 So, we were looking last time about the cascode part and one of the thing which I said last time about cascode amplifiers has the advantage and just recapitulating what we did last time and we said it can you can increase the gain without loss of gain bandwidth product points and that is the strength of cascode over cascaded system. And essentially this is what makes cascode very interesting. However you may find in many applications cascodes are not used in fact okay. I am not saying it will never be used. There are different variety of cascodes opams available telescopic or folded or otherwise each has some value but they are not the ones which are commonly used okay for every purpose. Specific purpose cascode is the best amplifier. In most cases the amplifier which are used is we shall see later. Simple one stage to stage amplifier which is good enough for many applications but in an integrated circuit you will be suddenly actually told to match with something else in a digital domain or its inputs is very different and then you need specific amplifiers to drive that okay. So why we are teaching this because these are the real life systems in which any odd applications may require any odd kinds of amplifiers okay. However we will first learn basics and when the application comes okay now we know this you want so this is better for you okay that is what the designer should know. If this is an application this is the input it has to then what amplification I should have so that it can drive better. So this fact last time I said that cascode amplifiers are superior to cascades because in cascade the gain bandwidth of a single stage remains constant for any cascading stage and therefore as you increase the gain by cascading the bandwidth keeps going down. If you have three stages the bandwidth will further go down whereas in cascode the gain bandwidth product does not change even if you boost the gain and that is something fantastic about cascades. Penalty we shall see soon what is the big penalty we pay at least an integrated circuit that is a big penalty we are paying for all of it okay. So this is what last time I was talking so I will start again basically I start now my amplifier a typical cascode amplifier shown here is driven by biasing current of IDS fixed current current source as you can say and we want to find the gain of this amplifier and we also want to find this gain bandwidth product okay. Now the way I write that we have been already designing amplifiers what is the typical amplifier last time I showed very simple amplifier just few seconds a simple amplifier we have discussed already can have current source load or whatever so biasing you wish and this is your V in this is your V out okay and we are interested in this amplifier is V0 by V in and we know typically this will come minus gm times sorry r0 we have derived this earlier we just know this so what now I am going to do is I have a cascode which has more than one transistor sitting there in series there and therefore I will say can I get gm effective and ro effective so that the gain remains gm ro okay so instead of just gm the for the cascode I am saying gm effective into ro effective and if I derive these two values for this cascode case I still have the gain which is minus gm effective into ro effective now this is what follows and therefore I thought before I start you should know that I am basically trying to replicate what I already know and convert that into my earlier knowledge is that okay all of you so the after all this when I will show you the actual solving circuits is the best way of doing things using Kirchhoff law so some other amplifier I show okay you do not know anything fair enough you can always put equivalent circuits okay so I will show you some examples of solving with any kind of 3 series 4 series or n series or n parallel anything can be connected and then we can have equivalent circuit which may become quite complicated sometimes but can always be solved using Kirchhoff laws so it is not necessary that this technique which I show you is the ultimate or something but this why we choose chose this kind of is actually trying to compare with the earlier amplifier so if I do you equivalently that then you say okay in comparison what is going on okay so is that clear as I say I am this method which I am showing you is only to show you comparisons this need not be always used by anyone but can also be used if you feel nothing very great about so okay then we say if it is equivalent transistor situation is i is equal to gm effective times v in plus go effective times v0 as we did for normal amplifier instead of gm and a geo I will now say maybe if you I should do something like this so we can write that ac current through the for this equivalent amplifier stage will be i is equal to i is the current to the amplifier so gm effective times v in plus geo effective times v0 we already done for common source normal so this is the extension of that so what is the game I am trying to say I will figure it out what is gm effective for the cost code and what is RO effective or geo effective for the cost code and continue to use this same method of solving further is that clear what is the method I am suggesting I am only trying to why I did I repeat I want to see how much that gm effective is different from single stage amplifier gm and how much RO effective is different from RO of the single stage amplifier does that change because you said gain is boosting okay so that means their gm RO product must be increasing so it is possible that gm effective may be higher also RO effective may be higher or one of them may be higher other may not change or vice whatever it is there are possibilities we like to see which one is really pushing the game okay and why then in spite of increasing gain GBW or gain bandwidth point does not move okay so that is this so okay to say it let me put it this is a capacitance see which is load capacitance this amplifier is driving a load capacitance this load capacitance is such that all other capacitance are can be neglected except the output capacitance which takes as a net load capacitance is that clear remember bandwidth can be a function of input capacitance says series capacitance says an output capacitor I say this is giving you what we call dominant pole as of for the case of solving so we start with the case one I say if v0 is 0 in this expression the I is equal to gm effective then gm effective is I by vn when v0 is 0 just substitute 0 so gm effective is I by vn okay when v0 is 0 look at this expression put v0 0 so I by vn is essentially gm effective when v0 is 0 now similarly one can say if vn is 0 and then we figure out I by v0 is nothing but go effective is that okay just either this 0 or this 0 so I want to solve these two cases when v0 is 0 and vn is 0 and this equivalently then I get gm effective is I by v0 and gm effective is I by vn so now I want to find I and vn ratio or I and v out ratio for the two conditions which I have written is that okay as simple as that nothing great to make v0 0 I say it is a fixed v so no change so v0 is 0 there let us say I flows through this boat transistor should have okay please remember for an AC circuit any DC point is at 0 so we reference which is a VDC for this is essentially grounded for the AC purpose so for this if I say this point is 0 and then I solve I by vn ratio for this circuit and I will give effective if I say v in is 0 and I measure I by v0 to here then I will say I will get due effective so this technique is simple and this is only try I say keep trying to equal the gm on RO values of earlier this so that I see comparisons is that okay so the method is trivial to some extent but it does give you some physics behind what is going to happen is that okay everyone okay to start with of course in reality this may not be necessary but in most integrated circuits unless specified otherwise transistors are always made equals okay for variety of reasons we shall come to see actually if they are not identical it creates hell of an issue okay so we will try to make transistors identical in many cases but if needed we can modify what that change will be normally thresholds are rarely changed only it is the W by ratio which may be different for two cases okay so now right now I for the simplicity I assume M1 M2 identical that is the thresholds are same the W by also same as everything else is also same see off the same everything same okay for the transistor M1 this is the lower one the current flowing in this transistor Vgs is nothing but V in Vgs for this M1 is V in so beta V in minus VT square into 1 plus lambda V01 is the current flowing in M1 is it clear to you beta times Vgs minus VT square into 1 plus lambda VDS so I just I am not actually using I am just trying to show you what current is flowing there however if I only look this transistor from the small signal point of view then I say here I is equal to gm1 times V in gm1 times V in is the current source going out current going in the output plus gm1 of this into V01 okay now for this transistor M2 slightly the current is similar now I is equal to beta 0 minus V01 minus VT square into 1 plus lambda minus VDS is 0 minus V01 0 minus V01 so it is this can be a current in M2 please remember these two current should be identical because in a circuit only one current can flow one circuit point however I write similar expression for I in for M2 so I is equal to gm2 and what is Vgs for this transistor 0 minus V01 is that correct what is the Vgs for M2 0 minus V01 so it is minus V01 I repeat what is Vgs for this M2 gate voltage is 0 source voltage is V01 so 0 minus V01 so gm2 is into minus V01 plus go to now how much is VDS 0 minus V01 0 minus V01 so 0 minus V01 which can be written as minus gm2 0 into V01 is that correct is that okay so I have an equation which gives me relationship between for M2 and M1 and what is the ultimate time I am looking for what is first I am trying to derive or gm effective so what value I am really looking for I by win okay so this equation is what is important okay so I want to find gm effective so I want to find I by win all small things is always valid it is we had just told you other day be small okay I think you should now go back and do some little more calculations yourself the way small signal analysis is done in any IV characteristics let us say this is the character any idea what we fixed is this DC point is that clear and superimpose on this some signal assuming that the slope here does not change small signal analysis currents are similar except that the DC value has been taken out of it essentially saying I total is I AC plus I DC V total is V AC plus V DC so anytime we substitute the DC value together and subtract so the difference is only AC you subtract total minus this you will get the AC so the expressions are valid the current I is so adjusted that both M1 and M2 are in saturation is that clear so I has that that is what I say bias point is fixed by IDC this value so if I fix that value I am making device saturated if it does not occur yes I will have a problem because there is no gain then possibly I may not get enough gain or 0 gain sometimes but I assure you as of to solve this problem that I is so adjusted that the device both devices can enter into saturation is that clear the only word which I have not stated which I soon I will say there is another word which I use going to use is called headroom okay what is that word headroom that how much maximum I can shift that device still remains in saturation or go below till remains in saturation is called headroom I have that play available okay so I assume that my device has sufficient headrooms so that device remains in saturation this is assumption which can be proved otherwise it is something saying that first you assume and then prove that is all mathematicians do so why not we do it small as I say you add total VDC plus view everywhere DC and subtract DC value out of it you will get similar equation is that clear to you as long as you are in a small signal this is all small signal circuits we are solving as long as you keep doing this it is fair enough though no man either you may say if the slope slightly goes away or headroom is so small then you actually will never remain in saturation so where do you bias is a very crucial point for us okay right now we will come back to this issue in case it is needed but let us move quickly so from the first that first second equation which I just derived VO1 can be derived as minus I divided by gm2 plus just now I wrote an expression I am now going to find VO1 and substitute this VO1 in that equation for M1 is that correct what did I do evaluate VO1 from M2 and substitute in value for M1 okay so if I do this I get I is equal to gm1 V in which was the term anyway there and I have VO1 VO1 was the product there so it is VO1 is now replaced by this minus I gm2 by g2 then I collect I terms so I is equal to 1 plus g1 upon g2 plus gm2 is equal to gm1 into V in then evaluate gm effect as I say better method service what put up equivalent circuit and solve nothing better than that but why I keep saying you why I am using this technique just to show you how much difference we will do if I go from single stage to Costco because that gives you some numbers immediately to see this may not be the method you should use for real circuit solvers okay even spice does not use this but why I chose this this is not even given in many books because if you do not believe that this is why I want to explain you how much Costco is really affecting you and this is the way of explaining that so I g into 1 plus g1 just collect this term on the other side and I is equal to 1 plus g1 by g2 gm2 is equal to gm1 V in and therefore gm effective is I by V in that is what you said for this condition we already said VO2 is 0 we already assumed and for that we are solved so gm effective is gm1 upon 1 plus g1 plus g2 gm2 if I connect the term properly I will get gm effective is gm1 times gm2 plus g2 upon g1 plus g2 plus gm2 is that okay now gm typically is trans conductance of an amplifier should be higher or lower for a good amplifier gains higher so typically and g0 which is 1 upon r0 that means g0 should be very small if I want larger gains so in general since the output resistance of a transistor is the order of mega ohms typically gm is the order of millimodes okay since this is million that is mega or 1 upon micro you can say gms are always much higher than GOS of course given a value please check it device is still saturation and this is valid if not use the full expression I am not saying but let us say gms are much higher than GOS which normally will be in two to three orders then you can leave GOS and figure it out this term if I leave G02 so gm2 plus this compared to gm2 plus G02 I leave G1 this by this is one so effective gm is nothing but equal to gm1 so after even cascoding I have not really modified my gm term it is modified I am not saying it is not if you substitute this value it will be slightly different from the just gm1 because this term is slightly larger than this geo is smaller but let us say it is 10 this is 10 or minus 3 minus 6 minus 6 so even if there is a slightly more value than the denominator but ratio wise it is extremely small point 99999 something kind of thing so we say practically gm is effective gm so one of the feature of a cascode was that it did not change gm then if gain has to boost we already said what is gm gain is gm effective times RO effective so obviously if your gm is not changing the other term must be getting boosted otherwise there is no way I can boost the gain okay so then we must now evaluate RO effective and figure it out how much it is more than the normal RO I have okay and if it is much higher that means my gains will be proportionately higher is that correct as the ratio of RO effective to RO is essentially times that much times the normal gain will be boosted okay without losing gm value is that correct now we will see why we are interested so strongly talking about gm gm so is that in cascode we have not gone out of our everything done okay we just say gm effective is same as gm is that okay everyone has written okay case 2 I want to find RO effective so what is the technique we said put Vn equal to 0 is that okay for AC everywhere this is anyway grounded and put a output voltage V0 which enters a current I this is a standard method of finding output resistance short all the independent voltage current open the current sources and short all voltage sources independent ones is that clear to you and then apply a voltage at the output and measure the current there for the circuit V by I there is the output resistance of same technique has been employed by us is that okay so okay for m1 transistor Vn is 0 so gm1 into 0 plus go 1 into V1 or to say V1 is I by G1 that okay for the transistor m1 Vn is 0 so Vn is 0 so V1 is I by G1 for the m2 transistor the upper one I think you are know you already have that but for the m2 transistor the current is gm2 times Vgs how much is Vgs 0 minus V1 so minus V1 and how much is Vds for this GO times Vds or this minus this so V0 minus is that okay I collect the terms again minus gm2 plus GO2 into V1 plus GO2 V0 just collect V1 terms and then substitute V1 from this into the second one last time what did I substitute I picked up from m2 and substitute in m1 now I picked up from m1 and substituted in m2 then I get I is equal to minus gm2 GO2 by GO1 into I plus GO2 V0 just substitute V1 from here here and you get these expressions what do I then collect collect I terms and then V0 by I is RO effective I collect I terms and then V0 by I is the output resistance as seen from the output terminal okay net so okay so I collect the terms of I and then figure out I by V0 is GO effective substitute everything and I get GO effective is GO1 into GO2 upon GO1 in plus GO2 plus gm2 and I invert it to make it RO effective so RO effective is GO1 plus GO2 plus gm2 upon GO1 GO2 divide this each term 1 upon GO2 plus 1 upon gm2 GO1 plus gm2 by GO1 into RO2 okay it will do some correction of terms and you get RO1 plus RO2 plus gm2 RO2 RO1 what is the RO effective RO1 plus RO2 plus gm2 RO2 RO1 that is something fantastic is happening okay but what is gm2 RO2 the gain of second stage amplifier individually is that correct so AV2 win can get if you are noted down I will just write down the expression okay so if I collect all this term and write gm2 RO2 as AVO2 then I get RO effective is precisely if you seek it I think there is some mischief here this will come RO1 into RO plus that RO2 term should not a AVO2 into RO1 it is up to so RO2 plus AVO2 RO2 RO1 and RO2 are equal why are they equal we said transistors are equal currents are equal 1 upon lambda id is constant so ROs are equal so essentially what is it trying to tell and gains are typically of the what order 100 to 1000 to 10,000 maybe million gains of amplifiers can be as high as 1000 10,000 a million as well okay 10 power 6 gains are possible not that every time use them but possible opams of 10 to power 6 or anywhere available is that correct gains are available so normally AVO2 will be very high compared to any other term there okay and neglect all ROs but AV into RO will be heavily increasing the RO value okay AV is 1000 and RO is mega ohm so from mega ohm it has gone to big ohm okay so now the output resistance of a cascode amplifier can be boosted by this AVO2 term okay and if RO effective is so high AV0 which is nothing but gm effective into RO effective is as much high though gm effective is same as gm1 is that correct gm effective is same as gm1 but RO effective is gain times the RO is now pushed there one of the stage gain AV2 into RO is boosting the gain RO and therefore the gain equally get now we were making a statement okay that if I would have done a cascade what would have been the gain instead of putting cascode let us say AV1 is the gain of first stage AV2 is the gain second stage and they are equal then it will be AV2 is that correct so here also if you see if you just look at this expression AV0 it can also be expressed in do not write all of it just write this bank AV0 is AVV1 plus AVO2 plus additional this since AVs are much larger square terms are larger than single terms so typically gain is gain square same as what cascade could have got it is that clear cascade could have got how much gain square of two of them here also I am getting same gain little more but almost same order of gain then what did I achieve I have you said gain is no gain product it seems to be similar for that case of cascade as well as cascode but what is changing or not changing is the gain bandwidth product is that point clear the effective gain probably in a two stage cascade is same as one single cascode it has not of course additional term is there so you may say yeah it is still boosting but in 10 to power 6 plus 10 to power 3 it is still 10 to power 6 so it is a gain is only AV square it is not really boosted as one thought but then what is that I keep saying boosted from the single stage I have certainly I have only one single stage there okay and I have boosted the gain from 1000 to 10th hour 6 okay which is what I did for the cascode because RO effective I could boost thousand times okay now the term which is most important for us as I say you write down this expression they are simply derivable and there is nothing great about the point I was trying to make is essentially cascode single stage is not very great compared to two stage cascade because we are comparing all the time with that why not connect to amplifiers in series which is always done many places so what is that cascode is actually doing so we say okay there what was the problem I said if I make AV1 and AV2 there or AV square the gain bandwidth will have new gain but the bandwidth will go down because GBW is constant okay so as I keep putting cascade stage my bandwidth will keep on going reducing by that number but that is what does cascode does the same and why do it cascode if it does and then we say I have beaten up please remember GBW to some extent is called figure of merit or technology constraint we always say GBW is constant kuchnikar something now with cascode can be beat this so called technology constraint and here is that explain everyone written down this is what it is saying gain bandwidth product of a cascode is GM effective by CL but what is GM effective GM 1 by C effective is that correct C effective means CL plus any other capacitance you put there is fair enough the gain of a cascode is AV square the bandwidth is GM effective by CL GM effective how much GM 1 for a single stage the gain bandwidth product is GM by C because GM 1 is this which is same as the earlier one so what is it trying to say gain bandwidth product of a cascode is same as single stage amplifier but from the single stage amplifier what has it changed gain how many times thousand times or whatever the gain of that amplifier initials this that many times I have boosted the gain but my gain bandwidth product has not changed is that correct so that is something I achieved that I have not missed the point of GBW but I still boosted please I last time showed you some figure maybe I will repeat for you you just write down this in the case of cascade what would have I lost in case I boosted the gain by that much amount I would have lost the bandwidth is that correct now I say no gain bandwidth is not changing even with the cascode stage but gain has been independently as if boosted by is that clear by increasing the this because GM by CL is no different from GM by CL for the two cases so the GBW point did not move that help that GM effective is GM 1 has helped me now to fix that point independent of which amplifier I use that so here is a figure will come back to it okay maybe I should drop better figure just a minute I will come back to this expression once again in case you have not written let us say I have an amplifier which is gain in DBs versus frequency so this is my gain and let us say it has a single pole as I have taken the case this is my GBW this is my AO1 or something if I would have boosted the gain okay so what and this is my bandwidth if I have boosted the gain for by cascading let us say I go here okay then this bandwidth point would have itself so what is the change would have occurred BW original and BW new so I would have lost the bandwidth I increase further I would do this gain this further bandwidth will go now if I do the same thing for cascode what I am doing is the following A versus omega this is my normal single stage amplifier okay now I am boosting the gain okay I am going from here to here okay I may still lose some bandwidth I am not trying to say because that ratio is not exactly 1 so some part I may but the slope is such that it still meets the same point so what has changed of course it should have been slightly this side okay so bandwidth reduction is still there it is not that it is not there but it is the way figure is not very good it is very close to the original bandwidth but what I shifted again I have boosted the gain to keep the same slope now my bandwidth likely is going down but marginally going down and I say as if I am not even worried about that small change so I say okay I am retaining bandwidth what I had but I have boosted the gain is that correct that is the strength of cascode okay we will solve an example give you the result I agree but it is not related only to that that is what I say because this geo effective has many terms so is GM effective so overall since you have to meet GBW if you extrapolate it for higher gains essentially what I am saying you keep extrapolating that value then you will find gain has been boosted but bandwidth is not reduced in the same ratio is that clear that is exactly the strength of cascode now catch word in this all game was which is why I said you all the figures and everything was this which you should realize what I did actually I just fooled myself okay and I said okay okay this last line you can read down that is most important the gain bandwidth of cascode is same as gain bandwidth of a single stage amplifier however AV0 is AV square clearly the technology constraint is now new is under root gain cascode into the old GBW is now constant is that correct because the new gain square again a cascode may gain square IR signal stage may gain if equivalently it is gain under root gains this of cascode multiplied by old GBW is the new constant which you have got this is called the new parameter figure of merit for a amplifier normally this would have been your figure of merit I have now cross that limit okay that is exactly what I was trying to achieve think of it how did I solve so gain square into bandwidth is now constant not gain bandwidth that is something is great we have achieved out of this is that clear to you because further increase will shift now you I mean it is not that infinitum it can go is that clear to you so it is no bandwidth is reducing it is not that but it is gain square time bandwidth is constant not gain into bandwidth now all this game which I did not show you or did not say specifically if you see your cascode amplifier once again very seriously I am just trying to fool myself I kept on telling this is a single stage cascode okay yeah it is called single stage cascode but in reality it is also cascading two things you have one amplifier which is common source okay output of a common source which is VO1 is fed to this amplifier which is common gate okay I am actually cascading two amplifiers even now only thing is instead of two common source amplifiers the second stage now is common gate and the advantage of when I now show you common gate this will be very obvious like in the case of common base amplifiers the current gain is unity same is in this case the common gate amplifiers do not give you any current because source current drain current same so it is almost unity current gains but what is the advantage it allows you to shift the let us say what is the first stage is doing as you look at it M1 is a common source amplifier so what does common source amplifiers characteristics converts voltage into current common gate amplifier improves the output impedance just now you have said so what I am still available this is the gives the same current so what has improved the current source equivalent voltage dependent current source has become far superior compared to single stage is that clear so I have not really achieved fantastic or something I was trying to boost every time but the idea was to show I have improved the current source from a normal stage amplifier which has now higher output which is a good current source how higher output resistance means good current source all that I have done is by this technique is to improve good current source which a normal amplifier we see normal voltage to current converters have relatively RO is not very bad even there 10 to power 6 or kind but this has actually how many how much I want for a good current source infinite so by boosting this I am trying to get better current source out of the amplifier is that equivalently so actually all statement the funder point is that if I connect the two which way can get me what really I am looking for so another design case 3 amplifier will discuss quickly from this other is book common source common gate source followers and I will see which ones have what property and when connected what it will do for you is that correct so as a designer I will do analysis because then I have to know value so I will have to do analysis substitute the values but I like to know a priory that which ones I should use for purpose sake of I am looking for and this is what I am trying to hint every now and then that this course is not just solving a problem but just to tell you that by doing this I can achieve something which a priori I was not aware I could solve a cascode I will show you RO increase but why I did all this because I figured out this actually can do this job so tomorrow there is some biomedical instrumentation requirement comes you want a constant current to be driven because the time taken for charging the input capacitor there has to be constant or ECG monitor at that time you may require a cascode amplifier because you need a constant current source as good as possible other it has a time constant will not switch quickly okay. This is an issue which in real life applications one can see which ones to use so as a designer I will not be told to go analyze karka dikhao but for doing analysis I will figure it out what is that characteristics I get out of it which when needed I can then imply so the difference is that clear analysis is always required to prove a point but you must remember the point is more important than the analysis at then we are doing all this to show okay if this need appears I will pick up this design now what people designers are expert about as I keep saying they are very good in copying all of you are but maybe all of us also so they try to use someone else's result but many a times the kind of value some spec is not same between the two and that may actually kill you the other way because that is what the analysis really if you do sprinkle on this this may also go like GM effective here remain GM you do something else maybe that also changes and the whole purpose would have actually lost is that clear so in future the difference between analog circuit course and analog design course is only this that we still solve the same circuit which we are done n times but we now learn from them what what is the advantage of using this so these issues you should keep in mind all the time because these are the only things which are required for a good designers given an application decide oh this spec so can I do this plus this this plus this so that must strike you and there is no solution for each case available in the market is that correct each application will require different thinking and therefore different design and therefore people say it is slightly difficult okay. It is always gain bandwidth product GM by C is a gain bandwidth product GM by C is a gain bandwidth product I will solve this right now take from me I am not solve for a frequency response so far so I have not come to it I will prove it what you are asking it is always gain bandwidth product and not bandwidth huh for a cascade yes but cascode that is what I proved you that the GMC does not change but gain still boosted correct gain increase which is in cascade did not happen because any single stage I follow another stage the gain if gain increase bandwidth automatically goes down GBW may still limit your point but the bandwidth reduction is too strong if I keep increasing the gain stages thousand times correct the thousand time bandwidth each other cascode does that since it actually allows you to slope to be modified compared to cascade I am able to reduce bandwidth I am still reducing bandwidth but marginal compared to the cascades we will solve a problem plot a bodice plot and show you where they match okay okay. Now before we go this cascode stage sorry sorry I am very sorry is this sorry just note down I will come back to solve the values of this and then I will show you all that I am trying to say I have changed the constraint now okay using cascode I have changed the constraint because gain square yeah then a gain square and under root gain gain into gain bandwidth is gain square bandwidth gain cascode is AV square under root of that is AV so gain into gain bandwidth is gain square into bandwidth is now constant this is the new constraint is that okay cascode is not a single stage amplifier that is what I have proved just now this is only true for cascode stages this has been derived for cascode otherwise gain bandwidth is a constant single stage we are done put second stage gain increases bandwidth goes down nothing you can do on that that is Sankrasek a go get the Dushram lehling no no no it does under root of that that is the what I am saying it will actually reduce by under root of that I show I will give a value I will actually plot the real values for that then you will see actually slope changing is how much okay okay I will check it but this is correct okay someone says is this gain increases enough can I do something more than that yeah I think cascode people came out with another circuit they say here is another cascode amplifier which is gain boosted cascode amplifier now already you have boosted the gain now I have an amplifier which further boost the gain what it should not change again being cascode stage the GM should not change but what should therefore it increases RO effective so you want even better current source now yeah another case in which I can have gain boosted cascode okay the method here is simple in the first cascode case you just first draw the circuit and then expression you start looking at my first and then you write I have a same as M1 M2 here I actually put a ground there the gate of 2 was referenced to a DC value or grounded for AC instead I have an amplifier sitting there which is receive an input VO1 and transferring – a times VO1 to the gate of M2 is that clear last all that I did that I have boosted earlier it was 0 – VO1 now a VO1 – VO1 okay so it is slightly a plus 1 times VO1 is what now I am giving to the gate of M2 this is called boosted gains or boosted ROs okay same way I can repeat the same expression for M1 and M2 okay and figure out by same logic instead of VDS what should I use for the first M1 VDS M as VO1 no change for this the VDS is how much now – VO1 – AGB times VO1 is that okay how much is this VDS this – this what is output of first amplifier here at the gate a so a – a times VO1 or AGB I call gain boost so AGB times VO1 – VO1 that is AGB plus 1 times VO1 is the VDS substitute this term does not change why does not change because that this remains same there is nothing change at the VDS is that okay only VDS has changed for M2 VO VDS has not changed for M2 so I am only changing the V please this voltage is the gate voltage which is – a VO1 is that correct this is still VO1 so VDS is – a VO1 – VO1 so – a plus 1 VO1 if I substitute this I get GM effective long enough expression I derived and if I say RO1 by RO2 is close to same so these are once these are larger these terms are larger this can be neglected if everything cancels so GM effective still GM1 so however if I look at geo effective now for the same expression instead of 0 – VO1 where you are using it should use 1 plus AG times VO1 terms so now you are going to get geo effective RO1 plus RO2 AV2 into 1 plus AGB times RO1 is not it so it was AV2 times RO1 so gain was resistance are only boosted by the gain stage of this normal now that itself is getting multiplied by the boost stage gain of this is it okay to you not clear I am only substituting where it was 1 1 plus AGB okay and rewriting the same so now how much is RO effective specifically saying AV1 times AGB times RO1 this into this into this this are small terms so what is the earlier one equivalent we said AV2 times RO1 now I multiplied by AGB now this AG amplifier could be what kind of amplifier it could be it could be even another cascode okay that cascode may be driven by another cascode and may be the minimum value 0 finally can be there at the input so 1 stage 2 stage n stages of cascode can bring a desired gain what boost you are getting is that point clear the amplifier I am using can itself be a cascode amplifier which may be driven by another cascode and another cascode and one and go till 3 stage 4 stage okay so that the net gain is what you are really looking for okay in that case you will be able to get whatever AGB you are looking for that can be attained without losing what bandwidths anywhere okay so you just boost it there no pulse will hit you there keep doing and then you will get heavily boosted gain in this case is that correct okay so what is the advantage of gain boosting possibility that is why I say I am not discuss about the stability yeah we will come if there are this is equivalent circuit assuming very clean stage but in real life yeah not only the instability that 0 will be issue okay not the pole as much okay the worry may actually come from a 0 okay the 0 may not be even on the imaginary exit may actually lie on the left which means it will oscillate what is that it makes this is a good current source I put there okay but this current source cannot be created out of nothing so it could have a P channel device could have an N channel device it could have a series of 2 devices even there whose output resistance will also be there is that clear to you the issue is now what I assume this is a ideal current source okay so it is infinite but in real life this may not be infinite where that resistance will hit you then at the output node whatever is RO here this RO will shunt it and if that RO is smaller all that this cost code people were doing is useless because of that resistance is smaller output is only seeing that resistance is that clear so that is why I say as an amplifier you may worry you should not use that because the actual RO may not be very high for normal amplifier it will be say already there then what do you do so already is the output resistance nothing RO effective RO effective is 1 billion ohms parallel to 1k okay so issue was that okay you have this you may have more than one transistor in series now my worry start something like this let us say for some reasons this is one example this may be many other this may be P channel may be grounded for the heck of it or may be I put VB here which is bias something now this be in here this is my output this is equivalent to some RO2 RO3 you may call now the one is of course as I say what is the major worry you have boosted RO here and if this RO3 is smaller so the output is RO3 so what should I do here then I should also cascode the upper ones may VG2 VGG3 and VGG4 if I do this yeah I have the RO of this stage may be as much as RO of the lower stage and may be half it may still be but at least it is in giga or higher values you can create out of this but this is the issue if you are boosting your RO and if you are three or four transistor in series that madam should be was asking me how do you guarantee this what is the guarantee of a saturation a transistor is in saturation if it is VGS minus VT is equal to VDS or smaller than only then devices in saturation if that is so VD1 let us call this VDSat 1 VDSat 2 VDSat 3 and VDSat 4 for this case for RO boosting you did for RO has got what you wanted goods current source everything you say achieved now you have a issue that there are 4 VDS in series is that correct the minimum you have VGS minus VT VDSat is VGS minus VT 1 value okay typically how much is the excess voltage I said 100 200 millivolts okay but now you have 800 millifold your supply is 1.2 volt or 1 volt at times keep it saturated your head room is practically so now you are worried that whether this amplifier will remain amplifier for a long any change in values of VINs or VGS then I have a problem this amplifier may not remain in such all transistor may not remain in saturation if they do not remain in saturation all the theory which I derived as she said how do you know yeah I also do not know but possibility there that will be lost so now I am worried that is this a good solution if I had to boost so much I must ensure that I am not using this essentially as an amplifier that correct but this is called telescopic okay so I can I will have shown an amplifier OPM which is telescopic OPM which has its own advantages at some cost so we like to see that all the time putting things in series may not be that advantages as it looks if you have 3 volts apply fantastic no problems 11.2 there is an issue and why I keep saying now that because now power supply for digital part is 1 volt and I have forced to work with them independently with a bipolar why I do all this I am doing all this because I am forced to work on CMOS and also on the technology node which is digital node so I figured out it was not so trivial as I thought I just thought everything is good everything is good is not really that good as I thought okay so do not use cascode as a very though as a concept it is fantastic no one can beat its concept okay but remember where to use that is a very crucial factor okay we will use in a cascode OPM show you so to avoid this to some extent I will quickly last few slides I will show you which is the word I used okay I will do this cascode amplifier with resistive load later there are two ways of doing this cascading one is called normal cascode the other is called folded cascode the right is folded left is normal what is the difference I did here both transistors were N channels the common gate amplifier transistor was also N channel and the driver was also N channel is that okay however in the folded cascode the driver or the input of an amplifier is N channel device okay but there is another device which is P type and it is look at it instead of going up it is folded down now you say what is so big about we will see what is it big about by some things is that point clear this is called folded cascode it has advantage of layouts and it has advantage of supply voltage requirements both yes of Chitra Razavi ke book me hey voice maker ke book me be okay so there is nothing very new I am teaching I am trying to bring those issues which at times they do not want to hammer on so I am hammering that nothing great really being told okay I will write give the sheets I will quickly read for you and explain please keep your figure right with you because I cannot put both together so I will use figure from my lower down sheet and I will show you what for the first normal cascode what is how do we define VoV Vg1- source is grounded so Vg1- Vt is VoV so Vg1 is VoV what is VoV Vgs- Vt is that clear so I just trying to be little funny on that what is Vd1 which is our Vo1 shown there what is Vd1 is the over voltage which is Vd saturation plus additional voltage which will allow transistor still remain so that is the limit which I want to go to so we say Vd1 is the over voltage which is Vd sat plus some head room voltage is that clear what is why this Vh has been added with this additional this also device should remain in saturation Vg2 look for the term for the Vg2 which is Vd1 plus VoV for the second transistor which is Vgs- Vt for that okay plus Vt because this is essentially Vds of the second transistor Vds of the second transistor so it is VoV plus Vtn then the Vg2 can be written as substituting here Vd1 from here into this I get 2 VoV plus Vh plus Vtn then I evaluate Vd2 which I get 2 VoV plus Vh plus V2 V0 the small V0 is a average value of a signal which may actually because of the capacitors there average value okay which can be treated 0 in many cases but just say swing aya because Vo is swinging on the output so what is the maximum value should be taken care in case it takes you out then so we want to keep it otherwise in many cases Vo can be treated as a 0 value but for the sake of evaluation the peak value I should know I will add that also in my calculations is that correct this small Vo is the peak value of the output which may not be very small at times in a small larger signal so I assume Vo is present which in normal case is very small and neglected but for the sake of clarity I say okay add that also so I get supply voltage should be 2 VoV yes Vd2 is the voltage at this node is that correct this essential is this plus this plus this or directly this please remember in a transistor whichever way go in a circuit if that is not happening then there is an issue this voltage whatever is V1 if this voltage is V2 and this voltage is V3 V1 must be sign wise V2 plus this is the game either you go through gate and from gate to the drain or source or you come from directly this if I want to evaluate this and I do not know something so I can go through gate and come to the drain is that point clear this is one way of doing it this is what I have used this method is very strong used in bipolar is that clear this is your Vbe this is your Vbc and this is your Vcb so we keep saying when the transistor is externally reverse bias but may become forward bias because this may become forward that is exactly saturation stage in bipolar this is exactly how they do so this technique please be clear at a given two nodes whichever path you come some total of voltage must be same because between two node this is the difference okay that cannot be changed to method is this or this must be same loops cannot be go through this or go through this voltage difference is respective okay so tricks of the trade is try games whichever you know you use that value to achieve the other one okay so I may be a small mistake here there but the point I am saying if I do similarly for folded cascode please note down that Vd2 there Vd2 sorry I am sorry she is right please remember this is Vd2 P channel device please remember this is P channel device drain is downwards this is Vs2 this is Vd1 okay is that correct the source voltage of P channel is same as drain voltage of N channels okay I1 I2 are the current sources which are biasing NNP device they may be same in most cases they will be same okay but yes V0 is essentially small one this is Vd1 so Vs2 is same as Vd1 Vd2 essentially is referenced to either this or this minus this plus that is what I am saying normal either find Vd2 is equal to this plus Vds of this but Vds can be derived from Vgs side also is that clear to you I repeat Vgs minus Vt is Vds saturation so I can go this way or I can go through gate and still arrive at the same equivalent values okay try it yourself up the point I am saying is this Pd2 is what is V1 large drop across current source V1 is the drop across current source so Vd2 small ac signal average value plus V1 is Vd2 is that okay I repeat this voltage plus whatever is ac on that it may be not be small peak of that so this value is drop across this plus small average extra value of Po is essentially the Vd2 what is Vg2 gate voltage I know this value I know Vov so I can reach Vg2 is that correct so Vg2 is nothing but Vd2 minus Vtp okay please remember Vs2 is same as Vd1 which is V1 plus Vov plus V0 then the V supply voltage is Vd1 plus V1 what is supply voltage this value plus the drop across source or this value plus Vs2 is same as Vd2 so I calculate V supply for folded cost code is 2 V1 plus Vov plus V0 I repeat V0 is a very small term but in a larger signal the peak value may not be that small therefore we kept it for the security why did we keep because you should not go that it is taking transition linear mode so we put that value so if I compare these two values what do you see really the V supply voltage of a folded cost code is slightly smaller than V supply voltage of full cost code of normal cost code so what does that did we achieve something better what was the problem I first they told you if I put everything in series I have that problem of use supply voltage requirement so partly I can save myself from that great trouble by putting folding okay now in this case folded cost code has used n channel as driver p channel as the other side folding you can do otherwise okay these are essentially therefore cost c mass folded cost codes is that correct the complementary word in digital they say gates are common in analog we never say they are common p and n channel if they are then we say they are we must please do not think at no time barry exceptions I should not say at no time most cases we only use independent p channel and n channel and never use they are the common so a common gate materials terminals in that sense it is not equivalent of a digital c mass it is always both p and n channel have been used and therefore it is c mass amplifier is that clear otherwise please do not compare it with the other side so the problem of power supply probably can be taken care by folding techniques okay and that is what we will use in the folded cost code opamp we will actually fold it to save some supply requirements see you then