 lecture and we are going to discuss wave shaping circuits, by definition wave shaping circuits are those which shape the waveform of a given wave in that sense rectifiers are wave shapers because they in the half wave rectifier they reject the negative half of the wave, they only retain the positive half and the waveform of the output is quite different from the waveform of the input, so rectifiers are wave shapers, a full wave rectifier flips the negative half into the positive half, so that is also a wave shaper and before taking other specific circuits I wanted to do a couple of interesting examples on rectifiers which are also as I said wave shapers and one of the examples is this, we have a circuit like this try to draw with me, there are 2 diodes and 2 capacitors C1 D1 and then the voltage across this is V1, then we have another capacitor C2 here and there is a diode D2 which connects like this alright, the input between these 2 is Vm sin of omega t and the voltage between these 2 points is taken as V2 V2 and a load that is the resistance across which you require a DC basically this is a rectifier circuit, a full wave rectifier, the load is here RL, you are required to find out V1, V2 and VL under the condition that RL C1 or C2 is much greater than the time period of the given sinusoid which is equal to 1 over F, is the question clear alright, now let us see how this circuit works, when we have the positive half of the sine wave between these 2 points that is this polarity is positive, this is negative then obviously it is this diode which conducts D1, current flows like this and then comes back comes back through the capacitor to the other point of the sinusoidal source, so the current flows like this, can it also flow through the resistance RL, no because if it has to then it can come back through C2 via D2 to the supply, now I have drawn it, so it can pass through RL and then come through C2, it can pass through RL then come back to through C2 to the supply, but in that case diode will also, no diode is connected here, if the current comes back to the same point where it lives it has to go to the other end of the voltage supply alright, so the only path is this, the other path is this RL then via C2 via this point alright, okay, let us consider the situation when it becomes negative, let us say this is negative and this is positive, then obviously current passes through like this through the capacitor C2, through the diode D2 and to this part, here also a part of the current might flow through RL and through C1, but it has to come back to the same point, why should it come back to the same point alright, so this the green and the red, these are the two directions of current, what actually happens is the following, that C1 charges to the peak of this voltage Vm and C2 charges to again to the peak of the voltage, but with a polarity this positive, this negative, that means that is V1 if RL was not there, forget about RL, if RL is not there then no question of discharge of the capacitors come, so C1 voltage is held at Vm and C2 voltage is held at Vm, so V1 would be equal to Vm, V2 would be equal to Vm and the voltage across the load would be twice Vm and therefore this is a voltage doubler, not only it holds the voltage at the peak value, but it doubles the voltage because we have two parts V1 and V2, two capacitors, so it is a voltage doubler peak detector circuit, now suppose RL is not there, then this diode charges the capacitor C1 to the peak value Vm, as soon as the voltage falls below Vm, C1 holds the charge diode fails to conduct, so C1 is charged to plus Vm, similarly then the negative half, that is when this terminal becomes positive with respect to the upper terminal, current flows like this and C2 charges to Vm, alright, now when one of the diodes is not conducting the corresponding capacitor may find a path through RL for discharge, if RL is there, if RL is inserted then either capacitor when it is not charging may try to discharge through RL through RL and through C2, but this discharging tendency shall be restricted because of this relation, if the time constant is large then the decay will be very small during one time period and therefore this voltage Vl is approximately equal to twice Vl, this is a practical voltage doubler circuit. Sir while C1 is charging in the first half of the cycle, sir why cannot the current go through RL side, RL is present in the circuit, yeah but where does it have to come back, so it has to come back here, but is there a finite potential difference across C1, there is, yes this is Vm, so then why cannot the current come through RL, so the current can come through RL as I said it can come like this, it can come like this but the tendency is restricted because the time constant is large compared to the time period, that is in one time period the decay of the voltage either V1 or V2 due to discharge through RL shall be negligible, if RL is not there then you see the voltage shall be exactly through twice Vm, with RL this twice Vm passes a current through RL and therefore both C1 and C2 have a tendency to discharge, but this discharge during one time period shall be very small because of this relation and whatever charge is lost during one cycle shall be regained again during the next half cycle alright and therefore this voltage is approximately a constant equal to twice Vm, this relation should be valid, suppose it was a short circuit, suppose RL was 0, so that RLC is negligible, suppose RL is very small so that RLC is negligible then obviously both C1 and C2 when C2 is not charging it will discharge through RL alright and if it is short circuit the voltage will be 0, so you see VL depends on the value of RL precisely the product CRL because it is this product C1 RL and C2 RL which is going to determine the decay during the period that either capacitor is not charged, the second problem which is also very interesting is the following, suppose we have a DC V1 which fluctuates, a DC V1 which fluctuates may be it is a rectified power supply but the power supply itself fluctuates instead of 230 volt it sometimes becomes 240, sometimes falls to 210 and so on, so V1 fluctuates our purpose is to reduce the fluctuation of the DC that is in between V1 after V1 we shall put a circuit the output of which shall be less fluctuating than V1, such a circuit is known as a voltage regulator circuit, as you must have learnt as you must have seen in the in the domestic applications the refrigerator does have a regulator, the TV it is advisable to run it from a regulator, what basically it does is whatever the fluctuations in the power supply is it could be an AC regulator or DC regulator, well 230 volt fluctuates and if I rectify this 230 volt and supply to an electronic circuit for example a public address system, we want that the DC supply should not fluctuate, so fluctuating DC is required to be made as constant as possible and one of the possible circuits is this is what the problem is about, that there is a 1k resistor here called this RS, RS could actually be the internal impedance of the DC supply and then to reduce the fluctuations what we do is we use a diode here, a battery of 10 volt and the resistance of 20 ohms alright and then we have the load let us say RL which is equal to 2k this is what is given, you are required to find out the variation in VL that is delta VL if V1 increases from 16 volt to 16 volt to 24 volt which means a 50 percent rise okay, a 50 percent increase in V1 what is the corresponding delta VL, you will see that the corresponding delta VL would be much less than 8 volt and therefore in that sense the circuit acts as a voltage regulator, let us see how to analyse the circuit, you see that if V1 is positive then this diode always conducts, so let this current be I1 and let this current be I2 alright then obviously we can write 2k VL equations, one is that V1 is equal to 1000 I1 plus I2 alright the drop across this plus let us say plus I2 times RL, well RL is 2000, this is one of the equations around this loop and if you write around this loop then you shall get 10 plus I1 into 20 okay should be equal to this voltage drop, I am assuming the diode to be ideal, so 10 volt plus minus then I1 times 20 should be equal to I2 times 2000 alright from which you can eliminate I1, what you want is I2, you want to find out VL, you want to find out I2, I shall skip the algebra, if you do the algebra that is eliminate I1 from this equation then you can get I2 in terms of V1 alright, you can get I2 in terms of V1 and therefore VL, VL is simply I2 times 2000 in terms of V1, the relationship as I worked it out is VL equal to twice V1 plus 1000 divided by 103 which means that delta VL is equal to twice delta V1 divided by 103 alright, if I increment this by delta V1 then obviously delta VL should be equal to this and therefore if delta V1 that is V1 increases from 16 to 24 then it is 2 times 8 divided by 103 and delta VL is 16 divided by 103 which is less than 0.16 volt, you see how things change okay, so this circuit basically acts as a regulator when there is a 8 volt fluctuation, 50 percent fluctuation, now what is the percentage fluctuation, what is VL when V1 is equal to 16 volt you can calculate that out, we have the relation the VL is approximately equal to 10 volt, so 0.16 in 10 multiplied by 100 so much percentage approximately 1.6 percent, so a 50 percent fluctuation in the input causes only a 1.6 percent fluctuation in the output which in effect then is the circuit for a voltage regulator, we shall consider other more sophisticated kinds of voltage regulator later in the course but I could not check the temptation of bringing to your notice a circuit which is a useful circuit and which uses a diode, yes you had a question. A filter circuit is not necessarily a regulator, a filter circuit basically improves the ritual content, that is a filter circuit makes a fluctuating quantity into a more or less constant quantity but if the supply changes, if the supply suppose from 230 drops to 200 then the corresponding DC will also change, given the constant supply, the supply is constant a filter circuit improves the DC content that is it rejects the AC, so it makes the voltage more constant but it does not regulate, yes. 10 volt and for the battery and This is a specific circuit, we shall know, we shall discuss the design of voltage regulator circuits later, this is a specific circuit I just wanted to illustrate this with the help of the example, yes. Sir what was the use of the diode in the previous circuit? If there were no diodes we would still have got the same demand. Yes that is a good question, what is the use of the diode in this circuit? Yes, this is because if the fluctuation such that the polarity changes, then the Very diode will become effective, ineffective, it will become effective, there is another use for the diode, you see the 10 volt here, this is the use of the diode, the 10 volt here is not effective in setting the current through either 100 ohm or 2k, is it not right? The 10 volt supply till this voltage rises above 10 and this is the purpose of the diode, if the diode was not there this equation was not at the value, you would have to take, you would have to apply superposition, the effect of 10 volt also you will have to apply, it is that this branch allows current only in this direction, that is actually current goes through the battery, the battery does not such a any current here, this is the purpose of the diode, if the diode was not there this voltage regulation action would not have been there, you understand the purpose? This circuit is a human lateral circuit, it passes current only from top to bottom not from bottom to top, as it would have been if the diode was not there, if the diode was not there the battery would have set to 1000, out to 1000 no matter it is 2000, okay, any other question? Now let us consider some more weight shaping circuit, the first circuit that you consider is a clip-off or a clipping circuit, a clip-off circuit, the popular meaning of clipping is that if you have something long you cut it off, chop it off or you clip it off by means of a pair of scissors, that is what a clip-off circuit does, a clip-off circuit is a circuit in which the output V0 versus Vi is a straight line, is a straight line of let us say unity slope, it did not necessarily be unity but let us simplify the thing, of unity slope till the voltage, till the input voltage reaches let us say a positive value of capital V, beyond capital V the output is a constant, alright, in other words, V0 is the output and Vi is the input, pardon me, V0 is the output voltage, V0 is the output voltage, V0 is equal to Vi, when Vi is less than capital V, when Vi is less than capital V, okay, this is what we have observed earlier, okay, Vi is less than capital V, this is equal to V when Vi is greater than V, alright, this is the definition of a positive clipper, positive clipper, that is when Vi goes more positive than capital V, then it simply chops off, for example, if you have a sine orb like this, if you have a sine orb like this as the input and this is the level let us say capital V, then the output of the circuit shall be it will be chopped off, the peak will be clipped off and then negative half it will go like this, then again it will come like this, it will be chopped off and so on, is that clear, this is called a positive clipper, the top of the waveform is clipped at capital V, it cannot go beyond capital V, on the other hand you could have a negative clipper also, a negative clipper by definition is 1 in which the output during the positive values is not affected but the output cannot go beyond let us say minus V, this is a negative clipper, this is Vi and this is V0, so that if you have a sine orb like this, if you have a sine orb like this and this level is let us say minus V, then the output of the of the circuit shall be this, positive half is not affected, in the negative half it cannot go beyond minus V, then it comes back, so this is a negative clipper, if you want to shape the wave such that part of the positive portion is clipped off and part of the negative portion is also clipped off, then obviously the characteristics shall be like this, maybe it clips off at this level and perhaps it clips off at this level, where this could be VA let us say and this could be minus VB, so when this is VO and this is VI, when VI is between VA and minus VB, when the input voltage lies within this range, the output is exactly equal to the input, when the input exceeds VA, the output saturates at VA or clips off at VA, when the input tries to go below minus VB, the output saturates at minus VB, so this is a positive as well as negative clipper, so it is a clipper for both halves, positive as well as negative halves, if VA is equal to VB, then it is called a symmetric clipper, that is if VA is equal to VB, then the chopping off shall be symmetrically done alright, on the other hand if VA is not equal to VB, it is called an asymmetric clippers, a clipper circuit has many applications as you shall see later and the circuit implementation of a clipper circuit will do for the most general circuit, for the most general clipping that is an asymmetric clipping on both sides, then you can specialise it to any particular positive clipping or negative clipping and the circuit is simply like this, but you have a input VI which is to be clipped in the positive as well as negative halves, so what you do is include a resistance R, then you have a diode and a battery VA, a diode D1 and the battery VA, this ensures that the voltage across this cannot exceed plus VA, is not that right, whenever the diode conducts it acts as a short circuit and therefore the output saturates of levels of at VA, on the other hand if you want to level of at minus VB all that you have to do is to reverse the polarity of the diode and the polarity of the battery, V sub B, this ensures, this path ensures that the output voltage cannot go below minus VB, when this diode conducts the voltage across these 2 points shall be minus VB and therefore you can now specialise this to a positive clipper, negative clipper, symmetric clipper or asymmetric clipper, for example if this branch is not there then it is simply a positive clipper, if this branch is not there then it is simply a negative clipper, if both are there and VA is not equal to VB then it is an asymmetric clipper, if both are there and VA is equal to VB then it is a symmetric clipper. Let us take an example, oh suppose suppose we consider the positive clipping alright, that is VI, suppose VI exceeds VA, then you know V0 remains at VA, I am going to explain a little bit about the function of R, why is R there, okay, if VI exceeds VA then V0 is equal to VA, so what happens to the difference VI minus V0, obviously that has to be dropped and that is why this resistance is essential, alright and the value of the resistance is determined by how much current you require, alright, if you need a large current, if you need a large current obviously this resistance has to be small because this determines the current, alright, okay. Let us take an example, this example is a resistance R, a diode with a plus 10 volt D1, 10 volt plus minus and in the other branch the battery is absent, no battery, this is D2 and we wish to find out V2 when V1 is a sinusoid, V1 is a sinusoid Vm sin omega t, alright, you see the characteristic of this circuit is VB equal to 0 and therefore the characteristic is simply, simply this, forget about this line, it is like this where this voltage is 10, this is 10 and in the negative part it is simply the negative axis, so this is V0, this is V2 and this is V1, is that okay, now suppose I supply, let me draw it again, my characteristic is like this, this is 10 and this is 10, suppose I apply sinusoid at the input, I must mark my V1 and V2, suppose V1 is a sinusoid going like this whose peak value is greater than 10, peak value is greater than 10, then what shall I have at the output, what am I plotting here, I am plotting V1 versus omega t, okay, this axis is omega t, then at the output what I shall observe is, it will go like this, level of a 10, come back and then and then remain 0, then again it will go up, level of come back and go to 0, is that clear, the negative halves are chopped off totally, the positive halves are chopped off at 10, if this voltage, if this peak value was at 20, then at the middle point it is chopped off alright, this is the waveform, now you can calculate its average value, effective value and so on and so forth, this is about clipper circuits, the next we consider clamping circuits, clamping circuit, okay, a clamping circuit is a DC is alternatively, can be alternatively looked upon as a DC shifting circuit or average level shifting circuit, for example, you had a, you had a sinusoid like this, suppose I want to shift the sinusoid up so that the minimum value is 0, then obviously my sinusoid would be like this, okay, if I want to shift it up, I could also shift it down so that such that the peak value is 0, okay, maximum could be 0, do you understand what I am doing, so what I am doing is I am simply adding a DC, adding a DC either positive or negative so that the waveform is either shifted up or down, the circuit which performs this job is called a clamping circuit, let us take a few examples and a clamping circuit also is an application of the simple diode, suppose we take an example like this, I have a V1, a resistance R, then I have a capacitor C, alright, I have a capacitor C and a diode across this, now what do you think, if this is V2, this is a sinusoid, okay, sinusoid or any other waveform, I do not really care, let us say it is a sinusoid, like this, alright, now what do you think, what do you think will be the maximum positive value of V2, when the diode conducts, when the diode conducts the voltage across this is 0 and therefore what will happen is this voltage, this waveform of V1 shall go down such that the maximum positive value is 0, now why does it go, suppose V1 equals to Vm sin omega t, let us look at the action of the circuit, what happens is during the positive halves, during the positive halves C charges, alright, C charges to what voltage, Vm with this polarity plus minus, agreed, when V1 goes negative then the circuit does not conduct and therefore C remains at Vm, C remains at the peak value, alright, it is very important that you understand the simple circuit and you see by KCL, by KVL that V2 is nothing but V1 minus Vm, is it not right, V2 is equal to V1 minus Vm, alright, so when during the positive halves, well, if this is V1 and you subtract Vm from it, this is Vm, what happens this, the waveform goes bodily down so that the maximum value is Vm, the maximum value is 0, is that clear, by KVL V2 is, pardon me, about the capacitor, during the positive halves cycle the capacitor charges, it charges to the peak value, then when this voltage goes down the diode does not conduct anymore, so the capacitor does not get a path to discharge and therefore it remains at Vm irrespective of what goes on in the external world and by KVL V2 is this voltage V1, no current in the circuit minus Vm and therefore what happens is V1, this waveform comes bodily down by Vm, in other words its maximum value now becomes 0 and the minimum value becomes minus 2 Vm, that is all, there is no distortion in the waveform, the waveform does not get clipped, there is no distortion, all that happens is it goes bodily down, now if this diode was reversed then obviously what we would do is, the minimum would be lifted to 0, the minimum would be lifted to 0, the maximum would go to twice Vm, all right, now suppose you say, yeah, or it would, the first cycle, in the very first cycle the diode would conduct, then it will remain at Vm, all right, it would not budge, let us put it that, it will only budge if you connect the load here, then the capacitor gets a path to discharge, all right, can it be replaced by a constant voltage source which is equal to Vm, which one, which one is a constant voltage source, V2, in place of C, in place of C can it be replaced by a battery, battery equal to Vm, why say the, you know why you do like to replace a capacitor by a battery, oh, is it not, yeah it is equivalent, yes, because this is constant, yes, that is quite correct, but as a practical engineer you will never do that, the battery is much more costly affair than a capacitor, also it has to be maintained, a battery requires maintenance, all right, what you can do by a passive element, you will never use an active element for that purpose, now question is, does it have to be clamped at 0, does the maximum, can I clamp the maximum at let us say plus 5 volt or can I clamp the minimum at minus 5 volt, all that I have to do is to add a battery, now we will have to add a battery to the diode, all right and look at this circuit, let us take a specific circuit, let us say V1, now I am omitting that internal resistance and so on because it is only the voltage that we are concerned with, there is a capacitance and suppose I have a circuit like this, for bringing variety into experience the diode is in the other direction and we have a negative battery here, let us say 5 volt minus plus, all right, this is my V2, now what is the relation between V1 and V2, V1 is again Vm sin omega t, let us look at the working of this circuit, obviously, obviously when V1 is positive during the positive half cycle, the diode does not conduct and therefore there is no current through C1, when V1 is negative during the negative half cycle, this diode conducts and current flows like this and therefore there is a voltage V sub C developed across C which is of this polarity minus plus, is that okay, is that okay, now what is the value of Vc, Vm, Vc would be equal to V minimum, Vm okay, Vm minus 5 because Vc plus 5 should be equal to Vm, is that okay, now what is V2, V2 is equal to V1, V1 plus Vc, why plus Vc, minus plus minus plus, V1 plus Vc, that is equal to V1 plus Vm minus 5, all right, is that clear, now if you draw the waveforms now, this is the original waveform, where this is minus Vm, now where would this, how would this be, would this be raised or lowered, it would be raised by from minus Vm, it would be raised by, yes, 5 volt, it would be clamped at what, the negative peak would be clamped at, minus 5 volt, do not you see the battery is across the output, so if minus 5 is this level, then all that will happen is, this voltage will go up and this will go up, I do not know, you can draw it, is that clear, okay, so your clamping can be at any level, yes, no, if the battery was not there, then clamping would be at 0, not the capacitor, the total voltage across this cannot exceed Vm and therefore Vc plus 5 volt should be equal to Vm, pardon me, why, it cannot be less than that because the capacitor will charge till the diode conducts and the diode will conduct to a value such that the voltage across the capacitor plus the battery voltage is equal to the minus, negative maximum, in practical case yes, but if everything is ideal then one cycle is good enough because the voltage will go up to minus 5, minus Vm and the whole thing will go up to minus Vm, the capacitor will charge and then the capacitor will refuse to budge, it will remain at minus Vc, whatever the voltage is, is that okay, so this is an example of negative clamping at an arbitrarily set level minus 5 volt, similarly you could get positive clamping at let us say plus 2, is it possible to get clamping at both levels, both positive and negative clamping, obviously not, because it is simply a DC level shift, there is no distortion in the waveform, so if you raise it, you raise it by a certain amount, you cannot raise it simultaneously lower it, then effectively it will be lowered, alright, okay, we next consider a circuit which you might have encountered earlier, a differentiating circuit, differentiating circuit is also a wave shaping circuit, that is whatever the wave is given, it is differentiated and the wave shape is changed, if you differentiate a sine wave, what is the shape of the resulting wave? Cosine Cosine, does the waveform change? No, there is only a phase shift, that is not the case if the waveform is non-silusoidal, for example if the waveform is like this, a sawtooth, have I introduced a sawtooth? Yes, if the waveform is like this, then what would be its differentiated wave form? It would be a constant up to this, then 0, hold it, 0, derivative is not defined, you will get an impulse function, have you had an impulse function? No, we will come to it later, let us say for the time being that it does not drop instantaneously, let us say it drops with a slight slope, all right, what happens is that from here it goes negative over a certain range, then again it goes positive and so on, yes, my figure unfortunately has got distorted, so let us, okay, let us make a triangle, okay, then obviously its differentiated waveform shall be like this, over this portion it will be positive, then it will be negative, then positive, then negative and so on. Now you see from a triangular wave we have produced a rectangular wave, from a triangular wave by differentiating we have produced a rectangular wave, this is indeed wave shaping and this is performed by this simple circuit that is a capacitor in series with a resistor where the capacitor value and the resistor value are so chosen that CR product is much less than, CR what is the unit? Time, is much less than the time period of the periodic wave which you want to differentiate which is equal to 1 over F, all right, under this condition what happens is the following, you have a C and an R, let us say this is VI and this is V0, let us say this is V sub C, then VI is equal to V sub C plus V0 by KVL and if V0 is much less than VC which will be satisfied if the condition CR is much less than capital T, all right, then this is approximately equal to V sub C, all right and therefore V0 which is equal to VR is given by I times R, I is the current in the circuit and I is nothing but C DVI D phi, all right and therefore the output voltage is proportional to the differential coefficient of the input voltage and this is why it is called a differentiating circuit, this is slightly hazy as to why this is needed, you should remember that the price that you pay for a good differentiator is to have an output voltage which is very small, if you want to increase the output voltage by increasing R then the quality of differentiation also goes down because of this approximation that VC is much greater than V0 and this will be satisfied if CR is much less than T, in a sense it means that this current, this current is approximately held, the physically what it means is that this current is approximately held as a at a constant value, when will this so happen, it will happen that irrespective of, irrespective of positive or negative polarity of VI, C gets very little time to discharge through R and this is what this means, CR is much greater than T, okay, let us take an example, let us suppose that we have, we have a square road like this, now let us suppose it is a perfect square road, what you expect is, what you expect is in the practical circuit if I have a C and an R like this, in an ideal circuit the DDT is infinity, DDT here is minus infinity alright, in a practical circuit what happens is the capacitor charges alright, the capacitor charges almost instantaneously if R is very small, in other words the CR product is very small, the capacitor charges almost instantaneously like this, then when the voltage becomes a constant, the capacitor input becomes a constant, the capacitor tries to discharge, it discharges to 0, when it comes here the capacitor charges in the other direction, other direction almost instantaneously and then it discharges like this and so on, this is what we shall absorb if you feed a rectangular periodic pulse at the input, these are pulses of large height and very small width and therefore they are called spikes, you shall get a spike like waveform, this you can very easily absorb in the laboratory by taking a capacitor and a small resistor from the laboratory assistant, simply wear it up like this, take a pulse generator which produces rectangular pulses, feed the input and see what happens at the output, observe the output in an oscilloscope, I believe this is one of the experiments in 110m alright, we shall consider next time tomorrow integrating circuits and their implementation by open differentiating as well as integrating circuits.