 This is the 34th lecture on sinusoidal oscillators, an example of positive feedback. We have so far discussed only negative feedback and we should see how positive feedback can be utilized. As I mentioned, the concept of feedback started with positive feedback. One wanted to increase the gain of an amplifier but then this is people soon found out that if you increase the gain of an amplifier the circuit becomes unstable. It tends to oscillate and then this principle of positive feedback which gives a tendency of oscillation was utilized to generate sinusoidal waveforms and this is the subject that we are discussing today namely oscillators. In the general feedback configuration you saw that the feedback gain is given by AF equal to A divided by 1 minus A beta and if the feedback is negative then this sign changes to plus A beta. Now we are talking of positive feedback so we want A beta to be positive, A beta to be positive and I also mentioned that if A and beta are frequency dependent, if this quantity A beta depends on frequency then at the frequency at some frequency omega equal to omega 0 if A beta is equal to 1 then AF tends to infinity and the amplifier shall require no input to produce an output which means that it oscillates. It oscillates at the frequency omega 0. It may so happen that besides omega 0 there are other frequencies at which this condition is satisfied and all those frequencies shall be produced and therefore you shall get a non-sinusoidal waveform which can be decomposed by Fourier series into a sum of sinusoidal waveforms. Even if this condition is satisfied only at one frequency it is logical to conclude that this would be a critical condition, it is a critical condition. In fact what we want for starting of the oscillations is that A beta at omega equal to omega 0 should be slightly greater than unity so that the oscillations start and stabilize. How do they stabilize? Our analysis would be so far, negative feedback is based mostly on the linear equivalent circuit of the transistors and the active devices. Now as you know the transistors are highly non-linear devices. It is only for incremental operation that we can consider it to be a linear device. Now because of the non-linearity once the oscillation starts there is nothing to stop the oscillations if the device was purely linear and therefore oscillation amplitude should continue to increase. It requires no input and therefore output is fed back to the input, amplitude will increase then again amplitude and it goes on. Ultimately the power supply and the non-linearity of the device restricts the oscillation amplitude. So it is non-linearity which restricts the amplitude of oscillations and this non-linearity is very difficult to analyze because it involves non-linear differential equations with amplitude stabilizing non-linearity which cannot be modeled exactly and therefore we cannot find out the amplitude from a linear analysis. The other question is who starts the oscillations? I mean how does the device know that it has to oscillate? The gain becomes infinity. Now what happens at the input noise? Noise usually is white that is it is a white band noise from amongst this noise components that particular frequency is selected at which A beta omega equal to omega 0 is equal to 1 which as you know we said this is a Barkhausen criterion, Barkhausen criterion that is A beta equal to 1 is the Barkhausen criterion for oscillation at the frequency omega 0. Now this non-linearity which stabilizes the amplitude has another effect, the non-linearity produces distortion of the sine wave. Any sine wave which is not pure sine but a distorted sine may be slightly clipped off at the top and slightly clipped off at the bottom. As you know Fourier series says that this can be decomposed into a fundamental component and it is harmonics. So any oscillation that you generate in the laboratory through such circuits will have harmonics in them and there are ways of reducing the harmonics which we shall go into a little later. Obviously the common sense says that if you want oscillation at omega 0 you cannot generate subharmonics that is you cannot generate omega 0 by 2. Its harmonics will be Fourier series says omega 0, 2 omega 0, 3 omega 0 and so on. And therefore common sense says if you have a low pass filter which passes omega 0 and cuts off others obviously you can get a pure wave form. This is very simple means okay but let us look at the circuits first then we shall then we shall look at this phenomenon. Now if you generate oscillations at omega 0 okay and suppose the temperature of the room changes you know transistor parameters are heavily dependent on temperature and other environmental conditions and therefore omega 0 may change. Omega 0 is usually determined by a passive circuit, resistance capacitance and inductance and all the 3 parameters all these 3 passive circuit elements also change with environmental conditions they change with aging and therefore omega 0 may also change there can be drift in omega 0 as time proceeds. If you keep the circuit on for a very long time the transistor heat dissipation may be enough to generate to change the circuit elements and therefore a measure of frequency stability is used with the usual sensitivity parameter. Sensitivity of omega 0 with respect to X I have already defined this will be equal to X by omega 0 partial of omega 0 with respect to X partial because X may be a multiplicity of parameters it could be resisted capacitor inductor temperature humidity and many other conditions. So this is a measure of frequency stability any oscillator can be characterized by this frequency stability criteria. So let us look at some specific circuits and then we shall go into consideration of this. Now if the oscillations are to be generated at low frequencies up to about 100 kHz up to about 100 kHz I must say F0 it is convenient to use resistors and capacitors as the frequency determining elements because inductors at low frequencies not only we require very large values but inductors are difficult to make without losses. You make an inductor it always has a resistance associated with it there are losses. In addition inductors may have eddy losses if you use a core okay and there are magnetic non-linearities you know the hysteresis curve and therefore inductors are not very favored elements at low frequencies and low frequency for RC oscillators it goes right up to 100 kHz. On the other hand if the frequency is beyond 100 kHz then we usually use an LC oscillator because inductors are easy to make you can make you wind a couple of turns on a pencil and take the pencil off that becomes an inductor okay these are the simple air cord inductors at high frequencies so and I must also tell you that there is a limitation on inductors it can go up to about 100 megahertz beyond 100 megahertz of course the problem of it is difficult to make an inductor beyond 100 megahertz which shall have sufficiently high Q okay because the inductor is small the resistance is also small and the losses take over. Number 2 beyond let us say 1000 megahertz 1 kilo megahertz okay the conventional circuits the concept of lump circuits that loses its meaning so you have to go to some other devices we shall consider for this particular class RC and LC oscillators only you must also realize that in integrated circuits we have no other alternative but to use RC oscillators because we cannot make an inductance in that small space so even for high frequencies we shall have to use RC oscillators or LC oscillators where L is simulated by RC elements and active devices okay the one of the popular RC oscillators is the so called phase shift oscillator and the basic device that is used is a positive feedback device in which you have an amplifier you have a basic amplifier which is magnitude A angle 180 that is a phase inverting amplifier the ordinary single BJT common emitter amplifier is a phase inverting amplifier or a common source amplifier FET is that or you can also use an op-amp for example you can also use an op-amp in which the input resistance is let us say R1 and this resistance is R2 then the gain as you know because of the op-amp if the op-amp is assumed to be nearly ideal then the gain is minus R2 divided by R1 and the input impedance is R1 so this can be modelled as this can be modelled as if this voltage is Vi then you have Vi the input impedance is R1 and then A Vi which is let me take this polarity the output impedance is approximately 0 okay output impedance approximately 0 and if the amplifier is ideal and A is negative okay A is negative or we could say A Vi with this with the polarity reversed if I use negative here and positive here then I use A as positive okay I could also use a common common source amplifier for example let us use a JFET R sub D plus VDD and then R sigma shunted by C sigma depending on the type of the active device that you are using the analysis shall be slightly different you see if you use this as the as the basic amplifier V sub I then you know that the equivalent circuit is V sub I the input impedance is 0 and then you have GM V sub I because this is shorted GM Vi and in parallel with R D in parallel with the load resistance R D whose equivalent circuit is the following Vi if you convert this into a Thevenin equivalent then you get GM R D Vi with this polarity minus plus so there is a phase shift of 180 degrees but here you have an output resistance R sub D okay so the basic device can have an output resistance can have an input resistance as you saw in the case of the op amp the inverting op amp inverting configuration if it is a BJT amplifier basic BJT amplifier the output is taken here if it is a BJT then both input resistance shall be there input resistance and output resistance output resistance would be R sub C and the input resistance would be approximately R Pi beta plus 1 RE will come if this is if the emitter is not shorted but the if if the emitter is not shorted you know you get a deduction in gain because of negative feedback all right so it is your choice if you can do with RE unbiased go ahead if you cannot then you bypass this okay so the the amplifier the basic amplifier the basic inverting amplifier in general shall have an input resistance RI let us call this minus A or I will simply say A Vi with this polarity and a resistance R0 okay this is the basic the A circuit A circuit the beta circuit is usually a passive circuit now since this is an inverting amplifier let us connect this since this is an inverting amplifier we want to connect between this point and this point a network a beta network for positive feedback now if it is to be positive feedback obviously the beta network well let us say this is grounded this is grounded okay I have connected a beta network the output is being feedback we shall assume in the analysis that R0 is approximately 0 we shall assume that the basic network basic amplifier is an ideal one that is RI tends to infinity and R0 tends to 0 or or that the input impedance of the beta network is much larger compared to R0 and that the output impedance of the beta network is much smaller compared to RI and therefore in an actual circuit in an actual transistor circuit you shall have to take care of these imperfections okay and the frequency oscillation that we shall derive shall not be valid this will be modified frequency of oscillation and the condition for oscillation they have to be modified in accordance with the imperfections of the basic amplifier okay now in the RC phase shift oscillator the beta network is 1 which produces which has to produce a 180 degree phase shift because you want a beta to be equal to 1 the angle of a beta should be equal to 0 or 360 or multiples of 2 pi okay so the beta network must produce a phase shift of 180 degrees now since the beta network is passive it shall also introduce an attenuation that is the gain of the beta network in this direction shall be less than 1 which is the purpose which is being compensated by the basic amplifier that is the product of a beta should be equal to 1 so if beta is let us say 1 by 29 then a has to be 29 to be able to generate oscillations and the phase shift network that is usually used is the following I am starting from the output end of the amplifier so this will be my V0 okay then CR CR and CR a 3 section phase shift network question why are 3 sections needed I want to produce 180 a single section RC network can produce at the most 90 2 section can produce at the most 180 but that happens at very high frequencies okay or very low frequency that is at DC or 180 this is the limit and therefore to produce a phase shift of 180 degree in between 0 and infinite frequency we require 3 sections okay we cannot do with less than 3 sections and this is the simplest network that is used since I am connecting this to the input the input of the amplifier the output of this CR network should be VI agreed this is the condition for oscillations okay or if you so desire let us decouple them let us call them V0 prime by VI prime so we have to find out for this passive network the transfer function VI prime by V0 prime and this will be my beta and I have to find out the condition under which beta has a phase shift of 180 degrees the analysis of this network can be done by various ways but the if you have if you have been faithful to SCDR and his 204 you should now know that there are simple techniques of analysis you do not have to write loop equations and node equations you cannot you can you can take Thevenin's theorem for example if if I apply Thevenin's theorem here then I get V0 prime SCR divided by SCR plus 1 D is missing this is the voltage source and then we have an impedance which is yes R divided by SCR plus 1 you should know this R and C in parallel produces this then that comes in series with C 1 R then another C and another R this is V0 prime I am sorry VI prime okay at the next step what you will do is apply Thevenin's theorem here there is no chance of making a mistake unless unless you are very keen on making a mistake on the other hand in loop analysis node analysis you miss orientation or you miss a term then you you you get into problems also you have to invert matrices it would be a 3 by 3 matrix inversion nothing is needed here you can you can do it almost pay inspection if you continue this up to this end up to this end I will I will skip this analysis you get the transfer function beta of S let us do in terms of the Laplace transform variable the result becomes 1 divided by 1 plus you can verify this 6 divided by U where I do not want to write SCR again and again U I have used for SCR okay plus 5 by U squared plus 1 over U cubed this is the result that you get I have put it in a particular form for convenience of analysis what is the does this check is this result correct or is an obvious defect in this result you see at DC what is the transmission at DC for this network 0 so if U equal to 0 you see that the denominator goes to infinity and therefore at DC response is satisfied at U equal to infinity the denominator is simply 1 so the transfer function is 1 at U equal to infinity all the C's act as short so the transfer function is 1 you must do this checking and I mean automatically and spontaneously it does require any effort okay now since I am interested in generating pure sinusoidal oscillation I should look at beta of j omega and you see that this is 1 minus 5 by omega CR squared I have taken this term and this term I put U equal to j omega SCR then plus yes j 1 by omega CR whole cubed plus j or minus j minus j okay this will be minus minus 6 upon omega CR whole squared by me no squared okay let us not make a mistake now if you look at this expression obviously if the phase is 180 degrees then the imaginary term should be 0 phase is 180 degrees means that beta will be a real quantity and the negative quantity that is all and therefore for 180 degree phase shift we require 1 by omega CR whole cubed should be equal to 6 divided by omega CR which means that this is satisfied at omega 0 equals to omega 0 squared is equal to 1 by 6 CR which means that omega 0 would be equal to 1 over root 6 CR or f 0 the frequency of oscillation would be 2 pi root 6 CR C squared R squared of course okay so if at all the circuit oscillates this will be the frequency of oscillation as you see the frequency of oscillations are determined by C and R the circuit the passive circuit elements the beta circuit elements okay you also see one difficulty that if f 0 is to be varied if it is to be a variable frequency oscillator then either all the C's have to be varied simultaneously simultaneously because this depends on C and all C's were equal assumed to be equal or all R's have to be varied simultaneously and the usual usual instrument that you get in the laboratory are C decayed oscillators their phase shift oscillators in which the C's are varied simultaneously the 3 capacitors are ganged together so that one dial variation varies all the 3 C's these are usually air capacitors large air capacitors mounted on the same shaft and if you rotate the shaft all the 3 capacitors vary identically and simultaneously this requires very huge mechanical precision and that goes into the cost of the equipment the components themselves are not costly but this mechanical precision nevertheless such oscillators are very popular and they are available in the laboratory yes yes you can but it is not according to a very pleasant rule I mean you do not know how much to be varied you vary one capacitor you do not know what is the rule for variation if you analyze again now in a laboratory you just want to vary a dial 1 kilohertz 1.1 that is it and this is the simplest way to do it of course then you will not buy the instrument it is so inconvenient you have to calibrate every time and so on nevertheless this is the usual thing of course resistors can also be varied 3 3 potentiometers can be put on the same shaft and varied simultaneously but C variation is preferred is preferred because it requires less mechanical precision than resistors also resistors as you know there is a contact problem there is a potential meter in a capacitor you have the inter-digital type so it goes in or out on the other in the potential meter there is a contact if you go on varying this there is a wear and tear whereas in a capacitor this wear and tear is not there so capacitors are preferred however if omega 0 is equal to 1 by root 6 cr then you go back to the transfer function the imaginary term is 0 so you get 1 minus 5 divided by omega 0 squared C squared r squared and as you can see omega 0 squared C squared r squared is 1 by 6 and therefore this becomes minus 1 over 29 and that is what gives rise to the fact that the gain required for the basic amplifier A has to be 29 and then as I said to start the oscillation 29 does not suffice you have to make it slightly greater than 29 to start the oscillations otherwise it would be off and on off and on it may oscillate it may not oscillate because it is sitting on the border line okay you have to make it slightly greater at the cost of a slight distortion in the generated waveform because A equal to 29 is the is the Barkhausen criterion A equal to 29 satisfies the Barkhausen criterion exactly that means if A equal to 29 you will generate omega 0 if A equal to 29 plus delta you might also generate harmonics because of the non-linearity that comes into effect but this non-linearity this harmonics can be taken care of otherwise. So if I draw an oscillator circuit if I now draw an oscillator circuit let us say using the FET we have r sub D r sigma and C sigma this is plus VDD this is the basic circuit then we draw we bring in the feedback circuit C r C r no further coupling capacitor is needed because this capacitor itself serves that job C r and then another C r and this now has to be taken to the input this is the circuit of an FET FET r C phase shift oscillator 180 degrees phase shift could also be produced if the C's and r's are interchanged if capital R one this would be a leading phase shift or lagging phase shift no from here to here would it be a leading or lagging phase shift leading oh take a simple C r the phase shift is it leading or lagging leading the output leads the input but 180 degrees is 180 degrees whether it is leading or lagging it comes back to the same negative axis okay. So I could even take a simple R C R's and C's could be interchanged but you see the problem if I interchanged I will require an extra blocking capacitor extra coupling capacitor that is correct so this is preferred but then there is a problem with this one also life is always a mixture of pain and pleasure okay the cost that you the price that you pay is that the output impedance of the of the FET is R D and therefore at frequency omega 0 R D must be much less compared to 1 by omega 0 C C comes in series with R D if you recall the equivalent circuit equivalent circuit is GM R D VI that is the with negative polarity in series with an R D and therefore R D unless it is ensured that R D is much less than 1 by omega 0 C the frequency of oscillation shall be affected the gain of the circuit shall also be affected this transfer function beautiful transfer function that we got in terms of you shall no longer be valid R D shall show its tip and you can make an analysis on the other hand if I use a resistance here if I use RC then this resistance can absorb R D that means the first resistance the first resistance instead of using R I will use R minus R D you understand this using that coupling capacitor this is the price that you pay but R D can be absorbed in R the point clear no okay suppose suppose I have an oscillator like this R D I will not draw the rest of the circuit and then I have RC RC RC okay suppose I produce 180 degrees phase shift with this then obviously the equivalent circuit would be minus plus GM R D VI whatever that is in series with R D and then I have R okay I want this to be R so instead of R I shall use R minus R D then the formula shall remain valid all the three R shall be identical all the three C's shall be identical now obviously in the if it is an FET common source amplifier then this R is not disturbed this R is not disturbed it is not shunted by anything agreed on the other hand if you had used an open and inverting open did I use R2 or R1 R2 and R1 okay then using a CR network for example here okay let let us use a CR network this is grounded CR CR and CR okay this I am going to connect here the input resistance this is my VI the input point input resistance of the op-amp is R1 and therefore R1 shall come in parallel with R isn't that right again the frequency of oscillation and the condition of oscillation will change because one of the registers has become different okay can you suggest a remedy instead of this instead of I simply don't use this and I use R1 equal to R don't I get the same result because this point is virtual ground I get the same result so even in an op-amp it is easy to take care and the frequency of oscillation shall still be given by 1 by root 6 CR and R2 by R1 will now be required to be 29 1 by 29 minus 1 by 29 or plus 1 by 20 no it has to be exactly equal to 29 the gain of the the gain of the stage has to be minus 1 by 29 so R2 by R1 must be 29 is the point clear on the other hand if I have a BJT if I have a BJT I have a problem okay I have the biasing resistor equivalent biasing resistor R sub B and from here I connect the RC or CR circuit then you see RC affects the frequency of oscillation and the input impedance which is the parallel combination of RB and R pi also affects the frequency of oscillation you might say that I will make RB parallel R pi equal to my R I can do that then the frequency of oscillation would still be given by that simple formula provided 1 by omega 0 C you have made much greater than R sub C otherwise R sub C will show in the formula okay it would be instructive not to make any of these adjustments and find out how the frequency of oscillation and the gain requirement changes okay yes right so A has to be plus A has to be minus 29 yeah but we assume that we assume A beta equal to 1 beta is minus 1 by 29 yes A has to be minus 20 A is minus R2 by R1 and R2 by R1 has to be 20 sure sure right oh we assume the of M to be ideal yeah R is same as infinity not 0 then it will be shorted but this point is virtual ground and what we wanted is an R going to ground and therefore this is a very simple and neat way of designing an RC phase shift oscillator often is the best best choice because it does come in series no I have converted this way voltage so my gain is now minus GMRD okay now let me also mention that 29 is not a very sacred figure 29 is not a very sacred if you taper the circuit the phase shift circuit like this C R first one then you make the second one as C by A and AR so that the time constants are the same but you see the loading of the first stage by the second stage would now be less and therefore this will make less attenuation and we go to the third one C by S squared and S squared R if I do this tapering then the gain requirement can be increased and it is instructive to do this analysis and show that if A can tend to infinity if A is very large then the gain requirement A can you guess what the gain requirement would be I am putting with a big question mark you verify this that with increasing A with increasing A the gain requirement decreases and if A tends to infinity obviously A cannot be infinity because then this resistance is not there this capacitance is not there is that right so there is no feedback but A can be between between one and infinity and you can see in fact if A is 10 capital A the gain requirement is 1 point it is less than 2 I am claiming this but you verify this the next question is suppose I have designed an op-amp based this is R op-amp based RC phase shift oscillator how much should this be now how much should this be oh this is very 29 times R because you want a gain of 29 minus 10 okay then you have CR CR and finally you have a C only which connects it to this okay where do you take the output where do you take the output at the say this point it is one one choice the other is which are this are across this then you have no grounding no no physical ground here here here or no that is a Himalayan mistake this point potential is 0 you can take the output anywhere because throughout the except this point and except the ground because everywhere there are sinusoidal oscillations all signals in the circuit shall be at the sinusoidal at the frequency omega but this choice is a gem of a choice I should have used a different color this choice why because anywhere else you take the output suppose I take the output here I am going to connect it to some device I am going to use this oscillator and that will create a loading of this resistance which means the frequency oscillation itself shall change on the other hand if I take the output from here the output impedance of the op amp is approximately 0 and therefore there is no loading you can take you can take to any load except when it is a short circuit you cannot take it to a short circuit obviously because you cannot short this then nothing goes alright so this is the point at which you should take the output next even if you take the output here okay let us bring the output here as I said there always be some amount of harmonic distortion that is there will be higher frequencies so what you can do is one of the simple ways of reducing distortion or reducing the harmonic content is to use a low pass filter and you could simply use you could simply use a low pass filter like this simple RC circuit such that Omega C what is the cutoff frequency of this if this is R1 and C1 Omega C is the 3 dB cutoff 1 over R1 C1 not RC do not confuse with these elements 1 over C1 this has to be where should you put the cutoff frequency slightly greater than Omega not slightly greater than Omega not such that at 2 Omega not the attenuation is let us say down by at least 20 dB that should do it as it is the proportion of harmonics to fundamental be very small you reduce it further another 20 dB 20 dB means attenuation by what factor for me 10 attenuation by factor of 10 and multiplication by a factor of 0.1 so it should bring it down if you want if you want well one of the problems is now you are going to take your output here right and any loading of this any loading of this is going to change your cutoff frequency you should be problem so what should you do you should use a buffer here you should use a unit again amplifier if you use a unit again amplifier here plus 1 whose output impedance is 0 obviously there is no loading of this but if you are using an amplifier if you are using a buffer a unit again buffer another op-amp why do not you use the this circuit from the output of the oscillator you use let us say R4 I beg your pardon C4 R4 R3 do you know that this is a low pass filter no very simple to see R4 divided by SC R4 plus 1 this is the impedance of this divided by R3 so the gain of the circuit would be this obviously this is a low pass filter with a cutoff at what is omega C4 this 1 over C4 R4 these are very simple things to do you see from the output of the last op-amp you apply this filter and then take your output here this is usually incorporated inside the oscillator circuit and if you want better rejection then use a second order filter instead of first order use a second order filter or use a higher order once much higher order if you want very fussy if you are very fussy about the waveform the distortion content of the waveform okay this is about the phase shift oscillator as I said even if you make even if you couple a low pass filter to reduce the harmonic content who stabilizes the amplitude how do you stabilize the amplitude anything changes temperature changes for example the amplitude will either rise or fall on the other hand in the in the laboratory if you are making let us say if you are testing a communication equipment you want the input or testing the frequency response of an amplifier let us say you are going to vary the frequency you do not want the input voltage to change okay you only want to monitor the output voltage you might like to record it on an on a recorder the frequency response then the input voltage should be fixed you plot output versus frequency well you can say even if the input changes I will take the ratio fine but then you require a calculation you do not require it if the input voltage that is the oscillator output voltage is a constant how do you stabilize the amplitude there are amplitude stabilizing circuits which are which use non-linear elements like temperature sensitive resistors negative temperature coefficient or positive temperature coefficient or even Zener diode or ordinary diodes which are very temperature sensitive the main problem in oscillation stabilization of amplitude the main problem in stabilizing the amplitude of an oscillator is temperature heat the heat the very circuit generates it other circuits in the neighborhood generate it the temperature changes from season to season and therefore temperature stabilization is one of the one of the crucial things that has to be done with an oscillator the second kind of RC oscillator is called the so-called wind bridge oscillator the wind bridge oscillator the the simplest circuit using an op-amp is this I will draw the circuit and then carry on the analysis later there is a resistance this is also be basically the wind bridge oscillator requires a gain a positive gain amplifier instead of a negative gain amplifier if the gain of the basic amplifier is positive well this is a this is obtained by this R2 R1 and this is grounded and the input is applied here okay obviously between this point and this point the gain is 1 plus R2 by R1 which is positive that means the phase shift is 0 therefore the beta network that you want should produce a phase shift of 0 instead of 180 you can say if you can do it 180 why should you bother about 0 because it uses less number of components that is the circuit that is used is a series RC and then a shunt RC and the output is taken from here and applied here this is the so-called wind bridge oscillator why it is called a wind bridge we shall see later but you should realize that this circuit is capable of producing a 0 phase shift now 0 phase shift obviously can also be produced by a simple potential divider why don't you use that no frequency who will determine the frequency a potential divider is insensitive to frequency if the gain is point 1 it will be point 1 at 0 it will be point 1 at infinite frequency also so who will determine the oscillation frequency that doesn't mean that it will not oscillate it will oscillate it will and get stuck either at the plus power supply or the negative power supply we don't want that we want to generate sinusoidal oscillation so the beta network has to be a frequency sensitive network all right how does it how does it produce a 0 phase shift let's see qualitatively we will make the analysis later qualitatively can you see what kind of a circuit is this we have a C R R and C what kind of a circuit is this if this is my input V1 and this is my V in and this is my output what kind of a circuit is this does it favor low frequencies is it a low pass filter no at DC obviously the transmission is 0 does it favor high frequencies no at infinite frequency this is short and therefore this is a it could also be band stop no it is a band pass somewhere in between somewhere in between it produces a maximum like this starts from 0 goes to 0 at infinity somewhere in between it produces a maximum and one can show by simple analysis that this maximum occurs at the frequency 1 by C R and that at this frequency V out by V in at omega 0 is equal to 1 third you see the advantage of this oscillator what is the advantage that the gain required is only 3 instead of 29 so the resistor spread what is the spread that I want here if I want a gain of 3 only 2 R 2 by R 1 has to be equal to 2 so the resistor spread is 2 there is no reason why we cannot make these resistors identical to these 2 we can do that so the resistor spread is only 2 not quite resistors spread is only 1 tell me how common sense I will make R 2 as R 1 in series with R 1 I can use identical resistors everywhere we will start from this next time