 okay. So we were looking yesterday about the pole zeros we will come back quickly for some few minutes about that again okay. If you take please if you take an arbitrary function x plus j y the phase of this function will be tan inverse y by x please remember what I am saying I am taking an arbitrary function f is equal to x plus j y okay then the phase for this function is tan inverse y by x and we call that as a phase angle ? and we can see substitute some values if y is x tan inverse 1 is 45 if y 0 tan inverse 0 is 0 and if x is y is very large in closing to 10s and 100s tan inverse infinity is 90 so it approaches 90 degree it does not reach 90 unless it is infinite but to say it that may one can save y is large the phase is 90 if y is 0 phase is 0 and if y is 1 y is equal to x essentially what I mean then it is 45 okay. So you can see the phase will vary from where to where from 90 towards 0 degree as the omegas vary okay as the frequency let us say similarly we can write omega by omega 0 so as the frequency changes and you can see for every 10 10 times the frequency the phase shifts by how much 45 degree you can do a 45 for this 0.1 is very small so 0 10 is very large so 90 so you can say what was the gain function was changing with frequency 20 dB per decade and what is the way phase will change 45 degree per decade so every times the frequency changes by 10 times the phase will go down by 45 degree and gain will also go down by 20 is that okay so this is something you should realize that every gain fall okay will be associated with phase going down okay so from 90 to 0 it will actually run down and typical figure which for a what we call as a phase diagram phase response is this is I am plotting theta versus frequency so different frequencies your phase so what are these points these points are the same points at which gain was changing from say 20 dB to 20 plus 20 40 dB plus 40 every time it follows 20 dB per this phase also will follow is that clear and that is this together and same corner frequencies as they were together this amplitude diagram that is the magnitude in dBs and phase in theta is called Bode's plots okay these two are together called Bode's plots please remember we are we actually should do as isentropic values real life but since we are not interested in how much exactly at this frequency otherwise we want to know from where it falls where it becomes 0 those two or three major frequencies we want to know so these were called corner frequencies and Bode suggested that for any stability as we shall see later the things we want to know what are the corner frequencies by Mathematics the day one what are these corner frequency we said they represent poles the transfer function that is transfer function becomes infinite at that value of frequency okay we did not talk in this so far anything about 0s okay we did not talk it simple case can be thought let us say I have a pole let us say magnitude some function and it has a pole somewhere here at some frequency we call omega p1 and let us say at the same point I have also a 0 that is the numerator also becomes 0 at the same frequency some reason then from here up to this it will be same and 0 will be actually going plus 20 dB because it is numerator so the value will increase 20 dB for a decade this is minus 20 dB per a decade so after omega 1 p1 equal to omega up or which is I am using the same what will be the actual gain actual gain will remain is that idea clear to you 0 means the numerator and if you do same thing as we did for pole every decade of frequency the gain will increase by 20 for a pole it will decrease 20 dB down so from this frequency onward plus 20 and minus 20 will keep canceling per decade is that clear is that clear but let us say there is only single 0 here and there is a another pole occurs here which is a omega p2 and there is no other 0 then so what is the slope of this can anyone tell me is the second pole down minus 20 for this plus another 20 from the other pole so it is minus 40 dB per decade but that means at this frequency gain will start because this 0 will actually put it to 20 dB per decade all the time but what is the gain falling now here minus 40 dB so the together if I do it then from this point onward you will have minus 20 dB per decade is that point clear I repeat I have one 0 and one pole at the same frequency so the compensate the word is compensate okay 0 compensates pole or also in the other word 0 nulls the full null means cancels okay it nulls the whole this is called pole 0 compensate this is why I am showing you this initially because I want to tell you we want to actually increase gain for higher bandwidths so there are there are possibilities in which the pole what is the bandwidth essentially means what is the definition of a bandwidth I give till which the gain is constant from this frequency down the gain is not if there is no 0 gain would have fallen so this is our initial band but if you have a 0 this will continue bandwidth is increasing so if I cancel a pole I have actually increased the bandwidth but at this point since the two poles and one 0 are occurring onward the gain will start falling from here itself how many dB minus 20 dB per decade so what is the new bandwidth now Omega P2 because beyond that only gain will start falling so by circuit way if a device is fixed I do not have play on that but I can have now a circuit if I can create my choice 0s and poles okay then I can tell her my frequency response of an amplifier is that point here by deciding where should I have my poles at 0s of course they will be limited by the intrinsic property of the transistor how much is GM how much whatever biasing you did but apart from them whatever is normal amplifier I had I can monitor I can add some RC is somewhere some way so RC is pole 1 upon RC is the frequency of a pole or time constant so if I add us do something a RC time additionally my external part on mine I will be able to change the bandwidth is that point clear so something you have to understand that circuit people are always smarter enough okay even if they are working in an adverse circumstance they may still be able to give you something which you are looking for okay is that clear this is something tricks of their trade in all analog designs now question is why you have to solve so many problems in analog because for every circuit you are given their poles will be different their gyms will be different so for every one of them you will have to think redo it again is that correct this does not happen in digital hardware why because once a gate is a XOR gate or a NAND gate or a NOR gate do whatever it will remain that function is that clear in analog this does not happen and therefore it is challenging in a way that every problem you have to take it independently there is no some force theory is correct rules are known but you will have to solve for every case to do something what you are asking for so there is no standard design available okay I want an opium of this kind so just buy it from some it is not like that for that require a lot to redesign all of it you say something change allowed to redo it again okay that is why analog designs or analog circuit are much more interesting if I say otherwise than the digital so is that why we are doing all this because we should be able to tailor our bandwidth for the use someone wants to use that 100 mega what do I do I have only one circuit and I have to play games on that okay so there are tricks which you should be able to know and we will see they are not one there are three kinds but when we do all this there is another term which will add later which is called stability this word we shall come little later after feedbacks and we will say if you do some mischief otherwise the way I shown you it may lead to unstable situations okay and we will see why they are called unstable so it is not that you can do a randomly anything and get away you know you have to be suddenly you find amplifier is not giving constant gains of signals are shooting like left and right okay so this stability is another parameter and that time we say we will use another theory which is feedback to stabilize a circuit I may use feedback at the cost of something and I may use some zero pole compensation technique or something to improve the bandage even then but something I mean moves in doing all of it is that correct so please remember there is nothing there is no free lunch as we say there are no free lunches same in circuit you cannot achieve something greater unless you give somewhere so at every point what is at stake that is asked for and what then I can give you okay if you do not mind here then I will give you this okay so this is what analog designs are of interest to everyone because they they are you have to keep thinking this place should not go to low which is demanded by them or it should not still you want to improve I say okay if you give me 10% here I will give 50% accepted like this so there are design issues which this course as I say we are not really looking into but I just want to give perspective for perspective of this course because why are we doing this analysis so many times because at then I may have to design somewhere a hardware for them okay that is electronic engineers job that we should be able to design a system may go on finally on a single chip that is another issue but design has to be first formed okay and therefore the tricks of analysis must be known to actually design a hardware that is why we are actually looking into theory must seriously okay so having told you we are discussed first day one pole circuit here is another circuit okay so how much how much capacity how many capacitances were there in our first circuit one capacity so how many poles it gave one because one RC time constants gives me one pole is that clear we will we are not going to discuss 0 we will see how 0 appear but right now we are only looking into poles now I have another circuit in mind in which we have two poles why two poles I say how many capacitance should I have for two poles to capacity two time constants I must be able to create to have two poles so here is a circuit which is called a two pole circuit I have an input source which has a series resistance or S and I put another capacitance CS in series to this this S really series word okay then there is a capacity resistance RP which is in parallel shunted by a capacitance is that okay a simple circuit there in one time what we do either we had this or we had this now I have put both of them CS and CP are both present simultaneously is that okay is the word clear both capacitances are simultaneously present with two resistances okay this is a simple network once again so if I solve V0 by Vn I can solve as we did earlier how do I solve take a parallel combination of this take a series combination of this then what is V0 this upon this plus this into Vn is that correct with what is it equivalent to saying Z1 because you did not say resistance Z1 Z2 this is V0 this is Vn so V0 a Z0 as a 2 upon Z1 that okay so I calculate this I calculate parallel combination of this and then just take a ratio of that is called potential divider just divide it and you get V0 by Vn as the transfer function and what is the transfer function here we are looking it is a voltage gain why it is called voltage gain because V by V output by input voltage so the transfer function here is AVS and why I am talking of this S word because a frequency dependent term capacitance into S is the 1 upon CS is the impedance of the capacitance or rather CS is called admittance of the capacitance okay so if I do this V0 by Vn and if I define two time constant tau s is Rs plus Rp into CS this S is subscript otherwise no no it is not subscript sorry no no it is okay not it is a subscript okay or maybe I should rewrite CS and the other is RS parallel Rp times do you get the point why this how this parallel combination will come if I see a resistance from the CP side what is the resistance Cp will see when I see from output what should I do Vn should be grounded so this RS CS will come into parallel and for the resistance that will occur if you separate the terms it will be essentially coming from that circuit okay so this is the two time constants I have defined one is Rs plus Rp into CS the other is RS parallel Rp into CP why I define from there because I am seeing those functions here so I define it from there okay so then I write this as a function so how many to do you see this is a quadratic S term is appearing this is SS square this is S plus this is that correct this into into S 1 plus SS square tau p tau s plus 1 plus all this into S tau s so there SS square plus SS a SS square plus BS plus C is in denominator is that clear can you write such functions which I did earlier for example if I have a function SS square plus AS plus 1 let us say I can always write this as S plus S 1 S plus S 2 of course this may become some different value because when you connect it or not right now it will be one but if you want to separate what is the way you can separate these two functions by partial fractions if necessary you can remove put a partial right now one can find this is equivalent of this can see SS square into S 1 S plus S 2 S into S 1 S 2 okay and divide S 1 S 2 so you will get some kind of similar expression so maybe maybe be here plus so I can see any such quadratic term in denominator can also be represented as this if I take S 1 S 2 outside then I can write S upon S 1 plus 1 S upon S 2 plus 1 you can also write that please remember S 1 S 2 are constants so I can take them out and then I get I can get a function which is 1 plus S is omega S 1 will be also in terms of another omega so 1 plus omega by omega 1 into 1 plus 1 plus omega by omega 2 kind of function I can create if I have a trans function denominator of this point is that clear so whenever there are two poles two capacitance what how many poles it will give two poles is that correct SS square term appearing will always give you two poles is that clear so this fact has to be understood whenever there are two time constants involved there will be two poles please remember it may also create a 0 which in this there is I am I am avoiding there is no equivalent source I have put to create a 0 for that but I can see I can add a 0 also by some other but right now assume that there are no so this fact has to understand as soon as you write the trans function and you get a SS square terms in the denominator SS square plus S plus something you can always represented this as a product of two such S plus kind of terms and they say there are two poles S 1 and S 2 may occur is that clear this is the method of solving any circuit any transfer function now why I say any because it is voltage by voltage tomorrow it may be voltage by current third time it may current by voltage or current by current or it can have a mechanical system velocity force connected it can always this is a control system part anything has a transfer function output by input what is the stimuli stimuli you understand any energy source you put any stimuli is immaterial transfer function has nothing to do with voltage current or something it can be true for every one of them however in our case we are only looking into electrical functions is that okay so this technique is universal is that correct this technique is yes okay if S is j omega so j omega by S 1 will be constant so 1 plus j omega by S 1 now you can find its magnitude and pole very is that correct when you want to find a magnitude it will be much easier for me to do that is that this is a no compulsion you can do otherwise but then also that turn will come out finally when you do that so instead I did initially okay I am not telling that this has to be done but this will come anyway so I showed you it is much easier to start with like this please do not take it that this is has to be done as long as you have a function you can break into any of this such form and what automatically terms will come out or come in okay so you do not have to if you have seen him really you will get everything correct this is the way I am explain you how to do okay okay so having shown you this transfer function I have done this I am not going to solve it fully but just to give you some idea on that so with one can see tau S which is essentially RS plus RP into CS is called open circuit time constant is that word why it is open circuit what does it mean 1 upon CPW is infinite that is the impedance at the output from the capacitance P is open much larger so where it can occur assuming CP's and CS are equal values are not very different where it can occur at higher frequency or lower frequency much lower frequency 1 upon CPW will be much higher whereas if you see the other time constant which is RS plus this comparatively so this is I say is called open circuit time constant by similar logic we can say tau S represent time constant associated with CS and the is called short circuit time constant sorry tau S is also called short circuit time constant why it was called short circuit if you short circuit the output and see what is the time constant is RS CS related only so it is related time constant only to only to RS and only to capacitance CS and not to the seat so this fact that I can define sorry oh I am sorry oh sorry that was similarly tau P is called short circuit I am sorry I am I wrote it correctly and then in hurry I said something wrong similarly by similar argument I can say tau P is called short circuit constant because at that time the series capacitance I want to short so where this can occur very very high frequency is that clear to you 1 upon CS Omega will become close to 0 only when Omega is very high Omega is very high so what does do you get now two separate frequencies from this frequencies one is very very low frequency other is very high frequencies so one can see if they represent some way the poles the one pole is very close to 0 side the other pole is very far away from this others it is going towards infinite side is that clear to you so whenever there is this capacitors available depending on of course the value of CS and RP and the frequency RS RP frequency one of the pole may be dominant over other what which is the dominant will come from here 1 upon tau P will decide upon tau s I just said you know if s is infinite only then this will start dominating and this will short circuit that means the pole which is decided by much higher frequency the other term dominates at much lower frequency CP is going away it is only CS which is dominant is that correct so one can say the poles and zeros are normally not at same value they will automatically get split this is called splitting of poles what is it called splitting of poles this is a theory I am trying to why I brought this survey so another technique which I am using going to use by proper choice of RS RP CS CP what can I do say initially let us say some simple figure I can draw again for the same thing which I am doing again again let us say honest I will come back to it this is my omega P1 and this is omega P2 this is dominant this is lower this is low and this is high but I can change the both side what can I do I can bring this on this side and can bring this on this side I may actually bring them together or I can also shift the first pole towards lower side a second pole and first one I may shift out this is called pole splitting I can actually decide which pole is dominant for me is that correct how do I do that by proper choice of ours and since this is another method what we will see later ever stability of a amplifier can be achieved by splitting poles okay one method what did I say null zero nulling something the other way I use the splits okay and how do I split by varying the values of typical RC networks which I put there so that I can change my pole positions is that correct so these two techniques there is another one will come when you finally so is that technically you understood that these are not constant per se for a given circuit in analysis we will just solve it whatever values come but in real life if someone asked me know I want here then I will think what should I do now additionally that I can do what he is asking so to do this this is called design what is design the customer tells me I want this and I am asked to design for him but to design I must know what should I do that I can meet his specs okay so our course is all the time talking about analysis but you must be remembering every time that this is not analysis interesting course and we are not doing for the sake of interest because tomorrow you may be designers and you must know how to design so you must know which way one can design to get the specs someone is asking and specs will not be your hand because someone else will tell you I want this okay so a priori you must know of a this is our time we did it but he wants this so what should I do I have learned one professor taught me like this I use this technique and let us see what happens okay this is what is all about in real life okay no one actually tells you to solve problems in the exams as we did here in the real industry anytime but they will come oh this product will be marketed this is the date at which a market value is highest so please put this product in this time then you are told this is what I want and that time of course this you will do in graduation or PhD much more detail way but right now to bring you why we are doing it you know few years ago everyone used to ask me for it my analysis number number Karthik you so I am now trying to every class I am now telling no it has a relevance in the end of your career at the start of your career because there at that time I will not be there books will only give analysis okay so you you can see my even the book which part I am even it is 400 500 pages every time you cannot carry two three books everywhere at that time only this helps and therefore the tricks are I am okay yes because that will be very small yeah but that is the that means that will start dominating one upon RC of that will start dominating smaller the value okay one upon RC is smaller is the first time gain will start falling the amplitude will start that is called a dominant is that clear no I won't say it is 0 0 and infinite are the extreme value there will be a finite value of CSS also there will be a finite value of CPP also okay so it is a way still will be a ratio but which one will dominate numerically by magnitude is that clear you do not get that word 0 or infinite as the actual value because then there is no system left is that clear that is the many time I show you an error to you tending to infinity okay tending to 0 otherwise if it is shorted and open then there is no there is no component there what to show on that okay is that clear okay having told you if I show you this two frequencies which I just talk one is called FL one upon 2 point tau S the other is called FH one upon 2 point tau P the first one since being lower it is called the lower corner frequency FH is called the upper corner frequency or higher corner frequency and if I plot the gain versus frequency by Bode's technique we are not plotted full Bode right now I only I figured out what are the corner frequencies so one at here one at here in between we say the value of V0 VS is constant below FL again will fall beyond FH also gain will fall V0 by gain means V0 have been both sides it will fall so what is this frequency from FL to FH should be called where V0 VS is constant is called mid band or essentially the band width or essentially called the band width in most cases FL may be very small close to say 1 hertz or 10 hertz is that clear in many cases the lower cutoff frequency may be as low as few hertz okay at that time FH itself is your bandwidth because FH 100 megahertz minus 10 hertz is as much as 100 megahertz is that clear 100 megahertz minus 10 hertz is as much as 100 megahertz so if this values are very small then the normal bandwidth is FH value now can you think little more from this kind of figure which I drawn you can see from here if I have a network from anything below FL the gain is falling sharply 20 dB per decade okay and any free just forget about FH right now anything ahead the gain is higher constant so what is the equivalent I could say again maybe we use fresh paper what I am saying I have two functions I am now drawing this is my FL okay this is gain versus frequency so below FL gain is very small or 0 I made step function kind and beyond FL it is large and constant by same argument up to FH gain is constant and below beyond FH gain is 0 so can you think what is it what is this actually doing I have a circuit I do not know what is it has a many frequencies F1 F2 Fn okay what it will do at the output from input to output if it passes through this it will not allow frequency below FL okay so which what are the frequency it is passing higher frequency so it is called high pass filter it is called high pass filter what is this up to FH it will pass all the in frequency with the gain and beyond FH it will not allow any this so it is a low pass filter so this is high pass filter and this is low pass filter can you think two filters out of this before we go to solve two more filters from the same figures can be created what will be band pass this is what was the band pass this is essentially your band pass okay from FL to FH is the band in which this is there beyond that roughly it is 00 okay so the band pass filter has something like this F1 and F2 so it is F2-F1 is the band in which it is passing the rest all frequencies so what will be the fourth one is called band reject so what should be done there you can see same combinations now I will fit okay I would say one is here which one will be this filter this is HP this is LP the only thing is LP go ahead or HP go Picheli of this frequency can that band reject over see FH that is FL is larger than FH I told you by circuit I can change FL and FH value here FL is large smaller than FH here FL is smaller than sorry F1F2 is same FLH is however here by design I made FL higher than FH is that correct so in between the frequency now what is it gain is how much or signal is 0 so this is called low power high pass this is low pass sorry low pass is here first and high pass is ahead and this is called band pass this is called band reject is that correct so from the same low pass band high pass theory by adjusting time constants I can create four filters low pass high pass band pass band reject okay all that I have to adjust is the value of FL and FH and they are functions of what RNC so by proper choice of network values I will be able to design any filter which will be of kind which you are looking for I will not give an answer but just think where do you think this will be useful a core filter I bring F1 FL FH very close okay I bring FL FH very close maybe here so I have transfer function versus very very close to what this filter should be called notch filter since all of this game what is important what is that we are talking adjust your transfer function value with RNC and you can create all kinds of filter you are looking for is that correct there is another filter which will study properly it is something like this what is it called all pass now question arises why such a filter is required at all okay is that clear all pass all frequency path is still we put that filter so I am right now I am not talking of I am only thing network I know gains right now I am only talking transfer motion V0 by V in there is no amplifier there is no transistor right amplification can only come if there is a active device in this anyway did I show you any transistor no these are only transfer functions passive networks what are these passive networks along with them we may put the amplifiers are something to get something more out of these filters because you need driving you need many things but basically filter has nothing to do with active device okay we can always create a filter using passive networks this is also required all pass something will show you in the using an opamp we do need an all pass filter in opamp circuit sometimes we need all pass but still we want there okay till we want that to be there so is that clear filter theory filter does not get basically it is only RC network combinations which can allow certain frequency signal to pass and certain other request signal not to pass okay or picture a combination of the two this is the technique again why I am doing it this is a technique I may use it later in my filters okay so I just told you today itself that all these tricks I use an actual circuit theory anywhere I use I will just follow this okay this I have to adjust I will put so next time if I show you some network I have said this is a low pass do not ask me why I told you how it will do is that clear right now I show you how it works as a low pass high pass band pass band is that okay so something this basic thinking which I thought I hope that network theory people should have done that so at least my course since I need this later so I do not want to miss it later though sudden this was a low pass kasey so I will tell you how she is a secure okay filters John banana is passive new you will take active filters that means there will be transistors or opamp or some active device will be there okay not that without them we do it but basic network is a filter filter means something is pass something is not pass is filter okay let us revisit I have a normal amplifier except CC rest is same which I have discussed there is an input capacitance seen there is an output capacitance CL okay this is a series resistance of the source there is a load resistance or D okay which is the drain resistance or D and this is a unless it otherwise it is NMOS transistors okay have you seen what is the circuit first forget about CC the all other is a standard amplifier okay and now I add a capacitance between output and input what is the output point this drain point what is the input point this is the gate point so I introduced a capacitor between drain and gate what should be called feet feedback gate drain to get I have connected now okay now this word is why I brought this again you can see there are two ways the signal can go which are the two ways signal can go from drain and gate it may go from gate to the drain then it is called feet forward what is it called feet forward if the signal goes from here like this it is called feet forward but if the signal returns through the same path to the gate it is called feedback is that correct but capacitor to do no side you can success so how do we in real life we will prefer feet forward be stopped and feet back be allowed okay what is that trick we are going to play later I want to reduce feet forward problems because feet forward is a big problem will show you and I may allow feedback to occur is that clear so but in this circuit do not you see that the signal can go this way or it can come this way who can stop it okay now who can only stop it provided this voltage as are such that signal can go from one side to the other only is that clear to if there is this potential is smaller than this may go so we must now look for the potentials at the two ends so that signal can go either way only is that here this is the trick again we are going to please remember between two nodes signal can go from one to two or two to one feet forward feedback either is possible and as I say if there is a feet forward what is the problem is just okay we can say right now by associate time constant one of the poles at the input okay without CC is 1 upon RS seen so what is the trick I am now showing you a quick post they may not be accurate once if CC is not there what is the time constant that they input RS seen so that it should give a one pole okay at the input side okay okay now I have a problem which I say there is a famous theorem which is called Miller's theorem okay this is a famous theorem which is called Miller's theorem it is essentially says if you have a two port network and on the one of them is the impedance Z okay then this Z can be split into and let us say what is the input if your input to signal is V1 at the output you have a signal AV1-AV1 it could be plus AV1 also it does not matter but AV1 is the gain function if this is your two port network can you think where it can be Abhijab Bihai circuit is may a high slim and using this you have a signal V1 at the output you have amplified signal AV1 okay and in between output and input there is an impedance Z is that okay it can be capacitance it can be resisted it can be RC together or it can be even inductance okay or it can be LCR anything Z according to Miller's theorem what is Miller is saying this could be split into a Pi kind of circuit non-Pi kind of circuit two port kind separation the first component Z1 is essentially equal to 1-A times that is this Z can be split at the input side as 1-A into Z and is minus please take it is minus so it is 1 plus a magnitude wise is that correct and if a is large you can even neglect one so it is game times the Z is seen as the input please remember where it is between output and input you are an impedance Z so it will reflect at the input side as game times that value please remember this minus is plus actually is minus so 1 plus a have magnitude wise is that clear a is minus so 1-a is 1 plus magnitude wise e okay so this is larger or smaller a is larger is a gain function so larger than at least 10 20 hundred times okay so this impedance at the input will come much larger at the input side however if you see equivalent on the right side of that is at the output side this Z2 is a upon this minus are removed now is positive 1-a is 1 plus a both minus a is minus a is minus no no a is positive here means this with minus a is the gain a is equal to minus V2 by V1 assume the in a there is a phase shift in all transistors okay this minus sign is only taken care for opposite sign V0 is minus GMR is that correct so it is that minus a is positive but that is the so that is what I say okay if you feel so much worried okay first up okay a is positive is that okay now so essentially it is game times the Z will reflect at the input but gain upon 1 plus a is how much one almost one that way into Z will be reflected at the output this is called Miller's theorem so any such network connecting between output and input can be split into part in the input part at the output please remember the condition of Miller's theorem and if that is not true then the Miller's theorem is not how do I how can I derive this can you do me how can I derive this do not use this values this is Z1 this is Z2 you evaluate from write an equation for this and find what will be impedance seen here and what will be impedance here equated and you will get this expression Kirchhoff law like I mean okay please try okay the most important part in Miller's theorem which most books avoid telling you is this the Miller's theorem is only and only valid only and only if there are two paths from input to the output is that correct Miller's theorem is only valid if there are two paths from input to the output if there are no two paths the Miller's that means if this is not a time something which I showed V1 and V2 this cannot be split is that clear if this is because of the transfer function of the some active device if there is no such active device sitting there there are no such two paths created there the Miller's theorem is invalid so what I am saying do not say this is V1 this is V2 so this will be split this is not possible okay so what is the equivalence I am asking for Miller's to be valid there should be a gain term or some function through which input can go one path and this is the other feedback path or feed forward path is that clear to you so if you have a two parts only and then the Miller's split is allowed is this condition clear to you if this kind of circuit is not possible in this that is why I have put it here indirectly is that correct that a means essentially gain function coming from if that does not exist no such split is possible this theory if you move someday I will give you a some one hour lecture on Miller you know so this is only nutshell I brought you actually theory is very interesting long issue not to prove my all points right now I am assuming something and saying it is proved but I can show you without assuming so some other day so is the Miller theorem clear so if you look at my this is what I am going to use very often so can you see for where I can use now Miller's theorem for which component yes which component I can use Miller theorem CC output to input is the only component is CC how many paths you can see from here one through here one through here is that clear so there are two paths available and that is why CC could be split now is that correct yeah so CCs come to equivalent circuit your pointer is understood but let us wait few seconds one is through transistor get to drain gmvgs-gmvgs a current or transistor there I know gate say gate may be just like I have a brain current minus times I know that a part Mila BOOTRA external part via your capacitors a job that I so dope dope part key theory job only over the key kina is a here Miller left that if you go hata lia transistor was a hata lia so there is no Miller theorem this fact has to be understood you cannot split otherwise is that clear so only when you see such things then you say I split is possible okay okay yeah it always correspond to external feedback in transistor even internal feedback is always there which is the component which is always connected between gate and drain CGD CGD so that that is also there but this CC I put it additional to CGD so I add CGD to CC the total capacitance between drain and gate is CC externally put by me okay so I have now two paths one the transistor other through CC is that correct and this will give me gain with a phase of minus 180 degree is that clear and this will be another one so I am now looking into this situation and therefore I can say now CC is splitable what is the amplifier gain in our case just look at it leaving all capacitances what is the amplifier gain bina kuch so che bolo now is amplifier again kithna hata R0 is infinite man liya bhoj jada hai so GMRD minus GMRD parallel R0 if you wish to the gain of this amplifier is minus GMRD please remember R0 is parallel with that but since RDs are much smaller normally I leave R0 but in real life what should you write RD parallel R0 okay okay so it is minus GMRD is the gain okay so what is this value I have got that minus a word which you are seeing is this GMRD is the value which I got for the gain function okay so can I now write an equivalent circuit using I know the gain I know CC so I can now write an equivalent circuit what I said so far is the exactly what I do there is no difference in what I said is that okay so Miller's theorem is valid for our case so at the node 1 that is at the input side what will be the capacitance seen now one is seen one please look at your circuit what is the first capacitance I have given input capacitance of a transistor seen one so that exists as it is please what are the node 1 I am talking the gate terminal I call node 1 the drain terminal I call node 2 okay so first is seen one and shunted by it how much is the capacitance will be 1 minus minus GMRD times here 1 plus GMRD times CC plus seen one is actually the input capacity is it becoming larger or smaller earlier seen may be order of say 0.1 puff how much 0.1 or even less than that femto farad maybe 10 to 50 femto farads let us say GMRD is 100 okay let us say CC is 0.1 puff so how much I am adding to some femto farads how much initial capacitance I have few femto farads seen one I have added 100 times CC even if it is 0.1 puff I say or 100 now femto farads even then it will add now tens of pups so what is the value capacitance has increased enormously if there is a resistance here per say equivalent something what will be the time constant will increase or decrease increase so the 1 upon time constant will increase or decrease decrease is that correct because larger 1 upon R C is larger 1 upon R C is smaller so we a lagakha ki a just feedback may capacitance if you put the pole on the input side will go lower value or higher value for which side it will move C has increased 1 upon R C C has increased so P will become lower value so the first pole has if I would not have put this let us say there is no CC where would have been my pole R into C in 1 if I put a CC where would the pole shifted 100 times on the left or what is I value is that correct that means smaller is that correct so do you see that I am now tailoring my FLFH values by just putting a CC there I have brought down one pole to the very left side okay how much I can do it that will be decided by which values by the GMs by the RDS and by the CC I connect generally I may not play too much on GMRD why I will not play too much on GMRD because that is the gain I do not want to play too much on gain so I will only change the value of CC so CC may tell me where should I put my pole please remember seen one will be very small compared to CC 1 plus this time CC so essentially it is the CC which may decide the pole at the input is that correct unless of course there are additional capacitances available at C1 is that point clear C in 1 is larger than you may see how much is parallel total sum of that but in case it is only CGD CGS that may be smaller than 1 plus GMRD times CC okay and in that case we say CC may now start deciding the pole at the input side is that correct so have I tailored it I changed this is what design is all about I tell her it I say okay with the hold all down okay see the next side at the output side there is a load capacitance and how much is the capacitance it will reflect that the output I said a upon 1 plus a means 1 only so it is CC only so the output capacitances CL plus CC so the second pole will be 1 plus RD times CL plus CC please remember 1 plus GMRD times CC is very large but CC may not be that large as small I mean that large so is load may be external I do not know it may be large also is that clear so this pole will be generally smaller or larger than the earlier P1 CL and CC will be comparatively smaller some of them is that clear RD is typically few kilons are tens of kilons RS is in few hundreds of homes there sees are increasing there so which pole is earlier than the second one P1 is earlier or P2 is P1 is much earlier compared to P2 this did you see I split it so I moved P2 far away or I brought P1 away from the P2 is that clear if I do not put CC then what will happen this CL and see in may be opposite please tell it what can happen if C1 is comparatively higher smaller value RC in may be larger 1 upon RC in may be larger compared to 1 upon RD CC CL so then what will happen the output pole may start dominating over input pole now what did I do I made input pole dominate over the output pole by putting CC is that I remove the CC then it may possible value of CL and RD as such that this pole may come earlier than RS CN 1 okay that we say okay output pole is dominant now then the input one but by putting a CC what did I have confirmed you that the input pole will always be dominant over the output is that clear so CC is going to decide where should I have my bandwidth is that correct why it is called bandwidth point because beyond that the gain will start falling 20 dB down pole means gain will start 20 dB down so this is the value is that clear your bandwidth I am controlling through what CC external capacitance I am controlling is that clear these are the tricks which all amplifier design is used where do I put my pole is that correct see pole 1 is 1 upon RS times CN CN I have increased by CC because gain time CC is appearing RS is let us say 100 ohms and this is say gain is high enough GMRD time CC is also high enough so this value may be let us say gets to set one key 1 kilohertz okay this is say 10 kilo ohms and this is CL is equal to something like this and this goes to let us say 10 kilohertz because of RD or this combination so this pole has started dominating which is this one-tenth of this thing but if this CN is seen one only okay which is 100 times less than that and RS is smaller than the P1 they become larger than the P2 in which case I would said output pole may dominate but by putting CC I guarantee it said something that the input pole will dominate is that clear to you so by why I did not want to try on the output side can you tell me why because the CL is not known to me many times so I do not know what bandwidth I will use is that clear to you why I am always looking at the other side is that point clear because CL is something unknown to me in my analysis I put some value okay but in real life where it will get connected I know idea okay and then that should not decide my bandwidth is that some unknown quantity where my this amplifier is going to drive should not decide where my bandwidth is so by putting CC what did I do I decided my bandwidth independent of what is ahead is that clear to you this is called what you want can you get without the output telling us what I should but in analysis we need not do all that we just calculate for all capacitance all resistance what are we get and find which is higher or which is low is okay that you have P2 right is that clear so in analysis which will do one you need not worry all this which I am telling you why I am telling you all these tricks because this is how I will design is that clear CC designs deep bandwidth which is external external means I am putting it though I do not know but this amplifier I am designing I am adjusting CC value for my bandwidth the assumption is here will not be so small or something which will start that may still happen otherwise then okay so some conditions I must meet but generally if the gains are larger and CCs we put around 1 puff point 1 puff to 1 puff this is always on the left side input pole will dominate but please do not take it that every time input pole will dominate over output poles is that you have to calculate and find but by trick by this roughly will always occur for general techniques is that okay this tricks I am telling you because I have figured out over the year that people always feel that why do we do so much of this so I am now explaining why we do that is that correct so we will start a circuit in which this has nothing to do with actual knowing where the poles are what we will do is we will have a network will find all are these and calculate whichever is omega 1 omega 2 omega 3 then we may substitute the given values and find oh where is omega 1 where is omega 2 where is that clear so in real analysis we are done bother which dominates but in design I do want to know where I control myself last but not the least before we go yes okay I made a mistake he may be right put why okay do not say that because he I think he was right I did not want to that time agree it is an admittance essentially capacity is in admittance run okay you are very good thank you for correcting Z by Y sorry that there is a for RNC RNL Y is for admittance because it is a one upon that line so I made a mistake thank you very much okay there is another important issue we have seen poles so far okay now people say why should 0 occur what is the transfer function 0 means at the output V0 should go to 0 where is the 0 in a transfer function when the output goes to 0 now output can only go to 0 if there are two signals opposite phase and equal magnitude a asset current a if we see not the opposite current are equal is not with the potential 0 there is same two parts opposite polarity gives the same magnitude then I will create output 0 is that good I into R I into R both cancels minus IR plus IR R1 R2 may be different but I1 into R and I2 into R2 opposite fade R1 R2 may decide they may cancel at some point is that clear this is exactly what 0s are all about just now I showed you there are two parts equivalently we say one is the path through a transistor that say an input V into and the other path is let us say an input V in one which actually V in one as both are same but even if they are different I do not care the other path is this a concept problem the current may be feed forward output per two path millenium is a X signal say CCK tariff Sarah do Sarah transistor can do Sarah GM time Sarah but transistor always give me how much phase shift minus miss 180 degree shift but this is always given me 180 shift this is direct is that correct if the values of IR for here IR from this are such that at this potential output is balanced because of this or because of this then we may say because they are opposite polarity there is a possibility of those values canceling each other is that point clear X a path a 2 180 out of path a do not a value same opposite polarity with the net AC current is that clear so we will become 0 so output will become go to 0 is that now clear why 0 can occur we are created a CC and you say okay I shifted the poles you did very great thinking isn't it when I put a CC I was thinking how great look I shifted poles and all that but this problem is suddenly seen by me now I say Baba if such magnitudes occur then I may actually get a 0 is that clear and 0 means 20 dB per decade gain enhancement pole means 20 dB per decade going down now where is this 0 and where is this pole if that occurs at the same first frequency then nothing has happened actually it has cancelled each other so now I have to start looking into where this 0 falls is that clear where this 0 is occurring at what value of RC and GM this can occur is that correct what value of these three 0 can occur and if that those values still the dominant pole is Omega P1 fine otherwise I have to say or air then I am not gaining anything is that clear so this was a pole and at the same frequency if the 0 starts okay same frequency then this is the pole 0 has cancelled each other so the whole purpose of shifting that pole has lost is that clear you can see then only it will start so the purpose of a 0 may not be as good as we think it may be actually bad oh so V in 1 V into only to show you two paths V in can be seen they are same you connected here if V in 1 is same as V in 2 it is only to show you two parts it may mean a electricity okay what is saying that sir you are like I mean yes I mean in the code this is just to show you two paths yes ma'am no no see this node as I say if they are equal this node and this node are same that is what I said you only show true path I made it separate what you are asking is correct if they are same like this noise okay so the point I am trying to say that whenever there is a feedback capacitor between drain and gate not only you will get a different pole but you may also land into a 0 is that correct there will be a 0 associated with this feedback capacitance every time and those values may actually change your board a plot anyway is that correct so you thought only pole I can adjust then you realize that not only pole but now this 0 is also coming okay so in your analysis you must also figure out where is that 0 now okay and how it will come into expression in the numerator somewhere term must go to 0 that will be 0 is that correct if that at a given frequency numerator must become 0 okay if that occurs then we will say there is a 0 present and the value of that 0 and value of the pole you are shifting should not be same is that okay this is what we are as I say I am not done so far amplifier this is what we want to do okay actually but all this today we did is general theory how do I bring 0s high band poles what do they mean okay is that okay so today we only did a general theory okay for poles and 0s next time we will do an amplifier and find poles