01- experiment diode j09

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EXPERIMENT 1.a +

Clamp circuit *

Concept

Sometimes it is desirable to clamp the signal to a certain voltage level (i.e. prevent it fromtransitioning below or above a certain voltage limit). Diodes are used in

implementing these clamp circuits. An application example is in processing analog

video signals before analog to digital conversion.

Procedure

1) Get a 1 F capacitor and a diode 1N4001.

2) Connect them in series with a function generator, as shown on Fig.1.1.

3) Adjust the generator output to square wave with frequency 200 Hz (or other

waveforms), and amplitude 2 Vp-p.

4) Get an oscilloscope, connect channel A to the function generator output andchannel B to the two terminals of the diode.

5) Turn on the function generator and the oscilloscope.6) Observe the output.

The circuit configuration is shown in Fig.1.1.

Fig.1.1 The clamp circuit drawn on the simulator._______________________________________________________________________________________________________

* Refer to Chapter # 3 Section # 3.8 Page # 194in Micro-electronic circuits By: Sedra/Smith

+ Simulate /Build


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Results

The output waveform will be identical to the input waveform, except that it is shiftedupwards by 1 V (i.e. the output limits are 0 V and 2 V). The output waveform is shown

on Fig.1.2.

Fig.1.2 the input and output waveforms of the clamp circuit.

Summary

The clamp circuit is used to limit the output signal voltage from exceeding certainlimit; in our experiment it prevents the output from going below zero volt.

Practical work

1) Repeat the above experiment after reversing the polarity of the diode. Sketch the

input and the output waveforms.


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2) Replace the capacitor C1 with a 10nF capacitor, and report the difference in the

output waveform.

3) Sketch the input and the output waveform for the circuit in Fig.1.3.

Fig.1.3 circuit for question 3


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4) Add a resistor 1k in parallel with the diode, observe and sketch the output,

explain your observation.

5) Repeat question 4 with a100 K resistor, and show the difference between the

output in this case and the previous case.


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Clipper Circuit *+

Concept

Clipper circuits are sometimes called limiter circuit, because they limit an input voltageto certain maximum or minimum values. Although this is a very simple diode

application, limiters find application in a variety of signal-processing systems.

Procedure

1) Get a resistor 10 K, and 2 diodes 1N4001, 2 constant dc sources 3 V.2) Connect the circuit as shown in Fig.1.4.

3) Get a function generator and connect it as shown in Fig.1.4.

4) Adjust the generator output to sine wave with frequency 1 kHz, and amplitude 10

Vp-p.

5) Get an oscilloscope, connect channel A to the function generator and channelB to the output.

6) Turn on the function generator and the oscilloscope and observe the output.

Fig 1.4 the clipper circuit

___________________________________________________________________________________

* Refer to Chapter # 3 Section # 3.6.1 Page # 184 in Micro-electronic circuits By: Sedra/Smith

+ Simulate /Build


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Results

The output waveform will be clipped from the positive region by about 5.7 V and fromthe negative region by about 5.7 V, thus the output signal is limited between 5.7 V and

5.7 V. The output is shown in Fig.1.5.

Fig.1.5. The input and output waveforms of the clipper circuit

Summary

The clipper circuit is used to limit the upper and/or the lower limit of a signal voltage,

these limits are determined by the dc supply value in any branch plus the voltage drop

across the diode in forward bias conditions.

Practical work

1) Set one of the dc sources to Zero and observe the output, sketch and comment onthe output waveform.


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2) Open one of the branches, and sketch the output waveform. Describe the

operation in this case.


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Voltage Doubler *+

Concept

As the name of the experiment reveals, this circuit is used to generate a voltage that

equals double the peak of the input voltage. The technique described here can beextended to provide output dc voltages that are higher multiples of input voltage peak.

Procedure

1) Get two 1F capacitor, two diodes 1N4001.

2) Connect the components as shown below on Fig.1.6.3) Connect the function generator to the input.

4) Adjust the generator output to a square wave with frequency 100 Hz, and

amplitude 10 Vp-p.5) Connect oscilloscope channel A to the function generator output and channel

B to the output capacitor.6) Turn on the function generator and the oscilloscope and observe the output.

Fig.1.6 The voltage doubler circuit drawn on the simulator._______________________________________________________________________________________________________


+ Simulate/Build


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Results

At the peak of the negative half cycle, D1 is forward-biased and D2 is reverse-biased.This charges C1 to the peak voltage of the input signal. At the peak of the positive half

cycle, D1 is reverse-biased and D2 is forward-biased. Because the source and C1 are in

series, C2 will try to charge towards twice the peak voltage. After several cycles, thevoltage across C2 will equal twice the peak voltage.

The output voltage will be a dc voltage with amplitude that is double the volt of the input

peak voltage. The output is shown in Fig.1.7.

Fig.1.7 the input and output waveforms of the voltage doubler circuit.

Summary

The voltage doubler circuit is used to produce a dc output signal with double the peak

voltage of an ac input signal.

Practical work

1) Change the input to a square wave with frequency of 100 Hz and amplitude 10 Vp-

p, report the results.


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2) Describe the output waveform in case of reversing the polarity of D1 & D2.

3) In Fig.1.6, does changing the input waveform from ac to dc yield the same output

waveform.


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EXPERIMENT 1.B +

Half Wave Rectifier *

Concept

One of the most important applications of diodes is in the design of rectifier circuits. Adiode rectifier forms an essential building block of dc power supplies required to build

electronic equipment. We use a power transformer to get a (5-20) V dc output from a 220

V 50HZ ac output. This transformer consists of two coils wound around an iron core thatmagnetically couples the two windings. The primary winding, having N1 turns, is

connected to the 220 V ac supply; and the secondary winding, having N2 turns, is

connected to the input of the rectifier circuit. Thus an ac voltage of 220(N1/N2) volts rmsdevelops between the two terminals of the secondary winding. The appropriate turns ratio

N1/N2 for the transformer is selected to step the line voltage down to the value required

to yield the desired dc voltage output. The half wave rectifier circuit converts the input

sinusoid to a unipolar output. It utilizes alternate half cycle of the input sinusoid.

The output of the half wave rectifier circuit is pulsating and therefore undesirable as a dc

supply for electronic circuits. A simple way to reduce the variations of the output voltageis to place a capacitor across the load resistor. This filter capacitor serves to reduce the

variations in the rectified output voltage. In this case the circuit will be called the peak

rectifier.

In selecting diodes for rectifier design, two important parameters must be specified. The

current handling capability required of the diode, determined by the largest current thediode is expected to conduct, and the peak inverse voltage (PIV) that the diode must be

able to withstand without breakdown. PIV is 50% larger than the largest reverse voltage

that is expected to appear across the diode. It can be shown that the peak inverse voltagein case of the half wave rectifier is equal to the peak of the input voltage from thesecondary windings. the average and maximum diode current of the peak rectifier is

given by:

Iavg = IL (1+ pi * root(2 * Vp /Vr))

Imax = IL (1+ 2* pi * root(2 * Vp /Vr))

And the ripple voltage, which is the difference between the upper and lower limits of the

output voltage, and is given by:

Vr = (Vp * T) / (C * R)

Where T is the period of the sinusoidal input voltage.

_____________________________________________________________________* Refer to Chapter # 3 Section # 3.5.1 Page # 172 in Micro-electronic circuits By: Sedra/Smith+ Simulate/Build


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Procedure

1) Get the following components: diode 1N4001 and a resistor 1 k.2) Connect the diode in series with the 1 k resistor.

3) Get a transformer with turns ratio 220/12.

4) You may simulate the circuit in either of the following configuration:a) Connect the primary coil terminals to the 220 V, 50 Hz supply and the

secondary coil terminals in series with the diode and the resistor.

b) Use a function generator as an input instead of the 220 V, 50 Hz supply,adjust the output to 50 Hz frequency and 10 Vp-p.

5) Get an oscilloscope, connect channel A to the function generator output and

channel B to the output resistor.

6) Turn on the function generator and the oscilloscope and observe the input and theoutput waveforms on the oscilloscope screen.

The half wave circuit configuration is shown on Fig.1.8

Fig.1.8 the half wave rectifier circuit.

Results

The input sinusoid is half rectified and the output wave consists of the positive half

cycles as shown in Fig.1.9. Notice the absence of the output corresponding to the

negative input half cycles.


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Fig.1.9 the input and output waveforms of the half wave rectifier.

Summary

The half wave rectifier is used to generate a half rectified wave from the input sinusoidal

signal.

Practical work

1) Replace the resistor R1=1 k with a 100 resistor and sketch the output signal.

Explain the results.


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2) Explain the difference in the voltage levels between the input and the output

waveforms.

.

3) Put a 1uF capacitor in parallel with the 1 k resistor and sketch the output,calculate the ripple voltage, the average and the maximum diode current.

4) Repeat question 3 with a capacitor 100 uF.


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Full Wave Rectifier (Bridge Rectifier) *+

Concept

An alternative implementation of the full wave rectifier is the bridge configuration, the

circuit is known as the bridge rectifier because of the similarity of its configuration withthe Wheatstone bridge, does not require a center-tapped transformer, a distinct advantage

over of the full wave rectifier of the circuit of Fig.2.4. The bridge rectifier requires four

diodes as compared to two in the previous experiment (2.a). The bridge rectifier circuitoperates as follows: during the positive half cycles of the input voltage, the current is

conducted through diode D1, resistor R1, and diode D2. Meanwhile, diodes D3 and D4

will be reverse biased. There are two diodes in series in the conduction path, and thus theoutput voltage will be lower than the input voltage by two diode drops. This is somewhat

of a disadvantage of the bridge rectifier. Next, consider the negative half cycle of the

input voltage. The secondary coil voltage will be negative forcing current through D3, R1

and D4. Meanwhile, diodes D1 and D2 will be reverse biased. The important point to

note is that during both half cycles, current flows through R1 in the same direction andthus the output voltage will always be positive.

The maximum value of the peak inverse voltage (PIV) of each diode is given by:

PIV = secondary coil voltage diode drop voltage (forward biased)

And the ripple voltage, which is the difference between the upper and lower limits of the

output voltage, and is given by:

Vr = (Vp * T) / (2 * C * R)

Where T is the period of the sinusoidal input voltage.

Observe that here the PIV is about half the value of the full wave rectifier with a center-

tapped transformer. Fig.1.10 shows the bridge rectifier circuit, note that PIV is themaximum value of the peak inverse voltage between node 1 and node 2.

Fig.1.10 the bridge rectifier circuit_______________________________________________________________________________________________________


+ S/B


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Note that the bridge rectifier differ from the full wave rectifier which uses only two

diodes and a center tapped transformer as shown in Fig1.11

Fig1.11 full wave rectifier circuit

Procedure

1) Get four diodes 1N4001, a resistor 1 k. Connect each two diode togetherforming two branches.

2) Connect the two branches together in a bridge configuration, and connect the

resistor between the two branches of the bridge (i.e. between node 1 and 2).

3) Get a transformer with turns ratio 220/12.4) You can simulate the circuit in either of the following configuration:

a) Connect the primary coil to the 220 V, 50 Hz supply and the input of the

bridge (the two connecting nodes of the two branches) to the secondarycoil terminals of the transformer.

b) Use a function generator as an input instead of the 220 V, 50 Hz supply,

adjust it to 50 Hz frequency and 10 Vp-p.

5) Get an oscilloscope, connect channel A to the input and channel B across theresistor.

6) Turn on the power of the generator and the oscilloscope and notice the input and

the output waveforms on the oscilloscope screen.

The circuit configuration is shown on Fig.1.12,

Results

The input sinusoid of the secondary coil is rectified and the output waveform will be inthe positive region as shown in Fig.1.13. Observe that the output waveform has twice the

number of the positive half cycles as the inputs.


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Fig.1.12 The bridge rectifier circuit.

Fig.1.13 The input and output waveforms of the bridge rectifier


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Summary

The full wave rectifier converts sinusoidal AC signal to DC with average value of 2/ the

peak signal of the input sine wave.

Practical work

1) Replace the resistor R1=1 k with a resistor 100 k and sketch the results. Andshow the difference, if exist?

2) Explain why there is a difference in the voltage level between the input and the

output waveform. How much is this difference?

3) Calculate the peak inverse voltage across the diode D1.


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4) Do you think the signal average in the full wave rectifier is higher or lower than

the half wave rectifier.

5) Modify the circuit shown in Fig.1.11 so that you can generate the followingoutput. Measure the ripple voltage for the shown waveform.

01- experiment diode j09

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