 problem session, basically our aim in this session would be to see the limitations of the assumptions that we made during the analysis of the oscillators. You recall that our active devices were all very ideal devices, infinite input impedance, no capacitance inside, zero output resistance for some of them for LC oscillators, we could permit a little, a little R0 there. So our problems would be directed towards practical oscillators in which the effects of the transistor capacitances and transistor imperfections can be taken into account. Before we take up a problem there is a small, small circuit that is left to be, to be discussed that was a crystal oscillator. Let us quickly do that then we will go to problem. Crystal oscillator of the Colpitz type, Colpitz type. You know that a crystal oscillator, a crystal is used in when its reactance is inductive, okay, between omega s and omega p, between omega s and omega p. So Colpitz type oscillator uses two capacitors and one inductance and to bring variety into experience let us say we aim at a FET oscillator. Then in an FET you have the R sigma C sigma, the input impedance is no problem, the output impedance you have an R sub D plus VDD. Then what you require is from here to here what you require in Colpitz is a capacitor in shunt then an inductor and then another capacitor. This is the Colpitz circuit, this should go here and this should come here. Now since the crystal is to be used and the crystal is used as an inductor it is this inductor which will be replaced by the crystal. Now where does this inductor appear? Between the drain and the gate and therefore you connect the crystal between these two points. You connect the crystal between these two points, okay. Then obviously you have to use a capacitor, the first capacitor, well you use it to ground, the first capacitor. The second capacitor is to be between this point and ground, okay, between this point and ground. So use another capacitor and to be able to terminate the gate you use a resistance here. Is this resistance essential? Would someone tell me why this resistance is used? Is it necessary for biasing? No, it is not necessary for biasing. It is yes and no. It is necessary because for DC you have to establish this point as zero potential. Now if you leave it open it is anybody's guess what potential it would go to. It might pick up straight charges and go to certain potential. Even if it wants to go, if you provide a resistive part the charge will pass to the ground and therefore this R G is essential in theory no, but in practice yes, okay. So the V G S, V G S is zero minus V S. This is the circuit of a crystal oscillator using the Colpitts type, Colpitts type architecture and this circuit was independently discovered by a gentleman by the name Pierre Sey and therefore this is also known as the Pierre Sey oscillator. Pierre Sey, Pierre Sey oscillator, okay. Pardon me? Oh we can, instead of R D, instead of R D we can use a chip but you know in LC oscillators my source can have an output resistance. It does not matter. It does not affect the operation of the oscillator and therefore a resistance can be used but if you do not want dissipation you can replace this by means of a chip, fine. You can replace this by means of an RF chip. Now these are all very simple circuits that we have discussed in theory. Now let us complicate our life and you will see life is indeed complicated. Even with a single transistor the imperfections cause many problems in the analysis and design. The first problem that we take, question 1, draw the circuit with me. I have not done this circuit in the class because there are hundreds of varieties of circuits. I thought some of them I will use to illustrate, illustrate how a practical oscillator is analyzed and designed and this is the first problem that I take. Draw with me. There is a transformer, 2 coupled coils and the turns ratio is N1 is to N2. This is N1 and this is N2. This, the primary of the transformer is used to form a tuned circuit with a capacitance, let us say C prime. There is a reason why I am using C prime. You will see later. This acts as the load of a transistor and therefore this is plus VCC and this terminal goes to the collector. The transistor is a common base, common base circuit instead of a common emitter. You see the variety of oscillator circuits that exist. The emitter is connected to the secondary and in addition for biasing you have an R sub B and C sub B which goes to infinity. It is very large. It is a bypass capacitor and in addition what else do you require to bias the transistor? Q. I have shown a PNP. This is not necessary. You can use it an NPN. What else is necessary for biasing? Pardon me. I must have a resistance between the base and the power supply. Let us call these resistances R1, some R1. Is that all? That is all but usually a resistance is connected between the emitter and ground also. RE which is usually chosen to be much higher compared to RPI. You see RPI will come in parallel with RE, is not that right? RPI from base to emitter will come in parallel with RE and therefore in order that it does not load unnecessarily we usually make RE much greater than RPI. This circuit does oscillate. It is not very easy to see how this circuit oscillates but obviously feedback is applied through a transformer, through magnetic coupling, through a transformer and the transformer winding, you know the meaning of the dots that the potentials of these 2 points rise and fall simultaneously. In other words if this goes up then this also goes up. So there is positive feedback. There is positive feedback in the circuit and it does oscillate. Let us see how it oscillates. The question is to analyse this circuit for the frequency of oscillation and the condition for oscillation. Now obviously the first thing we do, I have told you that in oscillator analysis you do not have to identify the A circuit and the beta circuit. You can simply draw the equivalent circuit and see what condition is to be satisfied in order that the input voltage is V pi when the output voltage is g m V pi multiplied by the load. So you just draw the equivalent circuit and oscillator analysis is eased out. Let us see what the equivalent circuit is. Now you see let us take the point E, C and B. B is grounded for the transistor. Between E and B there is a resistance R pi and this voltage is V pi. Do not make a mistake about the polarity. This is V pi. Polarity is B side is positive and emitter side is negative. Between C and D we have g m V pi. It is a common base circuit so the g m V pi is drawn horizontally. There is no reason why you could not draw it otherwise. Then you have at the collector between collector and ground you have the capacitance C prime then the inductance L with this polarity and you have another secondary whose dot is here and this couples to the emitter. In addition you have a resistance R E to ground. Now we have ignored several things in this equivalent circuit. R B C B combination is short so we ignored that completely. We have ignored the transistor capacitances. Since this is a high frequency oscillator transistor capacitances usually cannot be ignored. In other words I must include a C mu first, C mu from collector to base and therefore C mu can add to C prime. Is that clear? C mu is from collector to base therefore it comes in parallel to C prime. So C prime plus C mu alright. C mu is taken care of. Then R pi, R pi appears between this point and ground. I am sorry not R pi, C pi, C pi. This also should have been taken into account alright. Now then is then is the question of analysis. You see whatever the connection here whatever the connection at the emitter terminal, the voltage output V0 appears across this parallel combination of C prime and L due to a current g m V pi passing through this and therefore I do not care where g m V pi goes. I do not care. My V0 would be simply the this current multiplied by this impedance which is not not quite correct because the because of the transformer there is a reflected impedance and that is how this termination comes into effect alright. So let us say let us say R E parallel R pi. Let us not neglect completely. Let us call this as R pi prime. We will neglect it later whenever necessary. Since we can take care of it it comes in parallel with R pi why why ignore the poor fellow right at the beginning okay. Let us call this R pi prime. Then you see as far as the collector as far as the collector is concerned, this ratio was N2 is to N1 alright. N2 is to N1. It was N1 is to N2 from the other side. So N2 is to N1. Now what does this the primary sees? What does the primary see by looking into the secondary? There is a reflected impedance and the reflected impedance would be resistance would be multiplied by N2 by N1 squared and the capacitance would be divided by N2 by N1 squared and therefore as far as the collector is concerned, my equivalent circuit is g m V pi. Let it go to ground. I do not care wherever it goes. Then a capacitance. Now what will be the total capacitance? C prime plus C mu plus the reflected capacitance. What is the reflected capacitance? C pi by N2 by N1 whole squared okay. So C pi N1 by N2 whole squared. Is that clear? Let us call this capacitance as C. Then we have the resistance. What is the resistance? R pi into N2 by N1 whole squared. Is that clear? R pi prime. That is right. N2 by N1 whole squared and you have finally the inductance L which is coupled to another inductance with this polarity. Now what should be the condition of this terminal? Once you have taken the reflected impedance, this should be open and this voltage would be V pi with what polarity plus minus. Is that clear? Okay. What I did was I wanted to calculate V0. So I look at I look at this transformer. This transformer reflects a certain impedance into the primary. I have reflected that okay. So the the amplifier circuit, the amplifier circuit consists of g m V pi passing through C prime plus C mu L and the reflected impedance. So I have been able to calculate V0. Then V pi appears across the secondary. V pi appears at the secondary and the secondary should now be such that there is no reflected impedance. Obviously the secondary is to be open because I have taken care of the reflected impedance alright and it is with this polarity. Pardon me? Which one can be short? If it is short then there will be short reflected here which means that V0 will be also be short, agree? It is only infinite impedance reflected as an infinite impedance. Yes. How is g m V pi going to ground? How is it going to ground? As I said I do not care where it is going. It has to come through this impedance. I am calculating as far as my calculation of V0 is concerned this circuit is good enough. I do not care where it goes. I can leave it empty if you so desire. It goes somewhere but it has to come through this combination okay. Now I also know because of the transformer ratio, N2 is to N1, I also know that V pi should be equal to this is V0. V pi should be equal to minus because the polarities do not agree. N1 by N2 times V0, agree? What about the coupling? I have taken care of that. Oh we assume that it is ideal, 100 percent. That is right. Sir but that is well only when the current through this load, ZL? Once I reflect the impedance the current can be put equal to 0 in the second wave. No but that is well only when the current is only because of the secondary current. Here the current is not going only because of the secondary current. It does not matter. There is a current here. There is that through the load. It does not matter. This current goes through the load. Now the load I have reflected into the primary. Therefore the secondary can be left open. No, no current in the secondary because the effect of the secondary current has been reflected already. I am not saying that. I am saying that when we reflect this impedance and multiply by other turns ratio square. Right. That we do only because the current through this load is because only because of secondary load. No only. You see so far as if there is a load R here RL and the turns ratio is N is to 1 then this is equivalent to a simple resistance N square RL in parallel with a transformer whose output is open. As far as this side is concerned this is a perfectly valid circuit. Sir what if there is another DC or say AC connected to RL secondary? Oh, that is a different story. So here that is what is happening because... There is nothing connected. The current g m d pi is also adding to this load. Oh, that does not matter. How does it matter? Between this point and this point whether a current is coming or not the impedance is that of Rbe, Rpi, Cpi. Impedance does not change. Whatever the current I do not care. Okay. Think about it it should be clear. Now therefore my equation my equation for oscillation should be this is Vpi and my V0, V0 should be equal to minus g m Vpi divided by the admittance of the total combination that is j omega C plus 1 over j omega L plus Rpi prime N2 by N1 whole square. Agreed? That is my V0. g m Vpi passes through this. That is my V0 and Vpi, Vpi is minus N1, N2 by V0. So substitute Vpi here then cancel out V0. You get a certain relationship between the coefficients. And obviously that can be satisfied. Let me write this relationship. 1 would be equal to g m N1 by N2 divided by j omega C minus 1 by omega L plus Rpi prime N2 by N1 whole square. Okay. I have made a mistake. Alright. Now obviously this can be satisfied. The right hand side should not have any imaginary part because the left hand side does not have one and therefore the frequency of oscillation is 1 by square root LC. This C you must remember is not simply the external capacitance. It is supplemented by C mu and Cpi N1 by N2 whole square. Okay. External is the external inductance. That is used in the primary of the circuit. Now if this is so then obviously what we require is that g m N1 by N2 multiplied by Rpi prime N2 by N1 whole square. This would be the condition. This goes to 0 and this comes up and therefore this is the condition that I mean. Which means that g m Rpi prime should be equal to how much? N1 by N2. Now g m Rpi prime is RE Rpi divided by RE plus Rpi. This should be equal to N1 by N2 and if RE is much greater than Rpi by design then obviously beta should be equal to N1 by N2. In other words the terms ratio of the transformer should be equal to the beta of the transistor. Yes. Oh analyze the circuit to find the conditions of oscillation. Then I can give some numbers in terms of the circuit parameters. We require to calculate the frequency of oscillation. Okay. Now this problem was not too difficult. Let us next come to a difficult problem. More difficult problem. Although it looks simple. Simply says analyze the Colpitt's oscillator circuit shown below. Taking account of transistor imperfections. The circuit, this is also revealing what kind of circuit one makes in practice. Instead of a resistive load one uses an RF choke, RFC so that there is no DC dissipation. Then you have the transistor RE CE. CE goes to infinity. It is a Colpitt's circuit. So you have let us say a C1 prime, then an L and a C2 prime. You will understand why I am using primes because they are supplemented by things and then this goes to the base. Do you require a capacitor? Yes you do. You require a CC which is very large. You require, otherwise this DC will pass to the base. The circuit is not complete. You require the base biasing and the base biasing is done by 2 resistors, RB1 and if you want to conserve, no, if you want RB1 and RB2 not to load the transistor then you use an RF choke. Otherwise you do not. Usually RB1 and RB2 parallel combination is much greater than R pi and therefore that can be neglected. RB1 and RB2. The question is to take care of all possible imperfections of the transistor and analyse the circuit. In the usual analysis in the class we had ignored all those imperfections. Okay. So the first thing we do is to draw the equivalent circuit and now the real problem starts. RB2 should be grounded. I beg your pardon. This should go off. I beg your pardon. Please do correct the circuit. RB2 should be grounded. It is not connected to CC. Okay. Now let us draw the equivalent circuit. The equivalent circuit starts with RB which is RB1 parallel RB2. Then an R pi in parallel with C pi, C pi then you have this voltage is, I beg your pardon, V pi. This voltage is V pi plus minus, a gm V pi goes to ground. The RFC poses no problem. It is infinite impedance. We ignore that. Then we have the C1 prime L and C2 prime. This goes to this point and one thing that we have not drawn in this circuit is this minis of C mu even though it is a small value. Now C mu complicates the problem of analysis because this is a coupling between, if we call this as V0 between V0 and V pi. We cannot isolate the 2 circuits. There is a connection from here. There is also a connection from here. Now can we use miller effect? Please do consider. Please do listen to me very carefully. Can we use miller effect? You see the load here is not resistive. So that C mu 1 plus gm RL that will not work. Agree? The load here, the load that it sees is not resistive. In the miller effect we always had a C mu, the miller capacitance reflected was C mu 1 plus gm RL. It is not RL. It is no longer RL here. But there is a saving grace. The saving grace is that at the resonant frequency, at the frequency of oscillation, you see this is not a wideband circuit. It has to work only at one frequency. And at the oscillation frequency, at the oscillation frequency these will resonate and what will be left is a parallel combination of RB and R pi. Is that clear? If the circuit oscillates then between this point and ground the gm V pi source shall see a resistance because the condition for oscillation is that the inductor should resonate all the capacitances that are present in the circuit. Otherwise it cannot oscillate, okay. So approximately, approximately you can replace C mu. This is approximate. It will not give you an exact solution. I will come to the exact solution later. But you can you can ignore the effect of C mu by including a capacitance here which is equal to C mu then 1 plus gm R pi prime, agreed? And in addition C mu shall also come in parallel with this. So there is a C mu here. Yes? At the oscillation frequency, what this source sees is an anti-resonant circuit. Anti-resonant circuit C1 this capacitor in parallel with an inductive reactance which cancel each other, okay. And therefore what will be left is what will be left is this resistance R pi parallel R B. This is what it will see. You agree? No. This is not correct because gm V pi does not see the resistance across this. It is not a pure C L R. If it was so it could have. Is the point clear? No sir. I have been arguing wrongly and I am waiting and pausing after every sentence nobody objects, okay. From here to here what this current source sees is not a pure resistance because the resistance is here. It is not a parallel LCR circuit. If the resistance was put between this point and ground, yes I would have agreed. But it is not so. It is after the inductance. You see what I am seeing is the following. What the current source sees is a capacitance, then an inductance, then a capacitance and a resistance, okay. So what it sees here at resonance is not simply this resistance. Agreed? It is not this resistance only. It sees C in parallel with L plus this RC combination. That has to resonate. It does not mean that this input impedance will be this resistance. Nevertheless the input impedance would be resistive and that resistance has to be calculated. It is not simply this resistance, okay. So while in theory the procedure would be correct, the reflected Miller capacitance is not this. It would be something different. C mu 1 plus gm let us say r pi double prime, effective impedance seen by gm r pi, gm v pi. Do you see the problem of analysis? And then C mu has to come here. If I do this, mind you this is approximate. If I do this, then my equivalent circuit becomes simple. What I get is the following. We have to include C mu, C 1 prime, C 2 prime and C pi. All of them, okay. Then my equivalent circuit becomes the following. I get gm v pi. You understand the simplification. Even with the simplification it is an it is an approximate analysis. I will come to the exact analysis a little later. What I get is gm v pi, C 1, then an L, C 2. Now I have removed the primes. C 1 now is C 1 prime plus C mu. C 2 is now C 2 prime plus C pi plus Cm, the Miller capacitance, okay. And that comes across r pi prime and this is my v pi with this polarity plus minus. Now you analyze the circuit. You analyze the circuit and obviously you do not write loop equation or node equation. You make a Thevenin source here. You make the equivalent of a Thevenin source here and then it is a one potential division which gives you a single equation giving the frequency of oscillation and the condition of oscillation. I would skip this algebra. It turns out that omega not squared. Can you tell me what omega not squared would be? 1 by L, C 1 C 2 divided by C 1 plus C 2. It is the series resonance of these 3, okay. And the condition for oscillation is that gm r pi prime is equal to C 1 by C 2. gm r pi prime equal to C 1 by C 2 and if r pi prime is approximately r pi then you see that beta has to be C 1 by C 2, okay. This is approximately equal to beta. I would skip the algebra. If it is not so, if Miller effect is not acceptable, you see these results obviously will be quite away from what the exact results are because it was based on the assumption that the shunting effect or the bridging effect of C mu could be removed by using 2 capacitances, one at the input and one at the output which is not quite true because the gain is not purely resistive. We consider the gain only at omega 0 at that frequency. How does the exact circuit differ from this? Isn't it only in the inclusion of a capacitor here across L which is similar, right? So all this approximation was not really necessary. It does not, it does not give you more complication than what you did in analyzing this circuit. Instead of L, instead of L, it would be g omega L 1 minus omega square L C mu. Is that clear? Yes. Can it be said that I have evaluated minus of C mu in parallel to this? Oh, let us go back to this circuit. Where was C mu? C mu is between this point and this point. Now this point is the same as this end of L and this point is the same as the other end of L and therefore all this fuss about C mu and all this care about C mu was not really necessary but it creates complications. Having a C mu here, having a C mu here, obviously my C 2 will now be, can you tell me what is C 2? C 2 prime plus C pi, that is all. Nothing else, okay. And what is C 1? C 1 is simply C 1 prime. I am not including C mu, all right. If I include C mu here, then obviously these are the values. And in the analysis, all that you have to do is, first you apply Thevenin's theorem here, then you write a single potential division equation where you wrote J omega L, now should be replaced by J omega L divided by 1 minus omega squared L C mu. So it is not too difficult. What is difficult is that the condition of oscillation will now be of what order, what would be the condition of oscillation, what order? In the previous case, there were 3 reactances, L C 1 and C 2, so the degree of the denominator would have been 3. Now the degree would be 4. So you have a 4th degree, okay, the denominator Oh, let me draw the Thevenin equivalent, minus plus G m V pi times J omega C 1 and in series with C 1. This is the Thevenin equivalent. It does not matter whether it is resistance or capacitance or inductance. Thevenin equivalent is multiply the current by the impedance and put that impedance in series, okay. So the condition of oscillation now shall be, shall contain in the denominator a polynomial in omega, to be precise polynomial in J omega and in the numerator a polynomial in J omega. And what you have to do to find the condition of oscillation is to make the phase of the numerator equal to the phase of the denominator, okay and then find out the magnitude. After you cross out the phase what remains? That would be equal to 1 and this is a little bit of algebra. It might prove slightly difficult but if numerical values are given you should be able to solve it, okay. I would not go into that discussion. The final circuit, final problem that we discussed in this session is the following. The question is can, I do not know if you know the answer to this question, can an emitter follower be used, a single emitter follower be used to generate sinusoidal oscillations? You know the answer to this? Is the answer yes or no? The answer is yes, okay. Now how? A emitter follower has a gain slightly less than unity, okay. Emitter follower has a gain slightly less than unity. Emitter follower has a gain slightly less than unity. It is beta RE divided by, it is 1 plus beta RE divided by r pi plus 1 plus beta RE and since 1 plus beta RE is much greater than r pi it is approximately unity but it is always less than unity and therefore the beta network, I should not use beta HFA, the beta network that you should use should give a gain slightly greater than unity but a phase shift of 0 degree, okay and that network can be derived very simply. You recall this network, okay, R, R, R, C, C, C. You know that if this is VI and if this is V0 then V0 by V0 by VI is equal to minus 1 over 29 at omega 0 equal to 1 over, yes, what? Route 6, RC. This is the phase shifting network. Instead of CR I have drawn RC, okay. If that is so then you consider this voltage, V0 prime. What is V0 prime by VI at this frequency? You see VI is equal to V0 prime plus V0. Therefore V0 prime by VI is equal to 1 minus V0 by VI which means 1 plus 1 over 29 which is 30 by 29 greater than 1 and the angle is 0 degree, agree? Is this okay? So all that we need now is to use an emitter follower. Now let me draw the circuit. Without any more ado, VCC you have to have a blocking capacitor. No, I beg your pardon. Do you require a blocking capacitor? No, you do not. That is what it is. The voltage is from output is from here to here and this output is 30 by 29 times this output. The gain of the emitter follower is less than 1 and therefore the product can be equal to 1, slightly less. Do I require a blocking capacitor? I do not because there is a capacitor here but I do require a biasing arrangement. Do I require this resistance? No, 3R should take care of it. Yes, I know you are going to ask that question. Which minus sign? Yeah, what is your question? No, this is 180. V0 by VI is 180 but I am taking the output here which gives me 0 phase shift. It is a very novel way of designing and you see in an emitter follower many things, many imperfections are taken care of. R pi C pi and many other things. You can include R pi C pi and see that it does not change substantially. You can always choose your capacitors and resistors in such a manner. Even this resistance is not required because 3R takes care of it. It is a very simple oscillator. You can quickly wind it up in the laboratory and see. By putting the power supply on, you see beautiful oscillations. We will meet in another 10 minutes. We will take a break now, meet around the 10 minutes.