 about the cascode there are some issues which may have somehow brought to my notice in advertently or advertently I might have said wrong or right I already said a 6 page problem about what is the bandwidth issue in cascode which is now available on website should be yesterday night I sent to TAS they will put it on the website and also may be on Moodle depends on them so please look for issues about the maybe there is an issue which I did not secretly or maybe I said it wrongly either way in any case whenever gain increases the bandwidth will go down so there is no issue on that what I was trying to only say that if I have a cascode then the change in bandwidth reduction per say from the simple to cascode will be not that large compared to the increase in the gains is that point final word is clear that you may have enhanced gain and you will not have in the same ratio reduction in bandwidth is that clear that is the all that cascode does okay there is no issue in saying that cascode has the same bandwidth if I said it per say I do not think I have I have many times said bandwidth is not same but in case I have said it I stand corrected okay so we start for cascode so I have already given there so you can read that may be other class I will do today let me finish my because I have I like to use something of today's class and tomorrow's the Friday's for the test okay so three of the common amplifiers which all of us use in variety of applications is common source common gate common drain or what essentially now popularly known as source followers so today we shall like to do common source normal amplifier we already done day one we have been doing that often that I will not do it but I will use one of the interesting common source amplifier which is very interesting because it gives you lot of understanding as well as it has lot of features which we will just talk to you there are three types of loads which we can use in the amplifiers of course fourth one I have not written in the sense that resistor is always possible so I am talking of non resistive loads that is active device which is acting like a load so three kinds one is called diode connected the other is called current source loads and the finally one can have a linear load essentially we are trying to replace resistor by transistor in different modes of operations that is the basic idea each kind of load has some advantages and some disadvantages so we will see at least few of them if I am going today or at least on Friday these are very relevant because we also will like to at the end of this all three gate all these amplifiers why do you really need a difference amplifier at all if everything was so good with all of them so why did we look for different or defense so I will at the end hit these all amplifier some way and say that they are not good enough in many applications or most applications and therefore we are looking for a differential amplifier so I am not trying to say these amplifier will never be used I may show some applications but generally why defense are used probably will be obvious if I limit them oh this is does not give this is not good enough oh so look for something which all in all can I get little better in one and that is the difference so all these applications or all these amplifier I am talking because I want to show that they are also gain stage stages but they are their own limitations and they must be overcome if you want a good amplifier and that is the way we will come to it so this is the pedagogy of this why I am looking into this many of you are done this so there is nothing new in that I may rush sometimes if you have someone does not follow stop me but otherwise there is nothing extraordinarily I am talking about which you have not done earlier for example one of the thing which I may show at the end of this which this particular part may not show if the major worry was biasing in most of these amplifier which we normally never talk so I will show you that why this biasing is an issue for us and why defam actually solves that problem okay so the kind of things which I am trying to show in an integrate circuit what are my problems if I use this and what will be solution if I use defam so that is why I am looking in such pedagogy fashion okay so let us start with the first and the foremost amplifier common source normal amplifier I have discussed this is an interesting amplifier which shows that there is a resistance in the source and it is very popularly known as source degeneration okay now why this is relevant at times may be shown when the output appears for this amplifier this is the kind of equivalent circuit which I draw from the Razavi's book this is not mine but this is what taken from Razavi's book method I do not mean this specific but Razavi's technique and I like that technique simply because it explains from each node to each node what is happening in real circuit okay that actually explains it very well and because of that I thought Razavi's technique of putting an equivalent circuit is far superior than many other book in which people just say equivalent of this put replace by this I do not like that this is a straight forward equivalent circuit of a MOS transistor along with other resistors and nothing great about so here what I am doing is I have a load RD which as I say will replace later with active loads right now keeping RD there then there is a series resistance RS to the source and of course in this case also there are issue which may be at the end I may show you normally for all simple solutions we say there is source resistance is 0 the input source cannot be without source resistance so if there is a source resistance it may create problem not at the low frequencies but at high frequency so right now since I am more worried about the gain low frequency gains I am not showing you RS values some what is called R with the source itself but otherwise that will also appear in circuits right now I am assuming that there are all effects of capacitance only come at very relatively larger frequency and I am not looking into those frequencies I am in a mid band low frequency area where gains are independent of frequencies this is my assumption we will see where it starts so that is what our frequency response when we do for this we will see that actually. So the way as Razavi defines you have an input voltage V in to the gate to the gate between gate and the source there is a Vgs which is essentially he defines V1 as the voltage across gate to source and from source to the ground there is a source resistance RS at the output side rain side equivalent current source due to the input signal is gm V1 shunted by R0 and shunted by the body bias effect gm gmb times VBS finally there is a V0 output here I should have said this is a V0 output across RD and please remember source is not grounded so RD should not come into parallel to R0 or any of the current source this fact is clear RD is not in a cross R0 or sources it is to the ground so essentially it is if at all I should complete this circuit because that is what essentially the way I have shown is that clear so please that is the only difference from earlier work and this work now if I want to figure out the gain I know gain is nothing but gm times the output resistance and so we will figure out gm effective for this circuit which is I out divided by V in which is delta if you are looking into delta kinds of the earlier analysis which we showed there is another simple method I showed you I is equal to gm V in plus go V out if you use that then it is delta I out by delta V in is essentially same as gm. Now if this method what I am what Razavi does or I like to do is we figure out gm effective for this circuit we figure out R out for this circuit you can need not do all that also but this also gives you a feature that how much gm is changing and what parameters gm is actually getting influenced by because as a designer I want to know if I want to boost something what parameter is in my hand and that is something which this kind of analysis does and that is why I say I like Razavi not because of course he is a prolific writer great analog circuit man okay so first thing we do is we want to find gm effective so we will make V0 0 okay and put V in and find I out by V in that is the standard technique to find gm so I put V0 equal to 0 okay is that correct I is equal to gm V in plus go RO or GO V out I gm is equal to I by V in is that what we said it so I make V 0 0 so I ground this output is that clear as soon as I ground output there is no current in the RD so I just forget about RD so I have gm 1 V1 gm VVS plus R0 so now I look at the relationship between input side V in V1 and since this current I out how much I out will be current through R0 plus current due to the body bias current source between this source plus current source appearing due to dependent gm V1 okay these three currents must be summing to I out okay but the same currents is this is open circuited the same current must flow through RS to go to the ground there is no other path once the current comes through this there is no way it can go there so it has to go through this so essentially drop output current is flowing through even the source resistance which is true in a normal amplifier anyway current goes through the train to source okay so this is not very great however if you can see since there is a RS here and there is a current here there is a voltage drop that we call VX so we write what is VX the drop across RS therefore it is VX is equal to I out RS is that okay the drop across RS is IX RS which is nothing but my VS signs are correct because current I have now chosen is downwards so this assumption of plus minus is fair enough I am putting current downwards is that okay so VX therefore it is not my if I would have chosen this outer I out like this then I would have put a minus sign on this right now so you can choose either way it is not it is a matter of only your choice I have put I out in you may put I out out also so it does not really matter very much because if I out is outside I RO RD finally which will come will be taken positive but I out itself will come minus somewhere so minus GM RD will come finally okay so does not matter if I choose either direction of the other things will take care but this current sources must go from drain to so there is no other direction for them so this is what it is GM V1 into GM VBSR 0 so if I write this this is the expression if you see V in and if you see this loop starting from this ground to this ground what is the way we are we are telling you go from this ground to this is a loop so drop plus this plus which signs what are it is this plus this plus this must some out to be 0 with proper signs okay in a mesh that net voltage is 0 okay so we write then VN is equal to V1 plus because this is I use this plus minus plus minus so they add so VN is equal to V1 plus I out RS VX is I out RS I out I substitute this values in I out GM V in minus I am replacing V1 from here okay what is V1 V in minus I out RS is V1 will be V1 is V in minus I out RS I just substitute that GM V1 is GM into V in minus I out RS that is GM V1 plus GM V which is minus I out RS because VBS source to bulk and we are looking from bulk to source it bulk is grounded so opposite polarity is that okay source is that positive and this but we want the opposite so it is minus signs so minus I out RS is VSV how much is the current through R0 0 minus VX divided by R0 and what is VX I out RS so 0 minus I out RS by R0 this is Kirchhoff law and nothing very great on and this why I do this as I say I like I sense as I say I like all three books as much or four books but this method is very straightforward there is no assumption there is nothing we are missing in terms sometimes are going to be smaller or this automatically numerically will I mean lose the term to smaller so you do not have to worry which term is stronger or which term is but as a designer we show you that this is not the ideal way every time because then you solve everything so from there we will try to do that if this means what and then during design we will use our conceptual thinking for that so first to get that I think we should do analysis so if I find out G effective from this I get GM R0 RS plus 1 plus GM plus GM V times RS into R0 just collect the terms and figure out GM effective as I out by vain I repeat collect the terms of I out collect the terms of vain divide and you get your GM effective so this is the expression I got it now I start looking at these terms if you say RS is not very large few kilo ohms even then this term is of the order of 10 or 1 to 10 even if RS is in kilo ohms it is at least between 10 or something this may be one case fine but this into R0 that means it this denominator is now getting 10 times R0 value equivalently so if I just leave RS R0 cancels I also say right now GM is larger than GM be so I leave neglect that so why I got it is actually GM upon GM RS which is 1 upon RS is that correct equivalently saying it is just 1 upon RS that means the value of RS essentially is going to decide the game how much is my RS value is that point clear and this is very important okay is that okay so this is how you should look having seen an expression which terms dominate okay and GM is decided by what currents but right now what I am getting is essentially GM effective is not very much function strong functions of GM itself it is only a function of the load I am going to put there or source degeneration value which I am going to put this makes something interesting because that means if GM effective is not a function of technology parameters like thresholds or environmental parameter like temperatures I am getting an GM which is very very much constant and to my value I fix RS and I get my GM so that is designers thinking oh so I want to fix my GM which is relatively constant relative word has to be thought we will see why but otherwise you say relatively constant for most parameters of working then I say oh I got a good this GM value which I can fix now okay so as a designer now I say okay this degeneration help me to get something a constant value for my choice that is designers output from this as such otherwise there is nothing great in this expression as a designer I started looking oh it is only RS function RS is my value which I must decide in my designs is that clear so this is the learning part of that the first part is just deriving the second part is to learn okay what is that I got out of this expression that is the design issue so design issue says fix GM by RS that is simple the second term in the gain function was allowed and please remember R out is essentially the net output resistance okay but right now we first calculate the output resistance other than the load because as seen from the drain side so we calculate RO from here what is the condition there to calculate output resistance put a voltage source at the output and short all independent sources okay the standard technique of circuit analysis short all in independent sources please short means if the current source is there it should be open voltage source it should be shorted so do not try to show me something sir I shorted something and it got funny it may okay so okay in my case I say okay V in is 0 so from the same circuit I use the actually I never change my circuit because I just want to say I do not want to go out of circuit from which I started with so I grounded this I still believe since I am putting a V out here okay V out by I out is essentially what I will call as output resistance seen from the drain end for the transistor so I say okay what is V1 then since in this case this is grounded Vx please take it this is ground this is ground so what is the condition you are saying V1 plus Vx is 0 V1 plus Vx is 0 or to say V1 is minus Vx or to say V1 is minus I out or as is that this is opposite of this is grounded now you bring this terminal ground so it is minus I out or as yes V out is at the drain terminal whenever you calculate the output impedance or output resistance short independent sources at the input or anywhere in fact then actually apply a voltage source at the output which I am calling V out which is entering a current I out then the resistance seen there is V out by I out or why should call resistance if there are capacitance it will call impedance of that. Now RD is outside see this is the output resistance seen into the device the arrow you can see then I will shunt it you see next line the actual output resistance is RO parallel RD okay because why I do not want RD to be part of my calculations because RD is an external parameter it has nothing to do with device per se or circuit I am designing initially that can be decided by the next system as well okay. So I do not want to a priori decide what is RD so whatever it will come I will put it anyway there okay so right now I say I am only calculate is that should be clear now so RO is seen in the device at the drain side okay so Vx is I out or as V1 is minus I at RS so I out again same current this current this then this is Vx which is Vx minus V out you have to now remember the current is going from V out minus Vx divided by R0 what is the current in R0 V out minus Vx divided by R0 substitute all of this here V out minus Vx by R0 is the current through R0 minus GMI out all are minus I taken it because I mean I can understand actually there is plus all are same okay this current is some of this plus this plus this but the minus sign I am trying to say is because can you tell me why because in real life the phase out will come so I initially added that minus but you need not put you can put it plus here and still solve and there is nothing goes wrong okay so having done this I did all analysis and finally I get RO which is R0 plus 1 plus GM plus GMB times RS plus RS by R correct terms of V0 V out and IO divide and you get the output resistance R0 which is this and the final RO as I say is this output resistance R0 shunted by RD no because that I out see the problem I what I removed is the okay I will come back to it but let me first finish this talking this minus sign I did not put initially okay so I just took it this minus sign which I am going to get so I initially itself I used it so that finally I will get a minus sign but then I kept using to make it that plus with an additional minus sign the gain is GM minus GM RO and there I started plus GM RO so I just thought I will make I out the other direction so it will automatically phase out okay just come to it just wait okay I have point I understood your point just let the major feature in this R0 is quite large why do I say so because again GM plus GMB RS is larger than 1 if RS is kilo ohms may be tense RS is normally much smaller than R0 so that this term is negligible so what I this is larger than 1 so this is around 10 let us say so 10 times the R0 so output resistance is much higher right now so if you shunt it with the load and if the load is smaller than RO then it is the RD itself because whichever is smaller that will dominate however in some cases RD may be larger than RO which will be this case current source or cost code for example if I use a current source or a cost code at the load end that will be even larger gain larger than this gain because there it is GM times RO times other RO so that time the RD value equivalent will be even larger than GM RO this GM RO okay there will be an additional RO coming in the cost code or in the current stage there and therefore at that time instead of RD what will be the RO whatever is the RO you got that will be your output resistance so please take it do not always use RD as a smaller or larger RD could be larger if it is a current source which is replacing RD if it is normal resistance or RD can be smaller in transistors also when if I use transistor as a resistance and I say RD is small so which is the value which way I can get one of the active load I said the device is in linear mode or non saturated mode the resistance is very small 10 less than 10 K in that case RD will be always smaller so the bias how you did it how do you put an active load on that may decide RD smaller or larger than RO and corresponding parallel combination one or the other may in case they are equal will have half of it but otherwise one or the other will start dominating okay so typical value if I write GM RO effectively all that here I get this expression and now I again do the same thing which terms is smaller this term is larger GM this is larger this is smaller this this cancels so I figure it out roughly this becomes gain is equal to minus RD by now this ratio has very interesting features what is the importance of a ratio any parameter which changes the resistance like temperature let us say it has a positive thermal coefficient okay PCR is plus positive but since both are resistances made out of same silicon okay there PCR should be same so if of course in some sense you cannot say ratio will have same you know R plus DX is same not necessarily same RS because it is depend on the value itself also but to some great extent we say it will be always temperature independent it is independent of all transistor parameters nowhere W by L has appeared nowhere beta dash appeared music of I am not even looking at it okay is that okay to you so in essentially now I have a gain of an amplifier which is controllable by me independent of the device okay. So this fact has to be remembered that whenever you are looking for a good amplifier which is temperature independent or device parameter independent you should use but what is the limitation I got out of this typically RD can be few kilo ohms 120 kilo ohm 30 kilo ohms because larger I cannot put that silicon it will eat 10 will this may be 1 to 5 kilo ohms so what is the kind of gain I am looking for is 1020 30 so for all the advantages I got I saw that my actual gains are come down heavy so at the cost of living gains I have now achieved constant see is that okay so if your requirement in a circuit is that at any cost I want gain should not change for anything okay whether it is sensor based device which is in some environment audio systems where back noise is too high you prefer that gain should be independent of all other other inputs okay in that case this may help but the idea is gain is smaller so do not put this as a stage in which 1000 above gains are expected it never gives that okay the first amplifier which we discussed was common source amplifier and that is the word degeneration do you get the point why it is called degeneration the gain has degenerated from its high value GMR this lower value now and make it as constant as possible and therefore it is called source degeneration okay and that is the one of the major stabilizing factor in any amplifier is that point clear so anytime I put a resistance in source I am stabilizing something okay now this issue will use it in the feedback systems where actually we will see does this gives you a feedback it does essentially it is the first feature of a feedback it is a series feedback and we will see this stabilizes things okay in stability criteria the poles will move away okay and then we say system is becoming more and more stable this is the feature we want to now bring to you that why RS in series okay of source is that clear so these are we have learned all these but start looking from the design perspective I am given this what should I choose and that is how I am trying to explain okay so second amplifier of my choice stay common gate stage amplifier now this amplifier we already looked into somewhere where did we look into it in cascode the upper transistor was acting like a common gate so let us do it as this now this is first time I am trying to show you if I use other people's like boys or Paul Gray's book they sometimes show a very different kinds of equivalent circuit okay so I just thought I should copy from them and show this is also one way of putting the advantage of this circuit is it actually mimics the circuit drawn by normally our equivalent circuit looks like horizontal so we do not mimic the kind of circuit we are drawn the way they feel it and that is why I showed you that this circuit actually mimics the way you draw okay is that okay point nothing great it is only to show you that it looks like this so here why I am showing you this because in layouts when I do chip layouts for the transistor everything I actually mimic this and therefore if I have this I have much easier to lay out okay so I just showed you why such trade circuits are many times shown simply because the that is the way I will actually implement on chip during layouts so just for the heck of it nothing very great okay here is a this amplifier whose gate is grounded grounded yes just draw it and of course as I say I will go back to my standard that is equivalent because much easier to solve with that kind of thing but I just thought I should show you in some books the equivalent circuit could be shown in this fashion this includes all capacitances this I in RS is essentially like a V in RS is that correct equivalent it is a it is a if you converted from thevenin to Norton or Norton to thevenin I in parallel RS is equal to equivalent in voltage is V in series to RS some other RS not same series resistors gate is AC grounded AC grounded DC value you have to put for bias okay you have to keep it in saturation but for AC I think I have made it clear that dot dot is AC ground okay so if I now put if you are drawn please do yeah that is what I said this is a current source no no no which is DC there is no DC so it is all AC how that is just to show bias that is this these are current input current source you have input V in in series RS I can thevenin convert to Norton and say it is I in parallel to of course this RS and that RS is not same okay so right you may be right yes so gate is grounded so look for the lower actually Rada we technique I have gate grounded AC wise there is a potential between gate and source which I call V1 there is a okay so instead of this I used as because as I have used this I used the V in please remember this RS and this RS are not same current source output resistance is very high compared to the series resistance of the voltage source which is comparatively very low 100 ohm 500 ohms okay so do not confuse the same RS okay so this is a current input through series resistance RS this is V1 this is the output current GM1 V1 GMB VBS R0 shunted RD which is RD is brown to the ground okay now I assume IX the current at the output which flows through already is that okay your sign is taken care now but the same IX can now in a circuit it can go only like this is that okay should be if current is moving like this it should move up okay so that means this voltage IX RS is anyway negative is that correct so that sign which we are telling is taken care through the sign of IX see earlier case do you remember I used IX entering now I use IX coming out okay so if I use this equations this have you drawn the circuit equivalent so let us do max which is trivial what is IX actually this is grounded this is V0 so V0 by RD is IX so the first equation is IX is V0 by RD please check it sometimes because you know I do not check it myself when I am writing I just come here and show the slide at times I may miss sign wrongfully so be please bring as he said we will also check again yeah okay essentially I am trying to say current source is still downwards this is minus source to bulk that is exactly is the value which I am looking for but the VSB is always taken like this source to this is minus this is VBS so essentially it is opposite direction it goes down okay what I am showing you as far as sign is concerned I am correct on the sign okay so if I use the please do not write before this if I use this loop input loop ground V1 and this then I say 0 is equal to V in minus IX RS plus V1 is 0 is that okay just take a mesh from gate ground to the ground through the RS okay V in minus IX RS plus V1 is 0 why I want to do this because from this I can write V1 plus V in minus V0 by I replace this IX by RS with 0 I also see V0 just a minute I will come back this voltage please remember what I am looking for is the drop across this this which mesh I am looking I am going from here to here which loop I am looking I am going like this is that okay same common point now if I substitute this if I use this bin drop across let us say IR0 is the current in R0 so which is IR0 R0 minus V1 must be 0 because that loop is completed to the gate ground is that clear drop across R0 is IR0 times R0 minus V1 must be V0 okay which mean and now what is IR0 IR0 is this current minus these two currents because these three currents must sum up to IX is that correct so this current must be this current minus this current so if I do that I substitute this here I substitute this back here and then figure out the relationship between V0 and keep substituting nothing very great what is our aim to get the game which is V0 by V in all that I have done it I removed all IX terms all V1 terms to get a relationship between Vx and V in and V0 this is that expression okay at any given node the net current must be 0 is that clear Raj at any node net current so whatever Khrushchev law says put science accordingly okay so having done this all analysis I get the expression of gain which is V0 again GM plus GMB into R0 plus 1 R0 plus GM plus GMB R0 RS plus RS plus RD into RD why this is what is RD that the output voltage GM times RD okay now if this expression you see very carefully this term RS and RD are smaller than R0 is that okay RS plus RD are always smaller than R0 so I leave them GM times R0 this is anyway more than GM RS is more than 10 so 10 R0 maybe I add if you wish but even that term I can neglect I neglect one from here so I cancelled here this RD by RS is all that I get so even in common gate I would not really achieved any gain in fact it is it is very close to one only as that common gate amplifier show not voltage gain voltage gain can be ratio of RD and RS but the current gain is how much in common gate what is the current gain always unity source and drain currents are always equally independent of what you do is that clear so it is always unity gain for currents but the output impedances and input impedances can be separate so the gains can be different for different loads is that clear because currents are same so input current drops will be different and output current drops will be different and therefore the gains voltage gains can be higher but current gain always will be unity that is the importance of this however as I say this is not very frequently used so much as a good amplifier what did I talk about common gate as an advantage that it acts like a good current source so what parameter I should figure out for this amplifier output resistance because that is what I am right now looking for so what is the method I should follow for output resistance evaluation short all input sources independent put a output some in source of V out or VX entering current let us say I out or X VX by X or V out by X same method which we apply for anything we should keep using the same techniques okay now if I do R0 what is the method I suggest short V in but what is this case looks like if I short only V in this is like evaluating for a common source amplifier with degenerated RS okay a degenerating source okay so I solve this again and I get same expressions which I did earlier which is R0 into 1 plus GM GMB times RS this is the same analysis which we did earlier its repeats here okay now if I get this R0 what is R out again this R0 shunted by RD okay just write down and then is it larger or smaller anything multiplied by this factor to R0 the output resistance is higher that is what Costco we were trying on the output resistance I want to boost so the common get what did it do boosted the output resistance okay what is the requirement of a good current source the source shunting resistance should be as high as preferably infinite what is the third parameter I should look into gain I have looked into output resistance I which is the third one I should look for an amplifier the input resistance do you believe it will be higher or lower lower that is the fun part in that okay we will calculate so what is it trying to do a transform a circuit which has low output resistance to a higher output resistance okay so it is like a pseudo buffers okay that you are having a connecting at a lower input resistance this but putting to the higher output for the next stage okay so that the next stage does not get loads get loaded by this so this is interesting but of course I repeat if RD is smaller the output resistance will not be very high it may be a few kilo ohms 10s of kilo ohms 20 ohm 20 kilo ohm 30 kilo ohms okay but RD can be smaller also or larger than RO also if I use current sources okay is that okay expressions are everyone okay so same expression same statements if R0 is greater than RD less than RD R out is either R0 or RD now the next parameter of interest to me is the input impedance okay input impedance is seen from the input side which is the if IX is the current here and Vx is the voltage I actually calculate conductance IX by Vx IX already I can find from here GM Vx GMB Vx minus Vx by R0 okay which is GM plus GMB by 1 upon R0 since let us call this GM dash so if GM dash R0 is greater than 1 one can see this value is larger than 1 this is by divide by R0 so RN is effectively very low many because you know RS is external again this has to be always understood this is whenever we calculate any input or output resistances or impedances it is seen in the device output from the drain side if it is a output at the drain then seen from the drain side input always seen from the gate or source side but the additional resistances will actually modify that value corresponding but if this is very low even if RS it will further reduce it down okay so it does not really matter so what is the advantage of common gate amplifier that input impedance is very low impedance word is not clear because there are no capacitance as used but I will come back to it this is the issue where in frequency response that capacitance when they come which will have a dominant pole may decide from whether R is lower or higher because 1 by RC so this understanding that R is lower or more may help to even see oh this may be dominant is that clear so observation based on what we are evaluated is that okay so I keep I just now said the current gain in the case of this is always I X by I X therefore unity I 0 by N now one interesting feature which I did not say this was all under assumption of low frequency set equivalent circuits it will remain unity for a very large frequency and our thinking was that even up to FT it should remain one but in real life it does not remain one FT lefty okay so maybe some other day or some other time what frequencies actually it will not be one we will see that but as of now you can say for almost all frequencies the current gain is almost and this you must put in quotes almost not necessarily all times okay. So in a CG amplifier R in is low but R out is high or to say we can convert a normal current source into a decent or great current source which is also voltage controllable okay because if you change V in current value changes so it is good VCCS VCCS now let us do the last amplifier among them never is last but last but one you may say is that okay this part anyone but this is what you should last part is important for us the last amplifier of my interest is common drain okay which is popularly known as source follower there is little different between common emitter follower and source follower bipolar source followers are different in some sense to the MOS source followers or sorry common emitter emitter followers in BJT are not identical to common source followers of MOS transistor how much they differ and why do they first live it to you okay what is source follower what came from it was found that if V in and VGS of this transistor is such that V in means V capital V in so that VGA VGS plus input signal if that exceeds threshold voltage the transistor is on okay and once that transistor is on the current is flowing across through the RS and drop across RS is the output voltage is that correct the word source followers so output current is proportional to input and that current flows through RS so output voltage is directly proportional to input so source follows gate or inputs however one can say common drain because drain in this case is grounded there is no RD there since drain is grounded so all voltage are referenced from the drain point that is ground point so therefore it is also called common drain the drain is grounded means reference voltage has gone to the drain okay so everything is measured from drain ground potential therefore it was also named common drain amplifier. If you are looking for all capacitive mode equivalent circuit as shown here gate to source CGS gate to drain CGD CDB plus external load is CL GM VGS is this is the VGS across CGS so GM VGS is the current source GMB VSV is the another current source back by as if it is R0 is the output resistance please remember RS is the load across which V0 has been picked up okay and drain is grounded we will remove this capacitance next time before we start the differential amplifier I will like to maybe if time term no not will permit today I will see that whenever I have a series component in an amplifier which is connecting output to input I can put into two parts input side output side using Miller's theorem when is Miller theorem valid that is very important you cannot use Miller theorem randomly they are limitations or there are compulsions with which Miller theorem is valid because that will use again in the case of defam designs so we will like to show you what is the limitations in Miller's theorem okay is that equivalent circuit clear so first thing what will do is as usual remove all capacitances and solve for low frequency I am going to do one or two frequency response for me all these amplifiers one time okay and then the same procedure can be and one of the technique which I taught this batch of in second year which is 0 time constant value or open circuit 0 constant systems I will be able to show you which poles are dominant okay so you do not have to evaluate all of them you can figure it out which is the dominant pole which decides the band because once I know the bandwidth at damn care much of the otherwise okay so I must know my bandwidth so which need not solve all of it all the time so typical low frequency circuit is remove all capacitances put an equivalent circuit V and input voltage at the gate same gate to source voltage V1 this V1 word as I say I am using it from rather his book if you want to put some other VGS value or something is fine I know I repeat I am saying this I am using from I like this rather his techniques why I normally I have been using his terms this is my source drain is grounded now in this case I assume that R0 is very high and I just thought about it let us say this is my V0 the current in RS is V0 by RS there is nothing there is nothing else if this is my V0 the V0 by RS is my current in RS since this is downwards this voltage with reference to will become now minus V0 and then I substitute as usual all the terms V0 by RS is the current which is GMV1 GMV VSV substitute them again here and solve okay you that writing symbols I think what I meant is the current source going down in series in parallel to GMV I think I think I am made a machine writing sometime VBS sometime VSV please keep it as if the direction shown corresponding to that VSV this VSV is minus VBS in real terms so I at time this sign probably gets into wrong mood so please always assume whatever sign I am showing you assuming correct signs I am pushing the current down okay so if I get first equation is V0 1 upon RS plus GMV GMV1 my method like Rada is look for input side this is my input side is that correct this is V in this is V1 this is V0 is that correct this technique is very simple so you get Vn is equal to V0 plus V1 okay or V1 is minus V0 plus Vn okay then this to V1 I substitute here I repeat what is the method first I actually look for these current sources going into this then I look from this side input side I say Vn must be equal to this is that clear this is my Vn this must be equal to this nothing very machine done only voltage here at the same node same voltage must appear so if I write that I get this plus this is equal to this should be is okay this is equal to this plus this both ground is node K over a donor sum must be this same equation I wrote from there V1 is minus V0 plus Vn substitute V1 here I get V0 1 upon RS GMV minus GMV0 GMV1 collect the term for V0 collect the term for Vn and I get V0 1 plus GM GMV RS is equal to GM RS Vn or gain is V0 by Vn which is GM RS divided by 1 plus GM RS plus GM VRS 2 things you must remember denominator is always larger than the numerator marginally larger because the other terms are smaller okay you can say the numerator is marginally smaller than the denominator so the voltage gain of a source follower will be always less than 1 if there is no back gate bias then you may you may say it is very very close to the so what is the importance of back gate bias this is what I want to bring to why I actually use this can you think if there is no back gate bias what will what is the gain will become then even close to 1 because GM RS by 1 plus GM RS is very close to 1 okay but as soon as I add the term GM VRS I am actually increasing the denominator but that means gain will further go down because of back gate bias is that point clear so please remember why spice does not show your initial values because if you are not using the back gate bias your analysis it shows you are slightly away from your actual result so this is why also in important fact you must remember that the gain is positive in the sense the output and input are in phase okay because in that loop current in the same sense from input to output okay and therefore it is always in phase outputs okay therefore it is always plus if you look at the input input impedance of such amplifiers it is almost infinite except if there is no capacitance is open circuit okay if there is a capacitance it is decided by the capacitances typically input capacitance are very low CGS plus CGB these values are very low so unless it is because 1 upon Omega C unless Omega is very high the input impedance will be always very high at very high frequencies it may not be because then 1 upon Omega C may not be that high and then it will start shunting and that will be some kind of a pole it will start coming through so I am not very much worried about input impedance in the case of output impedance what is the technique we use remove the RS remove V in put VX there put IX going through there evaluate VX by IX are grounded V in wherever V in was there that I have grounded because RS is external impedance is always measured in the device at the output I never took already so yeah RS already is that okay load is not my hand is that okay that word was used there because in the source it was there but essentially it is an output load okay so if I solve this simple this current are same currents here mine this you can say sign taken properly then R0 is 1 upon GM plus GMB what does this trying to say R0 is higher or lower 1 upon GM plus GMB GMB is same as GM 1.6 so 1.6 GM GM get now that roughly millimodes millisiemens so R0 get now a few kilo ohms are lower because if number is little more millimodes then it will be even lower okay so the output resistance of a emitter follower is very low or lower and input impedances or input resistance is very high so where do you think it can be used in a buffer stage where input impedance you want to be infinite or very high and you want to match the load which is a lower value therefore output impedance should be matchable to that what is why it should be matchable what is the purpose of match loads power transfer is maximum okay so I want to put equivalent there so that I get maximum power transfers is that okay I had drawn this is very trivial nothing very serious anyone who feels that I had done machine something please go into Razavi's book and correct your signs or otherwise though I have not copied from Razavi but since I have taught Razavi for 10 years now I almost know the way I must have done okay so since I do not keep books anytime so there may be a sign in case you feel it please look into Razavi's book this is their technique so it must be correctly there quickly I show you and I will finish this this is something which you now remember and note where do you want to use source for followers the first thing I just now said if you need a system in which high input impedance and low output impedance then you should use buffers and source followers as buffers now here is an issue here is a designer problem. What was the first part circuit issue circuit analysis what did we learn we learned the problems now since VSB is related to V0 is that clear VSB is related to V0 VSB also changes VT you know VT is equal to VT0 to 5 plus VSB to the power half minus to 5 that means if V0 changes VT changes VT changes EM changes so VT is now a function of output through this body bias my VT changes the current will change current changes or V0 may change either remain which means EM may change the worst may come if this series resistance RS is even low think of it why I say not tomorrow I will tell Friday if RS is lower the effects are even worse so at least reasonable RS so source followers may not be that bad but if the RS values are too small then it will create this ideas variation or VOV variation very strongly okay and then all purpose of good buffer may not be actually seen this is first issue which common source source for the second problem which I see in buffers is you know at the end of please remember input signal may be small but normally what is the purpose of an amplifier output should be larger that is why gain we call it is not the problem with this source followers are that their output and input swing both gets lower example I give some values let us say I have an amplifier which I am using with VDD 1.2 volt thresholds of 0.4 volt and over voltage these are not very accurate value but roughly okay so we see the headroom now available to me is 0.4 plus 0.2 which is 0.6 which is almost half of the power supply that means just to turn on device on I have now so much voltage lost by me is that clear so I have very small swings now available for output or input is that point clear VGS minus VT keep device on below this you cannot turn on above this that means if this value is 0.6 or even more in some cases then the available swing to you is very small is that correct so only when a very very small signal amplifications you are doing this will be fine otherwise this may create a problem so source followers should be always thought of very low signal amplifiers however low power low voltage applications and you need many times in biomedical application larger swings for ECG for measurement for example such amplifier may not be used because the swings are very small so accuracy of measurement becomes very small so these are amplifiers which should not be used when you are looking for higher swings okay so where do you use them we use them we use them as small swing applications are in the case of RF for example front end which is low noise amplifiers or in a normal other not necessary RF or other amplifier systems the first stage amplifier which is called preamplifier you can have source the reason behind the input may not come from very high impedances particularly if like in case of RF system and antenna has 50 ohm or 75 balloons okay now you are inputting that signal okay which has that very low impedance so the first amplifier which is going to amplify signal and the signal is very low you are pushing it higher at that time this may be good for you is that point clear to you so do not think that these are not useful but do not arbitrarily put any time source follower thinking what is in that gain the one and no great problems okay it does create problems in actual designs the last two applications will not explain maybe I can but you ask then only the other application of a source follower is as a level shifter what is level shifter the other level Rajs in amplifier name what is level DC point so the output Q point that is DC value is lower than input DC point why VGS minus VT like again I repeat what I am saying if I want this amplifier to work this value a Vov plus VT is VGS so the voltage is voltage say at least Vov plus VT DC may is that clear it now voltage as a level shifter we can use this because the input level to output level can be brought how much shift I can do Vov plus VT not very large but at least it can go down you want further to put another stage for that okay you will reduce further then the last part it can be used as a load it can be lose used as a load if I am an amplifier say the n channel device this is my gate this is my this this is like a source follower okay and this can do a good job for you to give our RO high you are looking for RO high enough so this can give you good RO okay this are some techniques so this is why I brought this to your notice when to use what is most important in designs so source followers are only used for level shifting for LNA preamp and for finally as load device do not use that as an amplifier of any other use okay it has no good features we are not talk of frequency response it may have further impacts on them when I go frequency response I will explain more which one is even worse okay among them okay is that okay thank you for the day.