 38 lecture and we are going to discuss today negative feedback, negative feedback. As I have already told you, negative feedback reduces the gain. If it reduces the gain then why should one use negative feedback? Why negative feedback? And the answers are many. First, negative feedback desensitizes the gain of an amplifier to variations of device parameters, aging, variations of temperature and tolerances of parameters. The point is it desensitizes, that means it makes negative feedback, makes an amplifier gain less sensitive to all these variations namely device change that is if you substitute a transistor by another or change of temperature or aging or tolerances of components, something a capacitor which is required to be one microfarad, suppose you do not get 1.95 while it makes the gain less sensitive to this, if you use negative feedback. So desensitizing gain, then we have already studied stabilizing the operating point of a device, stabilizing the operating point of a device. You recall that in the BJT, BJT biasing, this resistance RE, all the other resistances, this resistance RE for AC it is to be, it is to be shorted, virtually shorted. Why was this needed? Because otherwise the gain is reduced. Why is the gain reduced? Because if this capacitor is not there, then RE acts as negative feedback and this is very easy to see. If you apply a signal between these two points, if you apply a signal between these two points and this is short, then the same signal appears between the base and the emitter. If it is not short, then the signal that appears between the emitter, between the base and the emitter is the applied signal minus the voltage developed across RE and you know that this voltage and this voltage are in phase, alright. This voltage and the collector voltage, they are out of phase but this voltage and the emitter voltage, they are in phase and therefore the actual signal appearing between the base and the emitter shall be less. If RE is not bypassed, if it is bypassed then the total signal appears and therefore RE applies negative feedback. Now at DC, at DC it is not shorted. At DC the capacitor is open and therefore at DC there is negative feedback through RE and it is this negative feedback which stabilizes the operating point and you know that RE is to be so chosen that RB is, what is the relation? Which one should be greater? RB should be greater, much greater than beta RE by which we mean RB should be greater than beta RE divided by 10, alright. This is the design consideration and this design, if this is satisfied then the operating point is stabilized and this stabilization occurs because of negative feedback, DC negative feedback. At AC we are short circuiting this. Because of DC negative feedback, the operating point is stabilized. We shall look at desensitizing of gain a little later. The third reason why negative feedback is used is widebanding. That is, if you have an amplifier, if you have an amplifier with let us say midband gain of A0 and an upper cutoff frequency of let us say omega 2, alright, if you have an RC coupled amplifier for example with a midband gain A0 and an upper cutoff frequency of omega 2, if you apply negative feedback, if you apply negative feedback then what happens is the curve modifies to something like this. If you apply negative feedback, naturally the gain reduces, the midband gain reduces but the upper cutoff frequency that increases. Omega 2 prime is greater than omega 2 and we shall show that due to negative feedback, due to negative feedback, what you give up, what you give is in gain. What you gain is in omega 2 that is the upper cutoff frequency. So the bandwidth is increased due to negative feedback and we shall show that A0 prime omega 2 prime that is the new gain multiplied by the new cutoff frequency is equal to A0 times omega 2. That means the product is a constant. You can increase one then the other shall decrease. We shall show this accurately, rigorously. Then the fourth reason why negative feedback is used is that negative feedback can be used to reduce distortion and noise in the output of an amplifier. Negative feedback reduces distortion and the fifth reason amongst many others, fifth major reason is negative feedback helps to increase the input impedance and decrease the output impedance. Negative feedback helps to increase the input impedance and decrease the output impedance. We shall look at each of these separately, all right. The first point was, first point was desensitizing gain, all right. In order to look at that, in order to appreciate how the gain becomes insensitive to variations of the device parameters or tolerance of circuit parameters or aging or temperature, just look at this G by 1 minus GH, this is the general formula, all right, 1 minus GH. Now if GH is much greater than 1, if GH is much greater than 1, then you see GF becomes approximately equal to minus 1 over H. In other words, it becomes independent of G, the internal gain of the transistor of the amplifier and therefore even if the transistor is replaced by another one or the circuit tolerances are large or the temperature is increased by 50 degree, the gain shall be determined by H which is the feedback network. As long as you keep the feedback network stable, everything, the gain shall be determined by the feedback network only. And this, one of the examples we have already seen in the op-amp network, for example, if you have an op-amp with an input at the non-inverting terminal and R1, R2 and R1, and the gain simply becomes 1 plus R2 by R1 which is independent of the op-amp parameters, op-amp gain, op-amp temperature or whatever it is, you can increase the temperature by 50 degrees, nothing will happen, all right. So it desensitizes gain. To take a more concrete illustrative example, suppose you have an feedback network, a feedback amplifier whose gain is 1 by G by 1 minus G H and take some specific example, let us say H equal to, if it is to be negative feedback then H must be negative. Suppose H is minus 0.01, very small fraction of the output, 1 percent of the output is fed back to the input. And suppose G is equal to let us say 10,000, can you tell me what GF is? GF would be 10,000 divided by 1 plus 100 which I can write as 100 divided by 1 plus 0.01 and it is approximately 100 times 1 minus 0.01, 0.01 and that is equal to 99, all right. Suppose the gain due to a change of temperature or device, may be the op-amp is replaced by another op-amp, suppose the gain changes drastically, suppose the gain becomes 2500, that is it drops down by 75 percent of its value. How does GF change? The new GF shall be 2500 divided by 1 plus, yes 25 and that you can see is 100 divided by 1 plus 0.04, all right. I can write this as 100, can I change the page? 100 divided by 1 plus 0.04 is approximately equal to 100, 1 minus 0.04 and you see this is only equal to 96. So gain change from 10,000 to 2500, a drop, a change by 75 percent makes a change of the feedback amplifier gain only to the extent of 3 percent from 99 to 96, is that clear? 75 percent change here is reflected only as 3 percent change and therefore the gain becomes insensitive to the gain of the internal amplifier. Whatever cause is internal amplifier gain may change due to change of temperature, due to change of device or whatever it is, all right. And you also notice that 1 by minus H, if G had, if G H was much greater than 1, then the gain would have been minus 1 by H which in this case is 100 and you see 99 and 96 are very close to 100. It approximately becomes equal to 1 by H. That is its control only externally, all right. Is the point clear? Okay. Now let us see, we have already explained the advantage of negative feedback for stabilizing the Q point, stabilizing the operating point. Let us look at widebanding. How does negative feedback cause widebanding? Suppose your internal amplifier G is given by A0 divided by 1 plus J omega by omega 2. Then what is omega 2? Omega 2 is the upper cut-off frequency. You see when omega is equal to omega 2, the gain becomes A0 divided by 1 plus J1, the magnitude becomes why A0 divided by root 2. Therefore omega 2 is the upper cut-off frequency. Now suppose you apply negative feedback. You apply negative feedback and suppose capital H equal to minus H, a constant where H is greater than 0, all right. You apply negative feedback. Then your feedback gain becomes A0 divided by 1 plus J omega by omega 2 divided by 1 plus A0 H divided by 1 plus J omega by omega 2. Is that clear? I have simply written this as G, this is G and this is GH, well minus GH, all right. Now if I if I clear this of the fractions, what I get is the following. You can very easily do this algebra. You can show that GF is nothing but A0 divided by 1 plus A0 H, 1 by 1 plus J omega divided by omega 2, 1 plus A0 H. If you clear the algebra, this is what you will get. If you notice this is of the same form as the previous gain expression, we can write this as A0 prime divided by 1 plus J omega by omega 2 prime where A0 prime obviously is the mid-band gain of the feedback amplifier, amplifier with feedback and omega 2 prime is the upper cut-off frequency of the amplifier with feedback. And you notice that omega 2 prime is equal to omega 2 multiplied by 1 plus A0 H and therefore omega 2 prime is greater than omega 2 which means that we have wide-banded the amplifier. The amplifier mid-band range has been expanded. You also notice that the mid-band gain A0 prime multiplied by the upper cut-off frequency omega 2 prime is nothing but A0 times omega 2, all right. This is A0 prime and this is omega 2 prime and therefore the product is the same. In other words, by applying negative feedback you cannot increase both. You increase one then you decrease the other, all right. Now this is a fundamental law. This is a fundamental law. It is a reflection of the uncertainty principle. The product remains a constant. It reflects in many engineering situations Heisenberg's uncertainty principle. So you understand why we say that negative feedback helps us in wide-banding. Next we look at distortion, how it reduces distortion. Distortion cannot happen in small signal amplifiers. In small signal amplifiers, so the signal excursion is around a small region around the Q point. There is no distortion. We have linear amplification. But suppose we take a large signal amplifier, like a power amplifier where the excursion is between the extreme limits, then obviously we reach non-linear regions and therefore there would be distortion. Now if I have distortion, if I have an amplifier with distortion, I can represent it like this. By an equivalent block diagram like this, we can represent a large signal amplifier or an amplifier which produces distortion as 1 having a distortion less amplification that is linear amplification. And then the distortion component, we can simply model it as another distortion voltage VD added to the perfect amplifier, all right. Is that clear? We are looking upon the output as being due to the output of a distortion less amplifier added to a distortion, all right. Can you do that? It should be obvious that we can do that, all right. That is you take the distorted waveform. For example, if you have a waveform like this, can you tell me what kind of distortion is this? There is a dip here. This is due to the second harmonic. You see, the fundamental is like this. What will happen to the second harmonic? Second harmonic will go through 0 somewhere here. And that is why there is a dip here, all right. This is due to second harmonic and I can always write this distorted waveform as the fundamental waveform plus the distortion due to the second harmonic. So this is what I have done here. This part is the linear amplification, distortion less and then I add distortion. In other words, the percentage distortion, you see the output, output is GVI, this is perfect, this is linear distortion less part plus VD so that measure of distortion as a percentage is equal to VD distortion voltage divided by GVI, all right multiplied by 100 percent, all right. Suppose now to this amplifier, to this distortion full amplifier, we introduce a negative feedback. Let us see what happens. GVI plus VD, yes, this is the sum. Percentage distortion is distortion component divided by the undistorted output multiplied by 100, okay. If VD is 0, there is no distortion, okay. Now to this amplifier, we apply now negative feedback. This is V0, we apply a feedback through a feedback network H and a summer and this is now VI, all right. Then V0 as you see, what is this signal? This is VI plus H times V0, okay and therefore V0 is equal to VI plus H V0 multiplied by G plus VD and therefore we get V0 1 minus GH equals to VIG plus VD or V0 is equal to VIG divided by 1 minus GH plus VD divided by 1 minus GH, all right. Now we play a trick. What we do is, we introduce an amplifier here at the input whose gain is 1 minus GH, all right. That is, we produce the amplification, total amplification in two stages. One is G and one is 1 minus GH, all right and let this be my effective input. VI prime, is this point clear? Actually what we are doing is exactly what we do, for example is a stereo amplifier. The first stage is a voltage amplifier, it is a small signal amplifier. So it is fairly linear, fairly distortion less and then we have the power amplifier which has lot of distortion, all right. So what we do is, we introduce an amplifier of gain 1 minus GH here. Then you see, does this affect the distortion voltage? No, but it definitely affects the actual input VI prime and therefore my V0, my V0 then becomes VI prime G, why? Because VI prime 1 minus GH equals to VI and therefore VI divided by 1 minus GH becomes equal to VI prime. So VI prime G plus VD divided by 1 minus GH, agreed. What is the percentage distortion now? Obviously it is VD divided by 1 minus GH multiplied by VI prime G and you see if GH is negative then the percentage distortion is reduced. Originally the percentage distortion was this much, VD divided by VI prime G multiplied by 100. Now it is reduced by the factor 1 minus GH, is that okay? What we have done here is that the gain has been split into 2 parts. One is a small signal amplifier of gain 1 minus GH and then an amplifier of gain G and we are assuming that all the distortion occurs in the second stage and this is precisely what happens in practice. The output stage when you have to drive a loud speaker has to have a lot of distortion because it is a power amplifier. In a small signal amplifier, the pre-amplifier as it is called distortion can be kept very low and therefore negative feedback is a means of reducing the distortion. Finally we consider how negative feedback increases the input impedance and reduces the output impedance and we shall illustrate this rather than in terms of block diagram and general theory. We shall illustrate this by means of a simple example and the example that we take is that of an emitter follower, emitter follower that is emitter follower is also a common collector amplifier okay. What we have is this is plus VCC, this is the input, let us call it V sub i, V sub s we have been calling this with a series resistance R s and let us say this is my output voltage will be, these are root mean squared values okay. It should be quite familiar with this circuit by now, it is an emitter follower or common collector amplifier okay. This is an example of negative feedback as I have already explained, the actual input between the base and the emitter is equal to the input, applied input minus the voltage developed across R e and as this voltage and this voltage are in phase it is a case of negative feedback that means feedback reduces the actual input applied to the transistor. And to analyze this, to analyze this you draw the equivalent small signal equivalent circuit V s then R s then you ignore R b, R 1 parallel R 2 you ignore or include this in R pi, now you cannot include it in R pi here, why not? Because R b does not come in parallel with R pi, R pi goes to R sub e and that goes to ground alright. Then you have a current generator beta I b, if this current is I b, beta I b and where does this go? It goes to, it goes to, no there is no R c, it goes to ground, the collector is connected to the positive of the power supply and the power supply for A c is ground and therefore you can calculate if you want the output voltage V 0, obviously V 0 is equal to beta plus 1 I b times R e alright. This current is beta plus 1 I b because I b comes here and beta I b comes here and V s the input voltage is I b multiplied by R s plus R pi, R s plus R pi plus beta plus 1 I b R e agreed by K v l this voltage is equal to drop in R s, drop in R pi and then drop in R e, the sum of the 3 and therefore the gain is simply the ratio of the 2, V 0 by V s from which I sub b shall cancel and therefore the gain g f, which feedback gain is now equal to beta plus 1 R e divided by R s plus R pi plus beta plus 1 R e, which obviously is less than 1, is not it right? Obviously is less than 1. You also know that R s, R pi and R e are of the same orders of magnitude and beta is of the order of 50 and therefore this term dominates in the denominator which means that this is approximately equal to 1 and this is why it is called an emitter follower that is the emitter voltage follows the input voltage. This is why it is called an emitter follower, emitter follower. Now let us look at this circuit again alright. Any question on this point? The gain calculation, the voltage gain is less than 1. Is there a current gain? What is the ratio of the output current to the input current? Beta plus 1, input current is I b, output current is beta plus 1 I b and therefore there is a current gain and therefore there is a power gain. The power gain is approximately equal to beta plus 1 because the voltage gain is approximately unity. Therefore the circuit is indeed useful but its use comes from a different consideration. An emitter follower is hardly ever used as a voltage amplifier or current amplifier. What it is used as is now illustrated by the effect of negative feedback on the input impedance and output impedance. Now if you look, if V s looks into the circuit what impedance does it see? What is the input impedance R i? It is simply V s divided by I b. No it does not see R s. It sees R s plus R pi plus beta plus 1 R e because this is V s divided by I b and therefore you see the input impedance is R s plus R pi plus beta plus 1 R e and if all of them are 1 k, if all of them are 1 k and beta is let us say 49, 49 then 50 plus 51 plus 52. This would be 52 k, all right. On the other hand if R e was not there, if R e was shorted to ground what would be the input impedance? It would be simply 2 k R s plus R pi. So the input impedance has been increased 26 times. What does the input impedance increase the factor depend on? Basically it depends on what? Beta and you know the Darlington connection can multiply, can square the beta. If you use 2 transistors the effective beta can be beta squared or approximately beta plus 1 whole squared. So if instead of 1 transistor you had used 2 in a Darlington connection then you could have increased the input impedance to 2500 k which is 2.5 meg and this is the major use. That means it produces an input impedance which is very high compared to whatever source you are connecting whether it is a microphone or a loudspeaker or whatever it is you can make the input impedance very high and that is because of the negative feedback, all right. Now let us see what is the effect on the output impedance, output impedance and for that one has to proceed very carefully. How does one calculate the output impedance? Well obviously output impedance would be impedance looking back from the output terminals with V s shorted but beta I B left untouched not open. Beta I B is a control source, control source must not be touched, all right. Let us see how to calculate this. So we draw the circuit backwards. We have an RE, across RE we connect a voltage source V and we find out the current I. Then R0 would be equal to V by I and if I look at the circuit what I have is an R pi and an RS, R pi and an RS the current through this is I sub B and no voltage, this is shorted. The voltage source should be shorted and then we also have across RE a current coming to this node from ground and therefore we have a beta I B from ground, all right. So there are 2 relations now which can be very easily written. The 2 currents that are coming in, 3 currents are coming in I, I B and beta I B and therefore I plus beta plus 1 I B should be equal to V by RE, this is KCL, okay. V by RE is the only current leaving, this should be equal to this current plus this current plus this current and the other relation is V, V shall be equal to the drop in R pi and RS, right. So it is minus I sub B times RS plus R pi. Now we have achieved what we wanted to achieve and therefore we get, have you been able to write this down, okay. Therefore we get R0 which is equal to V by I that is equal to minus I B RS plus R pi divided by minus beta plus 1 R B I B, I am taking help of this relation, I am writing the expression for I. So this goes to the right hand side plus V by RE that means minus I B RS plus R pi divided by RE. Now you cancel out I B, so you get RE RS plus R pi divided by RS plus R pi plus beta plus 1 RE, I hope I have not made a mistake or have I? No, not yet. Now in this practical consideration say that RS plus R pi can be ignored, okay. So if I ignore this then you see RE and RE cancel and therefore R0 becomes RS plus R pi divided by beta plus 1 and what is beta, what is 1 compared to beta? You can also ignore that and therefore you get RS plus R pi divided by beta and if RS if the source resistance is negligible, if RS equals to 0 then you see R0 is simply equal to 1 over GM and this is independent of all other parameters of the circuit, it only depends on the transistor. Not just on the transistor, it depends on the I sub C, if the collector current is known then you. Now what is the order of GM? 40 times I sub C, maybe I sub C is of the order of millions then 40 times 10 to the minus 3. So this is GM, then what is the value then? 25 ohm which is indeed a very low resistance, okay. It can be used to drive the next stage which might have an input impedance, if the next stage is a common emitter amplifier then the input impedance is of the order of what? Order for a common emitter transistor, what is the input impedance? It is simply R pi, it is simply R pi, R pi is of the order of K, 1 K is 40 times 25 and therefore there is no mismatch, it will deliver the total signal to the next stage. And with this we close our discussion on analog circuits. We shall have a brief glimpse at digital circuits in the few minutes that is left today and in the last 2 lectures on Friday, digital circuits. As you know digital circuits are the most important circuits today, they form the heart, they form the heart of a digital computer, microprocessor, any practical system whether it is radar or sonar or a defence countermeter, ECM, digital circuits abound everywhere and therefore it is important that you know at least what the rudiments of digital circuits are. The rudiments are, the elements are again diodes and transistors. As you know if you take a diode, if you take a diode and the voltage across this is less than the threshold or negative then the diode does not conduct current so it is off. On the other hand, on the other hand if the input voltage is greater than the threshold then the diode is on. So it is a difference between open circuit and short circuit and if you consider a circuit like this, let us consider a circuit like this, this is 5 volt, this is input Vi and this is V0. If Vi is low, is lower than 5 volt then the diode conducts, diode conducts and if the diode is good, is a good one then V0 shall be equal to 0. So if Vi is less than 5 volt, V0 equal to 0. If Vi is greater than 5 volt then the diode does not conduct and therefore V0 becomes equal to 5 volt alright. So 5 volt input therefore is a transition region between life and death alright, between 0 and 5 volt, there are 2 levels, 0 and 5 volt and this level traditionally may be considered as the binary 0 level and this level is the binary 1 level and all digital circuits, synthesis analysis they are based on logical 0 and 1 of the binary system rather than the actual voltages. Now this 5 volt for example could be 4.9, there may be a small current in the reverse bias diode, it does not matter, 4.8 that will also be considered as logical 1 and therefore between logical 1 and logical 0, there is a tolerance and this is one of the major advantages of digital circuits. In analog circuits, wherever there is a small change in the parameter or voltage, everything changes, here it does not change. Even if it is not 0 but let us say 0.5 volt because of noise or because of pick up, it does not matter, it will still be interpreted as 0 alright. This is diode, a transistor also can act as a switch, consider a common emitter transistor plus VCC and let us say we apply the input here, if input is and the output is taken from here to here, if input is negative or less than 0.7, less than the threshold of the base emitter junction, if VI is less than 0.7, then the transistor does not conduct, right. If it does not conduct, then what is V0? V0 becomes equal to plus VCC. On the other hand, if VI is much greater than 0.7, may be 5 volt, VCC can also be 5 volt. If it is much greater than 0.7, then the transistor conducts very heavily and this voltage would be very low and if you recall the transistor characteristics, if the transistor conducts very heavily, then naturally you might go into this part of the characteristic which is saturation characteristics. In digital circuits, the drive at the base is either very large or very small and therefore V0 becomes approximately equal to VCE saturation voltage and this is approximately for a silicon transistor, it is 0.2 volt. Now this logical level is considered as 1 and this logical level is considered as 0, all right. You also notice in this case that if the input is high, then the output is low. In other words, the output is the complement of the input. It is the other way round, if the input is low, then the output is high and therefore if you call the input as A, logical A, then the output is complement of A, A bar and such a circuit is called a not gate or an inverter. And it is from this point that we shall start on Friday.