experiment 1

Exp. # 1: Diode Characteristics, Diode Limiter & Clamper

Philadelphia University

Faculty of Engineering

Dept. of Communication and Electronics Engineering

Second Semester: 2009/2010

Group Members:

1_________________________________)

2_________________________________)

3_________________________________)

Electronics Laboratory

Experiment 1

Diode Characteristics,

Diode Limiter and Clamper

1-1

Course Title: Electronics Lab Lecturers: Dr. Omar Daoud, Dr. Khaled DaqrouqCourse No.: (650326/227) Dr. Waka' Farman, Eng. Wafeiah Al-Shabani

Eng. Imad Al-Maqusi


Part one: The Diode

Objectives:

Measure the forward voltage across a diode, and determine if the component is faulty.

Demonstrate the forward current and voltage characteristics of pn-junction and zener

diodes.

Reference Readings:

1- Electronic Devices, THOMAS L. FLOYD, Fifth edition.

2- Electronic Devices “a design approach”, Ali Aminian & Marian Kazimierezuk.

Theory

The diode is a semiconductor device that conducts currents much more readily in one

direction than in the other. The voltage across the diode terminals determines whether or not

the diode will conduct. When the diode's anode is at a higher potential than is the cathode,

the diode is forward biased, and current will flow through the diode from anode to cathode.

Unlike a resistor, in which the current is directly (that is, linearly) proportional to the voltage

across it, the diode is a nonlinear device. When the diode is forward biased, a small but

measurable voltage drop, called the barrier potential, occurs across the diode. For

germanium diodes, this value is typically 0.3 V; for silicon diodes, it is approximately 0.7 V.

.


Experiment 1

Diode Characteristic , Diode Limiter and

Clamper

1-2


FIGURE 1.1: I-V Characteristic of Normal Diode

The following formula is used to calculate the forward dynamic or AC resistance of the diode:

)1)

Where: ∆V: The small change in voltage across the diode.

∆I: The corresponding change in current through the diode

Equipments Required:

Resistors (1/4 W): 10, 100, 1 k.

1N914 (1N4148 or equivalent) silicon, small-signal diode. 0-15 V dc power supply. Signal generator. Dual trace oscilloscope. DMM (Digital Multimeter). Bread boarding socket

Procedure

1- Set your DMM to the diode test position.

Electronics Laboratory 1-3


2- Connect the DMM to the 1N4914 diode as shown in Figure 1.1. Measure the forward voltage (VF) across the diode.

Vf = ………………..FIGURE 1.2: Diode Testing

3- Wire the circuit shown in Figure 1.3:

a) Adjust the dc power supply to give the voltages shown in table 1.1.For each voltage, measure and record the dc voltage drop (Vd) across the diode, The diode current (Id); which can be measured by dividing the voltage across the resistance into the resistance value.

FIGURE 1.3: Measurement of Diode

Characteristics

Table 1-1


Input DC

voltage

Diode

Voltage

Diode Forward

Current

0.0 V

0.2 V

0.7 V

0.9 V

1.6 V

5 V

8 V

1-4


b) Plot the diode characteristic curve on the graphical sheet, and graphically determine

the diode's barrier potential (VB) and the forward resistance (Rf).

VB = ………………

Rf = ……………….

4- To obtain the I-V characteristics of the diode using the oscilloscope, connect the circuit shown in Figure 1.4, set function generator to 10 , 100Hz sinusoid input signal.

FIGURE 1.4: Measurement of I-V Characteristics

Using Oscilloscope

5- Set the oscilloscope to the x-y mode, with the following approximate settings:Horizontal sensitivity (CH 1) at l mV/division.Vertical sensitivity (CH 2) at l0 mV/division .



6- Draw the curve shown on the screen of the oscilloscope and graphically determine the diode's barrier potential (VB), and forward resistance (Rf).

VB = ………………

Rf = ……………….

Note that: Channel 1measures the voltage across the diode.Channel 2 measures the voltage drop across the 10 Ω, and using Ohm's law, it can be used to represent the current through the diode

= 1 mA/div

REVIEW QUESTIONS FOR PART 1

1. When an ohmmeter is used to test a diode, a very low resistance (but not

zero) in one direction means that the diode is

(a) open (b) shorted

(e) forward biased (d) reverse biased ( )

2. In this experiment, the measured diode barrier potential is approximately

(a) 0.3 V (b) 0.6 V (c) 0.9 V (d) 1.2 V ( )

3. If the 10- resistor in Figure 1-4 is changed to 100 and the

oscilloscope's vertical sensitivity is 0.5 V/division, then the vertical

axis, in terms of current, is



(a) 0.5 mA/division (b) 5 mA/division

(e) 50 mA/division (d) 0.5 A/division ( )

4. For which region of your experimental diode curve does the diode look

like an open circuit?

(a) Diode voltages less than the barrier potential.

(b) Diode voltages greater than the barrier potential. ( )

5. For the region of the diode curve greater than the diode's barrier

potential,

(a) the curve is essentially horizontal

(b) the diode forward resistance approaches an open circuit

(c) the diode voltage increases rapidly

(d) the diode current increases rapidly ( )



Part 2: Diode Clipper

Objective:

To demonstrate the operation of a diode clipper.

TheoryDiode clippers are wave-shaping circuits in that they are used to prevent signal voltages from

going above or below certain levels. The clipping level may be either equal to the diode's

barrier potential or made variable with a dc source voltage. Because of this limiting

capability, the clipper is also called a limiter.

Reference Readings:1- Electronic Devices, THOMAS L. FLOYD, Fifth edition.



15-k resistor, 1/4 W 5-k potentiometer 1N4001 silicon rectifier diode 0-15 V dc power supply Signal generator Dual trace oscilloscope

Bread boarding socket.

Procedure



1- Wire the circuit shown in Figure 1.5-A, and adjust your oscilloscope to the following settings:

Channels 1 and 2: 1 V/division, dc coupling.

Time base: 1 ms/division

Figure 1.5-A: Positive Clipper

2- Using the oscilloscope (in the dual mode),

sketch the input and output waveforms, and

measure the level at which clipping occurs.

VO )Peak) = …………………………..

3- Now reverse the polarity of the diode in the circuit, as shown in Figure 1.5 B, and then

sketch the input and output waveforms and measure the level at which clipping occurs.

V O )Peak) =…………………………..

Figure 1.5-B: Negative Clipper

4- Now connect the circuit of Figure 1.5-C,

and adjust the potentiometer so that the dc



voltage (VDC) is + 1.5 V, and sketch both the input and output waveforms, with

measuring the output clipping level.

V O )Positive Peak) =…………………………..

Figure 1.5 C: Positive Biased Clipper

5- Vary the resistance of the potentiometer from one extreme to the other. What happens

to the clipping level? .......................................................................................................

………………………………………………………………………………………….

…………………………………………………………………………………………..

6- Now reverse the polarities of both the diode and the dc power supply in the circuit, as

shown in Figure 1.5-D. Adjust the potentiometer so that the dc voltage (VDC) is - 1.5 V,

and sketch both the input and output waveforms, with measuring the output clipping

level.

V O ) Negative Peak) =………………………

Figure 1.5 D: Negative Biased Clipper

7- Vary the resistance of the potentiometer from one extreme to the other. What happens

to the clipping level? …………………………………………………………………......

……………………………………………………………………………………………




1. For the positive clipper circuit of Figure 1.5-A, the positive peak

voltage is approximately

(a) 0V (b) +0.6V (c) +3V (d) +6V

( )

2. For the negative clipper circuit of Figure 1.5-B, the positive peaks

are not clipped because the diode is

(a) reverse biased

(b) forward biased

( )

3. In all the clipping circuits in this experiment, the 15-k resistor is

used to

(a) set the clipping level

(b) set the peak output voltage

(c) limit the voltage across the diode

(d) limit the peak forward diode current ( )

4. For the circuit of Figure 1.5-C, the potentiometer is used to set the

clipping level of the output's

(a) positive peaks (b) negative peaks

(c) positive and negative peaks ( )

5. For the circuit of Figure 1.5-D, the potentiometer is used to set the

clipping level of the output's

(a) positive peaks (b) negative peaks

(c) positive and negative peaks ( )

Part 3: Diode Clamper



Objective:

Demonstrate the operation of the diode clamper circuit.

TheoryThe clamper is a diode circuit used to change the DC reference of a waveform without

significantly altering the shape of that waveform. The positive clamper shifts its input

waveform in the positive direction; the negative clamper shifts it in the negative direction.

The negative clamper is identical to the positive clamper except for the polarity of the diode

and capacitor. Clampers are easily distinguished from clippers in that they include a

capacitive element.

Clamper time constant

10RLC »Tinput )1)

Peak output voltage

VO (peak) = VIN (peak-to-peak) - Vd )2)

Reference Readings:

1- Electronic Devices, THOMAS L. FLOYD, Fifth edition.



10 resistor, 1/4 W l0-F electrolytic capacitor, 25 V 1N4001 silicon rectifier diode Signal generator Dual trace oscilloscope

Bread boarding socket.

Procedure



1- Wire the circuit shown in figure 1.6 and adjust your oscilloscope to the following settings:

Channels 1 and 2: 1 V/division, dc coupling.

Time base: 1 ms/division.

Figure 1.6: Positive Clamper

2- Using oscilloscope (in the dual mode) obtain

both the input and the output voltages ,

and sketch their waveforms

3- Note that the clamping action is not perfect.

The negative peaks of the output waveform

are clamped not at zero volts, but at a small

negative voltage, why??................................

……………………………………………..

4- Increase the peak-to-peak input voltage. What happens to the output??

……………………………………………………………………………………………

…………………………………………………………………………………………....

5- Now reverse the polarities of both the diode and the capacitor as shown in figure 1.7, and

sketch both the input and output waveforms.

Figure 1.7: Negative Clamper

6- Again you should notice that the clamper action is not perfect. The positive peaks of the

output waveform are clamped not at zero volts, but at a small positive voltage, why??

……………………………………………………………………………………………

7- Increase the peak-to-peak input voltage. What happens to the output??



……………………………………………………………………………………………

…………………………………………………………………………………………....




1. For the circuit of Figure 1-6 to function properly, the input frequency

should be at least

(a) 1 Hz (b) 10 Hz (c) 100 Hz (d) 1 kHz ( )

2. For the circuit of Figure1-6, if the input signal has a peak voltage of

VP, then the output signal is

(a) shifted upward by approximately VP

(b) shifted upward by approximately 2 VP

(c) shifted downward by approximately VP

(d) shifted downward by approximately 2 VP ( )

3. For the circuit of Figure 1-6, the negative peak voltage of the output

signal is approximately

(a) VP (b) -0.7V (c) OV (d) +0.7V ( )

4. If the peak-to-peak input voltage is increased,

(a) the peak-to-peak output voltage remains approximately equal to

the peak-to-peak input voltage

(b) the negative peak output voltage remains clamped at

approximately - 0.7 V

(c) the output peak voltage approximately equals the peak-to-peak

input voltage

(d) all of the above ( )

5. In order to change the circuit of Figure 1-6 to a negative clamper, you

must

(a) reverse the polarity of the signal source

(b) reverse the polarity of the diode

(c) reverse the polarity of the capacitor

(d) all of the above ( )

Conclusions



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experiment 1

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