 Okay we are looking for amplifiers and we want to design OPAM but when I started looking OPAM design I figured out that we have not done frequency response and one of the features of an OPAM design is the bandwidth calculations for a given what we called as phase margins. So I thought that I should quickly first run into frequency response of amplifiers and we will also give you a simpler method of calculating dominant and non-dominant poles. So today may be or even tomorrow I do not know how much, we will just look into frequency response. All of you are aware that a transfer function like AVS can be given as DC gain AV0, this is S by WZ, this is not QL plus 1 divided by S by W1 plus 1, S by W2 plus 1, WZ stands for the 0 of the transfer function, essentially it means at that frequency the transfer function has a value 0 whereas W1 and W2 are called poles which essentially means at those frequencies the gain becomes infinite or unstable as such. So in a typical mass amplifier we have around typical of course it is not necessarily 2 only but normal amplifiers will have 2 poles sometimes 3 poles and if you have a defam plus driver plus buffer plus output amplifiers it may have more than 3 poles as well and more than 1 0s also. But normally for a simple amplifier which we will discuss first there will be 2 poles and a 0 is most likely event okay, the most passive circuits which we use as filters like a low pass filter and a high pass filter in RC network has 1 capacitance and therefore 1 time constant and therefore 1 pole okay, these are called single pole transfer functions. So we will first look into quickly those single pole transfer functions, these are all standard you are done in second year nothing great has been told I am all that I am saying they in a transfer function denominator gives poles numerator gives you 0s as can be seen from the function itself and AV0 is called DC gain I repeatedly tell you earlier also that DC does not essentially means DC okay, though please take it from me and I am opamp is also a DC amplifier is that clear to you opamp is also a DC amplifier. So even at 0 frequency it has a finite gain okay, before this is a book which I follow and you can see why there is the name somewhere is mine sitting there so obviously I will support this book, this is a very famous book by micro electronic circuit by Sedra and Smith, I happen to share some 150 pages of that, so that is why I am name appears, so typical this is taken from page 34, 35 of that book, this is there are 2 filters shown to you one the first one which is RC network R here and C here is essentially low pass filter and the second one is capacitor here and resistor here which is a standard high pass filter and if you want to make any other filters you just make a combination of these 2 it can be band pass or a band gap or a notch whatever way you wish the basic 2 transfer functions are good enough to create any kind of filters however these are passive filters in real life if time permitting this I hope so we will look into at least putting at least one LRC based filter using switch capacitors so that possibly is the last part of this course, so we will implement a filter for that sake okay, so standard if you see I do not know that you can read but maybe I can see for one of them you can see the transfer function TS for a low pass is some constant which is DC gain as you can say 1 plus S upon omega 0 where omega 0 is the pole value for this transfer function and if I high pass it will be S upon S plus omega 0 where omega 0 is again the pole for the high pass. If you write in T j omega form S is equal to j omega substitute there you get this value the magnitude response which is essentially taking the magnitude of any complex function a a plus j b is under root a square b square and it is phase relative tan inverse b by a okay so that is standard technique which we have been using for complex system for last 3 years at least in IIT and maybe any years earlier. So we are interested in 2 response of any transfer function one is called magnitude response and the other is the phase response which is the angle associated with those poles or zeros so that is given by tan inverse minus tan inverse omega of omega 0 in this case it will be opposite of that tan inverse omega 0 by omega and we can say at omega is equal to 0 which we call DC okay the gain is k the other of course is 0 and the transmission at omega equal to infinity the low pass gives 0 this and the high pass will give k k is the DC value of that we also define 3db frequencies essentially we say at that frequency the gain falls by 3db okay that is the frequency we define essentially we will say the word later I am going to use the bode and what really bode contributed so that will be few minutes will come to it and these bode plots can be shown so this transfer function is given in the book these are standards so can be looked into book and these will be available to you afterwards anyway. If I plot its frequency response for the gain and the phase this is called frequency response and it is named bode essentially for simple reason that bode was the very old control system man okay and he had a book of 1948 call feedback control systems so if you read that book fantastic so what bode did essentially if you plot a frequent gain versus frequency and this is log scale normalize gain TDA bode by k 20 log of that in debates and this is the frequency scale in terms that is decades so 0.1 this ratio again normalized to Omega 0.1 1 ton okay and you can now see from here if I plot the gain versus frequency as I increase the frequency for certain frequency the gain remains constant and then it starts following as the denominator magnitude starts increasing so obviously gain starts falling now this point essentially if you see the transfer function this actually at 1 the gain has already fallen to 3 dB beyond from the maximum from its 0 dB let us say so essentially this point was called the pole point where we can say the gain has started now falling below okay. However what bode said extended up vertically put a little line and wherever this intersects he called it as a corner point or corner frequency okay and then he started saying okay instead of using 3 dB points use corners okay because anyway you are not going to evaluate from this what you are going to see where is the frequency at which gain starts falling so he actually made a simplified figure of this standard transfer function frequency response and therefore they were called bode plot so nothing extraordinary but something thought that time so if you if you are asked bode plot do not draw 3 dB then just draw this okay that is what the bode said okay and at this point when omega is equal to omega 0 one says below that the gain will frequency higher than that frequency omega 0 then the gain will start follow and if you see the function this slope is 20 dB per decade of frequency change or 6 dB per octave means to the power 8 kinds okay octadecimal octonal octal systems some people believe it is because bode did 6 dB octave so I this book also retains that 6 dB per octave of late being all decimal systems we all have gone to decades otherwise we will have to also change everywhere in form of 8. So this is essentially 20 dB or minus 20 dB per decade is the fall of the gain and somewhere down here you can see at the frequency which is much enough it crosses some limit but since it is minus there is no point of calling 0 but let us say this has some function k would have been not there this would have been some positive dBs and then it would have cross a 0 dB line essentially when unity gain is observed and below unity gain means there is no gain really there is a attenuation going on. So this point was very important in the case of normal transfer functions of an amplifier what will be that point gain bandwidth point okay so that is the very important point for us to know up to where gain is possible if you get the t j omega plots phase of that tan inverse omega by omega 0 minus because of the opposite signs so one sees from here if you see tan inverse 1 is 45 and tan inverse 0.1 is very close to 0 1 10th okay close to 0 and if you make tan inverse 10 which is close to infinite or much larger it becomes closer to 90 so if you go with your frequency from 0.1 to 1 to 10 times the omega 0 the phase goes from 0 to 45 to 90 minus because minus tan inverse. So the slope down here therefore is called 45 degrees per decade okay that is the slope with which this will actually go down the since this was a low pass filter the similar response can be shown for high pass in this case the gain actually reaches maximum sometime at the say whatever the pole frequencies the pole is on this side and the gain is falling on this side you can say high pass filters this was a low pass filter this normal in amplifier response is closer to mixture of a high pass and a low pass initially it rises then becomes constant and then falls so this precedes this and you have a normal gain but for open this does not exist actually the gain even at 0 is finite okay because it can that is the difference between normal amplifiers and an opamp at 0 frequency also DC can be amplified by an opamp is that clear to you and therefore there normally the response is exactly like this it does not follow here but for some reasons if you do some machine putting some capacitor some other registers around this can show a pass filter frequency even at the lower side okay so depends on the like coupling capacitors at times forces you to get into a high frequency pass system and therefore then you may see even in opamps which are called AC couple opamps you may see a normal mid band gain with fall on both sides okay so this is something which please remember Bode plot and frequency response are no different all that Bode did of course there is a lot of theory which you need for the it is a control system course okay and there the all the worries are only stability this was only shown for non-stable I mean 100% stable system we like to see in real life when we look for opamp design if there is a instability how do you get rid of it and when the instability occurs okay so we like to see we will see in case of small feedback system this and we show you that instability starts when something something happens in a feedback and in opamp we want to avoid it because we want to make amplifier stable okay so that is something we will see when we come to opamp design or at least stabilizing opamp maybe initially we just look for the frequency response and then we will figure out whether it was stable or not if not we say okay there are 3 more techniques which can make opamp stable okay is that okay so this is basic feature which all secondary writes at least I have thought okay so let us look for a typical common source amplifier whatever is true for common source is true for common gate common this basic equivalent circuit once you drop everything is identical for us so example is only shown for a very trivial common source amplifier with the resistive load if there is a current load what will happen RD parallel or 1 GM or R0 parallel whatever it comes corresponding loads will appear as of now RD represents load whichever way you create first thing we say okay if this is your this and we right now neglect there is a capacitance between gate and drain in a normal transistor which is called CGD which is very important capacitance right now I not shown it but there will be capacitance here from the substrate there will be capacitance to drain capacitance to source right now I am only interested to see input capacitance CGS okay also there is a output capacitance CDB let us say this is N channel and this is grounded and there is a capacitance here which is CDB okay equivalent circuit of a mass transistor we have done many days ago so there is nothing great about the CGS sometimes include CGB because if the channel is not fully inverted then age of that may give you get to bulk work this capacitance so essentially you can say CGS includes all of it C ox plus whatever it is and this represents CGS as far as circuit is goes only the value matters so I said it to CGS which may include CGB okay so if that is so I am interested in the output pole first let us say since this is no connection right now with this CDB is not CGD is not taken care so no connection between input output except transformation because of the current source okay and I assume right now R0 parallel RD R0 is very high compared to RD and therefore it is RD so if you see the pole which if you look at the gain V0 by this the pole as an output frequency at which the gain will start following is nothing but RD times CDB so essentially this means if I calculate the time constants I will be able to get the pole value is that correct so basic idea behind all this is to evaluate corresponding time constants however if the equivalent circuit of amplifier also contains CGD then some transformations are required because CGD will then be a part of input also you can see if you see only input site RS CGS is the only transfer function of this time constant available so first input where a frequency related term is one upon RS CGS output is related to R this time constant however as I say if there is a relation between this there is a CGD available then is that okay three capacitance is of course there is a CSB also is possible but that is lumped with CGS so here is the equivalent circuit of a MOS transistor amplifier common source I am applying a input signal through a series resistance of the source RS this is my CGS this is my CGD this is my current source GMVGS please remember this is not VN this is GMVGS this is R0 RD and CDB we can always calculate VGS in terms of VN by this simple divider rule 1 upon CGS upon 1 upon CGS S plus RS and therefore I get VN upon 1 plus SRG RS CGS as the connection between VN and VGS so if I want to find V0 by VN I will replace VGS by this term and therefore I can always get the ratio of V0 by VN okay the assumption here is when I calculate this this does not because that impedance seen is much higher so I neglect that okay if you do not feel like doing this solve all of it network there is nothing it will automatically come what I am trying to say okay so these circuits we have already solved n times so I am not going to solve it again I just have using a case of law either in generally most cases is the nodal equations I solve okay so if I use nodes and solve this circuit there will be two equations and therefore always everything can be solved so I get AVS of is equal to V0 by VNS which is minus GMRD 1 minus SCGS CGD sorry this is CGD by GM upon 1 plus RS CGS into 1 plus RS times RD CGD so if you compare this expression with the transfer functions the first time I said S by Omega Z plus 1 so if you say when this term is equal to 1 that is S is equal to GM by CGD the output goes 0 which essentially means you have a 0 of the transfer function okay so the zero is occurring at frequency of Omega equal to GM by CGD okay and at that frequency the transmission has a value of 0 however if you look at this value what is this value if you leave all frequency terms this minus GMRD is the that means remove all capacitors from the circuit this is a standard DC gain not 0 frequency DC gain which is AV0 which is GMRD so now you can see from the denominator you have 1 plus SR CGS 1 plus SR RDCGD this denominator I explain what is the purpose of doing all this now I can see two poles here okay when this becomes 1 minus 1 on this become I have two poles anyway I am seeing them but I want to see whether they represent Omega in and Omega out which are derived at least to great extent if they do then my solution becomes much easier to think then I will say okay look at only input look at the output I have two poles okay to prove that I am okay I will actually use expansion of this you will get SS square term S term and constant and from there we will figure it out which terms can be neglected and therefore input and outputs can be represented right here okay this is the trick which you want to use and using that trick what the new technique I am not really new is what may be some of you this technique is some 50 60 year old new world is not true this is given in various books so you can go back and look into that is both D is equal to I took some this GM which is here I have brought it down okay so I did D is equal to GM times 1 plus S of this into this okay and if I connect the terms I get something of this kind and GM RD is minus AV 0 so I I get finally a term which is 1 plus S CGS 1 plus GM RD CGD into S and what is being neglected here we are trying to say SS square terms are much smaller okay far off right now I kept everything okay only thing afterwards when I expand I left SS square terms and if I get this term then then I get this 1 plus RS C in if this is your CN CGS plus 1 plus GM RD times CGD C input then the pole this D is equal to 1 plus RS C in yes if this is higher than 1 this is sorry this is essentially pole which is RS C in okay so input pole is nothing but the series resistance or the resistance available in the input circuit and the net capacitance available if CGD would not have been there how much would have been CG the CN CGS so that was this with CGD why I brought this all expression here this is giving me some idea that how to get transformation of CGD from the towards input side which method I should use that what is the theorem we use Miller's theorem so this is the precursor to Miller's theorem I figure it out that I can actually look into input side and see that the input pole is essentially same as what I would have got except that now additional term of capacitance is coming because of CGD and which is becoming very important and very large because typically GM RD will be more than 10 or 5 at least if not more and maybe 100 or 1000 sometimes so in which case this term may start dominating this term is that clear and therefore the input pole may become that is the frequency becomes smaller and smaller as the game is larger or the coupling is larger is that clear what does that mean if the frequency falls means what does that mean that the cutoff point is shifting to a low frequency so your standard DC gain is for a very short frequency range and in normal amplifier what is the bandwidth definition up to which gain is constant so essentially we are saying if this becomes the dominant pole as we will see later and if this omega n starts reducing because of this term or this term then your bandwidth starts fall when this increases what does that mean gain increases gain into bandwidth is the constant you increase your gain and bandwidth goes down so obvious reasons this is valid statement is that clear so essentially we are worried about how much gain at what bandwidth you cannot have or if you want to beat little bit of that what will you cascode it but the penalty you will pay for maybe some more power and some more hardware okay so is that point clear the point here is not really great only I am trying to say I can always look at input poles itself to see the poles value and I need not have to do all the except that to take care of feedback I have added this additional term okay that is that to care now this all of you said suddenly and here is the theorem which is very popular all of us have been using for ages now if I have a impedance z between two nodes a two port network okay where there is a gain function a which is Vy by Vx this is very very important term a is Vy by Vx is that clear to you then we can convert this to this was a lateral this I can break into two parallel 0 1 and 2 which still has the same potential Vx and Vy provided we do equivalence of this then z 1 is z upon 1 minus a and z 2 is a upon 1 minus a into z now this please remember if there is capacitive then correspondingly you must figure a 1 upon j omega c is the z please take it otherwise do not go by random z is impedance so 1 upon omega c will be the value of capacitive impedance okay okay is that okay so this divider will be then actually will become multiplier for capacitance that is why it is 1 plus gain times the c has in is that point clear 1 upon omega c is the impedance okay so c becomes lower and therefore net capacitance actually increase okay this is very important most of you know this well working well and all that but here is a problem which you should know all a Miller theorem cannot be applied randomly okay and which is why an example is given if I have a simple passing network shown here which is R1 R2 obviously Vy Vx are not same so Vy is nothing but R2 upon R1 plus R2 times Vx so there is a gain equivalently gain but however if I make a transformation okay by the 2 0 1 z 2 functions on the one side I will get R1 plus R2 which is as good as any input this is fair enough value but if you look at the output side you get R2 parallel minus R2 you substitute there those values and you get R2 parallel minus R2 which means minus R2 by 2 R2 parallel mine that means there is a negative resistance at the output impedance which is not true because in real life no passive device can create negative impedance is that correct so now you do do not worry A upon 1 minus A you substitute Vy by Vx it will be 1 minus Vx by Vy just substitute those values and you will get minus R2 okay do not I am not trying to add myself anything okay so this essentially means that if I have a passive network something like this I cannot apply Miller's theorem is that correct so what is that limitation Miller did not say but was imbibed by him unless there are 2 paths from input to output Miller theorem is not valid in this there is only one path is that clear is that work clear to you all of you please remember Miller's theorem is only and only valid if there are 2 paths from input to the output is that correct otherwise like this the Miller theorem will give you the output impedance absurdly value from that okay this is a very this so if you have a normal amplifier you have a feedback resistor here the gain amplifier itself is one path is that correct they are through this input will go to the output currents are going input will be transfer current to voltage transfer or current to current no amplifier the signal goes from input to the output through amplifier only in a gene vigorous is transferring okay so there are 2 paths one through the amplifier and one through the feedback network only and only then the Miller's theorem is valid so please take it hurriedly do not use transformation because you are T network because go by Kardia or Ulta pie Kardia do not try something like this because that may lead to a absurd situations okay is that point clear to you everyone yes no no but then at that node still same path no no no what you are saying is if I have this path no these are not too bad they are same resistance splitting and passive they are not too bad here the phase between this and these are not same they are 2 separate paths going on is that clear to you through a 180 degree out of phase you are going and one directly you are going feed forward as the word goes and therefore the Miller is possible is that clear because that in every case you say 1 upon 1 minus a upon 1 minus a it has to be finite in this way otherwise that just divided output by input is not the gain function is that correct and therefore Miller theorem is always valid only and only if it is a active device sitting there so apply Miller only when there is an active device with feedback available to you apply Miller and you will not make any mistake in actually valuations is that clear this fact is known to many unknowingly but now I thought you should be knowingly know that why we do not apply Miller elsewhere okay is that okay all of you is this issue clear okay okay since we are on the two path system here is something which I thought you should immediately connect to this is nothing to do with the Miller but the same concept I thought maybe I extend if you see a normal transistor or amplifier shown here and you are just now you said there is a capacitance cgd in our case or actual capacitance which you can be which is called cf feedback capacitance but whenever if you have done a signal system course or normal signal theory or signal flow paths we are going to assume there is no feed forward there is only feedback okay but in real life feed forward exist okay if you have input signal going from here there is one way going to the output the other is going through the transistors at the output this is give you minus gmbgs okay because of the sign I chose and this will be directly what is the probable ICF whatever it goes since this is a capacity effect that means the impedance is 1 upon omega c so currents are functions of omegas okay so it can happen that at this output node if this current and this current are equal at a given frequency in magnitude but since they differ in phase by 180 they may cancel okay at that time the output will go to 0 okay so we say a 0 is essentially occurring because of the feed forward currents are matched by the transistor current which is out of phase at that frequency equal to magnitude of each okay then only you say there is no current and therefore the output DC AC output goes to is that clear so this 0 word which people keep talking is essentially is using the same concept as we used in Miller's theorem so I thought right now I should connect to you here that two parts is required to create a 0 otherwise there is no way function can become 0 okay so this has to be understood that why people I mean it is not very big thing I am talking but this is not understood by many that why 0 suddenly appear 0 is occur because of so here also there is a active device which is giving a phase 180 degree out of phase this is giving 0 phase if the difference of this magnitude becomes equaled in a man at a given frequency then only the net current with 0 and therefore net voltage will go to 0 this is a principle of 0 is that okay so please remember 0s are coming not out of nothing they actually come because of the real situation which you seen amplifiers is that okay all of you anyone so returning to a problem of frequency response of a common source amplifier a typical common source amplifier is here equivalent circuit is shown here the DC gain is gmrd and using the Miller's theorem now I can replace that cgd is 1 plus gmrd is the gain so 1 plus gmrd why minus become plus 1 minus a now 1 plus gmrd times cgd is reflected with the input in shunting the cgs and a upon 1 minus a is close to 1 gmrd upon 1 plus gmrd is 1 so I wrote only cgd parallel cdd so the output yes equal I just removed that this cgd has two paths 1 through gm vgs one directly through cgd gm vgs is appearing from the input isn't it that is the path okay so if I now use Miller's theorem I landed in this equivalent circuit and then I have already said I can I if I am not very fussy about exact values the input pole and output pole can be directly seen at the output side and the input side okay why by finding the time constants is that clear so what is the time constant on the input RS time cgs plus cgd times 1 plus gmrd is the input Omega 1 upon that similarly I have been to parallel combination of cgd and cdv that is some cgd plus cdv is the output pole many case the cdv may be very dominant term why last many a time those cdv value may not be more than half a per for less but that value may be very strong because of something else what if there is external capacitance that is you are driving next system which has a large input capacitance this will appear here so that is why when you calculate poles the output of input of the next stage must be used as the output for you because they are in real life this is what it is going to see is that correct is that point clear so at that time the seal is very large the output pole become very small frequency and that may become dominant okay if the seal is smaller or comparable to these values then probably this may still dominant okay but in many cases if seals are actually put that may become so do not go by my statement every time that this will be dominant you check it whichever is smaller is the dominant is that equivalent circuit okay all of you these are three rules you have already done in second year I am just trying to repeat because I thought this method which is relatively standard and now going to compare with the new as I say new in a pachas alpha runner or new technique zero time constant technique which many of you have learned if you have done my fourth second day in a course I taught them there but there if I recollect well I did not teach what is the other method of me you can always calculate dominant pole and I did not show them how to calculate non-dominant pole today I will show you I can also evaluate non-dominant poles okay is that okay figures are okay everyone okay is that okay will be because the next device I do not know if it shows me what is that I am driving and if no load is known normally in all circuits you know you are connecting to some stage so you roughly know the input of that stage okay in digital what do you say at least once you have said should drive okay and to safety we put at least 4 C oxy drive okay so we are guaranteed that the next delay is not worse than this what we are calculated okay here we could assume same okay using our technique as I suggested I can find omega in and omega out which is RSCJS 1 plus GMRD times CG input pole and output is 1 upon RD please remember this is radiance expressed in radiance per second and actually if you want frequency divide by 2 pi and you only get hertz so right now omega and frequencies are talked in same way but they are not 2 pi f is omega please remember unless you divide by 2 pi this is not hurts okay maybe in example which I saw I will show you hurts okay so typical transfer function then can be written and right now I have given up that 0 because I am not very keen about that 0 but if you wish we can do that as well no problems AVS is AV 0 upon 1 because this I am using it which I am going to be using 0 value time constant method and the actual method is what that subversion was asking you are a network so all 2 nodes get exactly whatever you wish no terms are neglected no terms are you are not missing anything okay but if you want to see approximately equal then tricks have to be followed otherwise as I say if everyone does it is of law there is no mistake because spice does not do mistake simply you guys only solve that network okay so spice results and other results but then why are we doing this because even to tell spice what W by I should start I should roughly get values now this much bandwidth okay so first I would guess otherwise what happens step by step 1 million years cable output okay so this is typically what the transfer function looks like okay the statement I was making if you want to solve I use expressions like this will appear this may gives you 2 poles omega p1 square s1 plus partial fractions a s plus 1 plus s plus 2 jubi high square expand koro equivalent Nikal or you Nikalsak so it may be the poles Nikal a RS1 plus GMR is CGD RSCGS plus RDCGD plus CDB omega put is 1 upon into 1 upon RSRD this this is using this so someone now you can verify what has been done in the 2 cases under what conditions this is same as what we did and this is close to something which we are done so conditions like I said then if RDCGD plus CDB smaller okay then this terms then omega in is same as omega p1 is that okay I will just look at this term if I neglect this this is what my pole was this is very trivial just as I say solve 2 node situation 8 may current as a 8 may current as a 2 equations banjangi or subquits follow which one you are talking C equivalent sorry I should have written somewhere C equivalent is combination of Cs or seeds CGS CDB CGS CGD CDB whatever 3 capacitance are 2 combination low that is equivalent okay may be I will write I will not CGS CGD plus CGS CDB plus CGD CDB okay capacitance these are all radiance per second omega the poop I said divide the hertz okay okay you do not call it dimension it is square of square of rad square this is a product yeah this is all some representation you are right this does not give units this only say C okay you can see from here if no no no she is right 2 resistance are there so I need 2 capacitances these are 2 capacitances is that clear to you to make a time constants I need RSC RDC these are CCs so it is C square I could have written but I said anyway I am not using it so I use only C okay is that it is only representation please do not go away the basic idea you can see if there are 2 resistances in a square term appear obviously there has to be 2 capacitances there okay the condition which I wrote if I do CGD CDB is smaller than this time and then CGS is larger than this time this omega p1 and omega p2 reduces to the same values which we derived with using Miller okay now most cases in most cases and not all cases this may be valid okay and therefore in many cases using Miller the values are good enough okay good enough like let us say if it is a cutoff is 4 giga 4 megahertz we are overestiming to 4.5 megahertz okay okay that is all that will happen if you do not use accuracy okay but otherwise anyway start point then I had to fall care 4.5 now I am not saying that we should not if you are very accurately expression I mean no one is stopping you are substituting everything there okay do substitute get 3.63489 whatever decimal hertz okay your your choice no I am not making fun but this is an issue which one has to understand in engineering up to what accuracy one should look at okay the reason why we do not go too much on this is in our assumption right now when I actually put this design on a chip there are other parasitic capacitances available to me even parasitic resistances will be there which right now I do not know okay so anyway may estimating yeah fifth or estimate that is the idea in real life what we will do is we will actually do this simulation find says laid out as we shall show you then extract from the layout the circuit which then takes care of all the additional parasitic which layout will give and then we will find the first values which you started with did not match with the you got back then pick till you get as good as match that is the design you must then correctly find which will give me correct okay if you do second turn around you lost money if you turn around you will be thrown out okay is that clear under certain conditions the Miller theorem is really really okay values it will give and you can still start with these designs as a guess works okay. Now one of the thing which all Miller theorem assumes and which is what the other theorems will also it assumes as if there is a dominant pool we started with this assumption if the two points are closed back the Miller theorem may create a lot of problems okay so selectivity will become very difficult and those this theorem should not be then applied okay but in most of the amplifiers which we use for variety of other specification the there will be a dominant port and the next pole will be away far away and 0 may be even further away okay and therefore this calculation of dominant pole is not that as one thinks the reality you can always get the accurate poles okay. Now since we have figured out in every such calculation RC RC RC we were looking the frequencies calculated based on time constant so we said why do all this then solve the network also why not we use little different technique to get the time constant themselves okay. If you see the expression little carefully this will become RS okay this CDRX CDJ RD plus this into CDD at time constant okay RS into this plus this is another time constant so all that we are seeing is there is a sum of time constant is all that is you are getting through all that expressions okay Swede may individually a Nikali no and I do not have to do all that analysis I just figure out those three and add okay this is essentially the technique which is very popular in designers it is called Baki kushma Liko last line Liko last the line Liko this suggests a simple technique of validation of poles using a technique called zero value time constant also popularly known as open circuit time constant this always figures out the dominant pole is that 0 value time constant technique always evaluates dominant pole okay the other term which it appears okay is essentially we will call it short circuit time constant analysis we will show it later and that will give you non-dominant poles open circuit time constant analysis gives you dominant pole short circuit time constant analysis will give you non-dominant poles okay last my favorite balloon we are assuming the zero is far far away from the pole frequencies what does that mean they do not contribute to the till you reach GB okay so they have no choice chance of coming earlier than GB zero frequencies 10 to 100 times of the GB okay so they do not bother us in any sense okay okay is that you do not have to write this I already said basically I figured out that if you see poles time constant and a though individually why not I calculate okay and then how do I calculate is the technique which is shown here okay for each time constants take whichever capacitor you are looking into for which time constant you want leave that capacitor but all other capacitors this they should be open circuited or impedance should be open they should be open circuited means actually the impedance is infinite as if there so just open there is other technique which I will show you in which actually capacitors are shorted that is impedances one all like that actually see shorted she is made infinity for all in capacitors are now as if not existing okay so I am short all independent voltage sources jabbi code or nika nika so what is the technique apply source find the current short suppose same technique nothing great open all current sources independent one and short all so GM VGS is not independent source it is dependent on VGS so do not short it or open it okay now we want to calculate a resistance R1 corresponding to the chosen capacitor C1 and then find out the time constant which is R1 C1 give me C1 choice how much resistance it sees in the circuit using this that I will call it R1 and the time constant associated then I will call it R1 C1 repeat this like for all capacitors one by one C1 O and then add all time constants this essentially one upon sigma Tj and not sigma of one upon Tj because that is harmonic mean that will get into your problems I repeat this is not sigma of one upon Tj it is one upon sigma of Tj is that clear they are two different the other one will give you harmonic means and that will be absurd then that is some all T times and one upon times the net how you get okay which is essentially your first time constant related omega minus 3db point the first corner phrase called dominant pole okay this is the technique zero value for the circuit of common source amplifier we obtain minus 3db point using ZVTC technique okay even my cgd code we are going to be there now I want to see seen from cgs what is the resistance it is seen if this is not there this is not there yeah okay then I see how much is the resistance is seen by cgs only RS only resistance is saying RS so our cgs name if I get then our cgs is nothing but RS so the first time constant as seen by cgs is RS times cgs expression the course man RS cgs term is that one clear the first in the denominator one of the term is RS cgs that has now appeared you open this you open this capacitances and at this cgs what is the resistance seen by it it does not see anything here there is no connection here so all that it see this and short inputs independent source okay you see there is a square cross so RS is seen by cgs so time constant associated is our cgs or RS times cgs is that okay everyone now what else I should do which capacitance I should do now cgd this is called zero value that is why it is open circuited now I want to see cgd k cross kya hai to my current source lagaya or this kya cross jay voltage of a b by ii is the resistance I shot the source I open cdb and I open cgs is that correct so this is the circuit I get this is like using a kishchoff law I solve this this is just solving again kishchoff law but follow circuit I open cgs I open cdb I put a current source test current and drop across that is my b test or vgd ratio of that is the resistance seen by this capacitance so if I solve this I get RS plus RD plus GM RS RD so that is my resistance seen by cgd so what will be time constant or cgd times cgd is the second time constant now start looking in this expression this may be RS RD common there are one plus GM RD term are I one plus GM RD times RS I open cgs I open cdb and across cgd I want to find the resistance as seen by this how to get the resistance any point between two points I apply either voltage or current source and find the ratio there okay so I applied a high test okay I later on call it test so I will also call test okay so I apply current source I test and measure a voltage across these two points the ratio of the two is vyi is the resistance seen by this once I put this then solving network is should be simple so rcgd is RS plus RD okay first quickly so how to is rcgd times cgd so this is the expression for this then the step 3 which is the step 3 open vgs cgs open cgd and start looking from the cdb side what is the resistance seen so you can see since vgs is 0 you are shorted this this current source is not existing so resistance seen by cdb is only RD is that correct so it is tau 3 is already cdb so team time constant up to team capacitance came in there please remember every capacitance will give a pole is that correct so three time constant should be available if you have three capacity so tau 3db is tau 1 plus tau 2 plus and not 1 upon tau 1 plus 1 upon tau 2 it matters hell of it so sum total kia which pole I am going to get through this method dominant pole okay so you have no omega p1 that should come after this I I I have to be here I have to be here to prove karna ke liye itna sath kia me ki yeh method ii ki yeh khut klik kar sakta aur thode mne aap koi the cga cdb ke value pata ladde nahi to aap baaki isko bhi chod sakta yo kya yeh tika yeh itna yeh itna yo yeh yeh iska cutoff that is what designers do so first order quick thinking huna chahi yeh because they will specify you bandwidth most amplifier designs are most analog people will specify you bandwidth because that is up to which it should work so that number should immediately be seen what is that my up to which I have to do and that is very crucial for our analysis is that okay so f3db 2 pi 3db is this expression and you can see this expression is almost same as the expression we derived earlier for omega p1 using Miller's theorem so essentially zero value theorems are representing Miller's technique is that clear and this pole is dominant pole now there is other poles if you want this full stop edge ke ba for non-dominant pole to we must know another technique which then I will get this okay that method is first measure ka naam actually dhushra kya tha open circuit time constant technique us ka ulta hai short circuit aap capacitor ko jaha open kar rahe tha nahi usko actually shoddy karen yeh so instead of telling you the technique I have chosen may be 5 minute meh ho jayega I gave an actual example which will get dominant and non-dominant poles okay ab jay main a value is calculate ke maa maa kuch ho a final value likh lo ake shawl karna maa aabi aapko se if expressions ka data dikhaha jatam okay main a shawl ke hai aap final value likh lo okay here one example I want to find the dominant and non-dominant pole of this amplifier these values which are changed from my values you note down this and solve the same example with this new values and inform me next time at least one of you what are those values it is there I am you I am doing a separate data but I asked you to solve Rs 10k RL 10k Cjs 1 puff CGT 20 puff gm is 3 milli amp per volt jay data aapko calculate I am going to use ZVTC technique to find dominant pole and short circuit time constant technique to find the non-dominant right now I am I have not used CDB we can use it to simplify I have just in most cases CDB can always be this you have to solve anyway so you note down this this is a common source amplifier equivalent circuit CDB is neglected as of now nothing very great okay. However this technique is given in Gray and Meyer's book those who wish to know it Gray Meyer new names two more names are come now hers and something here is the data RS is 1k RD is 5k IDS is 1 milli amp beta dash is so much rather beta dash into W by L is 100 milli amp per volt square CGT is 0.5 puff CGT CDB be a chance CGT be the zero a CDB the Haney open here so I calculate GM to beta IDS okay abit a quick change nahi hua current source biasing equivalent mirror sedia so GM is 2 into 100 10 to power minus 3 into 10 to power minus 3 the other 14.1 milli amp per volt now we first calculate the things using Miller's theorem so in sub key values so you don't follow anything here you write down this and I will give the final answer time is running out okay I repeat omega P1 this you should write data GM maybe you can write I calculate the pole is dominant pole is 1 plus GM RD RS CGT RS CGT plus RD CGT this is by Miller's theorem substituted all of it has given there by writing those data which I calculate so the dominant poles appears at 3.68 megahertz the second pole from the Miller's theorem is IDCGD so I calculate that to be 63.6 megahertz okay these are Miller's theorem and what is the technique I am comparing it with zero value and short circuit okay if I use ZVTC the problem is RS CGS have you calculate the kind of function Tau 2 is RS plus RD CGT plus GM RS RD CGT so Tau 2 will become then 3 into 10 to power minus 9 plus 35.25 into power minus 9 so the dominant pole is 1 upon Tau 1 plus Tau 2 this comes 3.66 same because okay is that okay in our Miller expression we have neglected certain term to get smaller version but if you use exact one this value will be slightly different from the one which I will get from ZVTC I mean I got from this now what is the next pole I want the non-dominant this is the dominant lower frequency so which is not done Fp1 is 3.68 megahertz Tau 1 Tau 2 the next technique is for the capacitance you are really looking for okay that is why I want to find resistance seen by CGD so Baki sub capacitance go physically short so as soon as I shot this so CGD is only seeing this since there is no voltage here no current source here so CGD is only saying RD so the RCGD that is resistance seen by CGD is RD so Tau CGD is RD CGD which is 2.5 to power minus 9 is that clear all other capacitors physically short them and then from the capacitance you want to see how much resistance it sees that is this equivalent resistor what is the next capacitance I have CGS so what should I short CGD okay but now VGS is not 0 is that clear because CGS is higher drop now so GM VGS however you can see from here this is VGS and this is also VGS 1 upon GM so you can see RS parallel RD parallel 1 upon GM is the resistance seen by CGS CGS is RS parallel 1 upon GM parallel RD okay and if I substitute all these values for Tau CGS I get 0.325 into power minus 9 okay and therefore short circuit time constant is sum of this plus CGD which gives me 2.825 into power minus 9 so if omega non-dominant is omega P2 1 upon this which gives me a value of 56.3 MHz how much was the bimolar 63. something so we are slightly by this technique we are slightly lower this pole but you can still see this is 63 or 60 and what is the dominant pole 3 MHz so we are really far away from each other so the validity is as good as long as these are separated so the a point or the pole I we have technique failure so this may not be at then you must solve if anything is worrying some do there is that okay so I can calculate the dominant pole I can calculate also the non-dominant pole 3rd pole we are not because anyway that 40 degree down should be okay so we are not worried about that okay where these techniques few comments and few limitation comments these do not lead to finding zeros first thing you must see I am neglected in my there is no way I am calculating zeros what is the assumption in that case zero is far far away from poles okay so no worries it is accurate enough there exist abhije main bola phir sa bhaiya likha hua these are these techniques are good enough if there exist a dominant pole separated by larger frequency difference from the non-dominant pole as long as yeh door hai yeh technique valid hai in most amplifiers luckily for us view in most analog system this condition is normally made but exam mein main jaan kuske ulta kar sakta hai that is another issue in this technique remove another catch word this which is very important for actual designers in all the actual circuit there will be coupling capacitor to separate DC and AC these values are my micro farads okay so do not use them because of normal frequency of any higher they will be anyway shorts so do not use those capacitances in any ZVTC open short circuit calculation just remove short them out and they are not there is that correct so in this technique remove coupling capacitors that mean short them evaluate by providing shots at higher frequencies so this is a catch word otherwise actual network mein wo see dekhaya uske piche bhi lag jao yeh aap toh okay we have to subscribe wo dekhaya aap ko pole okay so he said to deal that whenever you calculate the bandwidths for any amplifiers only the p1 is called or p1 is called the bandwidth of the amplifier is that correct is this technique clear so please remember these are the designers way of looking the things kischa fly is always true for all circuit people irrespective whether you design or you do not okay is that okay so on Friday we will start with opamps