 here is a cascode amplifier which is similar to a normal amplifier. This is your driver amplifier M1 which has a all standard value of W by L's VT etc etc. Series to that M1 there is another transistor M2 please look at it this is very relevant and good concept to understand there is another transistor in M2 please remember this is common source M1 is a common source system input is at the gate source is grounded whereas M2 actually has a DC potential at the gate for AC what is it ground for AC it is ground. So what kind of this amplifier or circuit will be common gate it is a common gate we have already done common source common gate common drain so now I say this is a common source amplifier over which we are sitting a common gate amplifier and it is biased by a constant current source I now this is I told you other day there are different way of biasing circuits one is by resistance network the other of course is by active load like a diode connected loads or a current sources as shown here they can be also taken from Miller sources okay Miller and mirrors are not Miller see mirrors current mirrors so it is biased in such a way that both M1 and M2 are in saturation okay is that point clear so the first problem one can see from here if you are drawn the circuit rest you do not write just listen and then think if M1 and M2 are in series as shown here in normal case if M2 would have been absent only M1 would have been there so how much would have been supply this is our VDD let us say source require some drop but the most of the VDS would have come in M1 is that clear most of the drop would have come in M1 so VDS would have been guaranteedly larger to make it in saturation now I want two transistors M1 M2 also to be in saturation if they are to be in saturation this voltage should be sufficient enough that VDS to plus VDS one should be such that both makes it in such is that clear that means power supply needs to be slightly boosted because otherwise I most of the voltage goes here then you have nothing left for to make them both them in saturation so one of the major problem in cascades are you may require larger power supply voltages okay it is not very true what I am making but roughly larger than the single stage amplifiers I would not say large or something because also made in the chips it cannot be more than 1.5 also we can adjust VDS we can adjust VT and therefore VDS minus VT still can be smaller than previous but irrespective order number I choose I need larger drops for M1 M2 to be saturation this is one is problem one you should understand now right now assume this current source is ideal and we say its output resistance is infinite okay infinite is that okay this is a cost code amplifier is that clear this sentences have written just for the sake of it you need not because I already explained you what I am talking these are some okay basically I am saying this is driver this is the gate common gate transistor this is the essentially current source biased now I want to analyze what is the condition what is the criteria I said last time to make cascode better than cascade that normally I say FT or unity gain amplifier a frequency or the gain bandwidth product as we call GBW that is constant for a normal common source or any amplifier so if I increase gain bandwidth goes down if I increase bandwidth gain goes down is that clear and the maximum frequency is unity gain frequency is that correct you cannot go beyond that because why because beyond that gain is minus DB means less than 1 so we have to now worry that how can I increase gain without actually losing on the bandwidth or which are getting that cutoff point to be any different if I remain there and still boost the gain then I achieved something which was not possible in a single stage okay that first case gain bandwidth constant or you FT is constant is essentially called the figure of merit you cannot go beyond that now I want to say even at that frequency gain is higher if I do that then I have achieved something is that clear and this is exactly what cascode is trying to do in this circuit if you see carefully the gain there is no additional load resistance I have put here the load resistance is essentially due to M2M1 okay series some combination of this will lead to R RO and this R is also parallel to this R of this source is that clear why because this is ground for AC so this R is also parallel to RO of these combination but that we treated as how much infinite so essentially I say this RO equivalent what I am going to see here is the only load resistance I have and therefore gain is GM times R0 okay and this R0 we will call RO effective because without this whatever is RO for this with two of them whatever we will call it say we will call it RO effective so if I can make GM RO effective larger than GM RO with and what is the bandwidth frequency will just show you GM by C so if GM by C is not touched and I increase R0 then I am increasing the game with not losing bandwidth is that clear so are essentially what cascode is trying to do is to boot output boost output resistance without losing anywhere in GM factors is that clear that is exactly what we are trying and this is an achievement which is actually not device is not able to tell us directly it is we say device is always cut off cut off you cannot go beyond now I am circuit I am showing you I can beat the system this is the trick in NLR how to beat the system some cost will play somewhere which we are not actually telling but at that cost I will boost this gain without losing the band is that clear this is why cascodes are used where do you use cascodes where you want gain boosting okay without losing the bandwidth okay here is that we wish to find GM effective and R0 I had a tendency to earlier for this is my old papers and this since I used to use R instead of R0 conductance Geo is nothing but 1 upon RO okay Geo is nothing but 1 upon RO so I want to calculate for this cascode stage a GM effective what is GM effective if the M2 would not have been there there would have been only GM1 is that correct with M2 to present the effective GM is we call GM and we derive that let GM and GM Geo1 are the trans conductance and output conductance of transistor M2 and GM2 and Geo2 are similar numbers for M2 names GM1 1 is for M1 2 is for M2 is that okay so if I define that GM1 and Geo1 and GM2 are the for transfer M1 and M2 by a network theory the small signal current flowing in this boat transistor is I which is equal to GM effective times V in plus Geo effective times V0 this is a simple two port network if you are not doing done so far in your network course do it again otherwise the easiest way to represent the two port outputs okay I is equal to GM effective V in plus Geo effective V0 okay if you cannot I will show you how but otherwise you should be able to give right to any two port network in and out V in V out and you will get this explanation okay now in this figure I call this voltage as VO1 this of course I still call it V0 output of we are not taking it but the drain voltage of M1 we defined as output 1 VO1 is that okay different definition nothing great about however if you see carefully the VO1 which is the drain voltage of M1 is also acting like a source voltage of M2 is that point clear VO1 is the drain voltage of M1 but it also source voltage of M2 now do you get the point so VGS is still available if this is your 0 for example AC also 0- VO1 which is good enough because then VGS- it is still will be possible is that clear so this is the trick which I played I said okay I put a series there the drain voltage of this is same as source voltage of this and therefore I can say my VGS is still plus VO1 and if that is larger than this it is still conduct okay so this is what I did I trick on that for AC signal that is what I say VGS to is 0- VO1 is minus VO1 my now from the current equation which I wrote I is equal to GM effective into VOVN plus geo effective times VO if I make VO0 do you get the point what I am saying if I make VO0 I upon VN is GM effective is that clear I upon V is GM effective if V0 is 0 maths okay if V0 is tends to 0 I by VN is GM effective so I write GM effective is I by VN when V0 is 0 by same argument geo effective is I by V0 when Vn is 0 in equation do you understand you make one codes output 0 next time you put input port 0 is that correct and then you get GM effective and geo effective I by V is both cases is I by VN is GM effective I by V0 is geo effective is that okay assume we apply if now there is a game we are playing okay why say is this is DC so not AC I at this node I apply V fixed DC value at the output where V0 is I apply fixed DC value we fixed so what is the AC value I am applying 0 so VO is 0 is that point clear VO is 0 when I say I applied fixed DC bias it cannot change DV DV by DC DI is 0 so we say is V0 there is 0 so we say assume we apply a fixed we fix at the M2 VO terminal therefore VO is that is the condition what is the condition I was asking for GM effective that I by VN is GM effective when V0 is 0 so I made that condition true and then I write VO-VS which is nothing but variation of the VDS to is nothing but 0- please remember this is 0 this is VO1 so how much is change in VO1 AC value 0- VO1 which is – VO1 so VDS is how much small VDS that is AC is – VO1 now look back the two for the two transistors I write two currents though I same in both transistors is that clear I cannot have two in a circuit there can be only one current which is I flowing through M2 as well as through M1 so for a transistor M I is equal to GM1 V in same two port network so I is equal to this is what we say intrinsic values external in the resistances will put later if whenever we need in actual circuit we want to know intrinsically whether we can break that GM that is gain bandwidth limits okay we can now so I is equal to GMV in plus GO1 VO1 what is why VO1 VDS of M1 VDS of M1 is VO1 so VO1 for the transistor to GM2- VO1 because that is the input now 0- VO1 so this is VGS plus GO2 again VDS is – VO1 is that okay just substitute values for both transistors M1 and M2 and currents are same please take it in a circuit same current can only flow okay so the at the drain of the M2 I am only still looking at this that is why it is 0- VO1 is VO for the VDS for this is that correct that is what I put – VO is that correct okay if I do this I can now find out from the equation 3 VO1 3 or 4 I do not know which one 3 VO1 is I upon GM2 plus GO2 yes simple maths okay VO1 is – I upon GM2 plus GO2 please remember G's are conductances 1 upon G's are ours whenever you wish it you can convert to back to ours now this VO1 value which I got from equation 3 can be substituted where in equation 2 in that case I is equal to GM1 be in please remember what was there – GO1 VO1 so it is GO1 times I upon GM2 so if I collect items I is equal to 1 plus GO1 upon GM2 to is equal to GM be in under what condition I derived this VO is external VO with DC value therefore AC VO is 0 and under that case I I by V in is GM effective so GM effective is I by V in GM 1 upon this quantity is that correct so I have now derived expression for GM effective is that and condition I have made what was the condition they wanted me to say that VO is 0 so I have made it pure 0 and I saw the equation and I got my GM effective now I want to see all these values which I put if I do little extensions of this I can bring this denominator back to numerator so the GM effectively GM1 into GM2 plus GO2 upon GM1 GM2 plus GO1 plus GO2 just remove that one part and just put the denominator to this now typically lambdas of the transistor is how much close to 0.001 002 or close to 0 as equally saying so R0 are normally infinite or even if it is higher mega ohms tens of mega ohms okay or at least a mega ohm is that correct unless your current is very very high R0 cannot be reduced to kilo ohms if it is very high current then the power will actually take care of your device failures you do not worry on that okay so if I say R0 are very large what does conductance is smaller or larger very small normally GMs are the order of 10 to millions per volt GMs are typically of the order of millions this Geo's are the order of microns Geo's are the order of microns is that clear so I say GM2 is much larger than either Geo1 or Geo2 so I neglect Geo1 and Geo2 from the denominator same way I neglect Geo2 from the top is that correct what is the condition I said since R0 are very high GM Geo's are very small compared to trans conductances which is typically a one order or three two three order higher therefore I neglect Geo2 Geo1 Geo2 equivalently saying it is GM1 GM2 upon GM2 which is equal to how much GM1 is that correct so even by using a cost code how much trans conductance I obtained same as equivalent of the normal single stage amplifier transistor M1 if I had used only M1 I would have got GM1 with 2 in this also I got equivalent of GM so what is the if GM1 are equal you can say GM is GM1 equal to GM2 also if you wish and it still it is valid for it if the transistor identical you can always use otherwise you can always get accurate value please remember this is the net value I have but in real life number wise GM will be very close to or sorry GM effective will be very close to GM1 so if you see this expression again okay okay I will come back first let us keep this and we will use the term bandwidth part later so what value we have calculated GM effective what is the next value I must calculate Geo effective what is the condition in Geo effective I said vain should be 0 so if I say so I say V in is 0 that is the condition you wanted V in should be 0 then I say Geo effective is I by V 0 when V in is 0 is that circuit okay input is grounded that is what you said V in is 0 so I grounded it I use the same equations of current again for this as well as for this for the second transistor I is equal to – GM2 V01 okay what is VGS for this 0 – V01 please remember VGS is how much 0 – V01 that is this – V01 is appearing here so the first is – GM2 V01 plus Geo 2 times VDS how much is VDS V0 – V0 – V01 so Geo 2 is okay I made a mistake so I rewrote V0 is V02 which is same so V0 – V01 this is another equation now if I write for M1 I had written it for M2 now I write for M1 current I in this is GM1 times 0 why V in is grounded so GM1 times 0 plus Geo 1 times VDS which is V01 okay yes bias we are not bias is the capital I could saturate current I input signal ground curve this is all man small signal every parameter I am using right now is small signal so from this equation V01 is I by Geo 1 from this equation 5 V01 is I by Geo 1 substitute this V01 in this expression is that clear from 5 this term goes I by V01 I by Geo 1 is V01 use this V01 value here and get new equations now is that okay all that I did from the second fifth equation I calculate V01 and so we substitute back in equate yes the biasing DC bias is coming from I source and nothing to do with AC signals is that clear to you M1 M2 are biased I am not saying this is VGS is biasing greater than VT small VGS has nothing to do with bias is that correct this is signal what is input to this signal is for M2 0 minus V01 is the VGS for that is the signal it is seen is that clear it is a AC signal is that clear M1 AC signal is grounded but bias is not removed bias is coming from I I have already told you if I is equal to beta times beta by 2 VGS minus VT square if I keep fixed I am keeping VGS minus VT fixed which is biasing it for saturation is that clear I repeat you have not got my issue any time I say I is fixed please look at it the way I said now in once for all capital I is if the device is in saturation assuming lambda or 0 right now so if I is fixed VGS minus VT is fixed we make I sufficiently such that VGS minus VT is always less than VDS choice of I is such that VGS minus VT is smaller than VDS so when I fixed active load which is called current source I am actually biasing the device in saturation the choice of I is mine is that correct choice of I is mine which will fix my bias for any transistor is that correct same current is passing in M1 same current is passing in M2 so both will be set left in saturation is that correct VGS minus VT for both should be smaller than VDS for each transistor is that clear that I am adjusting externally which is called bias current okay so all that analysis is clear all that we are talking is AC analysis and not DC analysis DC we are assuming that I am keeping both both of the first slide I showed you I is so adjusted that M1 M2 remains in saturation that I externally I did it is that correct I am not saying that if the voltage goes to fixed value it does not pass a current who said you VDD minus V6 still I can be adjusted all that I say as if you AC is made I am only looking for view AC now your question is if I may ground itself there then the whole of all current will actually go to the ground so I must put some potential there so that current can keep flowing is that correct otherwise all of current will be grounded I want AC grounded I do not want DC ground there is that correct and that is why I say put some voltage there so that the whole current does not short circuit out it should it will not go through M1 M2 then is that clear that is why that condition was made is that clear to you why that voltage was put otherwise if I physically ground that then all of I will actually go to the ground physical ground so no M1 M2 will be on is that correct so these conditions had to be made so assumptions in through all we are saying future way I do M1 M2 will remain in saturation this is DC biasing nothing to do with AC signals okay AC I am calculating is that okay but that is DC that decides GNs to beta I under root of that decides GNs so whenever I fix I am fixing GN is that clear to you I am as far as please get confused remove from this AC signals theory is based on values of DC you fixed is that correct that is what we keep saying RO is equal to 1 upon lambda times I whatever I fix GM is equal to under root beta times I so once fix I I fix both GM and R0 yes but that is a DC current yeah yeah but that is what AC what is the AC slope at any point is the slope we calculate what is theory of small signal we discuss all this way that at that point the variation is linear enough that we do not have to say major value changes DC value does not change yeah we does IDC plus small IDC will come but that current is how much say it is I am putting 5 milliamps or 1 milliamp and this is micro amps thousand times less or even lower is that clear that on any small signal analysis this value is always going to change you cannot say it is that but how do I calculate slopes if anything for I take this value and I get a slope here and I say yeah within this small change I put yes two values are actually being changing there is nothing I can do about but this value being what I the change here and here is negligible that is what ACs are all about correct it is always possible power supply will adjust to that small AC variation will over it all the time on DC okay always that is what we are and when we want to get output we remove that DC part by putting a capacitor so that the DC does not go to the output is that clear that is what we have been doing all through so far is that okay okay okay sorry he is right and I think I must thank him for getting clarifications otherwise I will be doing and doing and then everyone will confuse myself and yourself okay so if I substitute VO1 into I second whatever number of that equation I have three to equation 2 then I rewrite is gm2 I by GO 1 GO 2 V 0 is that clear ma'am is that just substitute VO1 in upper equation and you get this equation I forgot to I forgot this term so I rewrote again so I is equal collect I terms collect I terms on the left the other terms is GO 2 V 0 so I some of this is 1 plus GO 2 plus GO 1 up plus gm2 by GO 1 is into I is GO 2 V 0 and then what is the definition of GO effective I said I by V 0 when V in is this equations were derived based on V in is 0 so GA effective is I by V 0 which is GO 2 upon 1 upon GO 2 by GO 1 plus gm2 by GO 1 again if I linearize it GO 1 so I get GO 1 GO 2 upon GO 1 plus GO 2 plus gm2 okay then if you wish I can write RO effective is GO 1 plus GO 2 plus gm2 upon GO 1 GO 2 or can be written as 1 upon GO 2 upon upon GO 1 plus gm2 upon GO 1 is that okay what did I get I get RO effective is that correct I got RO GO effective 1 upon RO effective so I just got RO effective which is 1 upon GO 2 what is 1 upon GO 2 RO 2 1 upon GO 1 is RO 1 now if I do this you can see from here RO effective is RO 1 plus RO 2 this 1 upon GO 1 I made it RO 1 into gm2 by GO 2 or gm2 RO 2 also you can write okay but what is gm2 by GO 2 or gm2 RO 2 gain of sec M2 gain of M2 AV 2 so I write RO effective is RO 1 plus RO 2 AVT times RO 1 is that correct if AV is large enough how I make AV launch by increasing the W by L of those transistors or by increasing the currents okay if I increase sufficiently AV which I can then RO effective is how much 1 plus AV times RO 1 plus RO 2 and if the gain is say 50 or 100 how much is RO effective 100 times RO 1 if I do not put M1 M2 how much would have been output resistance RO 1 if I put M2 how much I got gain times RO 2 I got is that correct plus RO of course this plus term is there but this is much higher so what I have boosted the output resistance got boosted gain times in this cascode then we say AV effective is gm effective by GO effective or RO effective essentially so I get gm1 into RO 2 plus 1 plus AV 1 if I expand this well please note down this AV effective is gm gm effective upon gm effective is minus gm1 RO this I put here RO 2 plus 1 plus AV 2 or 1 then I say this can be written as gm1 RO 1 is AV 1 gm this is our route assuming gm1 is equal to gm2 everything is same then AV effective will be equal to AV square if gms are equal that is sizes are equal same current is flowing gms will be equal so AV effective will be equal to AV 1 square so what did I both gain is that clear gain I boosted how much square AV square I could have boosted my gain of any any amplifier let us say I have two such amplifiers A1 and A2 then what would have been if this is my way in this would have been also output of this VO1 would have become input for this and next gain times it would have come like this okay and if they are equal I still could have got the net gain is a square this is called cascade output of first is given to the input of next stage so gain boosting is not very difficult I could have put two in series like this cascade them and I will have got the boost of the gains okay so what did I achieve out of that that is what the cost code is different what is that we achieve please remember gain boosting is not the only thing we achieved what did we held constant in that gm we even with cascade a cost code we maintain same gm1 is gm effective okay because of that thing one can see the gain bandwidth product gm effective by seal gm by seal because gm is same gm1 is same is that correct so since if all gms are equal one can say gain of a cost code is game single stage square but the bandwidth remains same is this would have been possible in cascade no because full system would have gained bandwidth constant so if I increase gain what could have lost in that the bandwidth so cascade stages actually improves gain but lose on okay so this has have you got the point advantage of cost code over cascade that the cascade amplifiers boost gains at the cost of bandwidths cost codes boost the gain without losing the bandwidths is that but what is the penalty I am paying off top all of it I may require larger power supply voltages is that correct at the cost of VDD I achieved this is that clear so is that point clear so what is the advantage of cost code word cost code is a very interesting device that it allows you to boost output resistance without changing the trans conductance okay and that is something which no one else can achieve this is normally not done so much in discrete amplifiers this is always done in integrate circuits because that is much easier to adjust that GM1 GM2 identical or everything is possible on single chip two different devices if you take on a breadboard they will never be identical try any number of take thousand of such devices major IVs and CVs of them nothing will be equal okay so and a discrete such experiments are very difficult to prove sometimes but in case of ICs since the areas are very small everything is almost identical therefore integrate circuit use almost all such tricks more so in analog ICs or mixed signal ICs since future is only on ICs not on discrete so why do we discrete because discrete tell you which is the way you should go for ICs okay unless you know discrete you can do IC today I showed you same methods can be extended for IC designs is that clear okay so next chapter for our course is very very important frequency response of amplifiers now our assumption so far that there is a frequency of signal which device does not bother and it amplifies output at the same frequency please take my word what I am saying let us take an amplifier and I have an input signal at 20 kilohertz okay and a gain of say 100 so I know 1 millivolt of 20 kilohertz will become 100 millivolts at 20 kilohertz at the output let us say I change that 20 kilohertz to 50 kilohertz I still believe that it may still come so many millivolts at 50 kilohertz if I increase to megahertz then I am not very sure whether the same 100 gain will remain for that amplifier even at 101 megahertz or about which means the gain is also not only a function of V0 of being normal magnitude wise but also is a function of frequency so I must for an amplifier what is the condition I must have I must know the region of frequency band as we call say F1 to F2 in which gain is constant so that I know any frequency in this range will get amplified normally but beyond that the gain may not be same or before that F1 also cannot be what you want is that this is what we want to know this is essentially called the bandwidth or mid band frequencies of an amplifier I want to know where is the mid band what is mid band where the gain remains constant so first thing to notice I must now first do a mathematics to some extent and see how do I evaluate the frequency response of any system okay amplifier is a system which has some input some output any system has input and an output why I am saying this generalized because this theory which we are talking has nothing to do with electrical networks it can be applied to any energy mechanical or any kind okay this is what control system is all about is that correct we are only looking electrical but this is true for any other signals as well okay so let us start with our electrical for example I hope that in your maths course if not so far must have done Laplace transforms if you are not done it do it so we do J omega is what we call S which is called S domain analysis and if you have a system which is shown here any typical system which can be electrical in our case then it has been which is a function of frequency that means frequency can S is J omega so omega changes being S changes please remember anything which is written is only right now imagine any quantity but there will be also real quantity S is equal to sigma plus J omega is that clear there will be a real quantity plus imagine any right now I am assuming everything is imagined but in future in real life we will use that real quantities also there will be some output and please remember there can be many possibilities V voltage input voltage output voltage input current output current input current output current input voltage output 4 possible outputs for 4 possible inputs this is called ratio of output to input function is called transfer function ratio of output to input is called transfer function so how many kinds of transfer function in normal this system could be 4 kinds voltage will be voltage gains voltage in and current out is trans conductance gains current is input and output is current current gains current is input and output is trans resistance gains is that correct so 4 possible gains can be possible and there can be 4 therefore different we are right now using voltage transfer function why because in most cases amplifiers we are using are voltage amplifiers but later we will see it can be also any one of them is that the typical transfer function of a network shown here is HS or which is a J omega is V0S upon V in S and can be written as arbitrarily in this form K is constant independent of frequency in numerator you have a series of terms which is S-Z1 into S-Z2 into S-ZM divided by S-P1 into S-P2 into –-S-Pn this Z is called 0 of the function Zs are called 0s of the function and P's are called poles of the function is that correct this is nothing to do with network per se as I repeat this is true for any transfer functions why are we looking into this because let us say okay maybe there is another slide here or just from here if S is equal to P1 how much is HS if S is equal to P1 how much is HS infinite HS goes to infinite if S is equal to Z1 or Z2 or Z3 how much is HS 0 so poll is essentially saying the transfer function value goes to infinite and 0 means it goes transfer and value goes to 0 this is magnitude wise what we are trying to say is that clear so poles and 0s are those values of frequencies at which the transfer function may become 0 or may become or 10 to infinite is that clear that is exactly what mathematically we are saying how does it circuit wise we see is very important for us because there is where we are worried about okay is that general idea of transfer function clear any and please remember for there in our system we may use either of the four in actual amplifiers I may use a current amplifier then what is the output will be I 0 and input will be I in so there will be AI AV RI R and G both possible in this case is that clear so we only right now look for voltage transfer but do not worry it is true for all kinds of transfer here is a simple circuit okay there is a input V in AC there is a series source series RS of course there is no mass transistor RS is the source resistance there is a capacitance of C which has an impedance of 1 upon CS and there is a load which is RL and let us say by Kirchhoff law if the mesh current is IS IS times RL is V0 is a function of S is that correct V0 IS is the current in this mesh or loop so I times RL is V0 but how much is IS there is only one voltage source been divided by the net impedance this plus this plus this RS upon RS plus 1 upon CS plus RL is the net impedance in the loop series impedance so V in upon this is IS we write in V in into CS upon RS plus RL CS plus 1 nuts nothing great just take CS like this bring CS above please remember this is not sees of S is small as this S is the Laplace transform S J omega term you can write J omega C as well if you wish so V0 S upon V in S is RLCS upon RL plus RS times CS plus 1 this is the transfer function of this simple what is where is the capacitor in this circuit it is in series to the source and the output what could have been otherwise across RL also I could have a capacitor it will be called parallel capacitance this is called series capacitance for namesake okay is that okay obviously if the frequency 0 S is 0 you can see what is the transmission RCS is the numerator if S is 0 HS is 0 so there is a 0 at frequency 0 okay so a 0 exist at omega equal to 0 or F equal to 0 however if you look at the transmission again okay this to become 0 RL plus RS CS plus 1 then we can say RL plus RS CS equal to minus 1 at that time HS will become infinite and this minus sign is avoided in Calcutta is always on left half plane but just for the sake of it is 1 upon RL plus RS times C and this is it is called is the pole for this transfer function please take it what is RC called R into C time constant is that correct so I define RL plus RS times C as the time constant tau S okay as the time constant this then is that okay 0 exist at 0 and pole exist at 1 upon RC of this value of an upon tau S however we can rewrite the transfer function again H is V0 please look at your function V0 S upon VNS I use that RL plus RS I bring it outside so I get RL upon RL plus RS and I multiply it by RL plus RS these are same terms okay I just multiply and divide so I get RL upon RL plus RS is RL plus RS CS upon 1 plus and use RL plus RS CS tau and this RL upon RL plus RS K then HS is K times tau SS plus 1 plus tau S is that correct all that I did because on the numerator I do not have tau gating so I multiplied RL plus RS and divide RL by RS RL plus RS so this also become tau S this is a constant why it is called constant it is independent of frequency so it is called K tau SS upon 1 plus tau SS so we now say if it is a voltage gain transfer function is a voltage gain then AVS is RL upon RL plus RS if I write j omega j omega tau is upon 1 plus j omega tau is that correct now one can see now you have three terms here what are the three terms you see one is this term the second is this term and third is three terms are you sure you are three terms independently can be seen one two and three what I am going to do I will actually see response of each function with frequency initial and then what I say if they are all together that is this one how will they look is that correct I want to find variation of gain with frequency that is my aim frequency response you said I want frequency response so I want AV as a function of omega I want to find so I made a trick to understand the future I may not do it but initially I say okay I see now three functions one of this the other is this is the numerator this is in denominator so I use my tricks and I say okay first I get the magnitude I want how many quantities I should really find for any complex function magnitude and phase so first there is a phase I want to know the magnitude of the gain as a function of frequency it is RL upon RL plus RS omega tau S upon 1 plus omega square tau this is just a plus JB than a magnitude is under root a square V square is that clear so I just substituted the magnitude values where omega of course is 2 pi f okay so if I take AVF I can write RL upon RS RS 2 pi f tau S 1 plus 4 pi square f square tau a square to the power of same function rewritten in the form of frequency is that clear the function which I wrote here is same as this only thing now written in 2 pi f forms okay I have omega is 2 pi f now I start taking this is gain and gain can be expressed in some numbers which is called decibels is what is the why we use decibel as a word because essentially the decibel was found for measurement of noise or measurement of audio signals initially but they define when T log P2 by P1 or other if I put one media is called DVM is some function power in but since in our case power can be IV IV and I being same we can write sorry 10 I am very sorry this will be V square by R this will be V square by R so it if I say gain voltage kind of this this this is 10 square means 20 D log V2 by so AV in DB is 20 log V2 by V1 which is gain okay is that correct so if I want to find any DB values all that I do is 20 log 10 of magnitude of this value which is AV J omega is that clear 20 log V2 by V1 is the voltage in voltage gain in DBs okay so and but V2 by V1 is the gain actually what we have found so I actually write this function as star you know in log what is the log A into B by C will be log A plus log B minus log C a log is like yeah okay 18 terms go a lot a lot there so the first term is 20 log RL upon RL plus RF the second term will be 20 log 2 pi tau s and the third term is minus 20 log 1 plus 2 pi f tau s square or 4 pi square f square tau s square okay so let us start plotting individual terms please remember if I want this total what is the theory I am applying when I say like this like this what is the theorem called this plus this plus this with a minus I what is it called super positions so I actually do individuals and superimpose them to get the net AVF in DBs so the first term which is 20 log RL by RS I plot as a variation of frequency whether it will be less than one or more than one because this is RL and the denominator is RL plus RS so it will be less than one therefore minus okay so I say from 0 DB it will be some minus value which is 20 log RL plus some value depending on the ratio of RL and RL plus RS please remember why minus because denominator is higher than the numerator but this is always constant because no there is no term in this which is frequency dependent so this remains constant is that correct why I do all this simple things first because I want to make it very clear that things are trivial in real life okay unnecessary do not get you know excess or a analog was both by here what is the value I repeat 20 log RL upon RL plus RS has a fixed value throughout any frequency it remains constant is that okay that okay now use the second term the second term is 20 log 2 pi TOS okay 20 log 2 pi TOS so I say at F is equal to 1 upon 2 pi TOS please take a frequency F is equal to 2 pi TOS what is the value of bracketed term 2 pi F TOS is how much one how much is value of log 1 0 log 1 is 0 so it is a 0 DB is that clear F is equal to 1 upon 2 pi TOS if that is equal to 1 so at F is equal to this frequency the gain is 0 DB this second term only I am saying not the net one in the second term if I multiply it by 10 frequency okay that is if this becomes 10 upon 2 pi TOS okay then what will be the value 2 pi TOS is 10 how much is locked in locked in age 1 so how many DB is 20 DB so at 10 upon 2 pi TOS how much is the value of gain this function 20 DB is it okay 20 lock 10 is 20 DB by I take another other side 1 upon 2 pi TOS that is one tenth of that value so what is the again one tenth means how much minus lock 10 means how many DB is minus 20 DB is that man clear a 0.1 means 1 by 10 1 by 10 means log may minus lock 10 so lock 10 is 1 so minus 20 DB will occur at one tenth of this frequency and plus 20 DB will occur when it is plus 10 times the frequency 10 times this frequency you are 20 DB 10 times less is minus 20 DB so if you go from 1 10 to 10 how many slow how many how much gain DB is we are increased 40 DB and how many orders of frequency how many orders 2 orders 10 and 10 okay so per 10 which is called decade how much DB we are increasing 20 DB so the slope of this transfer function gain for the second term is 20 DB per decade slopes are called 20 DB per decade is that kind of I decade because every 10 times this 20 DB will come into picture okay 100 if you do it how much it will be 40 DB if it is 1000 it will be 60 DB is that clear so every times you increase 10 times another 20 DB will appear is that correct so 20 DB per decade is the slope of this is that correct by same logic if I do the third function which is 20 minus 20 log under root of 1 plus 25 tau a square up to when that f is equal to 2 pi tau s is equal to 1 how much is the value of that value should be at this frequency minus 20 log of no no not to root 2 1 plus 1 under root is that correct so under root which in numbers it is 3 DB minus 3 DB is that correct this is minus 3 DB so what is it showing that at f is equal to 1 upon 2 pi tau s the gain has fallen from 0 to 3 DB down at this this frequency please remember initially how much of the gain when f is smaller than 1 what is this part I am showing you when f upon 1 upon 2 pi s a boy chota so 1 plus 0 as a for I so I am getting a 0 DB gain as I start increasing the frequency that 1 upon 2 pi s term coming into picture and then we may start saying the gain will start reducing at at that frequency the gain would have fallen by 3 DB minus 3 DB essentially 0 to minus now this further down if you say what will happen one may be negligible or this and again 20 log decade it will start start if you increase f 10 times again it will come the same way 20 DB so now you can say 3 functions I have plotted which ones first one is this second this one is this and this one is this so I have 3 functions which I created and I now plotted 3 different frequency response term for 3 terms but what is important for us I want addition to this okay so if I add all of them together and this value is 0 DB please take it for us I will serve a function they can I try please remember this is less than 0 some value okay from this frequency onward the gain is going towards this is 0 please remember if I am adding now all three of them this plus this and this is 0 okay so if I add all three there then what do I get initially of course this may be RL upon minus 20 this and it starts rising up to what value it was rising towards 0 DB 1 by 2 pi tau s but actually that value is not this is 0 from the third one is how much 3 DB down at this point it is not 0 but as minus 3 DB so it is somewhere minus 3 DB is that point clear up to this frequency at this frequency how much is actual gain minus 3 DB 1 is 0 the other is minus 3 DB so this is minus 3 DB okay then I if you see further the gain starts falling and the other one is gain starts please remember this is 20 DB per decade this is 20 DB per decade down so what is the sum total of these 3 to this is 20 DB this is plus this is minus how much is constant 0 so afterwards the function becomes constant plus 20 you can see in mathematics geometry wise plus 20 minus 20 is 0 0 okay per decade 20 DB it will rise 20 DB it will decrease so I essentially know it is offset yeah you are right I am I should have put instead of 0 but assuming that term is 0 small okay you are right what is saying is correct I should this term is not 0 what is saying is this term right now I neglected that okay you are right I fully appreciate so what is it trying to show you this means please remember this is the crux of it now this is the name which I am now going to give you this is the frequency response of the transfer function I used this is the 3 DB point and what is this point is occurring at where f is equal to 1 upon 2 pi tau s beyond this frequency gain is becoming constant gain is becoming constant if I draw two such lines one line straight way and the other from this 3 DB point straight way like this this point essentially represent 1 upon 2 pi tau s this point okay this is called corner frequency this is called corner frequency and actual curve is isentrically following these two curves is that correct isentrically this is your real car this is the car which I am now modeling to is that clear this is the real value is that correct this is what I am now saying equivalently two lines like this here it may be so this frequency is also called 3 DB frequency or minus 3 DB frequency but this is called corner frequency this thinking that such 3 DB points can be put sharply are like this was first suggested in 1940 by the famous mathematician or other control system man all control systems are mass people vice ever all mass people can do controls what is the name of that person Bode okay Bode or board some people say I do not know I call it Bode and here so I will call Bode Mr. Bode first time suggested that to look at the response I must know the corner frequency because I am interested in below which frequency gain is higher or lower or above which I am actual value I may calculate by actual function okay 3 DB down but I am not interested in the value of that I am interested which is that frequency which is called the corner frequency therefore that isentrically substituting that essentially was first suggested by Bode and therefore all frequency response plots were called Bode plots of course this is only amplitude but we will also see next time frequency phase part of the frequency is that clear to you is that clear so why it was called Bode are this first part is not Bode is people snars even now call this board is actually no Bode only suggested they should have a corner frequency concept which was first time suggested by because in real life I want to know what is the frequency up to which gain I should not use or beyond which I should use or vice versa I only want that value the actual magnitude may be 3 DB now who care I want that frequency is that and I will be always away from those corner frequency okay to be guarantee only on this side or on this side is that clear that is why these plots were called Bode plots so in future we may not draw the actual curves we may draw always Bode plots okay in but when you ask to do real you must do asymptotes on that both side for example something like this occurs you know it will be something like this is that clear it will be asymptotic here and showing the here okay that is actual values but we will only draw like this is that correct this is how Bode thought and that is what the Bode's diagrams are all about see you next time.