 I just want to show what we did last time we actually put an equivalent circuit of that all that is difference now between the earlier common source was that there is a source resistance which is not bypassed by the capacitor even for AC okay yesterday's question was asked something where where is the biasing that Rg were actually was Rg1 parallel Rg2 to the VDD and is equivalent for AC Rg is appearing okay is that clear so someone asked where is the bias so you should know that it must have been coming from the bias network okay those who are there are many people ask me what is it the word origin stand from from where it comes from okay so some parameter is in the equivalent circuit coming so from where it come from so for example a capacitance why it is there or how from where it comes okay in the circuit device side how it comes okay that was the word origin so I do not know why there was a confusion but if so please remember for the English may be different was okay so this is what we did last time we calculated the gain function last time the current voltage gain is roughly equal to RD parallel RL by RS and we said what is the advantage of this the advantage is that that none of the two parameters of the none of the two numerator denominator terms have device parameter dependence no beta no GM or whatever it is and therefore it is now even temperatures therefore it is very stable amplifier at the cost of reduction of gain okay so this is why we still use these amplifiers then we also did something about the common drain or source follower essentially the difference between the last one and this is that this there is no drain resistance RD is 0 in this case RD is 0 that is why it is common drain it is going to the ground drain is going to the ground and the output is taken at the source itself output is taken at the source and this is that source resistance RS this is additional load resistance if it is there this is that RG which is the bias equivalent resistor RG1 parallel RG2 this is your signal this is R signal or R series whatever you call as the source resistance that source word is coming common source in the signal source resistance okay and then we say we want to find the gain of this amplifier initially we assumed R0 is very large infinite we opened it then this is your RD I made it 0 that means drain is grounded so this RD even if you told I have been there it does not matter but essentially this node itself is grounded I am picking voltage at the source there is a RS and there is the external load RL okay and the please remember I am taking simplicity R0 is much larger what does technology wise it means R0 is very high two possibilities exist when R0 will be very high lambda is very small close to 0 or our IDS is very small you are operating at very small drain source current but at that case what is the risk you may have if the currents are very low what does that mean the device may not be in active mode or this may be just the age of active mode it may be closer to off state okay so to lower current is not possible and to high current will take the device out of saturation into the linear mode and therefore again we do not want to go up there okay so this is the equivalent circuit what is the advantage of by a mass over bipolar one of the interesting thing it gives you that there is no connection between gate and source okay DC current there is no DC current between gate and source because of insulator in the case of bipolar what was that base collector base emitter junction exist okay and that means IB flows both through the emitter as well as through the base this fact is not there in mass so it is much easier to evaluate because if there is a current common to two no two sides then you have a problem to hold okay so this is what we were doing last time a source follower and my word follower I said output must follow the input can you tell me what two things it should do when I say output follows input for any any such voltage or current gain or transfer gains what are the two parameters are associated with that one is the magnitude one is magnitude the other is phase okay phase so when I say output follows input it should follow in both sense what does that mean amplitude wise it should be roughly same and phase wise also it should be same that means 0 phase difference between input and output this is a major different from common source amplifier which we did what is the different phase difference there 180 degree so the first thing it will be different that it will be and therefore it is much easier because gain is unity and phase is 0 so now you can connect some inputs out to the outputs out without any great difficulty all that we must do is input resistance and output resistance should be tolerable so that we can use them as a in between buffer either say be or either say yes oh yeah yeah yeah you are right you are very I am sorry I fully apologize because you know that RS the same old this idea you are very right sorry you are correct okay actually that is how we did because then we said V in okay the next equation will come that then this V in is equal to how much this volt plus drop RS is that you what your point of view is taken care if this is not grounded then there is an issue will come VGS will be equal to V in impact okay but that is not true what what you are saying is very correct this input voltage then is an assuming R series is 0 this voltage plus this voltage is actually the V in is that correct do you understand the this no RG of course is across it so it does not have any problem across that is VGS across VGS is across RG so voltage across RG is VGS plus drop across RS must be equal to input voltage if this resistance is 0 is that okay so this his point of view very correct other than that equations will go or yours okay very good thank you okay having then all that analysis one can immediately say one can write RSL is the parallel combination of RS and RL R0 is treated infinite so GM VGS RSL upon 1 plus GM RSL and VGS is V in IG is 0 AB is GM RSL upon 1 plus GM RSL if GMR actually you need not say you an RSL we should say GM RS at least should be larger than one then one can be neglected the gain will be close to but it will always be less than one is it clear it will always be less than one because one plus something even if that quantity is much larger than one still numerator will be smaller than the denominator by small quantity if not greater so AV will be always closer to one but mean not need never reach one and since the gain is positive here what does that mean phase is 0 the same phase so output and input has gain of one means output follows input magnitude wise as well as phase wise this is therefore called source follower is that here in between equations I am not doing because I we did it please remember circuit solving is identical use this voltage equal to this this current is only flowing through this please remember R0 is only flowing through RS and RL okay that current is equal to GM VGS okay just write the equations as we did for common source and same equations similar equations can be derived yes yeah in case oh if I read somewhere it is wrong because essentially okay I think that that term should not occur this is not you are right but anyway this may not be correct but this other terms are correct okay no no it should be then V in here then you must actually replace VGS by V in in terms of please remember then I must replace VGS in terms of V in okay and then substitute V in assumption is I mean tell you RS GM RS is smaller RS is smaller drops are smaller will that this gain will come close to one otherwise this one will not be one it will be 0.95 0.93 or 0.98 maximum the value which I have calculated to show close to one so I wanted to remove things so that it becomes but in real life what she said we must solve the network as it is and get V0 by V in whatever terms you get is that correct this was done to show my only aim was to show you that is close to one so I made machines so that I can get close to one okay okay what will be the for example here is something if you have an RG and R series the VGS will be something related to that okay as we did earlier and then we must find R in an RG again as normal devices okay please look at it this solution I do not want to do it I just want to show you how do we worry about the solving a network is that clear what is the game just draw equivalent circuit put voltages as given to you and see which loops are you are getting to solve solve the two three equations if three unknowns come to unknowns two equations you should be always able to solve all parameters in terms of the other is that correct that is the technique will follow throughout the course is that okay that is the course for example even in a common source someone ask you to find this not common so this source resistance follower there is a RS here there is a R0 here and please remember that RS is the source resistance R0 is the load resist I mean output resistance and why did I show you the opposite sign drain is down and source is up please look at it drain is down and source is up common drain please look at the circuit it is a common drain is grounded so it is the opposite this from here this is current is entering the source side source from here the RS is going to the ground so RS and these two are in parallel doing RL RL is external so these two are in parallel because because this is now coming from ground to a source so the science which I use there if you see clearly it is showing upwards then this is RS how do we calculate the output resistance what is the method I suggest short or input side sources independent sources put at the output VX as the source assume IX is the current starting from here the ratio of VX by X is the output resistance we are looking into this side okay so what we do is we write we check this equivalent circuit this side I already grounded okay this side we already grounded so I get IX please remember where is VGS not this VGS is here please look at this is good this is the source is that correct output resistance is calculated where please keep looking looking at the source is that clear looking at the source this is my output so the output resistance is looking at the output side which is my output side this is my output side so I must look around is that clear source following where is the output taken from the source so the output impedance should also be seen into the source okay so if I do that which I did here the output this is my source this is my drain and this is my R0 is that okay so this is so this is doing this is current source this is our this is R0 so I write IX plus GN VGS is VX upon RS current through this VX upon R0 but VX is minus VGS 0 why this is grounded no current between any of them okay this is grounded so no current between source RG and R series so this is equal to minus VGS this is minus terminal this is plus terminal why it should be plus at the gate otherwise no current can flow VGS should be greater to make it current flow in the transistor so VGS has to be plus here minus here is that okay so if I put VX equal to minus VGS then I resolve these expressions I get R0 is VX upon R so you can see it is GM VX plus VX upon 1 upon RS plus GM can be written 1 upon 1 upon GM GM can be written as 1 upon 1 upon GM plus 1 upon RS plus 1 upon R0 what does this 3 1 upon 1 upon means the 3 are in parallel so 1 upon GM parallel RS parallel R0 are the this is the output of course here I actually converted back but otherwise these 3 resistances are in parallel that is the output resistance of a source follower can you now see is R0 very large compared to RS which one no no this is R0 will be this is R0 now oh sorry you are perfect perfect one upon GM parallel sorry no just like RS yeah you are perfect that that has to be say it is a parallel combination so 1 upon R is 1 upon R1 plus 1 upon R yeah yeah perfect for the calculation of low output resistance we should say load is outside because it is not our making it is someone else is going to connect but when I want to have a low voltage calculation I use that why because I want to know if that appears how much is I am going to actually get the output voltage is that clear but the measurement is R0 forward this device has an amplifier how much output resistance it is giving and with which it should get compared then your point of is well taken all out is parallel to RL and now we must see whether this is good enough or not good enough to change the output for the next input for the next stage is that clear to you that is exactly what we are looking how much is this about compared to the load which will be affected or not affected is that clear that is what we are really looking okay so we first find out what is the available R0 for us okay okay so this roughly finishes the source follower what else we have to do as it common source then we did common source sorry this is the X no current between these two this is grounded this must be grounded but the signs are plus minus is that okay so this potential is minus of Vgs this is 0 input is 0 no current between resistances this potential is therefore equivalent of a ground this is 0 and this is Vx but the definition is this is plus Vgs so Vx must be minus Vgs is that okay circuit wise that okay yeah I assure I am sorry you should very deep thank you very much keep doing this yeah always looking at the device side loads not external load leave it always look at inside the transistor okay always look inside the transistor okay if you someone may ask you actual output resistance as seen by the load external load then put RL parallel to that whatever it is okay RO dash you may call it and then parallel of actual load external work with this RO you have calculated this will be always in parallel but in common source the output is taken at the drain in now so we look at the drain and the output wherever here output is you look into the drain that point okay here it is in source so I looked into source is that okay always remember wherever is the output without external load whatever impedance seen there is the output resistance okay what is our actual game later we will want to do we are we are going to work on something called feedback systems and what are we trying to do there we actually want to change gains I mean we do not want to change again but they will change but we will be able to control this are in and all out that is exactly what feedback does okay and we say why other feedback feedbacks are more important for what we call stability we are not done multi stage so far and we will see what happens if there are more than two or three stages of an amplifier your first is amplifier output of that is given to the second stage amplifier output of that given to third stage amplifier maybe and stages how can we go in stages what are the maximum number of stages you can view and if there are odd stages and if there are even stages what will happen to the output input relationship that is exactly what we will do in multi stage okay and then we will see oh then it is becoming like a oscillator and not even oscillating what is the world I said what does that mean can anyone think I say it is not an oscillator but it is oscillating what does that mean oscillator is fixed frequency is that clear oscillating does not mean that it is a fix really say it is varying amplitude is varying and phase also is varying okay that is called oscillating systems what is oscillator it oscillates at one frequency that that is the difference so amplifier the different difficulty in amplifier designs is something like this when I try to make a good amplifier is start oscillating it is called unstable okay whenever I start actually design an oscillator that it should work as good this it will start giving amplification okay then we are worried it is amplifying so design is interesting that when you want to control this something goes away so control that this goes away so how do you match that is all the circuit okay here is another last configuration maybe we will do which is common gate configuration we are done common source common drain finally common gate without thinking too much you just draw this circuit and think why someone should do common gate at all this current source is given to V- this load sorry this is all D is going to be DD or this is IQ is going to minus V or call it to VSS okay it can be also 0 but I just as a generalize I say it goes to minus please remember it cannot go to plus why current direction each and a to positive potential what I say is current source direction is at the end current source is receiving V- voltage okay it can be 1 0 okay it can be also 0 but it should not be positive okay and I told you yesterday there was a query small one of the 5 periods the major thing in analog circuit which is different from digital it is all many a times are the most time dual rail 2 power supplies plus VDD minus VSS or the VSS with which is a minus of that okay this is always analog circuit uses the advantage has anyone found this what is going on why should have a 2 rail supply in the case of analog digital we do not want it because there is nothing called minus voltage detector as 0 okay so I do not want that okay but in the analog I will prefer if it is minus volt to plus volt that means 0 shifted why I say there must be some something I am getting advantage by putting 2 rails typically voltage difference may be same let us say I have 5 volt so I will put VDD 2.5 VSS minus 2.5 so the net difference is still 5 volt okay but I prefer instead of 5 to 0 which is called single power supply device circuits single power supply circuits they are also possible but our preference to then I will say instead use VDD VSS half-half or what are the difference you need not everyone half awesome there may be different values can be given okay is that okay so we like to see if not next measure first thing I will tell you why minus is advantage in analog blocks do you see any disadvantage of this having minus in the case of silicon circuits if I make on chip do you see any problem if I have this no no VTG a good one see there two power supply I have to create one is plus and one is minus either externally I must supply you minus and plus or internally I will have to create some minus by shifts something level shifter I will have to thirdly I may have another problem that I have two lines running everywhere one plus one minus in circuits and integrate circuit every micron or every nano now is of importance if I unnecessarily run power supply lines everywhere I am losing that much area I am not doing anything with them so I would actually prefer one VDD okay but still analog people say please Baba if you can give us two power supply two dual rails that is most important not that it is great advantages that I am trying to make do not think that both could be gone on your own nothing but there is still positive advantage if there is something problem of something I may give one word also reliability issue there is something happens if otherwise so think of it why we still go for minus okay no no no RS is the actual part of the capacitance is for the external load when the signal goes to the external side I want to know how much is the external load then I do not want DC to be connected at the output which is external to me okay there I do not want DC to go is that clear no here it is not so here this this CG is different from that no no no this is my device this is load actual load of the transistor this is external load so before I get external load here I want decoupled this VDD should not go to the output this VDD should not go to the output so I just decoupled it is that okay this VDD should not go to V0 so I stopped it is that okay at this point please remember there is a DC plus AC is that correct we are only calculating AC is fine but at this point both AC plus DC are available to that means on this whatever DC voltage AC is running over it and that DC I want to make it 0 so I pass through a capacitor and only AC is bypassed okay is that clear so that is what exactly what we are doing okay what is the purpose of this CG compared to this RG because whenever DC part is concerned I want this to operate but when I want AC I do not want that to be in the way I have considered in my calculations because if there is no RG this current source cannot provide that bias to you so I must show you there is an RG sitting there okay if you wish you can see the way circuit could be also looked into I saw with a sorry I mean I saw okay this is source side this is drain side and this is gate side is that correct this is source this is gate this is drain okay in a 90 degree minute I saw boom earlier circuit because I want to draw equivalent of that so just so please remember how do I draw equivalent circuits I actually see the main circuit and equivalently put the components down is that correct that is easier to do if I do something and then I start drawing 90 degree then that is why I always try to show you that the way I draw main circuit the equivalent should follow okay that is easier to draw no reason why you cannot show circuit like this you can always show like this there is absolutely correct okay but for my simplicity I always show you it he can I say input you want to give from here at this side so I input output you want to collect here this should be common gate is that okay drawing the circuit is identical only thing is 90 degree shift Karnia minus okay this is your signals and source resistance okay this is of course DC so nothing to be drawn here between source and gate is grounded please remember gate is grounded RG is bypass for this so gate is grounded source is this terminal Vgs exists between source and gate okay what should be polarity of Vgs J plus source minus this has to be always correctly drawn okay Vgs from source to drain there is a current source and how does it flow from drain to source so drain to source so this is gm times Vgs okay this is your Rd normally this is your external load which is also grounded is that okay so the equivalent circuit is that clear to how do I do why I draw this circuit like this because then I can draw this circuit almost identically done is that correct so please remember that this is not relevant per se as far as circuit and a board is concerned okay but for drawing this if you draw this you know much better how to draw equivalent we are not interested in this we are interested in this so we want to know what is it okay where yeah yeah we are right between source and drain there is a R0 right now our assumption is R0 is very high yeah absolutely no problem no no it it only proc creates problem of solving equations now you remember not only current path is this but also part in the R0 okay part in the R you are right what did I tell you in the class first day I will solve simpler problems and I will ask you higher difficult problem okay is that okay so let us do that quickly analysis now keep that circuit in front of you V in please remember V in if this is the ion current this drop plus this drop is that okay this drop and why minus sign I put because this is minus plus Vgs from here on the top it is minus Vgs so this drop minus Vgs is your V in is that okay last people is that okay this voltage is equal to this plus this okay both side go on ground okay okay from the input loop however this I in call actually I may be calling IIR I in a same is there any other source you see from here what is I in actually current in this circuit if she is her problem in and then little little different but let us say R0 is infinite that this current and this current are identical but in opposite polarity GN Vgs is entering here I in is entering here they are same currents there is no other current source here okay. So we say I I or I in maybe if I made I in call it I in everywhere okay I but I was minus GM Vgs and therefore I substitute this I in here I in is minus GM Vgs so we will is I in RSE I replace I in by GM Vgs GM RSE plus 1 into Vgs is that okay so how much is Vgs minus how much is Vgs minus V in upon 1 plus GM R is that correct so how much will be Vgs V in related if RSE is 0 they will be equal okay phase out but they will be equal by magnitude by they will be equal so that question is always there if RSE is made 0 just make RSE 0 now instead of initially doing right now you can say so generalize this okay so V0 is how much is V0 from the circuit can you tell me minus GM Vgs will pass through this is that correct my minus current is going out from here so the drop across this is minus GM Vgs RO parallel RL GM Vgs RD or not RD parallel RL Vgs I had just calculated Vgs I had just calculated is it okay third bench is it okay so Vgs is minus V in 1 plus GM RSE and B0 is minus GM Vgs RD parallel RL okay so what do I now do put this Vgs here then I have V0 in terms of V in is that current V0 in terms of V in is it okay what is the phase difference in the case of mass transistors between drain current and source current so V0 V in is that clear since the drain to source current in same direction if I input at source and color to the same side on the there they should be of the same science let us see whether it comes because it is minus here minus it may look hanging so that may become plus so V0 is minus GM RD parallel RL into minus V in upon 1 plus GM RC so the voltage gain is plus GM RD parallel RL upon 1 plus GM RC please remember once again V0 and V in are in phase so in common gate what is that we are achieved in phase outputs okay how much is the game really will be assuming RSE 0 it is still GM times RD parallel RL as normal amplifiers is that correct but what is important here where this will be used very often is this find the current gain for this I0 by I exercise all amplifiers though I did not calculate may be next some of them I will see we must calculate all four terms which are four terms I said can you tell me voltage gain current gain input resistance and output and if they are capacitance what should be new name should be input impedance and output impedance okay so we must calculate all four quantities in some cases we may do two more quantities also which can be two more in gains input current output voltage input voltage output current I0 by V in V0 by IN so 4 power trans conductance trans resistance voltage gain current gain there are four possible gains in an amplifier is that clear now at times we want to use them as a current amplifiers at times I want to use them as a voltage amplifiers at times I want to use amplifier the trans conductance amplifier and at times I want to use the trans resistance amplifiers each has a different applications then we will say one of them which dominates when so I want this kind so common gate is better or this is better or that is better which I should use for this they will be only three of this kind with of course so degeneration fourth one but any one of them should be chosen corresponding to the kind of gain maximization you are looking okay is that clear in analog what are we looking for possible gains and see whether it matches one of your four combinations okay whichever gives you better results use that for us whatever you are equivalent circuit we can always calculate anything to anything ratio of anything you tell us okay so we must evaluate all four of them and keep ready so such tables are available in books that is what all designers do are a Cali fan so which they mean a cali cali cali that is how 80% designers fail in the life actually because they never think that your condition is whether exactly same as what they wrote I0 is current through the load okay but that says the other resistance is parallel the net current at the output please remember you know she has a I think I we must honor her words we should not take this is something like this is that correct this is the output they may have a division is that correct the current output is here is that correct so you must take output at the output net current so it may divide into already it may divide into R0 it may divide many parts but the net current is I see is that okay no no no they are in parallel why should it only be there there will be three parts coming GM VGS will actually divide into three parts so the I0 is actually at the output terminal and not through the load resistance the output terminal you are you are you are very correct but let us look at it see the current in this whatever current coming from here will divide into this so this current is what is outputted from the transfer this ratio I am looking gains are always connected from the drain side voltage is everything we measure here so at the drain current coming out is how much is that correct so this is my I0 and this may divide so the net current if I put a one node here not current is essentially entering two parts what you are saying why not only this okay you calculate even for this does not matter this will be actual load current which you are looking this is also possible but when I say something I say I0 here okay yes for the in oh yeah it is called short circuit current I see okay that is another way of swing solving a circuit maybe one day I will spend on how to use only open circuit and a short circuit values that is a much easier method of solving larger circuits and I am avoiding all of them simply because in my opinion if you do this kind of analysis it never gets into an error okay that is the method I suggest you because in my opinion Christian law solving is the easiest on such systems is that clear I am not suggesting there are three more methods I can show you for example people do not use gms will use gm r0 as the output voltage sources that different ways of doing things I am showing you my simpler method to tell you that if you want to make zero error free solutions use this equivalent circuit and solve which will never get into an error is that okay for this the input resistance is Vgs by ion but I in is minus gm Vgs so Rn is 1 upon gm is that correct minus Vgs is being that is what we said gm Vgs is I in Vgs by I in is there are in so Rn is 1 upon gm how much is the output resistance output resistance method is what is the method I suggest short the input source since there is no current here this is grounded this is very small no Vgs therefore no gm Vgs so Rd is the only resistance there the R out is Rd okay Vx by Ix is Rd only how much is the gm value in transistors other than we calculated not in bipolar in the case of mass transistor what is typical value of gms order QIC by KT gm in millions man you know which 38 millivolt milliamps per volt I had so mass may get other or come over come over now so it is of the order of 10 to power minus 4 10 to power minus 5 and 4 volts is that clear so Rn 1 upon gm is of the order of 100 K Rn is of the order of 10 K to 100 K is that clear that value you should know typically of the order of 10 K to 100 K depends on the bias current you are using that it will decide on that and Rd of course is the Rd value used for biasing as well as for the load so whatever is Rd is your output is that okay so what is the advantage of common gate you saw from here the voltage gains how much was the voltage and we said gm Rd parallel Rn is that correct that was the voltage so which is like a normal amplifier but it is in phase its output resistance is low is input resistance is high is that clear what did we do by common gate the gain is like a normal amplifier it is in resistance is so relatively high and its output resistance is now controllable by me how much Rd I fix that is my R0 okay that is exactly what and it is in phase and therefore no problems of connectivity is that correct circuit can be connected directly from left to right okay that is the biggest advantage common gate gives another thing which is not showing here and we are not started which is called frequency response something else will happen for all three amplifier some may have larger bandwidth some are smaller bandwidth okay so when the bandwidth issue comes again we will have to make a choice which one to use so that is the way we will not look at it okay before we quit this area quickly I will not solve all of it there are different kinds of load I say in the case of mass amplifiers one of the mass amplifier will have okay by the way this terminology you should use their input is connected that transistor is called driver okay and the resistance is replaced by a transistor that is called load it is like a load okay this is called driver that is called load this is actually equivalent of this nothing great and nothing very big okay this is our load this is input this is my driver this is my output this is the symbol this is what we will do so how to make saturated load I say if gate is connected to drain Vgs- Vt will be always less than Vds by did by design therefore always in saturation so normally this source is when I say I shot this to get the resistance here this is like an infinite resistance here which means if this RS in a common source with source not bypassed is equivalent T so you do this analysis for this Vx by RX RX will be equal to 1 upon GM parallel R0 which one will be dominating in this this or this what I have you won a GM 1 upon GM how much will be R0 mega ohms so typically this will be always equal to 1 upon GM is that correct so equivalent replace karna kele 1 upon is that correct a shot saturated transistor load is equivalent of 1 upon GM of course if R0 is not very high then parallel whatever it comes okay but if it is R0 is very high then you can say it is almost 1 upon GM ha for us many are for us a Indian this driver act like a series resistance for this this amplifier of this transistor and says there is no current will flow because Vn is going to be 0 this is equivalent of infinite resistant TK for us watch the other why I do not say everything because I thought you should when I up a hints of the daily cook around for us okay this this is called equivalent resistance is the source of a load which is infinite I just tell her think of it why this is equivalent of an infinity this RS is tending to infinite what I am showing you is not given in a book this is what you visualize how do I visualize that means this is equivalent of that this is what I give a hints how to visualize let us go quickly on the gains of this here is very interesting equivalence of this GM VGS is the source city there is no source resistance so directly VGS is Vn GM VGS what is this ROD driver output resistance what is these two I am drawing from the same terminal this is the drain please look at it this terminal is the drain this is ground for AC so whatever this load resistance is is appearing here across to the ground is that clear is that point clear ground I those come each a fold ground each a layer okay so that is exactly what this circuit shows this is ground lower terminal is ground so GM VGS ROD 1 upon GML ROL I repeat this ROL can be larger or not larger compared to 1 upon GML check it before actually leaving or not leaving turns alternatively if you say ROD is much larger than 1 upon GMD ROL is much larger than 1 upon GM then one can show and this also I will not show you that V0 by V in is GMD or you do not have to say you remove these two RO and ROD then GM VGS passes to 1 upon GML is that clear if you open ROD and ROL this current passes through only GML is that 1 upon GML so the output resistance output voltage is 1 upon GMD upon GML okay what is GMD the transconductance of the driver to beta N dash W by a driver into the bias current IDSQ divided by GML to beta N by W IDSQ what is this IDSQ the operating which we are not sure DC bias part we are not shown that is the bias current okay so how much is the gain essentially in the case of NMOS these amplifiers under root of size of driver W by L of driver divided by W by L of loads so gain but I make a size of driver should be higher compared to the load is that okay so driver W by L is not a lot of now or load is not a lot of gain but that is how we do gain adjustment in silicon chips and NMOS amplifier on a silicon all that I will design an amplifier of this kind and change the sizes okay to get my gains is that correct do you also see interestingly here there is no device parameter device basic technology device from except W by L which is size which is controlling sizes of course can vary this also is the problem this W by L and these W by L may not be accurately put on the silicon then the actual game may not be exact but this under root may help you please remember under root reduces something but it also reduces the error that is the advantage you must look at something which is bad for you is that okay so this is called what are the other loads I can use this was NMOS enhancement mode what the other load I can use NMOS linear mode PGG logo I have already calculated R for linear circuits to put that R there 1 upon beta VGS-VT that value we know so we know what is our load for that what is the third possibility may correct depletion load it is called the R Nikola you can replace depletion loads and finally we can use P channels with N channel together and that is what CMOS inverters are all about we normally use CMOS amplifiers for full range we call it inverter if you only use the transition we say amplifier okay there is no difference between the two okay the other amplifiers please look into the book which has my name also okay okay before we quit for the day there is one interesting thing I said you earlier for a normal amplifier Omega T which is the figure of merit what is the M omega T means unity gain into bandwidth or it is called gain bandwidth okay Omega T is essentially gain bandwidth how much is that I said roughly GM by C CN or whatever equivalent okay this is what we said fair enough so what is the figure of merit I say Omega T okay so if I increase gain I reduce bandwidth if I increase bandwidth if I increase gain reduce bandwidth by increase bandwidth I reduce gain because this is fixed is that clear but that is limitation you said now essentially I said that is the maximum frequency up to which one gain at least should be seen can go beyond now this is creating a plot I want larger gain okay but I do not want bandwidth to be lost so something let us say we are not talked about right now gain versus frequency okay but what we are essentially we will see later when I showing for a open this is this is Omega T this is 0 dB this is called bodice plug which we are going to do soon okay frequency response of a gain function so whenever gain becomes unity we say it is the FT or Omega T what really I want is or it is not this is that correct what I am looking for I may have this frequency where fall occurs may be slightly reduced but the gain is higher but it should well down exactly at the same point is that correct so I want what is it called gain boosting what we did here is gain boosting but what we did not lose FT is that correct normal amplifier method I saw who is it cannot escape either we go to gain to come here jada who is I want to see can I do this mischief now or can I break this what is called as technology constraint what is constraint this is constraint I want to break this here is the circuit which does that okay please remember what the amplifier which is now going to be taught is essentially what is the going to do it it may boost the gain but will not affect FT is that correct if I do that then what I did it I have broken this technology constraint okay but some cost care they now is just roll over point a but if can I can still do that that may be still acceptable to me then I have gain boosting okay this is essentially done by an amplifier called cascode what is cascade I repeat if someone says cascade which we have to do but we are not done essentially a1 a2 a3 is ka input is ka output next ka input next ka output next ka input this is called cascade relevant range a Kpch cascading okay what is the gain in such cases a will be a1 a2 a3 is that correct in Costco but then as you boost the gain what will according to this will happen the bandwidth commotion okay now I want to see gain boost that is not losing balance okay so let us see whether different between cost remember why I showed you immediately this what is that I am achieving in Costco enhance gain is that correct here also I am enhancing the gain is that correct so enhancement of gain is my aim but in the first case I do not lose my bandwidth or I do not lose my WT but in the next next I will finally because we will start rolling down earlier and so finally it may actually become somewhere like this I do not want that to happen sorry I do not want to use that kind of things okay so cascade nahin karna jata main cascode karna okay hodatha introduce karthik next time calculate karna this bar hodaya a figure banan detain jisya aapka idea lag jaai ki main itna gap kyo maaram or why I showed you this common gate will be now visible here is the device okay another source another way of biasing mass transistor if you recollect I showed you by mirrors or what we saw last current sources so many a times bias I that okay this is also biased by current source I from this current source from where it can come from the mirror okay from the mirror okay so let us see I take a easier case first maybe I can replace transistor later P channel or N channel this is the current source IDS which is biasing the device and now I have two transistors in series first is driver and second I actually apply what we call be reference we reference is DC volt is that clear we reference is a DC bias for what create VGS for here okay which is for AC what is the terminal going this terminal ground common gate make a bullet up go gate is common to both gate is common to this gate is common to this is that we are common source lower transistor is acting like a common source call it M1 M2 for example M1 and M2 so M1 is sorry M1 is common source system but M2 is common gate is that okay actually digital circuit me is cool camp what is this called this is called transmission gate or a pass gate will logic depends on this pass gate so whatever is being will be transferred to V out when the 5 goes one if I 0 V out is the last V out is that correct there is a transistor than conduct okay this is called pass gate in digital this is what we use very often in digital here this is like now think of it if this so called M2 would not have been there this is like a normal amplifier a bias equivalent resistor RO of this series to this driver this is RO of this series to this okay so this is a normal and mass amplifier with a load of R0 now I put M2 in between so what is the way gain can be improved in this because there is no external load in the case of such circuits the gain I said you how much gain of such amplifiers are GM times are the please remember R0 is the resistance of source I mean current source okay so GM times are 0 is this is cool but I have to say I can increase by increasing sizes but if I increase sizes what will happen capacitance will go high up so the GM vice you will not improve on me okay so what I say okay I damn damn care about this but can I boost R0 please remember what I am trying I will boost R0 so that my this increases GM R0 enhances if this is R0 dash okay but GM I kept roughly same as what earlier without this with this together GM if what is the GM for this full of this delta ideas by delta VGS is still GM now to get that okay so if I maintain GM new equal to GM old what I am going to get bandwidth will be I mean WT will be affected or not affected no no area change GM is same CC so FTI fixed otherwise but what I boosted is R0 so what I boost I will boost gain so now I have broken it I improved the gain so this circuit essentially enhances the output resistance is that correct this circuit essentially enhances output resistance huge number gain times R0 this will be gain times R0 gain of this into R0 will be appearing as the new R0 okay so gain will be boosted but FT will not change is that clear the game if I change GM then I have a problem because then I have changed both okay but if I fix my GM then I do not change FT or do not change bandwidth for that matter but what I will change gains R0 but I mean cost scores enhances the output resistance without losing GM so this M2 allows what is called cost score allows you to boost the R0 of this is that correct next time we will calculate this okay