for illumination control, temperature control motor speed control, and as static switch to turn as power ON and OFF.

27. Write DIAC application.

DIAC is used as a Triggerly device in Triac phase control circuits used for light, temp, Motor control. DIAC is not a control device.

28. Draw Two SCR version of a Triac.


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29. Draw the triac characteristic.

30. What is Luminescence?

Light emitted from a solid when it is stimulated by the source of incident energy. This phenomenon is called lumincate. If the incident energy in the form of photons(Light) it is called Photo Luminenec.

31. Name two categories of Optoelectronic devices.

(i) Photoconductive device: When radiation is incident on a semiconductor, some absorption of light by the material take place and conductivity increases photoconductive effects.

Eg. Photoresistor – Photodiode, photor

(ii)Photovoltaic cell generates and voltage due to light incident (eg) solar cellPhoto voltaic cell generates a voltage while photo conductive requires an external source.


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32. What is Photo conductive Cell (PC) –LDR?

The photo conductive cell (PC) or detector is two terminal device which is used as light dependent resistor. It is made of thin layer of cadmium sulphide (Cds), lead sulphide (Pbs). It’s resistance decreases with the presence of light and increase in the absent of light.

33. Name few application of photoconductive cell.

Used as light detector for automatic street lighting Switching circuit, Introdor detector Camera shutter opening during the flash.

34. What about Photodiode?

Silicon photodiode is a light sensitive device also called photodetector which converts light signal into electrical signal.

35. Explain about photodiode operation.

When light falls on a window lens fixed on the junction more no of electron hole pair generated due to light incident. The movement of electron-hole pair in a properly connected circuit (Reverse biased condition) increases the magnitude of the circuit current. This current wholly dependent on the light incident. Thus light signal is converted into electrical signal.

36. What is dark current?

There is minimum Reverse leakage current flow in the circuit even when there is no light. This current is called dark current.

37. Write the application of Photodetector photodiode, phototransistor.

Light detector, demodulator, encoder high speed country light operand switches, punched cord reader, sound track films.

38. What is the advantage of phototransistor over photodiode?

The current produced by a photodiode is very low needs amplification for control application. Photo transistor is a sensitive semiconductor. Light detector which combines a photo diodes amplifier. It permits a greater flow of current. The current increased by a factor of

39. What is Photovoltaic effect?


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If the PN junction is open circuited, the light energy is used to create a potential difference proportional to the frequency and intensity of the incident light. This phenomenon is called photovoltaic effect.

40. Explain photovoltaic cell/solar cell.

Photo voltaic cell is a light sensitive semiconductor devices producer and voltage when illuminated which may be used directly to supply small amount of power. When sunlight is incident one photovoltaic cell, it is converted into electrical energy. Such an energy converter is called solar cell or solar battery.

41. Name application of Photo voltaic cell.

Photovoltaic cells are used in low power devices such as light meter. Solar cell are used to generate power in satellites.

42. Details of LED in brief?

LED – light emitting diode is a PN Junction device which emit light when forward biased by a phenomenon of electro luminescence. The excited electron in the higher state move back to its original level. This energy will be radiated as best in most of the diode. In some material Gelium phosphide (GSP), gallium arcnide phosphide (G2ASP), the radiation is mostly of the form light energy. in visible region.

43. What are different LED type and its material?

Gellium arsenide (G2AS) – Infrared radiation (Invisible)Gellium phosphide – Green or RedGellium arsenide phosphide – red or yellow

44. Write the application of LED.

Burglar alarm Picture phone Multimeters Digital meters Calculator Microprocessor Optical Communication devices Electronic telephone exchanges

45. What is ILD Injection Laser Diode?


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When the emitted light is coherent (monochromatic then this LED are known as ILD. ILD is used as optical source – more suitable for high data rate application.

46. Describe Liquid crystal display?(LCD)

The liquid is normally transparent but if subjected to strong electric field, disruption of well ordered crystal structure takes place causing the liquid to polarize and opaque. The removal of applied electric field allows the crystal structure to resign its original form and the maturings become transparent.

47. State the advantage – disadvantage of LCD

Advantages:- The voltage required are very small low power consumption –economicsDisadvantages:- Slow devices turn on – OFF time are quite large

(ii) Used on dc the life span reduced. Therefore they used with ac supplies with frequency less than 50 Hz(iii) Occupy large area

47. Comparison between LED and LCD.

LED LCDConsumes more power requires 10-250 mw per digit

Essentially acts as a capacitor and consumer very less power 10-200 W per digit

It requires external driver circuit due to high power requirement

Can be driver by IC chips

Good brightness level ModerateOperable within temperature range – 40 to 85C

-20 to 60C

Life time is around 100,000 hours Limited to 50,000 hours due to chemical degradation.

Emit light in red, orange, yellow, green and white

Invisible in darkness- requires external illumination

Operating voltage range is 1.5 to 5 V d.c. 3-20V acResponse time is 50 to 500 ns Slow decay time – 50 to 200 msViewing angle 150 Viewing angle 100

49. What is a alphanumeric display?

Display devices provide a visual display of numbers letters and various sign in response to electrical input served as constituent of an electronic display.

(i) Passive Display: Light controllers – they are modulator of light in which the light path pattern gate modified on application of electric field –LCD


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(ii) Active Display – Light emitter –they are generator of light –LED

50. Seven segment Display – Explain

The seven segment display which can display all numerals and nine letters. Each segment can be turned ON/OFF to form the desired digit.

51. What is Charge Coupled device? CCD

A CCD is an analog shift register that enables the transportation of analog signal (electric charges) through successive stages (capacitors)controlled by a clock signal. CCD devices can be used as a form of memory to store optical images. Eg: Image sensor in Digital camera.

52. What is Varactor Diode?

Varactor or variable capacitor diode is also a junction diode with a small impurity at its junction which has the useful property that its junction or transition capacitor and varied electronically.

53. Write various areas where Varactor diode used.

Varactor diode are used in FM radio, TV receiver AFC circuits self adjusting Bridges Circuits, bandpass filter.

54. What is LASER DIODE?

Light Amplification using stimulated emission of Reduction, Similar to LED, laser are used to convert the electrical signal to direct band gap material where high recombination velocity exist, optical gain can be achieved by population inversion of carrier by thro high level current injection by formly a resonant cavity.

55. What are the condition to achieve laser action?


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(i) There must be an inverted population (I.P) move atom in the excited state then the ground state.

(ii) Excited state must be metastable state.

(iii) The emitted photons must is stimulate further emission, This is achieved by the use of reflectry mirror citrine end.

56. Application of Laser Diode.

Telecommunication Fiber optic communication source Barcode Reader Infrared and red laser are common in CDROM, DVD Industrial application such as heat treating cladding, seam welding Image scanning Laser printing

PART – B

1. Explain in detail about Tunnel Diode.

TUNNEL DIODE

The Tunnel or Esaki diode is a thin –junction diode which exhibits negative resistance under low forward bias conditions.

An ordinary PN junction diode has an impurity concentration of about 1 part in 108. With this amount of doping the width of the depletion layer is of the order of 5 microns. This potential barrier restrains the flow of carriers from the majority carrier side to the minority carrier side. If the concentration of impurity atoms is greatly increased to the level of 1 part in 103, the device characteristics are completely changed. The width of the junction barrier varies inversely as the square root of the impurity concentration and therefore, is reduced from 5 microns to less than 100

(10-8 m). This thickness is only about 1/50th of the wavelength of visible light. For such thin

potential energy barriers, the electrons will penetrate through the junction rather than surmounting them. This quantum mechanical barrier is referred to as tunneling and hence, these high-impurity-density PN junction device are called tunnel diodes.

The V-I characteristic for a typical germanium tunnel diode is shown in fig. It is seen that at first forward current rises sharply as applied voltage is increased, where it would have risen slowly for an ordinary PN junction diode (Which is shown as dashed line for comparison). Also, reverse


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current is much larger for comparable back bias than in other diodes due to the thinness of the junction. The interesting portion of the characteristic starts at the point A on the curve, i.e. the peak voltage. As the forward bias is increased beyond this point, the forward current drops and continues to drop until point B is reached. This is the valley voltage. At B, the current starts to increase once again and does so very rapidly as bias is increased further. Beyond this point, characteristic resembles that of an ordinary diode. Apart from the peak voltage and valley voltage, the other two parameters normally used to specify the diode behaviour are the peak current and the peak-to-valley current ratio, which are 2 mA and 10 respectively, as shown.

Fig: V-I characteristic of tunnel diodeThe V-I characteristic of the tunnel diode illustrates that it exhibits dynamic resistance

between A and B. figure shows energy level diagrams of the tunnel diode for three interesting bias levels. The shaded areas show the energy states occupied by electrons in the valence band, whereas the cross hatched regions represent energy states in the conduction band occupied by the electrons. The levels to which the energy states are occupied by electrons on either side of the junctions are shown by dotted lines. When the bias is zero, these lines are at the same height. Unless energy is imparted to the electrons from some external source, the energy possessed by the electrons on the N-side of the junction is insufficient to permit them to climb over the junction barrier to reach the P-side. However, quantum mechanics show that there is a finite probably for the electrons to tunnel through the junction to reach the other side, provided there are allowed empty energy states in the P-side of the junction at the same energy level. Hence, the forward current is zero.


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Fig: Energy Level Diagram of tunnel diode

When a small forward bias is applied to the junction, the energy level of the P-side is lower as compared with the N-side. As shown in fig (b), electrons in the conduction band of the N-side see empty energy level on the P-side. Hence, tunneling from N-side to P-side takes place. Tunneling in other direction is not possible because the valence band electrons on the P-side are now opposite to the forbidden energy gap on the N-side. The energy band diagram shown in figure (b) is for the peak of the diode characteristic.

When the forward bias is raised beyond this point, tunneling will decrease as shown in fig. (C). The energy of the P-side is now depressed further, with the result that fewer conduction band electrons on the N-side are opposite tot the unoccupied P-side energy levels. As the bias is raised, forward current drops. The corresponds to the negative resistance region of the diode


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characteristic. As forward bias is raised still further, tunneling stops altogether and it behaves as a normal PN junction diode.

Equivalent circuit:

The equivalent circuit of the tunnel diode when biased in the negative resistance region is as shown in fig. In the circuit, Rs is the series resistance and Ls is the series inductance which maybe ignored except at highest frequencies. The resulting diode equivalent circuit is thus reduced to parallel combination of the junction capacitance Cj and the negative resistance – Rn. Typical values of the circuit components are Rs =6, Ls =0.1 nH, Cj =0.6 pF and Rn =75.

Fig: Equivalent circuit of tunnel diode

Applications

1. Tunnel diode is used as an ultra-high speed switch with switching speed of the order of ns or ps2. As logic memory storage device3. As microwave oscillator4. In relaxation oscillator circuit5. As an amplifier

Advantages

1. Low noise2. Ease of operation3. High speed4. Low power

Disadvantages

1. Voltage range over which it can be operated is 1 V or less.2. Being a two terminal device, there is no isolation between the input and output circuit.

2. Write in detail about PINDIODE.

PIN diode structure

The PIN diode receives its name from the fact that is has three main layers. Rather than just


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having a P type and an N type layer, the PIN diode has three layers:

1. P-type layer

2. Intrinsic layer3. N-type layer

The instrinic layer of the PIN diode is the one that provides the change in properties when compared to a normal PN junction diode. The intrinsic region comprises of the undoped, or virtually undoped semiconductor, and in most PIN diodes it is very thin - of the order of between 10 and 200 microns.

There are a two main structures that can be used, but the one which is referred to as a planar structure is shown in the diagram. In the diagram, the intrinsic layer is shown much larger than if it were drawn to scale. This has been done to better show the overall structure of the PIN diode.

PIN diode with a planar construction

PIN diodes are widely made of silicon, and this was the semiconductor material that was used exclusively until the 1980s when gallium arsenide started to be used.

PIN diode characteristics

It is found that at low levels of reverse bias the depletion layer become fully depleted. Once fully depleted the PIN diode capacitance is independent of the level of bias because there is little net charge in the intrinsic layer. However the level of capacitance is typically lower than other forms of diode and this means that any leakage of RF signals across the diode is lower.

When the PIN diode is forward biased both types of current carrier are injected into the intrinsic layer where they combine. It is this process that enables the current to flow across the layer.

The particularly useful aspect of the PIN diode occurs when it is used with high frequency signals, the diode appears as a resistor rather than a non linear device, and it produces no rectification or distortion. Its resistance is governed by the DC bias applied. In this way it is possible to use the device as an effective RF switch or variable resistor producing far less distortion than ordinary PN junction diodes.

PIN diode applications

The PIN diode is used in a variety of different applications from low frequencies up to high


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radio frequencies. The properties introduced by the intrinsic layer make it suitable for a number of applications where ordinary PN junction diodes are less suitable.

In the first instance the diode can be used as a power rectifier. Here the intrinsic layer gives it a high reverse breakdown voltage, and this can be used to good effect in many applications.

Although the PIN diode finds many applications in the high voltage arena, it is probably for radio frequency applications where it is best known. The fact that when it is forward biased, the diode is linear, behaving like a resistor, can be put to good use in a variety of applications. It can be used as a variable resistor in a variable attenuator, a function that few other components can achieve as effectively. The PIN diode can also be used as an RF switch. In the forward direction it can be biased sufficiently to ensure it has a low resistance to the RF that needs to be passed, and when a reverse bias is applied it acts as an open circuit The fact that the PIN diode has a low level of capacitance because of the additional intrinsic layer in the diode, means that it can switch more effectively than other forms of diode.

Another useful application of the PIN diode is for use in RF protection circuits. When used with RF, the diode normally behaves like a resistor when a small bias is applied. However this is only true for RF levels below a certain level. Above this the resistance drops considerably. Thus it can be used to protect a sensitive receiver from the effects of a large transmitter if it is placed across the receiver input.

PIN diode attenuator and switch circuit

The circuit above can be used as either a switch or an attenuator. This particular circuit is current driven, although by placing a resistor in series wit the inductor to the switched "+" line, the circuit becomes voltage driven with the resistor limiting the maximum current.

PIN diodes are particularly used in RF applications where there low levels of capacitance and also their switching and variable resistance properties make them very good in switching and variable attenuator applications.

3. Write briefly about Varactor diode.


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VARACTOR DIODE

The varactor or variable capacitor diode, is also a junction diode with a small impurity dose at its junction, which has the useful property that its junction or transition capacitance is easily varied electronically.

When any diode is reverse biased, a depletion region is formed, as seen in figure. The larger the reverse bias applied across the diode, the width of the depletion layer “W” becomes wider. Conversely, by decreasing the reverse bias voltage, the depletion region width “W” becomes narrower. This depletion region is devoid of majority carrier and acts like an insulator preventing conduction between the N and P regions of the diode, just like a dielectric, which separates the two plates of a capacitor. The varactor diode with its symbol is shown in figure.

Fig: Depletion region in a reverse biased PN junction

Fig: Circuit symbol of varactor diode

As the capacitance is inversely proportional to the distance between the plates (CT I/W), the transition capacitance CT varies inversely with the reverse voltage. Consequently, an increase in reverse bias voltage will result in an increase in the depletion region width and a subsequent decrease in transition capacitance CT . At zero volt, the varactor depletion region W is small and the capacitance is large at approximately 600 pF. When the reverse bias voltage across the varactor is 15 V, the capacitance is 30 pF.

The varactor diodes are used in FM radio and TV receivers, AFC circuits, self adjusting bridge circuits and adjustable bandpass filters. With improvement in the type of materials used and construction, varactor diode find application in tuning of LC resonant circuit in microwave frequency multipliers and in very low noise microwave parametric amplifiers.

4. Write in detail Silicon controlled Rectifier SCR and its characteristic.


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SCR (SILICON CONTROLLED RECTIFIER)

The basic structure and circuit symbol of SCR is shown in fig. It is a four layer three terminal device in which the end P-layer acts as anode, the end N-layer acts as cathode and P-layer nearer to cathode acts as gate. As leakage current in silicon is very small compared to germanium, SCRs are made of silicon and not germanium.

Fig: Basic structure and circuit symbol of SCR

Characteristics of SCR The characteristic of SCR are shown in fig. SCR acts as a switch when it is forward biased. When the gate is kept open, i.e. gate current IG =0, operation of SCR is similar to PNPN diode. When IG < 0, the amount of reverse bias applied to J2 is increased. So the breakover voltage VB0 is increased. When IG > 0, the amount of reverse bias applied to J2 is decreased thereby decreasing the breakover voltage. With very large positive gate current breakdown may occur at a very low voltage such that the characteristics of SCR is similar to that of ordinary PN diode. As the voltage at which SCR is switched “ON’ can be controlled by varying the gate current IG, it is commonly called as controlled switch. Once SCR is turned ON, the gate loses control, i.e., the gate cannot be used to switch the device OFF. One way to turn the device OFF is by lowering the anode current below the holding current IH by reducing the supply voltage below holding voltage VH, keeping the gate open.


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Fig: characteristic of SCR

SCR is used in relay control, motor control, phase control, heater control, battery chargers, inverters, regulated power supplies and as static switches.

Two Transistor version of SCR. The operation of SCR can be explained in a very simple way by considering it in terms of two transistors, called as the two transistor version of SCR. As shown in fig, an SCR can be split into two parts and displaced mechanically from one another but connected electrically. Thus the device may be considered to be constituted by two transistors T1

(PNP) and T2 (NPN) connected back to back.

Fig. Two Transistor version of SCR

Assuming the leakage current of T1 to be negligibly small, we obtain

also, from the figure , it is clear that


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Substituting the values given in Eqs(2) and (3) in Eq. (1) , we get

Substituting Eq. (5) in Eq(4) , we obtain

Equation (6) indicates that if i.e. the anode current IA suddenly

reaches a very high value approaching infinity. Therefore, the device suddenly triggers into ON state from the original OFF state. This characteristic of the device is known as its regenerative action.

The value of can be made almost equal to unity by giving a proper value of

positive current Ig for a short duration. This signal Ig applied at the gate which is the base of T2 will cause a flow of collector current IC2 by transferring T2 to its ON state. As IC2 =Ib1, the transistor T1

will also be switched ON. Now, the action is regenerative since each of the transistors would supply base current to the other. At the point even if the gate signal is removed, the device keeps on conducting, till the current level is maintained to a minimum value of holding current.

5. Explain in detail SCR Halfwave Rectifier.

SCR Half-Wave Rectifier:

Though the SCR is basically a switch, it can be used in linear applications like rectification. Figure shows the circuit of an SCR half wave rectifier.


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Fig: SCR half wave rectifier

During the negative halfcycle, the SCR does not conduct irrespective of the gate current, as the anode is negative with respect to cathode and also PIV is less than the reverse breakdown voltage.

During the positive half cycle of a.c. voltage appearing across secondary, the SCR will conduct provided proper gate current is made to flow current, the lesser the supply voltage at which the SCR is triggered ON. Referring to fig. the gate current is adjusted to such a value that SCR is turned ON at a positive voltage V1 of a.c. secondary voltage which is less than the peak voltage Vm, Beyond this, the SCR will be conducting till the applied voltage becomes zero. The angle at which the SCR starts conducting during the positive half cycle is called firing angle . Therefore, the conduction angle is (180-).

The SCR will block not only the negative part of the applied sinusoidal voltage, but will also block the part of positive waveform up to a point SCR is triggered ON. If the angle is zero, this will be an ordinary halfwave rectification. Therefore by proper adjustment of gate current, the SCR


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can be made of conduct full or part of a positive half cycle, thereby controlling the power fed to the load.Analysis Let V=Vm sin t be alternating voltage that appear across the secondary of the transformer. In SCR halfwave rectifier. is the firing angle and the rectifier conducts from to 180 ( radians) during the positive half cycle.

For Here the full positive half cycle will appear across the load. This is the

value of average voltage for ordinary halfwave rectifier.

When This shows that greater the firing angle , the smaller is the

average voltage and vice-versa.

If =0, then

6. What is TRIAC and Sketch it characteristic and describe its operation?

TRIAC (TRIODE A.C SWITCH)


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Triac is a three terminal semiconductor switching device which can control alternating current in a load. Its three terminals are MT1, MT2 and the gate (G). The basic structure and circuit symbol of a Triac are shown in fig. Triac is equivalent to two SCRs connected in parallel but in the reverse direction as shown in fig. So, a Triac will act as a switch for both directions. The characteristics of a Triac are shown in figure.

Like an SCR, a Triac also starts conducting only when the breakover voltage is reached. Earlier to that the leakage current which is very small in magnitude flows through the device and therefore remains in the OFF state. The device, when starts conducting, allows very heavy amount of current to flow through it. The high inrush of current must be limited using external resistance, or it may otherwise damage the device.

Fig: Triac (a) Basic Structure and (b) Circuit symbol


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Fig: Two SCR version of Triac: (a) Basic structure and (b) Equivalent circuit

During the positive half cycle, MT1 is positive with respect to MT2, whereas MT2 is positive with respect to MT1 during negative half cycle. A Triac is a bidirectional device and can be triggered either by a positive or by a negative gate signal. By applying proper signal at the gate, the breakover voltage, i.e. firing angle of the device can be changed; thus phase control process can be achieved.

Fig: characteristics of Triac

Triac is used for illumination control, temperature control, liquid level control motor speed control and as static switch to turn a.c. power ON and OFF. Nowadays the diac-triac pairs are increasingly being replaced by a single component unit known as quadric. Its main limitation in comparison to SCR is its low power handling capacity.

7. Explain DIAC is a bidirectional switch.

DIAC (DIODE A.C. SWITCH)The construction and symbol of diac are shown in figure. Diac is a three layer, two terminal

semiconductor device. MT1 and MT2 are the two main terminals which are interchangeable. It acts as a bidirectional Avalanche diode. It does not have any control terminal. It has two junctions J1

and J2. Though the Diac resembles a bipolar transistor, the central layer is free from any connection with the terminals.

From the characteristic of a Diac shown in fig. it acts as a switch in both directions. As the doping level at the two ends of the device is the same, the Diac has identical characteristics for both positive and negative half of an a.c. cycle. During the positive half cycle, MT1 is positive with respect to MT2 whereas MT2 is positive with respect to MT1 in the negative half cycle. At voltage less than the breakover voltage, a very small amount of current called the leakage current flows through the device and device remains in OFF state. When the voltage level reaches the breakover voltage, the device starts conducting and it exhibits negative resistance characteristics, i.e. the current flowing in the device starts increasing and the voltage across it starts decreasing.


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Fig: Diac: (a) Basic structure and (b) Circuit symbol

Fig: Characteristic of Diac

The Diac is not a control device. It is used as triggering device in Triac phase control circuits used for light dimming, motor sped control and heater control.

8. Draw the equivalent circuit of UJT & Explain its operations.

UJT (UNIJUNCTION TRANSISTOR)

UJT is a three terminal semiconductor switching device. As it has only one PN junction and three leads, it is commonly called as Unijunction transistor.

The basic structure of UJT is shown in fig(a). It consists of a lightly doped N-type Silicon bar with a heavily doped P-type material alloyed to its one side closer to B2 for producing single PN junction. The circuit symbol of UJT is shown in fig. Here the emitter leg is drawn at an angle to the vertical and the arrow indicates the direction of the conventional current.


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Fig: UJT (a) Basic structure (b) Circuit symbol and (c) Equivalent circuit

Characteristics of UJT Referring to Fig.(c), the interbase resistance between B2 and B1 of the silicon bar is RBB = RB1 + RB2. With emitter terminal open, if voltage VBB is applied between the two bases, a voltage gradient is established along the N –type bar. The voltage drop across RB1 is given by V1 =VBB, where the intrinsic stand-ff ration =RB1 /(RB1 + RB2). The typical value of ranges from 0.56 to 0.75. This voltage V1 reverse biases the PN junction and emitter current is cut-off. But a small leakage current flows from B2 to emitter due to minority carriers. If a positive voltage VE is less than V1. If VE exceeds V1 by the cutin voltage V, the diode becomes forward biased. Under this condition, holes are injected into N-type bar. These holes are repelled by the terminal B2 and are attracted by the terminal B1. Accumulation of holes in E to B1 region reduces the resistance in this section and hence emitter current IE is increased and is limited by VE. The device is now in the “ON” state.

If a negative voltage is applied to the emitter, PN junction remains reverse biased, and the emitter current is cut off. The device is now in the “OFF’ state.

Figure shows a fairly of input characteristics of UJT. Here, up to the peak point P, the diode is reverse biased and hence, the region to the left of the peak point is called cut-off region. The UJT has a stable firing voltage VP which depends linearly on VBB and a small firing current IP ( 25A). At P, the peak voltage VP = VBB +V, the diode starts conducting and holes are injected into N-layer. Hence, resistance decreases thereby decreasing VE for the increase in IE. So, there is a negative resistance region from peak point P to valley point V. After the valley point, the device is driven into saturation and behaves like a conventional forward biased PN junction diode. /the region to the right of the valley point is called saturation region. In the valley point, the resistance changes from negative to positive. The resistance remains positive in the saturation region. For very large IE, the characteristic asymptotically approaches the curve for IB2=0.


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Fig: Input characteristics of UJT

A unique characteristic of UJT is, when it is triggered, the emitter current increases regenerative until it is limited by emitter power supply. Due to this negative resistance property, a UJT can be employed in a variety of applications, viz, sawtooth wave generator, pulse generator, switching, timing and phase control circuits.

9. Describe in detail about UJT relaxation Oscillator.

UJT relaxation oscillator The relaxation oscillator using UJT which is meant for generating sawtooth waveform is shown in fig. It consists of a UJT and a capacitor CE which is charged through RE as the supply voltage VBB is switched ON.


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Fig: UJT relaxation oscillator

The voltage across the capacitor increases exponentially and when the capacitor voltage reaches the peak point voltage VP, the UJT starts conducting and capacitor voltage is discharged rapidaly through EB1 and R1. After the peak point voltage of UJT is reached, it provides negative resistance to the discharge path which is useful in the working of the relaxation oscillator. As the capacitor voltage reaches zero, the device then cuts off and capacitor CE starts to charge again. This cycle is repeated continuously generating a sawtooth waveform across CE.

The inclusion of external resistors R2 and R1 in series with B2 and B1 provides spike waveforms. When the UJT fires, the sudden surge of current through B1 causes drop across R1, which provides positive going spikes. Also, at the time of firing, fall of VEBI causes I2 to increase rapidly which generates negative going spikes across R2.

By changing the values of capacitance CE or resistance RE, frequency of the output waveform can be changed as desired, since these values control the time constant RE CE of the capacitor changing circuit.

Frequency of Oscillation. The time period and hence the frequency of the sawtooth wave can be calculated as follows. Assuming that the capacitor is initially uncharged, the voltage VC across the capacitor prior to breakdown is given by


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Where RE CE = charging time constant of resistor-capacitor circuit, and t = time from the commencement of the waveform.

The discharge of the capacitor occur when VC is equal to the peak-point voltage VP , i.e

If the discharge time of the capacitor is neglected, then t =T, the period of the wave.

Therefore, frequency of oscillation of sawtooth wave,

10. Explain about Laser Principle in detail.


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emission. This stimulated emission is the laser transition. Finally, a pulse of red light of wave length 6943 Å emerges through the partially silvered end of the crystal.


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11. What is charge coupled device CCD and explain its operation?

Charge-coupled devices (CCDs) are silicon-based integrated circuits consisting of a dense matrix of photodiodes that operate by converting light energy in the form of photons into an electronic charge. Electrons generated by the interaction of photons with silicon atoms are stored in a potential well and can subsequently be transferred across the chip through registers and output to an amplifier.

CCDs were invented in the late 1960's by research scientists at Bell Laboratories, who initially conceived the idea as a new type of memory circuit for computers. Later studies indicated that the device, because of its ability to transfer charge and the photoelectric interaction with light, would also be useful for other applications such as signal processing and imaging. Early hopes of a new memory device have all but disappeared, but the CCD is emerging as one of the leading candidates for an all-purpose electronic imaging detector, capable of replacing film in the emerging field of digital photomicrography.

Fabricated on silicon wafers much like integrated circuits, CCDs are processed in a series of complex photolithographic steps that involve etching, ion implantation, thin film deposition, metallization, and passivation to define various functions within the device. The silicon substrate is electrically doped to form p-type silicon, a material in which the main carriers are positively charged electron holes. Multiple dies, each capable of yielding a working device, are fabricated on each wafer before being cut with a diamond saw, tested, and packaged into a ceramic or polymer casing with a glass or quartz window through which light can pass to illuminate the photodiode array on the CCD surface.

When a ultraviolet, visible, or infrared photon strikes a silicon atom resting in or near a CCD photodiode, it will usually produce a free electron and a "hole" created by the temporary absence of the electron in the silicon crystalline lattice. The free electron is then collected in a potential well (located deep within the silicon in an area known as the depletion layer), while the hole is forced


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away from the well and eventually is displaced into the silicon substrate. Individual photodiodes are isolated electrically from their neighbors by a channel stop, which is formed by diffusing boron ions through a mask into the p-type silicon substrate.

The principal architectural feature of a CCD is a vast array of serial shift registers constructed with a vertically stacked conductive layer of doped polysilicon separated from a silicon semiconductor substrate by an insulating thin film of silicon dioxide (see Figure 2). After electrons have been collected within each photodiode of the array, a voltage potential is applied to the polysilicon electrode layers (termed gates) to change the electrostatic potential of the underlying silicon. The silicon substrate positioned directly beneath the gate electrode then becomes a potential well capable of collecting locally-generated electrons created by the incident light. Neighboring gates help to confine electrons within the potential well by forming zones of higher potentials, termed barriers, surrounding the well. By modulating the voltage applied to polysilicon gates, they can be biased to either form a potential well or a barrier to the integrated charge collected by the photodiode.

The most common CCD designs have a series of gate elements that subdivide each pixel into thirds by three potential wells oriented in a horizontal row. Each photodiode potential well is capable of holding a number of electrons that determines the upper limit of the dynamic range of the CCD. After being illuminated by incoming photons during a period termed integration, potential wells in the CCD photodiode array become filled with electrons produced in the depletion layer of the silicon substrate. Measurement of this stored charge is accomplished by a combination of serial and parallel transfers of the accumulated charge to a single output node at the edge of the chip. The speed of parallel charge transfer is usually sufficient to be accomplished during the period of charge integration for the next image.

After being collected in the potential wells, electrons are shifted in parallel, one row at a time, by a signal generated from the vertical shift register clock. The electrons are transferred across each photodiode in a multi-step process (ranging from two to four steps). This shift is accomplished by changing the potential of the holding well negative, while simultaneously increasing the bias of the next electrode to a positive value. The vertical shift register clock


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operates in cycles to change the voltages on alternate electrodes of the vertical gates in order to move the accumulated charge across the CCD. Figure 1 illustrates a photodiode potential well adjacent to a transfer gate positioned within a row of CCD gates.

After traversing the array of parallel shift register gates, the charge eventually reaches a specialized row of gates known as the serial shift register. Here, the packets of electrons representing each pixel are shifted horizontally in sequence, under the control of a horizontal shift register clock, toward an output amplifier and off the chip. The entire contents of the horizontal shift register are transferred to the output node prior to being loaded with the next row of charge packets from the parallel register. In the output amplifier, electron packets register the amount of charge produced by successive photodiodes from left to right in a single row starting with the first row and proceeding to the last. This produces an analog raster scan of the photo-generated charge from the entire two-dimensional array of photodiode sensor elements.

12. Explain about photoconductive cell (LDR).

Bulk type Photoresistor or Photoconductive Cell:

The photconductive cell (PC) or detector is a two terminal device which is used as a Light Dependent Resistor (LDR). It is made of a thin layer of semi-conductor material such as cadmium sulphide (CdS); lead sulphide (PbS), or cadmium selenide (CdSe) whose spectral responses are shown in fig. The photoconducting device with the widest applications is the Cds cell, because it has high dissipation capability, with excellent sensitivity in the visible spectrum and low resistance when stimulated by light. The main drawback of Cds cell is its slower speed of response. Pbs has the fastest speed of response.

The illumination characteristics of photoconductive detectors are shown in figure. (a). It exhibits the peculiar property that its resistance decreases in the presence of light and increase in the absence of light. The cell simply act as a conductor whose resistance changes when illuminated. In absolute darkness the resistance is as high as 2 M and in strong light, the resistance is less than 10 .


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Fig. Spectral responses of Cds and CdSeA simple circuit for a photoconductive detector is shown in fig. (b). The semiconductor layer

is enclosed in a sealed housing. A glass window in the housing permits light to fall on the active material of the cell. Here, the resistance of the photoconductive detector, in series with R, limits the amount of current I in the circuit. The ammeter A is used to measure the current I. when no light falls on the cells, its resistance is very high and the current I is low. Hence the voltage drop V0 across R is relatively low. When the cell is illuminated, its resistance becomes very low. Hence, current I increases and voltage V0 increases. Thus, this simple circuit arrangement with slight modification can be used in control circuits to control the current.

Fig (a) Illumination characteristics of the photoconductive detector , and (b) photoconductive detector connected in a simple circuit.

Applications The detector is used either as an ON/OFF device to detect the presence or absence of a light source which is used for automatic street lighting or some intermediate resistance value can be used as a trigger level to control relays and motor. Further, it is used to measure a fixed amount of illumination and to record a modulating light intensity.

It is used in counting system where the objects on a conveyor belt interrupt a light beam to produce a series of pulses which operates a counter.

It is used in twilight switching circuits. When the day light has faded to a given level, the corresponding resistance of the detector causes another circuit to switch ON the required lights.

It is widely used in cameras to control shutter opening during the flash. Twin photoconductive cells mounted in the same package have been used in optical bridge circuits for position control mechanisms and dual-channel remote volume control circuits.

13. Explain the Principle and working of photodiode.

Junction Type Photoconductive cell:

Photodiode Silicon photodiode is a light sensitive device, also called photodetector, which converts light signals into electrical signals. The construction and symbol of a photodiode are shown in fig. The diode is made of a semiconductor PN junction kept in a sealed plastic or glass


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casing. The cover is so designed that the light rays are allowed to fall on one surface across the junction. The remaining sides of the casing are painted to restrict the penetration of light rays.

Photodiode (a) Construction, and (b) Symbol

A lens permits light to fall on the junction. When light falls on the reverse biased PN photodiode junction, hole-electron pairs are created. The movement of these hole-electron pairs in a properly connected circuit results in current flow. The magnitude of the photocurrent depends on the number of charge carriers generated and hence, on the illumination on the diode element. This current is also affected by the frequency of the light falling on the junction of the photodiode. The magnitude of the current under large reverse bias is given by

where I0 = reverse saturation current IS = short-circuit current which is proportional to the light intensity

V = voltage across the diodeVT = volt equivalent of temperature =parameter , 1 for Ge and 2 for Si.

The characteristics of a photodiode are shown in fig. The reverse current increases in direct proportion to the level of illumination. Even when no light is applied, there is a minimum reverse leakage current called dark current, flowing through the device. Germanium has a higher dark current then silicon, but it also has a higher level of reverse current.

Fig: characteristics of photodiode


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Photodiodes are used as light detectors, demodulators and encoders. They are also used in optical communication system, high speed counting and switching circuits. Further, they are used in computer card punching and tapes, light operated switches, sound track films and electronic control circuits.

14. Write briefly about Phototransistor.

Phototransistor Phototransistor or photoduodiode is a much more sensitive semiconductor photodevice than the PN photodiode. The current produced by a photodiode is very low which cannot be directly used in control applications. Therefore, this current should be amplified before applying to control circuits. The phototransistor is a light detector which combines a photodiode and a transistor amplifier. When the phototransistor is illuminated, it permits a greater flow of current.

Figure show the circuit of an NPN phototransistor. It is usually connected in a CE configuration with the base open. A lens focuses the light on the base collector junction. Although the phototransistor has three sections, only two leads, the emitter and collector leads, are generally used. In this device, base current is supplied by the current created by the light falling on the base-collector photodiode junction.

When there is no radiant excitation, the minority carriers are generated thermally, and the electrons crossing from the base to the collector and the holes crossing from the collector to the base constitute the reverse saturation collector current ICO. With IB =0, the collector current is given by

IC = (+1) ICO

Fig: NPN Phototransistor (a) Symbol, and (b) Biasing arrangement

When the light is turned On, additional minority carriers are photogenerated and the total collector current is

IC = (+1) (ICO +IL)

Where IL is the reverse saturation current due to the light.


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Current in a phototransistor is dependent mainly on the intensity of light entering the lens and is less affected by the voltage applied to the external circuit. Figure shows a graph of collector current IC as a function of collector-emitter voltage VCE and as a function of illumination H.

Fig: Characteristics of Phototransistor

The phototransistors find extensive applications in high-speed reading of computer punched cards and tapes, light detection systems, light operated switches, reading of film sound track, production line counting of objects which interrupt a light beam , etc.

15. Explain the operation of Photovoltaic cell and solar cell.

Photovoltaic Cell

Photovoltaic cell, a light-sensitive semiconductor device, produces a voltage when illuminated which may be used directly to supply small amounts of electric power. In the photovoltaic device without any applied voltage, the junction generates a voltage depending upon the illumination and the load. The voltage generated is due to the accumulation of carriers produced by photon excitation.

The photovoltaic potential is the voltage at which zero resultant current is obtained under open circuited conditions. The photovoltaic emf is 0.5 V for either silicon or selenium and 0.1 V for germanium cell. The short circuit cell currents is of the order of 1 mA.

The magnitude of the current under large reverse bias is given by

The photovoltaic voltage Vmax which corresponds to an open circuited diode can be obtained by substituting I = 0 in the above equation. Hence,


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As IS >> I0, Vmax increases logarithmically with short circuit current IS, and hence with illumination.

Fig: Photovoltaic cell (a) Construction and (b) Circuit symbol

The voltage increases as the intensity of light. Falling on the semiconductor junction of this cell increases. A photovoltaic cell consists of a piece of semiconductor material such as silicon, germanium or selenium which is bonded to a metal plate, as shown in fig. (a). the circuit symbol for photovoltaic cell is shown in fig. (b).

The spectral response of silicon, germanium and selenium are shown in fig, indicating that photoconductor is a frequency-selective device. As the spectral response of silicon and germanium extends well into infrared region, its efficiency is quite high. Selenium cell has two advantages over silicon, viz, (i) its spectral response is almost similar to that of the human eye, and (ii) it has the ability to withstand damaging radiation environments, lasting up to 10,000 times longer than silicon.

Fig: Spectral response of Si. Ge and Se


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The characteristic curves of output voltage versus light intensity and output current versus light intensity are shown in fig. (a) and (b), respectively.

Photovoltaic cells are used in low-power devices such as light meters. Nowadays, with an improvement in the efficiency of these cells, more power is produced, as in solar cells which are photovoltaic devices. When operated in the short-circuit mode, the current is proportional to the illumination and photovoltaic cell is used to construct a direct-reading foot-candle meter.

Light intensity, foot-candies Light intensity, foot candies(a) (b)

Fig: characteristic of Photovoltaic cell (a) Output voltage vs. Light intensity and (b) Output current vs. light intensity.

Solar Cell:

When sunlight is incident on a photovoltaic cell, it is converted into electric energy. Such as energy converter is called Solar cell or Solar battery and is used in satellites to provide the electrical power. This cell consists of a single semiconductor crystal which has been doped with both P-and N-type impurities, thereby forming a PN junction. The basic construction of a PN junction solar cell is shown in fig. Sunlight incident on the gas plate G passes through it and reaches the junction. An incident light photon at the junction may collide with a valence electron and impart sufficient energy to make a transition to the conduction band. As a result, an electron-hole pair is formed. The newly formed electrons are minority carriers in the P-region. They move freely across the junction. Similarly, holes formed in the N-region cross the junction in the opposite direction. The flow of these electrons and holes across the junction is in a direction opposite to the conventional forward current in a PN junction. Further, it leads to the accumulation of a majority carriers on both sides of the junction. This gives rise to a photovoltaic voltage across the junction in the open circuit condition. This voltage is a logarithmic function of illumination.


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In bright sunlight, about 0.6 V is developed by a single solar cell. The amount of power the cell can deliver depends on the extent of its active surface. An average cell will produce about 30 m W per square inch of surface, operating in a load of 4. To increase the power output, large banks of cells are used in series and parallel combinations. The efficiency of the solar cell is measure by the ratio of electric energy output to the light energy light expressed as a percentage. At present, anefficiency in a range of 10 to 40% is obtained. Silicon and selenium are the materials used widely in solar cells because of their excellent temperature characteristics.

16. Describe with diagram the construction LED and explain its working.

Light Emitting Diode (LED)

The Light Emitting Diode (LED) is a PN junction device which emits light when forward biased, by a phenomenon called electroluminescence. In all semiconductor PN junctions, some of the energy will be radiated as heat and some in the form of photons. In silicon and germanium, greater percentage of energy is given out in the form of heat and the emitted light is insignificant. In other materials such as gallium phosphide (GaP0 or gallium arsenide phosphide (GaAsP), the number of photons of light energy emitted is sufficient to create a visible light source. Here, the charge carrier recombination takes place when electrons from the N-side cross the junction and recombine with the holes on the P-side.

LED under forward bias and its symbol are shown in figure (a) and (b) , respectively. When a LED is forward biased, the electrons and holes move towards the junction and recombination takes place. As a result of recombination, the electrons lying in the conduction bands of N-region fall into the holes lying in the valence band of a P-region. The difference of energy between the conduction band and the valance band is radiated in the form of light energy. Each recombination causes radiation of light energy. Light is generated by recombination of electrons and holes whereby their excess energy is transferred to an emitted photons. The brightness of the emitted light is directly proportional to the forward bias current.


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Fig: LED (a) LED Under forward bias, (b) symbol and (c) Recombinations and emission of light.

Figure© shows the basic structure of an LED showing recombinations and emission of light. Here, an N-type layer is grown on a substrate and a P-type is deposited on its by diffusion. Since carrier recombination takes place in the P-layer , it is kept uppermost. The metal anode connections are made at the outer edges of the P-layer so as to allow more central surface area for the light to escape. LEDs are manufactured with domed lenses in order to reduce the reabsorption problem. A metal (gold) film is applied to the bottom of the substrate for reflecting as much light as possible to the surface of the device and also to provide cathode connection. LEDs are always encased to protect their delicate wires.

The efficiency of generation of light increases the injected current and with a decrease in temperature. The light is concentrated near the junction as the carriers are available within a diffusion length of the junction.

LEDs radiate different colours such as red, green, yellow , orange and white. Some of the LEDs emit infrared (invisible) light also. The wavelength of emitted light depends on the energy gap of the material. Hence, the colour of the emitted light depends on the type of material used is given as follows.

Gallium arsenide (GaAs) – infrared radiation (invisible)Gallium phosphide (GaP) –red or greenGallium arsendide phosphide (GaAsP) – red or yellow

In order to protect LEDs, resistance of 1k or 1.5 k must be connected in series with the LED. LEDs emit no light when reverse biased. LEDs operate at voltage level from 1.5 to 3.3V, with the current of some tens of milliamperes. The power requirement is typically from 10 to 150 mW with a life time of 1,00,000+ hours. LEDs can be switched ON and OFF at a very speed of 1 ns.


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They are use din burglar alarm systems, pictures phones, multimeters, calculators, digital meters, microprocessors, digital computers, electronic telephone exchange intercoms, electronic panesl, digital watches, solid state video displays and optical communication systems. Also, there are two-lead LED lamps which contain two LEDs, so that a reversal in biasing will change the colour from green to red, or vice versa.

When the emitted light is coherent, i.e. essentially monocramatic, then such a diode is referred to as an Injection Laser Diode (ILD). The LED and ILD are the two main types used as optical sources. ILD has a shorter rise time than LED, which makes the ILD more suitable for wide-bandwidth and high-data- rate applications. In addition, more optical power can be coupled into a fiber with an ILD, which is important for long distance transmission. A disadvantages of the ILD is the strong temperature dependence of the output characteristics curve.

17. Describe the principle and Operation of LED.

Liquid Crystal Display (LCD)

Liquid Crystal Displays (LCDs) are used for display of numeric and alphanumeric character in dot matrix and segmental displays. The two liquid crystal materials which are commonly used in display technology are pnematic and cholesteric whose schematic arrangement of molecules is shown in fig. (a). The most popular liquid crystal structure is Nematic Liquid Crystal (NLC). In this type, all the molecules align themselves approximately parallel to a unique axis (director), while retaining the complete translational freedom. The liquid is normally transparent, but if subjected to a strong electric field, disruption of the well ordered crystal structure takes place causing the liquid to polarize and turn opaque. The removal of the applied electric field allows the crystal structure to regain its original form and material becomes transparent.

Based on the construction, LCDs are classified into two types. They are

(i) Dynamic Scattering type: The construction of a dynamic scattering liquid crystal cell is shown in figure(b). The display consists of two glass plates, each coated with tin oxide (SnO2) on the inside with transparent electrodes separated by a liquid crystal layer, 5 to 50m thick. The oxide coating on the front sheet is etched to produce a single or multi-segment pattern of characters, with each segment properly insulated from each other. A weak electric field applied to a a liquid crystal tends to a align molecules in the direction of the field. As soon as the voltage exceeds a certain threshold value, the domain structure collapses and the appearance is changed. As the voltage grows further, the flow becomes turbulent and the substance turns optically inhomogenous. In this disordered state , the liquid crystal scatters light.


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Fig: (a) Schematic arrangement of molecules in liquid crystal, (i) Nematic (ii) Cholestic and (b) Construction of a dynamic scattering LCD

Thus, when the liquid is not activated, it is transparent. When the liquid is activated the molecular turbulence causes light to be scattered in all directions and the cell appears to be bright. This phenomenon is called dynamic scattering.

(ii) Field Effect Type: The construction of field effect LCD display is similar to that of the dynamic scattering type, with the exception that two thin polarizing optical filters are placed at the inside of each glass sheet. The LCD materials is of twisted nematic type which twists the light(change in direction of polarisation) passing through the cell when the latter is not energized. This allows light top as through the optical filters and the cell appears bright. When the cell is energized, no twisting of light takes place and the cell appears dull.

Liquid crystal cells are of two types: (i) Transmittive type, and (ii) Reflective type. In the transmittive type cell, both glass sheets are transparent so that light from a rear source is scattered in the forward direction when the cell is activated.


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The reflective type cell has a reflecting surface on one side of the glass sheet. The incident light on the front surface of the cell is dynamically scattered by activated cell. Both types of cells appear quite bright when activated even under ambient light conditions.

Liquid crystals consume small amount of energy. In a seven segment display the current drawn in about 25A for dynamic scattering cells and 300A for filed effect cells. LCDs require a.c. voltage supply, A typical voltage supply to dynamic scattering LCDs is 30 V peak-to-peak with 50 Hz. LCDs are normally used for seven segmental displays.

Advantages of LCD

(i) the voltage required are small(ii) they have a low power consumption. A seven segment display requires about 140

W (20W per segment), where as LEDs require about 40 mW per numeral (iii) They are economical.

Disadvantages of LCD

(i) LCDs are very slow devices. The turn ON and OFF times are quite large. The turn ON time is typically of the order of a few ms, while the turn OFF is 10 ms.

(ii) When used on d.c their life span is quite small. Therefore, they are used with a.c. supplies having a frequency less than 50 Hz.

(iii) They occupy a large area.

Comparison between LED and LCD

LED LCDConsumes more power-requires 10-250 mW power per digit

Essentially acts as a capacitor and consumes very less power- requires 10-200 W power per digit

Because of high power requirement, it requires external interface circuitry , When driven from ICs

Can be driven directly from IC chips

Good brightness level Moderate brightness levelOperable within the temperature range -40 to 85C Temperature range limited to -20 to 60CLife time is around 100,000 hours Life time is limited to 50,000 hours due to chemical

degradation.Emits light in red, orange, yellow, green and white Invisible in darkness –requires external illumination.Operating voltage range is 1.5 to 5V d.c Operating voltage range is 3 to 20 V a.cResponse time is 50 to 500 ns Has a slow decay time –response time is 50 to 200

msViewing angle 150 Viewing angle 100

18. Draw and Explain about Seven segment Display.

ALPHANUMERIC DISPLAYS


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Display devices provide a visual display of numbers, letters, and various signs in response to electrical input, and serve as constituents of an electronic display system. Display devices can be classified as passive displays and active displays.

(i) Passive displays: Light controllers – they are modulators of light in which the light pattern gets modified on application of electric field, e.g. LCD

(ii) Active displays: Light emitters – they are generators of light, e.g. LED.

The optical devices described so far were capable of operating in an OFF/ON mode. LEDs are used as low consumption indicator lamps. Also, both LEDs and LCDs are potentially more useful as elements in alphanumeric display panels. There are two possible arrangements of optical displays, viz, seven segment and dot matrix, the choice being based on the display size, definition and allowed circuit complexity.

One way of producing an alphanumeric display is to make a seven-segment monolithic device, as shown in figure(a), which can display all numerals and nine letters. Each segment contains LED/LCD which can be turned ON or OFF to form the desired digit. Each segment of the array has to be switched in response to a logical signal. For example, fig (a) shows the response to a logic signal corresponding to 2, in which segment a, b, d, e and g have been switched ON and c and f remain OFF, Similarly, when all segments are ON, the digit formed is 8. If only the center segment, g is OFF, the digit will be zero. Common anode and Common cathode seven segment LED displays are shown in fig (b). Common anode type LED displays require an active LOW configuration, whereas an active HIGH circuitry is necessary for the common cathode type LED display.

The seven segment displays are used in digital clocks , calculators, microwave ovens, digital multimeters, microprocessor trainer kits, stereo tuners etc.

Another method of producing an alphanumeric display is to make a dot matrix of LEDs/LCDs in a monolithic structure. Commonly used dot matrices for this display are 5 7, 5 8 and 79, which can display 64 different characters including the alphabets, numerals and various symbols, by driving the appropriate horizontal and vertical inputs. A 5 7 dot matrix assembly using LEDs and the corresponding wiring pattern is shown in fig.

LED display are available in many different sizes and shapes. The light emitting region is available in lengths from 0.25 to 2.5 cm.


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Fig: (a) Seven segment monolithic device, (b) Common anode and (c) Common cathode configurations.


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(a) (b)

Fig (a) 5 7 dot matrix and (b) Wiring pattern for 5 7 dot matrix

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