Practice English Speaking&Listening with: Lecture - 34 Active Diode Circuits

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Hello

every body! In our series of lectures on basic electronics let us move on to the next. You

might recall in the previous lecture we discussed about the comparators; operational amplifier

used as a comparator. We discussed about different application circuits like the window comparator

where you have the op amp sensitive enough to detect the transition at two different

levels.

For example there is a low output between limits is one variation of the comparator

and the other one is the output will become high whenever the input voltage is between

the two limits. That is why it is called a window comparator. It is sensitive to two

threshold voltages of the input. When the lower threshold is crossed the output changes

state and remains in the state until you again cross the upper threshold by changing the

input voltage. Window comparators have got very nice applications and we also saw another

variation of a relaxation oscillator which you would have come across in transistors;

the astable, bistable multivibrators which are basically relaxation oscillators. Here

using comparator, a regenerative comparator to be precise you can use an additional RC

and make a relaxation oscillator out of it so that you get a square wave output from

the operational amplifier. You can generate the square wave and if you take the output

across a capacitor you might recall we also got a triangular wave. So you can generate

a square wave and a triangular wave. We also saw how to derive the expression for the frequency

or the period of such an oscillator. These are the things that we discussed in the last

lecture.

Let us now move on to the next one which is again a non-linear application of the operational

amplifier. But in this case it is a very interesting scheme where we are going to use a diode,

normal diode a pn junction diode made of silicon which is about 1 rupee in the market if you

want to buy and that is not an ideal device. It is not an ideal device because if we look

at the characteristics you have about 0.7 volts which is called the cutting voltage.

Only when you cross 0.7 voltage at the input the diode will get forward biased and you

will get sufficient current to the diode. That cutting voltage 0.7 volts is what you

have to sacrifice when you are using diode in the normal rectifying action. For example

I have an ac signal. How do I measure the ac signal? If I want to measure the ac signal

one way to do that is you can rectify. You can rectify the ac signal and give it to a

normal galvanometer or a voltmeter as a case and measure the voltage. The rectified signal

will either have half wave rectified signal or a full wave rectified signal. We know what

will be the average output of a half wave rectified signal or a full wave rectified

signal and you know what magnitude of the average voltage will be read by the voltmeter.

We can calibrate that voltmeter in terms of either the rms or peak voltage. This is a

normal procedure to measure ac voltages.

If my ac voltage is very large for example 10 volts or 20 volts if I want to measure

that I can directly put a full wave or a half wave rectifier and connect it to a normal

voltmeter, dc voltmeter and from the magnitude of the voltage I will be in a position to

calibrate the meter in terms of either rms. But unfortunately when my input signal is

about 1 volt or even less then if I use a silicon diode, to forward bias a silicon diode

you have to sacrifice about 0.7 volts which is a cutting voltage of the silicon diode.

If I apply a signal with a peak voltage of 1 volt after passing through the diode for

rectification you will be left with only about 0.3 volts or so which is a very, very low

voltage. If I say 0.7 if I say it in millivolts it is 700 millivolts which is a very large

magnitude of the voltage and you will get only 0.3 volts if it is 1 volt peak and that

is about 300 millivolt. You have lost more than 700 millivolt of your input voltage across

the diode. This is a very undesirable scheme.

How do I then detect the very low level signals how do I measure very low level signals? There

is one option available for this and that is you can use a good ac amplifier. You can

use an ac amplifier and amplify the voltage to a larger value so that 0.7 is very insignificant.

When you amplify the 1 volt by 10 times it becomes 10 volts. In 10 volts 0.7 volt is

not too much of a loss and you can calibrate using that by a normal dc. So you must have

an amplifier. But amplifiers can be very expensive or difficult to construct. Is there a way

by which I can use the same diode, pn junction diode and still rectify very low level signals

which are less than 1 volt? The point is that is indeed possible to make the normal diode

silicon diode as an ideal diode. For that all that you have to do is have an op amp.

It is not vey expensive. If you use an additional op amp along with the diode you can make the

diode behave like an ideal diode. That is the whole idea behind this active diode circuits.

When I have an op amp, the op amp has got an open loop gain of above 10 power 5; 100,000

as an open loop gain typically.

If I have a 0.7 volts as the cutting voltage in the case of a normal silicon diode then

if I use an op amp the cutting voltage for the closed loop will be equal to VK divided

by AOL where AOL is the open loop gain of the operational amplifier. VK is about 0.7

volts normally and AOL is 10 power 5 and 0.7 by 10 power 5 is about 7 micro volts. The

cutting voltage becomes reduced from 0.7 volts to 7 micro volts which is a very, very small

value and this is the greatest advantage that you get when I combine an operational amplifier

with a normal diode to obtain an ideal diode configuration. Having said that I will show

you a typical circuit. You can see in the screen a typical circuit which makes use of

an operational amplifier and a diode. This diode in the simplest case is a normal silicon

diode.

In this configuration the feedback loop from the negative input is connected at this point

after the diode and the diode comes at the output of the operational amplifier. This

resistor RL is actually the load resistance and this is a non-inverting buffer type of

thing except for the diode. When I give the sine wave input here I must get a sine wave

input here if the diode is not there. But because the diode is there something else

is going to happen. Let us see what is going to happen? When the voltage increases from

very low value, initially the diode can be in open condition. That means it is not forward

biased and there is no link. There is no feedback loop from this point to minus because this

is open. It is equivalent to almost connecting this point to the ground. It is in an open

loop condition. So when the voltages increases here this voltage will have to be amplified

by the open loop gain of the operational amplifier and the open loop gain of the operational

amplifier is 10 power 5. Whatever is the voltage here will be multiplied by 10 power 5 and

that will be the voltage that will be appearing at the output terminal of the operational

amplifier and even if it is about 100 micro volt, if the input voltage is around 100 micro

volts I have to multiply by 10 power 5 and this will be about 1 volt at this point. You

get 1 volt means the diode will be forward biased. Because at this point at the output

of the amplifier it is about 1 volt and on the other side it is close to the ground and

this diode will become forward biased. When the diode becomes forward biased there is

a closed loop here and when the input is going positive the output is also going positive

and it will be multiplied by 10 power 5 and it will be a very large voltage and it will

forward bias the diode. It will be much larger than 0.7 volts and so it will be forward biased

and I will get the same voltage as it is changing and when this will happen; whenever the diode

cutting voltage is more than 0.7 volts. To become 0.7 volts here I must have an input

voltage which is 0.7 divided by 10 power 5. That is 7 micro volt. Even when the input

is beyond 7 micro volt as it increases here the diode will be forward biased and I will

get the voltage here exactly in the same way as it is changing here at the input. I will

get all the voltages; anything above 7 micro volts I will get here at the output.

What happens when the signal goes negative? When the signal goes negative the output will

become negative and the output is open circuit. The diode is open circuit. It is no more forward

biased. If it is not forward biased the output will have to be the same voltage as this voltage

here at the ground. There is no current here and it will be zero. For all the negative

excursions of the signal the output will remain at zero for all the positive excursions of

the signal the output will go positive. Anything beyond 7 micro volt the output will follow

exactly the same as the input and what I will be left with if I look at the oscilloscope

will be half wave rectified with the amplitude starting even from 7 micro volts. You will

have no difficulties with reference to the cutting voltage of the diode which in normal

case should be 0.7 volts. But here because of the open loop gain of the op amp and because

of the configuration, the way I have connected it the cutting voltage of the diode is reduced

to 7 micro volts. This is the greatest advantage of using the diode along with the op amp so

that even very low level signals, peak voltages anything beyond 7 micro volts also can be

rectified and this output I can connect to a multimeter or a voltmeter and measure the

average voltage. In this case it is 0.31 times the peak, Em by pie, peak voltage by pie.

That will be about 0.318. If it is full wave rectified it will become 2 Em by pie. It will

be twice that value. That value I can now calibrate in terms of either the rms or whatever

that I want to do. This is a very, very simple configuration using one op amp and a diode

in the proper feedback scheme. I can get a half wave rectifier circuit using this combination.

Once I have a half wave rectified signal then I can have all application circuits of the

diode that we have already discussed at the earlier lectures when I discussed about the

diodes, semi conductor diodes. This circuit corresponds to a half wave rectified signal.

When the input signal goes positive the output goes positive also and turns on the diode

and the diode then acts like a voltage follower and the positive half cycle appears across

the load resistor.

But when the input signal goes negative the op amp output also goes negative and turns

off the diode because there is not enough voltage to forward bias. Since the diode is

reverse biased or open no voltage appears across the load resistor and the final output

is almost a perfect half wave signal because 7 micro volt is the only difference. Once

we can get an ideal half wave rectifier the question then is how to build a full wave

rectifier? We always require full wave rectifier when we want to amplify very small ac signal.

When we want to convert it to dc it will be nice to have full wave rectifier. In the case

of normal full wave rectifier you use transformer with a center tab so that with reference to

the center tab the two ends of the transformer will be out of phase by 180 degrees and I

can use two diodes to convert each half of the cycle and I can get a full wave rectified

signal. We have already discussed that.

But when it comes to an operational amplifier applied ideal diode full wave rectifier the

scheme is slightly different. You cannot simply use that scheme. But there could be other

clever schemes available and one such clever scheme I want to discuss. You can see in the

figure that I have shown on the screen you have the circuit of a full wave rectifier

and also the corresponding output wave form. In the full wave rectifier both the half cycles

should come in the same direction as it is seen on the bottom graph. This is a full wave

rectified output and this is the input signal which is going back and forth. Both positive

and negative excursions are available and even the negative excursion is pushed on to

the positive side in the output and you get a full wave rectified signal at the output.

This is what you want to give as the signal and this is what you want to obtain at the

output. What is the circuit? The circuit is what I have here. The first circuit here using

the op amp is nothing but a half wave rectifier. I have used R1 Rf here which has got a finite

gain when it is in the reverse biased condition. This diode is the one which is going to do

the rectification and I give the input signal which is in the form of sinusoidal wave at

this pane and what I am going to get here will be a half wave rectified signal. Now

that is given to the second stage. In the second stage what I have is a summing amplifier.

This is one input V1 here at this point and the other input is coming through this resistor

here. This input itself is coming as another input for the summing amplifier. The output

of the half wave rectifier is given as one of the inputs. These two inputs are amplified

by the summing amplifier and the result is what you get here.

The question arises how we get full wave rectifier when I have also sinusoidal input as an input

here. I will try to explain you with the signals that I have. Before that I should tell you

Rf is 2R3; twice R3. The gain factor here for the rectified signal is 2. R3 can be 10K;

Rf can be 20K and this R2 again here is 20K. That means the gain for the input ac signal

is only 1. But the gain for the rectified signal is 2. This is the important point that

you should remember. With this back ground let us try to look at this various signals.

You have a sinusoidal input and the half wave rectified signal will be something like this.

One half is cutoff, the negative excursion. Only the positive half cycles are there and

I still get the peak, V peak as the peak of the rectified signal.

Now when I give it to the summing amplifier, the summing amplifier amplifies the signal

to twice and it becomes 2Vp. This signal becomes 2Vp; 2Vp as the peak voltage and the other

signal is the input signal which is coming to the summing amplifier and for that the

gain is 1 and the same output is also there. These two signals will be added at the output

and what I get will be the sum of these two. If I add these two what is going to happen?

This is too week peak this is Vpeak in the opposite direction.

When I combine these two I will get only one Vpeak in the upward direction. If I come to

the second stage, in the second stage there is only one peak voltage in the upward direction

and there is nothing from the other input and when you sum them you again get a peak

voltage here in the same direction. Like that for each wave you get only a net output peak

voltage which is coming over here and when I monitor the output voltage of the summing

amplifier I get a full wave rectified signal. That is what is again shown here. This is

the Vin. This is immediately after the half wave rectification and after I have combined

the two signals using the summing amplifier I get the output which is a full wave rectified

signal.

There are some important points that you should remember with reference to a full wave rectifier.

This is a precision rectifier. It is called a precision rectifier because it can rectify

without that cutting voltage cutting in any of your voltage from the signal. It is only

7 micro volts which is very, very small value and the diode behaves almost like an ideal

diode. The circuit is very common and is pretty much found in many of the text books. The

most important point is the resistors that we have seen in the circuit R1, R2, R3, Rf

will have to be very, very precise resistors with tolerance 1% or even better. The tolerance

of these resistors should be very, very good.

We have to also worry about the high frequency response. The high frequency response of this

circuit is not good when you use very large resistors. If you want to improve the high

frequency response you have to reduce the resistors that you make use of in this circuit

and also you should try to use a low impedance output from a signal source, a sinusoidal

source.

All these things we will also see and you can have a precision rectifier having very

good linearity even at the millivolts range. Even couple of millivolts at very low frequencies

a full wave rectifier works very well without any great difficulty. We have seen one circuit

which is a half wave rectifier and the second circuit is a full wave rectifier which makes

use of a summing amplifier and a normal half wave rectifier in combination. Now I will

move on to the demonstration table and show you these two circuits and the working of

the circuits. You can see the circuit board here.

This is the same circuit which I explained to you. You have the V input here and the

negative feedback loop contains the diode and you measure the output voltage after the

diode and there is a 10K resistor as the load resistor. This is a normal 741 op amp. The

same circuit is wired here. You can see the op amp here and the normal silicon diode connected

here and this is a 10K resistor, the load resistor, Vcc and all are connected to the

dual supply and the wiring is completed. The input is given from a function generator which

you can see here. I can select the frequency here. I have now selected in the 100 hertz

range. This is the one which is the depressed pin and it is in sinusoidal mode. I can have

sine, triangular, etc. I am now in the sine mode and I am in about 100 kilo hertz range

and the frequency can be varied using these two knobs and the output voltage can be varied

in course 0.2 volts, 2 volts and 20 volts and this could be for continuous variation.

There are two outputs here. One is 600 ohm and the other is 50 ohms. I have actually

taken from the low impedance output the 50 ohm output and I have connected here. What

I connect at the input is a sinusoidal input around 100 hertz and the output I am monitoring

using an oscilloscope. This is a dual trace oscilloscope. We have got two traces and the

bottom one is actually the input signal connected to the circuit. The top one is the output

signal after rectification. There is a half wave rectification achieved. You dont have

any output here. Then the signal is going negative and when the signal is going positive

you get that alone coming here.

Now I will vary the amplitude and when I change that the output amplitude also varies and

if I change the frequency the output also changes. I am changing the frequency now and

if you look at the oscilloscope there again the frequency is changed. It is an output

corresponding to the circuit and it is a half wave rectified signal. It is a very, very

simple circuit. I have used only one op amp here and I have used one diode. That is all

in the feedback loop; a very simple circuit which is able to perform output rectification

even for very low signals. How do you know it is doing it for very low signals? You can

see from the amplification factor in the oscilloscope. It is presently in about 0.2 volts per division

and it is around 0.4. So even for 0.4 I do get a very nice signal here and that shows

that even for very low level signals you will be in a position to have the half wave rectified

signal.

I will show you the next circuit which is a full wave rectifier signal. I want you to

observe the circuit here. It is the same as what I discussed. The first circuit is an

ideal half wave rectifier, precision half wave rectifier which we have already seen

using an op amp and a diode and the second circuit is actually a summing amplifier with

one input coming from here, the other input coming from the input.

These two are summed and that is what you get and when you combine these two what you

get will be a full wave rectified signal. The gain of signal for the half wave rectified

signal is 2. This is 10K and this feedback resistor will be 20K in the actual circuit

and similarly this will be 20K, 20K. This gain is only 1. What you see here is an op

amp, first op amp in which I have used a diode here. This diode is 4148, slightly different

diode from what I have used. But this is again a silicon diode and this is the second operational

amplifier which is used as a summing amplifier. I have 20K in the feedback resistor. All of

them are precision resistors, 1% resistor and at the input again I have a 10K resistor

here. The circuit is exactly the same as what you see here and I have now given the input

from the function generator as before. You have the sine wave input and the frequency

is around 1 kilo and I can measure the frequency and the amplitude over here and now this input

is given as the input for the circuit. The input is given here at the circuit and both

the input and the output are monitored in the oscilloscope. The bottom one is the input

signal and the top one is actually the output signal.

The bottom one is a normal sinusoidal wave which is the input I give and at the output

it is a full wave rectified signal. Both the halves are coming in the bottom and if you

want to have the other way if you want to have the signal coming up to the positive

side all that you have to do is interchange the diodes that you have seen in the circuit.

For example if you invert these diodes, if you change the orientation then you will get

the full wave rectified signal corresponding to the other side. You do the full wave rectification

easily by using one half wave rectifier and another summing amplifier. By properly choosing

the gain you will be in a position to get the full wave rectified output as you have

seen in the oscilloscope. If I vary the amplitude the output also is changing here and if you

measure the voltage which is around 0.2 or 0.4 volts as you can see in the dial and it

is for very low signals that you get the full wave rectified output and this is a precision

full wave rectifier circuit.

We have seen the demo corresponding to two circuits. One is the precision half wave rectifier

using operational amplifier and the second one is the precision full wave rectifier where

we have used a half wave rectifier along with the summing amplifier to cleverly obtain full

wave rectified signal. This is all discussed in many of the standard textbooks.

Now we will go further and try to see how to make an active peak detector. The peak

detector is very useful whenever you want to measure ac signals. The one way to do that

is you rectify the ac signal and connect it across a capacitor so that the capacitor charges

to the peak of the input signal and if the time constant of the output is so chosen the

output will remain high at the peak voltage for longer period so that you will be in a

position to make the measurement using a normal dc voltmeter. How do we make peak detector

using the operational amplifier? On the screen I have shown the circuit of an active peak

detector using a precision diode here. The diode and the op amp are in the form of an

ideal diode by having in the feedback network and the output I connect to the capacitor

along with the RL. If the time constants of the R and C are very large then the input

signal will charge to the peak value of the capacitor and before it discharges the next

peak voltage will come and it will almost remain at the peak voltage except for the

small loss due to RC time constant.

Here again you can do the peak detection for very low level signals; less than 1 volt you

can comfortably convert them into peak voltage and then measure using normal multimeter or

dc voltmeter.

Now I will show you the peak detector on the demonstration board and then I will come back

and continue with the rest. of the demonstration Here you can see the same circuit which is

a peak detector. You have the precision rectifier here with the diode and the op amp and you

have a capacitor here connected for peak detection and this is RL the load resistor.

The corresponding circuit is shown here. You can again see the operational amplifier. You

can see the normal diode and I have put about 100 micro farad capacitor in place of this

and I have RL which is around 10K here. The input is again given from the function generator.

As in the previous case the output is connected to the input of the circuit and the output

of the circuit as well as the input of the circuit is monitored on the oscilloscope which

you see here. On the oscilloscope this is the input, a sinusoidal wave at the bottom

and the output you see is a dc; it is a straight line.

But how much is the dc you can measure by using the dc key here. I press ac and dc key.

Now it is in dc and it is here. If I now release that if I make it into the ground now you

can see the signal comes at this point if I release it it goes up so that much is the

peak. What I will do is I will bring it at the bottom or at center. Then you can see

it is actually going up to the peak value. When I put it into the dc it goes to the peak

value so it has been charged to the peak value and I increase the input amplitude by using

the function generator. I am going to increase the amplitude and at the output signal in

the oscilloscope again you see a magnified signal and if I again make it peak detect

it is going up to the peak, the corresponding peak.

It is indeed detecting the peak. Instead of using a multimeter I wanted to show you by

using the dc coupling here it goes from the zero line to the peak value here and this

is the dc output that I get across the capacitor. This is the output. The output is a dc. It

looks like a straight line but it is shifted with reference to the zero line up to the

peak value. It is indeed a peak detector. With a very simple op amp diode and a capacitor

we are able to make a peak detector and once I make a peak detector I can connect it to

multimeter and calibrate it in terms of either rms or peak voltage.

We have seen three different circuits so far. One is the basic precision diode or a rectifier,

half wave rectifier and then we also saw a full wave rectifier. How a half wave rectifier

along with the summing amplifier can be constructed to give a full wave rectifier. We also saw

a peak detector by using a capacitor at the output. Once I have a diode then I can have

different types of application circuits with the diode and a capacitor. We have already

discussed it when we discussed the pn junction diode; for example active positive clipper.

You already know of positive clipper or negative clipper and the active positive clipper makes

use of a precision diode. A diode which is along with an operational amplifier becomes

an ideal diode that is used for generating various clipping circuits and the clamping

circuits and that is what we are going to see now. Let me show you a very simple circuit

of a positive clipper where again I have used an op amp and the diode. The diode and the

op amp combined becomes a precision rectifier and I give an ac signal here with the peak

voltage Vp at the input and at the output I should get a half wave rectified signal

in principle. But then what happens the second input or the non-inverting input of the operational

amplifier now I have connected to a potentiometer, a variable resistor, wiper of the potentiometer

and the other end is connected to plus; the one end is grounded and when I move the wiper

I can move it to this extreme. Then it will become V plus and when I move it to the this

extreme it becomes ground.

I can vary this voltage at this terminal from anywhere between zero to +V. +V could be any

voltage 2V or 4V. I can vary this voltage here and I can change the bias here. When

I change the bias what is going to happen? If it is at the zero this circuit is nothing

but a simple half wave rectifier. Because this is now in this condition when it goes

negative, when the input is negative the output is also negative and the diode will be conducting.

So there is a short, there is a feedback loop and the output goes straight there. The output

goes straight at the output terminals. But when the input goes positive with reference

to the other terminal then the output will become positive. At that stage the diode will

be reverse biased. So diode will not work. It will be open and the output will be equal

to this voltage and that voltage here is what you get at the other terminal and initially

you should get a half wave rectified signal. But if I now maintain this terminal at some

voltage different from zero for example +2 volts then anything more than +2 volts if

I apply here this voltage is more than this voltage and I get a negative and then it is

shorted, the output goes straight there. If it is more than 2 volts this is opened up

and you get the corresponding signal and also the output is clipped at this V reference,

2 volts in this case. If I make it 2 volts the positive peak will be cut off at 2 volts.

If it is 1 volt the positive peak will be cut off at 1 volt. Depending upon where I

put the reference this output will correspondingly be clipped. The negative input will automatically

come because during that time there is a feed back and you get the exact output voltage

coming over here, the unity gain clipper type of thing. You get a clipper with positive

points clipped at this place depending upon what reference you have in the operational

amplifier.

If you want a negative clipper what do you do? All that you have to do is reverse this

diode and reverse this reference voltage. You can do that and have the positive cycle

completely and the negative cycle can be clipped wherever you want. In some cases you want

both sides clipped. When I take a sine wave and clip both sides I will get almost sort

of a square wave. There is one such circuit which I want to show you. You can see here

the circuit has got two zener diodes connected in opposition together and the output will

be clipped on both sides at the +Vz and -Vz.

That is what you get and actually when one Zener diode is having break down voltage the

other Zener diode will be forward biased. You will get about 0.7 volt in addition to

the Vz. That is what is shown here as Vz + VK. VK is 0.7 volts which is the cutting voltage

of the diode. Vz + VK will be point at which it will be clipped both on the positive side

and the negative side because one zener breaks down for one orientation of the signal, for

one excursion of the signal and the other zener diode breaks down for the other excursion

and what you get will be a clipped sine wave or basically something close to a square wave.

When you want to convert a sinusoidal signal into a square wave you can make use of the

clipper of this type which uses zener diode in the feedback and you can get a square wave.

I will show you a simple circuit. The same circuit implemented for a positive clipper.

The circuit here is the same as what I just now discussed as an active positive clipper.

You have an op amp, you have the diode and you have the signal input here. You take the

output and on this side you have got a potentiometer. The wiper of the potentiometer is providing

the reference voltage for the positive terminal or the non-inverting terminal of the operational

amplifier. This is the plus. I have used the power supply of the +Vcc itself here and that

is what is being used for the reference voltage here.

This is the circuit and the corresponding circuit is shown here. You can see the diode,

you can see the operational amplifier and the resistors and the input is provided from

a function generator once again. You can see the function generator here. You have again

at 1 kilo hertz the amplitude and the sinusoidal wave and the amplitude is adjusted using these

two knobs and this is given as the input in the board and the output is connected to an

oscilloscope. In the oscilloscope both the input and the output are shown simultaneously.

If I slightly move you can see the two signals.

You can see the bottom one is the input signal, the top one is the clipped signal. You can

see the clipping there. The clipping can be modified by changing the wiper position or

by changing the reference voltage here. That is done by changing the potentiometer here.

When I change the potentiometer I am able to change the point at which the signal is clipped by moving the wiper. The

reference voltage is what is being changed now and thereby I am able to change the clipping

voltage.

This is a positive clipper and if I want a negative clipper all that I have to do is

change the direction of diode and change the reference voltage on this side. Then you will

be able to get the negative clipping and you can also do a square wave clipping. That is

both sides clipping, dual clipper both positive and negative by using Zener diode or combination

of such circuits.

Having seen the clipper circuits we also know a diode can also be used for clamping. That

means dc restoration as it is also called whereby you can take the sinusoidal wave,

normal sinusoidal wave when you take the average of the sinusoidal wave you will get the symmetric

sine wave with reference to the base line or the common line, the ground line. For example

on the screen you can see example of the sine wave which is symmetric with reference to

the ground, the common point. The excursions are equal on both sides +V peak and -V peak,

maximum values and it goes back and forth on both sides of the ground. When I say I

am clamping what I actually do is I change this line with reference to the sine wave.

I can take it all the way to the top then I have clipped at the peak positive point

and the entire voltage will not be negative or I can bring this line to the bottom and

clip it at this point at the bottom so that the entire voltage is positively clamped.

It is clamped at this point, the lower point. So the whole thing is positive.

You can do either positive clamping or negative clamping and the circuit is very, very simple.

Because it is much easier to understand from our earlier discussion on positive clamping

I have shown the earlier circuits also here for recapitulation. You can recall what we

did earlier. I have used a diode here and I have used a capacitor and I give a sine

wave input at this point. What I get will be a clamped sine wave. How this comes about

also I explained to you in the previous case. When the input goes negative then this diode

can be forward biased. This is ground. This is going both positive and negative as the

signal goes and when it goes negative the diode is forward biased and the diode will

start conducting. When it starts conducting it will charge the capacitors in the path

to the full peak value. During one negative excursion of the input signal the diode will

start conducting and it will charge the capacitor to the full peak value.

Once that happens when the input signal goes positive the diode will become open circuit

after the first cycle and you can almost ignore the presence of the diode now. What is going

to happen is you have now a new situation where the input signal is applied in series

with a battery. The capacitor charged to +Vp and the diode open circuited acts like a normal

dc battery of a value Vp. You have to now add this Vp to every instantaneous voltage

of the input signal and that is what you get as the output. When the input is maximum V

peak you have to add another V peak here due to the battery here or the capacitor and the

total output will be 2V peak and when you go to zero you will have this V peak and when

you go to the minus V peak this +V peak will add with that and you will get zero. The output

will start going from zero to 2Vp as the amplitude and what I have effectively done is I have

clamped this zero line at the bottom. It becomes fully positive and it is a positive clipper.

The output signal has got 2Vp as the maximum voltage. This is a normal diode capacitor

based clamper. Wherever I use normal diode I will still have the same problem of the

cutting voltage. Because this is 0.7 volts when I want to clamp it using 1 volt signal

it will all be lost. Only 0.3 will be charging this and the clamping will not be done very

well. If I can make this into an ideal diode then I can even use very low signals and get

the clamping action completely done.

I have used here an active diode which is an op amp with the diode connected as in the

previous example. If you look at this as an ideal diode this is my capacitor. The circuit

is very similar to this. The RL is what I have at the output load resistor. This is

the input signal. When I now give the input ac signal the clamper will charge for the

first negative cycle. When it goes negative this diode will become positive because there

is an inversion and the diode will start conducting even for about 7 micro volts and the output

will come completely and it will charge the capacitor. Once the capacitor is charged this

diode will become open circuited and whatever input signal I give it will be in series with

this Vp and the output will be shifted with reference to average dc by a value corresponding

to Vp. The whole output signal will become clamped at the positive side. Zero will be

clamped at this point and the entire excursion will be towards the top.

The output has the same size and shape as the input signal. There is no difference except

for the dc shift. Only there is an additional dc shift which is provided by the capacitor

charging to the full value of the peak value voltage. In this case the output voltage is

sum of the input voltage and the capacitor voltage.

V output is equal to Vin + V peak and Vin is a sinusoidal voltage with the peak voltage

Vp. When I add it with this when this becomes +Vp the total becomes 2Vp. When this becomes

zero the output is Vp. When it becomes -Vp the output is zero. There is nothing which

goes below zero. Even though the signal goes below zero because you are adding with the

Vp the output becomes only all positive starting from zero up to a maximum of 2Vp. That is

the active positive clamper that we wanted to discuss. Now I will show you the demo of

the active positive clamper and then we will go further.

(Video needs to be edited here. Sir is speaking in Tamil).

This is the circuit of an active positive clamper which just now I discussed.

You have the input sine wave from a oscillator and you have a capacitor here and then you

have the ideal diode with the operational amplifier and the diode. This is the load

resistor and you take the output at this point. The positive terminal is connected to the

ground. The non-inverting terminal is connected to the ground. The same circuit is wired here.

The capacitor is here about 10 or 15 micro farad and this is the diode that I have here

and this is 10K resistor. The wiring is done exactly in the same way. The input is connected

to a function generator as before. It is in sinusoidal mode and it is about 110 kilo hertz

and the frequency, the output amplitude, etc can be changed using these knobs and the output

is connected to the input of this circuit which is an active positive clamper. Both

the input and the output waveforms are monitored using an oscilloscope. Here the bottom one

is the input signal, the top one is the output signal. You dont see any difference here.

Both of them are identical. That means the gain of the amplifier of the circuit is 1.

But there is a difference. What is the difference? That difference you can see by looking at

the dc situation. I have a knob here. This is now ac mode. This is dc mode, ac mode.

In the bottom sine wave the ac is with reference to the ground which is here.

For the top one when I make it into dc still the sine wave is there and it has moved up

to this line, the central line. It is symmetric to the central line and when I dc couple it,

it moves up to this. That means the entire output has moved with reference to this as

the reference line or zero line.

I have clamped it positive. I will show you again and again, number of times so that you

can see on the screen that initially it is symmetric. When I dc couple it the whole sinusoidal

wave moves up. That means it is positively clamped. That is what we see here. Whereas

the something if I do for the lower signal you can see that there is no dc there. When

I put dc coupling I get only a straight line.

That means there is no dc. Everything is only ac whereas in the top one you see all of them

are ac but with reference to a zero line which is at the bottom and it is a varying dc; sinusoidally

varying dc all with reference to this reference point. It is positively clamped. That is what

we mean. This is the way to identify the dc clamping. If I want a negative clamping all

that I have to do in the circuit is invert the diode here and then you will get a negative

clamping. That means the whole signal will move to the negative side the clamping will

be done with reference to zero and the two peaks will be on the negative side. Just as

we saw now on the positive side we can go to the negative clamping by just interchanging

the diode, turning through 180 degrees and then …… the circuit you will be able to

get the negative clamping. The positive clamping and negative clamping are also very useful

in different applications in electronics.

What we have done today is look at how a normal diode can be made into an ideal diode with

the combination of an operational amplifier and we also saw how different applications

of an ideal diode can be used for half wave rectification, precision full wave rectification

and then peak detection using a simple capacitor and then clipping circuit by giving some bias

and a clamping circuit. All these different circuits based on active diode circuits using

an operational amplifier, our precision rectifier can be considered and the applications based

on those diode circuits have been discussed. Thank you!

The Description of Lecture - 34 Active Diode Circuits