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Expt No.1 Fixed Bias or Base- Bias amplifier circuit using BJT

Aim:1. To observe the Waveforms at input and output without bias.

2. To determine the bias resistance to locate Q-point at center of Load line.

3. To Measure the gain of amplifier.

4. To Plot the frequency response & Determine the Bandwidth Product

Apparatus required:

S.No. Name of the apparatus Range Qty

1. BC 2N2222 or BC 547 - 1

2. Resistor 1K,100K 1

3. Regulated power supply (0-30)V 1

4. Function Generator (0-30)Mhz 1

5. CRO 30Mhz 1

6. Capacitor 0.1f 2

7. Bred Board - 1

8. Connecting wires - Req.

Theory:

The circuit (also known as Base-bias circuit) see Figure 9.1 is the most simple of all

Common-Emitter circuits. It provides a straightforward introduction to the DC bias analysis of

transistors. The Q-point, however, is extremely beta-dependent. Because of this, this circuit has

limited applicability.

Procedure:

1. The connections are given as per the circuit diagram.

2. The input voltage is set in the signal generator.

3. The readings are measured by CRO from the minimum frequency to maximum

frequency.

4. Draw the input and output waveform as per the readings measured.

5. Determine the bias resistance using RB=Vcc/IB

6. Locate Q.Point at centre of the load line by using RB.

7. Measure the gain using Gain=20log10(Vo/Vi)

8. Plot the graph by using frequency Vs Gain.

9. Determine the Bandwidth product using BW=fC2-fC1

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Circuit Diagram:

Frequency Response curve:

DC load line:

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Determination of Operating Point (Q): (VCEQ or VC, ICQ or IC):

Bandwidth Product:

The bandwidth of the circuit is found as the difference between the cutoff frequencies.

By formula,

Tabular Column: Input Voltage (Vi) =1V

S.No. Frequency in Hz Output Voltage (Vo) Gain=20log10(vo/Vi)

Result:

Thus the input and output waveform, frequency response for fixed bias amplifier

circuit using BJT are drawn and following parameters are calculated.(i) Bias Resistance RB = (ii) Gain = (iii) Band width product =

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5. Measure the gain using Gain=20log(Vo/Vi)



Frequency Response curve

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Thus the input and output waveform with and without bypass emitter resistor of the

common emitter amplifier using voltage divider bias, frequency response of the common

emitter amplifier circuit using BJT are drawn and following parameters are calculated.

Gain =

Band width product =

Expt No.3 Common Collector Amplifier using voltage divider Bias

Aim:

To Design and construct BJT Common Collector Amplifier using voltage divider bias (self-

bias), Measure the gain, Plot the frequency response and Determination of Gain Bandwidth

product.

Apparatus required:


1. BC 547 - 1

2. Resistor 100K,33K,4.7K 1



5. CRO 30Mhz 1

6. Capacitor 10f 2

7. Bred Board - 1


Theory:

The voltage divider is formed using external resistors R1 and R2. The voltage across

R2 forward biases the emitter junction. By proper selection of resistors R1 and R2, the

operating point of the transistor can be made independent of . In this circuit, the voltage

divider holds the base voltage fixed independent of base current provided the divider

current is large compared to the base current. However, even with a fixed base voltage,

collector current varies with temperature (for example) so an emitter resistor is added to

stabilize the Q-point, similar to the above circuits with emitter resistor.

Procedure:



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frequency.


5. Measure the gain using Gain=20log10(Vo/Vi)



Design:

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Frequency Response

curve

Bandwidth Product:

The bandwidth of the circuit is found as the difference betweenthe cutoff frequencies.

By formula,



Result:

Thus the input and output waveform of the common collector amplifier using

voltage divider bias, frequency response of the common collector amplifier circuit using

BJT are drawn and following parameters are calculated.

Gain =


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Expt No.4 Darlington Amplifier using BJT

Aim:To measure and compare the gain and input resistance of the Darlington

amplifier circuit using BJT, determine the gain bandwidth product and to plot the

frequency response.

Apparatus required:


1. BC547 - 2

2. Resistor 100K,33K,1K 1

3. Regulated power supply (0-30)V 14. Function Generator (0-30)Mhz 1

5. CRO 30Mhz 1

6. Capacitor 10F 2

7. Bred Board - 1


Theory:

The Darlington amplifier is a useful circuit and has the advantage of providing a very high

current gain, high input impedance and higher output power. (It is not necessary to use matchedtransistors here) and often you see a smaller signal transistor driving a larger power transistor. The

current gain is approximately the product of both Q1 and Q2 forward current gains. One point to

note is that as Q1 emitter is connected to Q2 base the bias voltage required is Vbe1 + Vbe2. This must

be taken into account when designing bias circuits for the Darlington amplifier.

Procedure:




frequency.


5. Measure the gain using Gain=20log10(Vo/Vi) and input resistance using

Ri1=R1II R2

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Design:

Input Resistance: Ri = R1II R2 = R1.R2 / R1+R2

Bandwidth Product:

The bandwidth of the circuit is found as the difference between

the cutoff frequencies.

By formula,

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Result:

Thus the gain and input resistance of the Darlington amplifier circuit using

BJT are measured and compared its values and frequency response characteristics curve is

drawn and following parameters are calculated.

Input Resistance Ri1 =

Gain =

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Expt No.5 Source follower with Bootstrapped gate resistance

Aim:

To Measure and compare the gain, input resistance and output resistance with and

Without Bootstrapping.

.

Apparatus required:

S.No. Name of the apparatus Range Qty1. BC 547 - 1

2. Resistor 100K,4.7K,47K,120K 1



5. CRO 30Mhz 1

6. Capacitor 22fd,1fd,47fd 1

7. Bred Board - 1


Theory:

Bootstrapping involves the use of positive feedback from output to input of

an amplifier, of nearly unity gain, in such a way that a particular point in the circuit is

"pulled up as if by its own bootstraps" The bootstrap technique is used to make a low value

gate resistor Rg appear to have a much higher value as seen by the input signal.

Procedure:




frequency.

4. Measure the gain using Gain=20log10(Vo/Vi), input and output resistance using

Rg= RD =

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Circuit without Bootstrap Circuit with Bootstrap

Input Resistance: Ri = R1 II R2

Output Resistance: Ro = RE

Tabular Column:Without Bootstrap

S.No. Input Voltage (Vi) Output voltage(Vo) Gain=20log10(vo/Vi)

With Bootstrap

S.No. Input Voltage (Vi) Output voltage(Vo) Gain=20log10(vo/Vi)

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Result:

Thus the gain, input and output resistance of the source follower with bootstrapped

gate resistance circuit using FET are measured and compared its values.

Expt No.6 Differential amplifier using BJT

Aim:

To Measure the CMRR of differential amplifier using BJT

Apparatus required:


1. Transistor BC 547 - 2

2. Resistor 3.3K,1K and 4.7K 2,2,13. Regulated power supply (0-30)V 3

4. Voltmeter (0-30)V 1

5. Bred Board - 1


Theory:

Gain (dB) = 20log (vo/Vi)CMRR = 20log (Ad/Ac)

Where Ad: Differential GainAc: Common Mode Gain

The function of a differential amplifier is to amplify the difference between two input

signals. The output signal is proportional to the difference between the two input signals i.e., V O=

Ad (v1-v2).

If v1=v2, the output voltage is zero. A non-zero output voltage is obtained if V 1 & V2 are not equal.

The difference mode input voltage is defined as V d[(=V1-V2)] and the common mode input voltage

is defined as Vcm=(V1+V2)/2.

Differential mode Gain Ad = Vo/Vs

Common mode Gain Ac= Vo/Vs

Where Vs = differential input voltage

CMRR: (Common Mode Rejection Ratio):CMRR is used to define the ratio of the Ad and Ac

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Procedure:

Differential Mode

1. The connections are made as per the circuit diagram.

2. The DC input voltage is set in base of the both transistor.

3. The corresponding output voltage is determined at collector of the transistor.

4. the value of readings are tabulated

Common Mode:

1. The Connections are made as per the circuit diagram.

2. The single supply voltage is applied to the both the transistor base.

3. Then the same procedure is followed in the differential mode.

Differential Mode:.

Tabular Column: (Differential Mode)

S.NO. Input Vd[(=V1-V2)] Vo Gain=Vo/Vd

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Voltage

V1 V2

Common Mode:

Tabular column : (Common Mode)

S.NO.Input

Voltage(Vc)Vo Gain=Vo/Vc

Calculation:

Result:

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Thus the differential amplifier was constructed and differential mode gain,

Common mode gain and CMRR also calculated.

Differential mode Gain Ad =

Common mode Gain Ac =

CMRR =

Expt No.7 Class A Power Amplifier

Aim:

To observe the output waveform, Measure the maximum power output,

determine the efficiency and Compare with calculated values of class A

power amplifier.

Apparatus required:


1. SL100R - 1

2. Resistor 100K,4.7K,47K,120K 1



5. CRO 30Mhz 1

6. Capacitor 22fd,1fd,47fd 17. Bred Board - 1


Theory:

Common emitter voltage amplifiers are the most commonly used type of

amplifier as they have a large voltage gain. They are designed to produce a large output

voltage swing from a relatively small input signal voltage of only a few millivolt's and

are used mainly as "Small Signal Amplifiers" as we saw in the previous tutorials.

However, sometimes an amplifier is required to drive large resistive loads such as a

loudspeaker and for these types of applications where high switching currents are needed

Power Amplifiers are required. The main function of the Power amplifier, which are

also known as a "Large Signal Amplifier" is to deliver power, which is the product of

voltage and current to the load. Basically a power amplifier is also a voltage amplifier the

difference being that the load resistance connected to the output is relatively low, for

example a loudspeaker of 4 or 8s resulting in high currents flowing through the

Collector of the transistor. Because of these high load currents the output transistor(s)

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used for power amplifier output stages need to have higher voltage and power ratings

than the general ones used for small signal stages.

Procedure:



3. The output readings are measured by CRO.


5. The maximum power output measured by using Pin(dc)=VCC.ICQ=2VCEQ.ICQ

6. The efficiency of this amplifier is determined by using o.

7. Compare with calculated and noted values.

Circuit diagram:

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Input and output waveform of Class A amplifier

Efficiency Calculation:

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Result:

Thus the efficiency, maximum output power of the class A amplifier circuits is

determined and measured. The output waveform, calculated values of the circuit is

observed and compared.

Expt No.8 Class B Complementary symmetry power amplifier

Aim:

To observe the output waveform with crossover Distortion, Modify the circuit

to avoid crossover distortion, Measure the maximum power output, Determine the

efficiency and Compare with calculated values.

Apparatus required:

S.No. Name of the apparatus Range Qty1. BC 547 & BC557 - 1

2. Resistor 100K,4.7K,10K 2,1,2

3. Diode (IN4001) 2



6. CRO 30Mhz 1

7. Capacitor 10fd,4.7fd 2,1

8. Bred Board - 1


Theory:

One of the main disadvantages of the Class B amplifier circuit above is that

it uses balanced centre-tapped transformers in its design, making it expensive to construct.

However, there is another type of Class B push-pull amplifier called a Complementary-

Symmetry Class B Amplifier that does not use transformers in its design therefore; it is

transformer less using instead complementary pairs of transistors. As transformers are not

needed this makes the amplifier circuit much smaller for the same amount of output, also

there are no stray magnetic effects or transformer distortion to effect the quality of the

output signal.While Class B amplifiers have a much high gain than the Class A types, one

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of the main disadvantages of class B type push-pull amplifiers is that they suffer from an

effect known commonly as Crossover Distortion. This occurs during the transition when the

transistors are switching over from one to the other as each transistor does not stop or start

conducting exactly at the zero crossover point even if they are specially matched pairs. This

is because the output transistors require a base-emitter voltage greater than 0.7v for the

bipolar transistor to start conducting which results in both transistors being "OFF" at the

same time. One way to eliminate this crossover distortion effect would be to bias both the

transistors at a point slightly above their cut-off point. This then would give us what is

commonly called an Class AB Amplifier circuit.

Procedure:



3. The output readings are measured by CRO.

4. Draw the input and output with cross over distortion waveform as per the

readings measured.

5. Modify the amplifier circuit to avoid the cross over distortion.

6. Draw the output waveform as per the readings measured.

7. The maximum power output measured by using Po(ac)=V2P/2RL or V

2pp/8RL

8. The efficiency of this amplifier is determined by using o(max)=78.5%

9. Compare with calculated and noted values.

Circuit diagram:
http://www.electronics-tutorials.ws/amplifier/amp_7.htmlhttp://www.electronics-tutorials.ws/amplifier/amp_7.htmlhttp://www.electronics-tutorials.ws/amplifier/amp_7.html

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Modified circuit diagram

Efficiency Calculation:

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Result:

Thus the efficiency, maximum output power of the class B complementary symmetry

power amplifier circuits is determined and measured. The output waveform with and without cross

over distortion of the amplifier circuit is observed. Calculated values of the circuit are compared

Expt No.9 Power Supply circuit - Half wave rectifier with simple capacitor filter.

Aim:

To Measure the DC voltage under load and ripple factor, Comparison with

calculated values and Plot the Load regulation characteristics using Zener diode.

Apparatus required:


1. Diode (IN4001) - 1

2. Resistor 1K,3.3K, 4.7K,47K 1

3. Step down Transformer 6-0-6 or 12-0-12 1

4. Capacitor 470fd,1000fd 1

5. Zener diode 9V 1

6. Bred Board 1


8.

Theory:

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The Half wave rectifier is a circuit, which converts an ac voltage to dc

voltage. In the Half wave rectifier circuit shown above the transformer serves two

purposes.

1. It can be used to obtain the desired level of dc voltage (using step up or step

down transformers).

2. It provides isolation from the power line.

The primary of the transformer is connected to ac supply. This induces an ac voltage

across the secondary of the transformer. During the positive half cycle of the input

voltage the polarity of the voltage across the secondary forward biases the diode. As a

result a current IL flows through the load resistor, RL. The forward biased diode offers a

very low resistance and hence the voltage drop across it is very small. Thus the voltage

appearing across the load is practically the same as the input voltage at every instant.

During the negative half cycle of the input voltage the polarity of the secondary voltage

gets reversed. As a result, the diode is reverse biased. Practically no current flows

through the circuit and almost no voltage is developed across the resistor. All input

voltage appears across the diode itself. Hence we conclude that when the input voltage is

going through its positive half cycle, output voltage is almost the same as the input

voltage and during the negative half cycle no voltage is available across the load. This

Explains the unidirectional pulsating dc waveform obtained as output. The

process of removing one half the input signals to establish a dc level is aptly

called half wave rectification.

Procedure:

1. The circuit connections are made as per the given diagram.2. The amplitude and time of transformer output is noted.

3. Then the rectified output with and without filter is noted.

4. Find the ripples and DC voltage of output of the rectifier.

5. Change the filter i.e., capacitor (470fd and1000fd ) and measure the output

waveform of rectifier.

6. The graph is drawn and to find the ripple factor value of these capacitors using RF.

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Circuit Diagram:

Model Graph:

Tabular Column:

S.No. Filter Vdc Vm Ripple factor

Ripple Factor

Ripple factor is defined as the ratio of rms value of ac component to the dc component in

the output.

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Ripple factor

Vav the average or the dc content of the voltage across the load is given by

RMS voltage at the load resistance can be calculated as

Ripple Factor:

Load Regulation:

1. For finding load regulation, make connections as shown in figure below.

2. Keep input voltage constant say 10V, vary load resistance value.

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3. Note down no load voltage VNL for maximum load resistance value and full load

voltage VFL for minimum load resistance value.

4. Calculate load regulation using, % load regulation = (VNL-VFL)/ VFL x100

Circuit Diagram:

Model Graph:

Tabular Column:

S.No.

Source

Voltage -Vs

(V)

Zener

Voltge -Vz

(v)

Load

Current (IL) mA

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Result:

Thus the DC voltage under load and ripple factor of the half wave rectifier is

measured and calculated values are compared. The Load regulation characteristics using

Zener diode is plotted.

Expt No.10 Power Supply circuit - Full wave rectifier with simple capacitor filter.

Aim:

To Measure the DC voltage under load and ripple factor, Comparison with

calculated values and Plot the Load regulation characteristics using Zener diode.

Apparatus required:


1. Diode (IN4001) - 2

2. Resistor 1K,3.3K, 4.7K,47K 1

3. Step down Transformer 6-0-6 or 12-0-12 1

4. Capacitor 470fd,1000fd 1

5. Zener diode 9V 1

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6. Bred Board 1


8.

Theory:

A full-wave rectifier converts the whole of the input waveform to one of

constant polarity (positive or negative) at its output. Full-wave rectification converts both

polarities of the input waveform to DC (direct current), and is more efficient.For single-

phase AC, if the transformer is center-tapped, then two diodes back-to-back (i.e. anodes-to-

anode or cathode-to-cathode) can form a full-wave rectifier. Twice as many windings are

required on the transformer secondary to obtain the same output voltage.

Procedure:

1. The circuit connections are made as per the given diagram.

2. The amplitude and time of transformer output is noted.

3. Then the rectified output with and without filter is noted.

4. Find the ripples and DC voltage of output of the rectifier.

5. Change the filter i.e., capacitor (470fd and1000fd ) and measure the output

waveform of rectifier.

6. The graph is drawn and to find the ripple factor value of these capacitors using RF.

Ripple factor:

The ripple factor for a Full Wave Rectifier is given by

the average voltage or the dc voltage available across the load resistance is

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Load Regulation:

1. For finding load regulation, make connections as shown in figure below.

2. Keep input voltage constant say 10V, vary load resistance value.3. Note down no load voltage VNL for maximum load resistance value and full load

voltage VFL for minimum load resistance value.

4. Calculate load regulation using, % load regulation = (VNL-VFL)/ VFL x100

Circuit Diagram:

Model Graph:

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Tabular Column:

S.No.

Source

Voltage -Vs

(V)

Zener

Voltge -Vz

(v)

Load

Current (IL) mA

Result:

Thus the DC voltage under load and ripple factor of the Full wave rectifier is

measured and calculated values are compared. The Load regulation characteristics using

Zener diode is plotted.

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