 We were looking for all these days, the difference amplifier, we have looked into the response, we also looked into feedback. The first and the foremost circuit, integrated circuit which is available in market for use of amplification is called operational amplifier and a typical circuit which shown here, of course the normal op-amp has three stages but in designing we also have what we call two stage op-amp, essentially a two stage op-amp is a op-amp without buffers or the output circuits. So there are two stages, the first stage of course is the very high gain stage because of the defam followed by high swing relatively lower gain stage which is the second stage of the op-amp and third of course is the buffer stage which drives the output, typical number of transistors required to create an op-amp is typically 22 and may extend a bit here or there up to 30, so it is not a small one transistor circuit, it has a large number of transistors even for two stage op-amp the number of transistor shown to you here are of course the transistor related to ISS has not been shown which may be more than two that is three one coming from mirrors for example, so you keep adding such transistor you find as many as 30 transistors or at least 22 transistors are required to create an operational amplifier. The two stage essentially says you have a defam, there are two kinds of two stage op-amp one is the two outputs Vo1, Vo2 the other is this is called double ended output and the second part which is shown here on the right is called single ended two stage op-amp, single ended means output is only one we out okay. Now one can see from here of course there can be different loads okay it can be mirrored loads or it can be constant current sources as shown here different loads can be given to defam as we did earlier, so there is nothing big about that right now I assume they are current sources okay with a whenever there is a current source what is the output resistance of that transistor always R0 okay in which case it is not R0 or R0 parallel something like diode connection shown in the next stage then it will be juan upon GM parallel R0s okay whenever there is a diode connection we have proved end times that whenever there is a diode connections you will get one upon GM shunting R0 and essentially it may be one upon GM itself okay. So here is as I say the output of the first two defam two outputs of a defam are given to a two amplifiers one on the left one on the right what kind of amplifier do you see this this is a P channel device so input is P channel input device is not N channel which is receiving input okay and what is the load for that for example for M5 M7 is acting like a load which can be a current mirrored load or any other load which you can create that clear now this means which kind of amplifier it will be single stage sources here so where common source amplifier this is a normal common source amplifier with P driver is that correct with P driver so not much different between analysis of N channel drivers or P channel drivers is that correct in CMOS normally what we do we have N channel drivers with a P channel loads here we have a P channel driver with N channel loads okay. So whatever gain of defam see VO1 will be proportional to VINs okay which is then given to the input of the next common source stage which will have some gain A2 okay so what will be the net gain at the VO2 or VO1 VO1 VO2 if A1 is the gain of defam and A2 is the gain of the output stages then the net gain is A1 into A2 that is what we essentially do in a two-stage open. So will this gain be then much higher what is typically gain of a defam do you expect what is typical gain of a defam GM times RO1 parallel RO2 or RO3 parallel RO4 ROs are typically order of mega ohms is that correct typically order of mega ohms GMs will be order of at least 10 to power minus 3 or mine in between minus 3 minus 4 amps per volt okay. So typical value will get gains of around 1000 or 5000 kind of things can be achieved through only defam is that correct the gains of a defam can be as high as 10 to or even higher sometimes depend on the GM value you choose RO is value on what these depend on GM and ROs both it is depend on the ISS is that correct so choose your ISS to get the possible gain values okay. Now once you decide the defam has this much gain and you have another gain stage which may have again GM times parallel RO of these okay and you adjust the if the currents are same for all the transistors one can still get some amount of gains may be 10 to 100. So typical opamp may have a gain of how much 10 to power 4 to 10 to power 5 typical opamp will have gain stage gain outputs of the order of 10 to power 4 or 10 to power 5 okay maybe and best design may be even higher but it may have some other problems if we gain too much gains on that so is that point clear so opamps are no different from amplifiers which we already designed already seen how they are solved okay this is the second part of this circuit is this is called single ended output and what kind of loads we are using here mirror loads mirror loads ISS bias current to beta dash W bi into ISS under root of that is GM one upon lambda ISS is ROs of that so one can evaluate what GM remember IS increase and IS decrease will have opposite effect on GM and RO so you may get reasonable gain if you still want higher make sizes double increase W bi okay that is the way gains are adjusted okay I will give expression I have written down those expression but before those I just thought what I am doing I showed you this is my standard defam with the current mirror loads this is my input between 1 and 2 terminal is my win this may be coming from the current source this may be your standard M5 or M7 number earlier we give and this is your common source output stage okay please remember again this has a P channel device which is receiving signal output of the first defam is given as the input to the common source amplifier with P channel drivers and N channel loads N channel loads so if we now calculate here are some expression which you do not have to think too much I just wrote down without either you were seeing for the defam for these defam first VO1 and VO2 will be GM1 RO1 parallel RO3 into be in VO2 will be minus GM2 RO2 parallel RO2 into be in if GMs are equal ROs are equal they will be correspondingly GM RO terms will appear then what is V out V out is the output of V out 1 is the output of VO1 as the input and that is GM of 5 into ROs parallel combination of 5 and 7 so that is what I did GM5 RO5 into RO7 parallel RO7 into VO1 substitute VO1 from here so you get GM1 GM5 RO5 parallel RO7 RO1 parallel RO3 into V in so the gain is V out 1 by N by same it will be V out 2 is to be same if GMs are same V out by V in so these are the 2 voltage gains please remember GMs are getting multiplied so our R0s are getting multiplied so essentially A1 A2 the 2 gains we are multiplying is that correct this into this into this and this into this is A1 A2 is that clear GM5 into this is A2 gain GM1 into RO1 parallel RO3 is the A1 gain so A1 into A2 is the net gain of a 2 stage of fan is that clear first stage of the defam as a gain stage of AM A1 which is GM1 RO1 parallel RO3 the second stage is a gain GM5 into RO5 RO7 so that is multiplied to the first stage gain so A1 into A2 is a final is that okay so without even evaluating all of it as we did equivalently circuit we can directly write these explain why how can we write these directly because we have derived all of them individually every case so in future we do not have to keep deriving everything we say okay for this this is what it is but to if someone said derive AB initial please go back right equivalent circuit and solve all of it okay but generally I mean for solving these we do not need that in the case of single ended what will change nothing much will change GM5 just look at this the output will be GM5 times parallel combination of 5 RO the 5 and 6 and into whatever output coming here which is nothing but same as what we did earlier GM1 into RO2 parallel RO4 into GM2 RO it is still the same only thing is it is single ended on the one side and what is the advantage there advantage there everything can be mirrored now from everywhere we do not have to put any current sources directly is that correct because mirror why I say mirrors are easier to create because once you create a mirror source current source you can always mirror any number of places how do I pass on that gate of that should get to the gate of that with a diode connection on the one side so it will just replicate the currents the next stage and if you change the size you can increase the double or triple or any number of amount of current can be pushed to the other side so it is much easier to control the currents and therefore the outputs by just using the good current source as my one current source I create and that I mirror everywhere wherever I want additional currents is that clear now the only thing is if P channel mirror has to be what should I do I last time given example this from N channel first you will have go to a P channel the output of a second stage should be a P channel load mirror which then can be connected to the next P channels is that point clear to you there was an example given to you first create an N channel mirror then the second part transistors this output or drain should be taken to the train of P channel which is mirrored to the next stage by again connecting source and sorry drain and gate keep then going to P channels and if you want back N channel what do you do bring that to the next end stage at the output and mirror it again for the end so how do we go from N to P transfer P to N transfer so any number of times any stages I can keep transferring the current to N channel to P channel which may be ratio of W by L which I can create is that clear that is the trick which we all follow so we do not actually use current sources normally why we do not do it because internally putting another bias is difficult you know VB you said normally what do how do we get that current source I need some bias potentials here this may not be same as we in that case I may have to put too many power voltage sources inside that means connection from the pin so many pin connections should get in for putting voltage of current sources creation so we normally put one VDD create a current source okay and then replicate everywhere whatever bias currents you need is that clear to you this is a trick we follow because otherwise on a board we do not mind there are power supply though power supply laying a 4 pins layer must a connector but in chip what is the importance of a chip compared to board what I said so many transistor I want to put in a smaller area that is the integration I am looking if I put additional voltage line somewhere I will have to leave some space for it that means I am not using that space for transistor so the trick and integrate circuit is avoid any extra lines if you can if you do not I mean if you need it even nothing else can be done is that clear so this is just to give a out why I normally use current sources and not so much normally current loads are this is just to that but in reality on a board I can put all kinds of loads resistive loads current source diode can I all can connect physically anything and get any outputs so the theory was shown to you what purpose what are the advantages advantages of any kinds and if it is a discrete device you can use most of it okay but in interior circuit the first major thing we do is to avoid areas giving not doing any silica any transistor there if transistor area is taken by something else as I forget it that is what I see the slightly different from normal discrete transistor circuits no difference basically in performance only thing why do we some things okay so yes no but on the like in this case I want the current to be taken from somewhere where from I get P channel currents so the gate of this must come from a P channel side if I want a current on this be fixed by me let us say I want so I have this P channel gate cannot be connected to N channel gates so they must be connected to P channel gates now P channel gate must be mirroring from other P channel device whose drain current must be the drain current of N channel mirror okay so you come from you create only N channel mirror but pass on to P channel on the top and then move to P channel when you want again N channel come down create to N channel driver go back keep doing any number of times so how much I therefore N channel P channel N channel P channel is that clear flip as I joined okay so this essentially what I was trying to say I may not do details about this open basic entire internal circuit as I said the three stage op-amp which is the real op-amp has the gain stage the single common source gain stage high gain stage common source gain stage followed by another circuit which is called buffer which is essentially for large driving of the output loads to dry loads larger loads we need larger currents and since we need larger current buffers are ones which provide larger currents is that correct buffer provides if you want larger current what will be there W by L's of transistors for larger currents larger or smaller if I want larger current transistor sizes should be larger or smaller larger larger W by L and larger current will that lead to smaller output resistance of that transistor 1 upon I is that clear so the larger the current R0 becomes smaller is that point clear so all op-amps are very low output resistance because of the buffer stages we keep there okay. However if you look at the input resistance can you tell me the input resistance of this op-amp is higher or lower either of them please look at it there is a gate connection which is insulated from source drain so the input resistance of a mass transistor is practically infinite but at least 10 to power 100s of mega ohms or at least 10s of mega ohms and above is that clear let us say it is a bipolar transistor will still be on I will be higher their base collector base emitter junction is connected there so it is still higher yeah it will be very relatively high why the resistance of source which will be RE RE or whatever we call that is emitter resistor how much impedance it will see at the input side on the base beta plus 1 times RE RE itself is 10 to power 5 or 6 okay so 100 times that so in either case op-amp has large input resistances is that point clear whether it is bipolar or mass intrinsically gives because a mass but even in bipolar because of RE your input resistance is extremely high and that is the property of op-amp what is the property of op-amp the input resistance of an op-amp is very very high output resistance relatively is very small it is not RO of the transistor it is the output resistance seen is very low compared to any other resistor on the chip is that clear this is the feature of an op-amp what is the feature of a op-amp large gains large input resistance and small output resistance is that correct this is the property of a op-amp okay just for the heck of it there is a op-amp which is also is an op-amp but it is not called op-amp it is called operational transistor trans conductance amplifier called OTA there are so Joe he is acting on the operational abhi a normally voltage op-amps are voltage amplifiers so there is a operational trans conductance amplifier are very popularly named as OTA okay so please remember OTAs are normally used in what we call GNC filters GM means trans conductance sees the capacitance these are continuous filters which are called what is continuous what is the other kind of filters could be digital filters okay which are not continuous filters so GMC filters are essentially continuous analog filters which can be created by a few capacitances and N OTA okay we will see that little later so is the issue clear what is the advantage of op-amp over normal amplifiers normal amplifiers unless it is emitter follower it is input impedance is very low how much as low as few kilo ohms okay H I H IB or RE is hardly QI by KT okay so that resistance is KT by QI so that is very small so two kilo ohms okay and now few kilo ohms is not what we are looking for okay so normal amplifier to op-amp his first thing we achieved is very very high input resistance and comparatively much lower output resistance is that clear that is the feature and the gain is very very high 10 to power 4 or 10 to power 5 will be the only thing is this gain should be called open loop gain what does that mean no feedback is right now available so from output to input there is no feedback connections so the open loop gain is 10 to power 4 5 may be higher than 5 sometimes is that these are the features which we are looking into a operational amplifier okay some of the characteristics which you all have done in the lab hopefully I just list them the few things which is of interest in knowing the characteristic of op-amp is the first and the foremost worry is the input offset voltage okay now what is the term we are talking so if you have an op-amp maybe which is a symbolize something like this plus minus V in 1 V in 2 are given names V in this so if you have V in 1 and V in 2 and this is V 0 we say in defam first stage if both inputs are 0 output should be 0 because no current output should be 0 is that clear an output of an op-amp is if GM is 0 output is 0 currents are 0 there but if we actually see inputs 0 and still we see some output in real life what happens that we thought that when V in 1 is made V into equal to 0 output should have gone to 0 but in real life if I monitor even if I say some voltage at the output okay that means the V 0 V in characteristics is something like this even when the V in is 0 here there is an output voltage is that correct there is an output voltage so that the difference between when the V in is 0 or V in is sorry V out is 0 to where V in is 0 this difference is called input offset voltage where should have been normally it should have crossed 0 it should have crossed from 0 it did not cross it crossed on the left so it required it when I went to V in I found some outputs okay. Now this occurs essentially we say this occurs because of the mismatch of the transistors okay there what can be mismatches there can be size mismatch there can be what else size is W and L mismatch what else can VT scan mismatch okay so if there are mismatches in the two transistors M1 and M2 or M3 and M4 then ideally we were expecting difference between them to be 0 which may not occur because one of them is not giving same outputs is that correct so there is some finite output potential you may see so how do I get rid of that let us say it is plus and minus if there is a offset here I should put in this or this plus some opposite offset voltage here which will actually cancel the other terminal offset on that is that clear let us say offset is plus 3 10 millivolts so I apply minus 10 millivolts at the input so 10-10 is 0 so always see in your open there is additional terminal given to you for offset cancellations all that we do to put a resist this resistive network with a power supply keep varying that part so that offset goes to output goes to 0 is that correct V plus you put a divider and change the voltage or change the resistance okay see that when V0 goes to output is 0 we say offset is cancelled but in real life offset may be present of course we will not look into this there are two kinds of offset which can be two kinds any random this is one of them can be random offset okay the other can be systematic offset so it is much easier to cancel systematic offset because what is I am trying to say oh every time it moves there I will actually put minus there as if it comes to 0 okay but sometimes goes there and sometimes goes there or do I do it so random offset which is essentially noise noise related they cannot be cancelled but systematic offsets can always be cancelled is that correct is that point clear so one problem which you can see why I shoot this the worst case offset could be VDD by the gain or VSS by of the gain because that is the highest voltage you know output divided by gain is the input so the plus or minus offset could be achieved the highest will be VDD will be maximum output of VSS minus VSS will be maximum lower side in output so divided by the gain is the possible maximum offsets possible is that correct this is the maximum possible outputs it can swing divided by gain will be the maximum minimum offset voltages cannot be because output cannot rise beyond VDD so VDD by AOL is the highest offset you can get what are reasons because you are now saturating there is no AOL beyond that AOL does not exist in fact well is 0 then once we did it it is AOL does not exist if at this point there is no gain okay so the maximum possible is VDD or VSS okay the second term of interest second worry of all of us is to get CMRR as high as possible which is 20 log AD difference gain by common mode gain if your mismatch occurs which is the first cause will also lead to CMRR because then ACM will not be 0 but will be some finite quantity that is GMs are not equal actually if you see the term it is GM 1- GM 2 occurs so if they are equal so it becomes 0 if they are not equal then there will be finite quantity will occur and ACM will be present then CMRR will be there which is typically what is the minimum CMRR I suggested you 80 dB there are opens I think some of the 725 723 or these opens may have a CMRR of 70 dB but they are gaining something else look at that some sheets and you see why they reduce the CMRR okay all 741s are always greater than 80 dB and typically the highest of 506 AD 506 will have a CMRR of around 120 dBs a third interesting parameter for this power dissipation please remember last time when I calculate there may be some this I did not show you correctly power dissipation is something like this VDD-VSS the maximum voltage available to you is that correct so when the current goes from here to VSS the maximum voltage swing is VDD-VSS but there are too many currents flowing one is of course this other is because of this then there is a mirror currents okay all vertical branches current going from VDD to VSS are parallel is that correct VDD to ground VDD to ground VDD so add all those currents because they are supplied by same VDD and received at the same VSS so the power dissipation of an opamp is all vertical path currents multiplied by VDD-VSS that is the power dissipation of an opamp is that point clear current in this path current in this path of course this together is in this path this path and mirror whatever any other hardware which is like buffer output to ground any current path going from power supply to the ground that current should be added to all other currents parallely going through because they are drawn from the same power supply reach to same VSS so VDD-VSS this is plus of course because it will add minus of minus VSS so let us say 2.5-2.5 means 5 4 multiplied by the net currents is the power dissipation of the opamp circuit is that correct the fourth quantity of interest is called PSRR now this is also very important there is a lot of research can be done on this but let us see only as a characteristic now this is one of the major design parameters for opamp but right now we will not look to in today's let us say the VDD and VSS are not constant which can occur why it can occur do you know why VDD may not remain VDD 5 volts or 2.5 volts it may become 2.5 plus minus something same way VSS can become plus minus something okay essentially say some noise is overriding on them and we know VDD will decide the current or VSS will decide the current DC currents so what it will change then all AC parameters are affected by the DC currents bias currents is that correct GM is proportional to root I lambda is 1 upon lambda is RO so you see any parameter any DC thing change the AC parameters are varying therefore gains may be that correct so we want how much should be how much we should tolerate the variation in power supplies this so that gain remains within the acceptable limits why we say acceptable limits let us say I am getting a gain of 1000 so I say if it is 995 does not matter if it is 1005 it does not matter but let us say I have a gain of 10 then I cannot say gain of 5 15 or same as 10 is that clear so the amount of tolerance at the gain stage you are outputs gain you are looking for within that way the variation in power supply be allowed is that correct as long as you are within that you may say it is fine okay this is called so what we are expecting how much is that tolerance allowed so that the output stage gain is not very much varying acceptable limit so we define here is the open loop gain divided by V out plus V plus because of this ratio whatever is the change occurring is decided by power supplier R R plus Y plus because it is on the VDD by same logic PSR or minus is a well divide you can see what I am what is this essentially I am talking variation in output due to variation in VDD or VSS these values are essentially called power supply the how do you really remove this much of the PSR are problems you are done is already the lab experiment on this if I want to see that VDD does not change very much but it may officially I do not want it but may occur then how do I remove that what do I do at the output you must have seen and if you are not simply sit now this is your VDD terminal I actually put a capacitor there is that correct which filters out those frequency components which changes the VDD is that correct have you seen any time or you are never connected all power supplies should be connected through a bypass capacitors to minimize the variation in VDD and VSS which will improve your PSR or values okay is that clear so we do not really do great things in VSS are also internally internally we can do lot of things but for external people also still you have a choice to put a capacitor but what is the problem with capacitor because it is only some frequency it will remove is that correct because if low pass filters so it will cut some of them okay depends on different frequency which you are going to get you may require n capacitor which will be very difficult to put so it is not that so 50 hertz certainly hum which is called will be easily removed maybe 100 hertz can be removed few kilohertz can but if it is a mega noise nothing can it will pass in fact so that is the problem in so PSR is not just for these it can also any other noise and therefore internally we do design to improve PSR is that correct so do not think that a capacitor is a sub problems all over okay now the next value for us which we already discuss is the slew rate what is the slew rate definition we give last time that how fast the output response to input okay so it is DV by DT is the rise of output to its maximum value which is essentially I said last time it is is divided by the load capacitance okay so this is also last time I already given you example in which showing one of the way the current ISS is fixed is requirement from slew rate requirement from power and requirement from the gain so all three should somehow get matched to everyone's requirement okay so this is essentially there are few more characteristics but these are essential characteristics which all must be knowing prior to we start working on opam in a lab okay now here is something the first though we are in our course this is but just to get to what he has taught and little differently from what he has taught you know why I say because I always want to show you little different from what others should be okay here is something what others this is a opam my ideal assumptions are always are very high because of this is that basically I am looking for MOSFET opam there is a input source with as an R1 as the series resistance there from the I define for opam two terminals one called inverting terminal the other is called non-inverting terminal it has nothing to do with minus and plus per se the idea is if I put input here the output gets inverted to the input if I put input here output does not get inverted and therefore this names are given minus and plus okay they are essentially V in one and V in two of a different okay in most non-inverting amplifier inverting on this is inverting amplifier the V plus terminal is grounded okay which means V in now occurs between V minus and V plus okay. So here is the ground from the V minus terminal I connect a resistance R2 to 3 I will not solve this problem directly but can you think which kind of feedback circuit is this output code V0 0 if there is no R2 me what does that mean what is being sampled voltage has been sampled so what is the kind of voltage sampling shunt sampling is that correct at this note what is being summed up essentially currents so what is the sampling mixing shunt mixing so this is shunt-shunt amplifier is that correct okay shunt-shunt amplifier me gains kasey bantah so this is the trick I did not want to show you but I thought you should see say feedback circuit setting right here okay this is no connection up please you either Galtis a connection the crowd will not be okay so equivalent circuit of a this is equivalent circuit of a op-am is something like this R I a current source if you wish across there is a resistance R 0 and this is your R is tends to infinity in case of Vgs in case of most on the stairs this is gm Vgs this source this current going down okay this is your drain this is your gate this is your source this is equivalent of that is that correct so we are saying gm V1 which is Vgs for us is equal to gm V1 shunted by R0 which is the output resistance of the amplifier and a feedback resistance of R2 connecting between gate and drain equivalently so if I solve this V in – V this is V1 voltage V in – V1 by R1 is this current IR1 is that correct there is no current here so where can current go only in R2 so this current R2 is essentially V1 – V0 divided by R2 so V1 – V0 by R2 so V in 1 by R1 plus V0 by R2 is Vi sorry V in V1 sorry V1 1 upon R1 plus 1 upon R2 is that correct it is because it will open key I have a kitchen a current or a current same Hona chair because there is no current here okay so whatever current is coming must be going through R2 so IR2 must be same as IR1 because what is the condition I put R I is infinite if there is a finite R I some current can flow okay then we may have to do little more calculations okay from the KCL you can see from here this equivalent the current V1 – V0 is going where assuming R0 right now is infinite for the sake of simplicity where this current will be going what is that equivalent current current GM V1 you look oh yeah I say sure who are current how path is go as such okay yeah path yeah current or a current same Hona chair a key path a loop may a key current Hona a loop may yeah I say sure Hona Hona current ground may as such that correct so GM V1 must be sorry GM V1 must be equal to V1 – V0 by R2 so I collect V1 terms V1 is into 1 upon R2 – GM sorry this cancer – GM is equal to V0 by R2 from here I calculate V1 why I am showing you all this I am trying to prove that that gain larger gain requirement you are I don't normally get an R2 by R1 – R2 by R1 but that I am okay only kiss condition may validate a condition up to the V1 is V0 by R2 1 upon R2 – GM was a pilot equation top car is equation may is V1 ke jagab value that do not remain a nickel so if I substitute Vn by R1 plus V0 by R2 or a V1 a term a into 1 upon R1 plus 1 upon R2 start collecting terms so I get V0 by Vn is – R2 by R1 or a though numerator or denominator pay though buddy terms are a 1 – 1 upon GM R2 and 1 divided by 1 plus 1 upon GM R2 we normally say in open the voltage gain is – R2 by R1 that is what we have been telling people earlier but I now derived the expression which is actually this R2 gain is this expression when this is equal to this only when GM is tending to infinity is that correct tending to infinity what does that GM tending to infinite means what what is the gain we are which gain we are talking tending to infinity open loop gain so if the open loop gain is very very high then GM times that value will be very high GMs will be very high and if that is very high these two terms can be 1 by 1 and then the gain is R2 by R1 so the condition under which this output voltage by input voltage is – R2 by R1 only exist if a value is very very high and what is the AOL I said you typically of a open stage 10 to power 5 is that correct 10 to power 5 since the AOL is very very high we can then assume that the output voltage is just – R2 by R1 times the input I will show you without doing this also we solve the circuit and we get it but actually this is the analysis which we should look under what condition I can assume that R2 by R1 is the gain function okay this is very important in understanding not that this is always be valid in the case of op-amps okay but to prove that my case is correct I showed you this is how I explained that things will happen automatically okay here is another parameter of interest to us is the output resistance how do I calculate output resistance of a network short input at the output apply a voltage source and the current starting from there Vx by Ix is the output resistance this is my R out okay. So I say Abhi have network Abhi sub is that okay the current Ix and again as of now I will say R0 is very high and I am neglecting current through R0 okay I can put everything there circuit is divided into one path is this and one path is this is that correct yes so Ix is GM Vgs okay and V1 is Vgs as I just now said so Ix is GM Vgs plus Vx please remember Vx minus 0 why it is 0. So Vx minus 0 divided by R1 plus R2 though resistance is a voltage minus a voltage divided by the resistance so Vx minus 0 upon R1 plus R2 is this current GM Vgs is this current so Ix is GM Vgs Vx minus 0 R1 plus R2 is that okay current divide a current either Gaia a current inner Gaia some of those two currents from here I calculate Vgs value which is R2 upon R1 plus R2 times Vx so I saw Ix I get Ix and from there I can get Vx by Ix so essentially what I am saying that will be also R1 I will come back to correct value but just see what value I am saying R1 plus R2 upon 1 plus GM R2 is the expression I got for R out is that correct from substituting I get Vx and Ix relationship so I find out that from here what is that you are again seeing that the output resistance if the open loop gain is very high if GM is infinite how much is RO if GM is infinite how much is the value of R out 0 so large okay the accurate value may come but in divided will always come 1 plus GM R if GM is infinity or very high then the R out will keep going lower and lower values is that correct you is that clear if the gain open loop gain is very high GM are very high R out will keep falling as the gain increases is that correct so one of the requirement of open was what that the R out should be very low R is very high already said VGSE like a open circuit okay R is high R out is low and how much is the gain I am talking very large open loop gains is that correct so by connecting open in the networks the way I have connected which is a negative feedback connection I am ensuring you that open loop gain is higher if then the output resistance is low and input resistance is high is that correct this is essentially accurate value you can still get it the point yes all in and saying that I must somehow guarantee to you is much higher open loop gains okay I think this is not correct okay now once I say then I can write this equivalent circuit very easily please remember how what is the equivalent circuit I am now saying a typical open I am now saying if there is my input voltage I say there is no current inside minus terminal why do I say so or no current in the V plus terminal why do I say so I can say so okay I will come back to it this is my input no current entering open at the output there is a voltage source which is V1 minus V2 times minus a and what is this a open loop gain please remember this is like a defam stage followed by a single stage followed by buffer minus al V1 minus V2 is your V0 is that correct I am assuming R0 very low and I am short removing it in voltage source where R0 will appear in a voltage or where resistance appears series or parallel series since the RNR out is 0 or very small this resistance is treated as 0 so this is my output terminal the voltage which is appearing at the output is minus aol times V1 minus V2 all that I said no current enters minus or plus terminals okay now we will prove this why did I say so if you see this amplifier once okay let us do this first if you do this if Ri is infinity what does infinity means essentially a open circuit but really there is nothing called infinity so that means there will be a very large resistance connection between this and this is that correct so how much will be current V minus minus V plus divided by R what is the current in this Ri V1 minus V2 by Ri is the current in I Ri is that okay current in this this voltage minus this voltage divided by the resistance in this if Ri is very very high how much is I I I Ri extremely small or less than nanowands less than nanowands is that clear to you if that is very very small we may say or practically 0 we may say this potential must be equal to this potential when the current does not flow what is the potential at the two end of any resistor say that is why we say no current so we may say now if this difference potential is V minus V plus the output is AV V minus V plus which is Vd and if our eyes very large no current flows through this means V minus must be equal to V plus I repeat is that clear why V minus is V plus because if we declare that our eyes very very high no current flows okay practically that means the voltage difference between V minus and V plus may be extremely small close to 0 so we say V minus must be equal to V plus that is why in opium we keep saying V minus is same as V plus we can see the same thing little differently again we have a V minus V plus separate into AV is the output is that correct that just now from this circuit AV into Vd is the output so Vd is a V0 by AV is that correct or V minus V plus is equal to V0 by AV if AV tends to infinity how much is Vd Vd tends to 0 so V minus minus of V plus tends to 0 so V minus is equal to V plus so either way if you look at it why this is either way coming because in deriving this circuit I assumed are infinite anyway that is why I derived this okay so it is not two different things but it is essential thing both ways you see V minus will be equal to V plus in an opium circuit provided the open loop gain is extremely high and therefore how much gain you expect at least should have 10 to power 4 and above or preferably 10 to power 5 and above okay that is the reason now let us say if I ground the V plus terminal I have physically ground this terminal then according to me the potential at V minus is how much what is the definition of a ground if the current sinks at that potential then it is ground sinks current must sink there yeah upper current ground pay Jara so we say it is physically grounded there is a potential zero Kia was up current sink over however in the open we said there is no really current flowing at V minus however this potentials are same because we say V plus is V minus V plus are physically grounded so V minus has a zero potential but there is no current sinking there is that correct I said V plus is V minus V plus I made it zero were physically by grounding so V minus I am saying has a potential zero okay however that does not allow current to get in we said it actually which means though the potential is zero no current sink is going on is that correct such a ground is called virtual ground is that correct why it is called virtual because current does not sink there is that clear to you current does not sink there otherwise it is ground but it is not actual ground it is virtual ground the difference is clear to you difference between actual ground and virtual ground the potential is zero one sinks the current other does not sink the current is that correct though potentials are same there this is very interesting and that makes opens circuit solving easiest because once I make V plus zero this terminal immediately I say at zero potential is that clear once I make it zero potential this current this current to solve that becomes extremely simple is that correct so this fact that the virtual ground exists only when defam has very high open loop gain and the input resistance is close to infinite under these conditions we say opam can say V minus can be created as virtual ground now here is a case which I must tell you what happens this is very interesting case this is question asked by many people in many interviews to many places including industry if you join okay if this is your VN and this is your V zero if you say this is that ground potential and actually show you will show me like this is that clear to you so this is a potential divider where center terminal is grounded which has those two properties then I may say the this is that virtual ground and the current is same here is that correct this fact must be understood that opens presents is a necessary to create a virtual ground and it is not a sinking ground is that correct by putting this this becoming a sinking ground is that correct so this is an essential feature we ask that is we always show us this question okay and as soon as I remove the infinite case this is the divider which has no input then and that means this will not act like a system which you are looking for is that point clear so please do not consider that this is physical ground this is only virtual ground once I say it is zero potential that is enough for me I am not saying it is sinking any current is that point clear once I have in my mind now I can say another net before I quit no there are few more networks we can solve okay here is a R2 of you Barbar Joe may be around here from here can you tell me if I want to increase the gain or output voltage or not gain output voltage what should I increase R2 either I should make R1 very small or make R2 very large so larger the R2 larger is the gain I am I know very well now from there but then typically R is let us say of 1 mega ohm sum Joe R2 10 mega ohm laga then chalega kya nahi akshwacha aya 100 mega ohm laga aapkote ratio chaiye aapto voltage vada na chate hain to aap R2 vada te chaiye 100 mega ohm kardee je toh 10 to power 6 gain mil sakta aapko okay now yeh sawal hain ki R2 kithna vada jai to a condition Joe aaproo kar sakta hain meh bhi kar sakta aapkar ke deke it should not exceed or at least not equal to it should be always less than the RI available to you that is the condition you must this is one requirement another requirement please look at it the real theory is what we go as long as the ratio is what you adjust the gain can be same I can have 100k and 10k for a game of 10 10 ohm and 1 ohm also I can have 10 so should I use 1 ohm 10 ohm 1k 10k 1 mega ohm 10 mega ohm kya values rakhna chaiye kispe dependent hain okay it does not cross RI that much I fixed what what values I should choose and what it depends on R2 by R1 to ratio same rakhna 10 ohm by 1 ohm is 10 100 ohm by 10 ohm is 10 1 kilo ohm by 100 ohm is 10 1 mega ohm by 100k is also 10 kya values as soon as I reduce the current I actually reduce the power dissipation is that clear to you so the choice of values of resistances are decided by the maximum minimum currents provided to you in the logbo is se jyada currents drama career okay is kya anuswari R choice karsai as an air ratio talk is capable of a point 1 over 1 ohm bilagasat is that correct so one has to accept but too high what is the problem I say the feedback problems will start then are I keep us near now is that clear so some way if you want to improve gains you will require R2 to be higher but you do not want R2 to be too high is that correct so what do I do here is the solution you replace your R2 by a T network is that clear to you what is my problem why I do not want to increase larger gain killer R2 me banana chata because if I increase R2 it will come very close to arrive I do not want that to happen so I said fine are we carrying a T network Daldian even now since it is a good open I1 is equal to I2 because V- because I put V plus at the ground equal to physically ground V- is a virtual ground so the current starting from input will be same as current in the resistor R2 this network is that clear let us say current in this circuit is I2 R2 me yeah so does a arrow many over liye automatically signs correct hojaini V by I cut this okay this potential I declare as Vx this is V0 this is Vx this is how much 0, 0 virtual round okay is that okay I2 I4 and I3 however I1 must be same as I2 yeah say nickel nevala current okay because we already said virtual ground is created no also please remember no current enters opamp input this is the reason why this is all possible okay now your potential Kipna hoga let us look at this potential Vx 0 please look at a 0- Vx divided by I2 or we can say I2 R2 is Vx with a minus sign is that correct 0- Vx divided by R2 is I2 so minus I2 R2 is Vx Kirchhoff law though no which may voltage difference divided by resistance is the current is that correct this potential minus this potential divided by R2 is I2 so minus I2 R2 is Vx however we say I2 is equal to I1 I2 is equal to I1 I am on a opamp requirement time so yeah I2 I2 is minus I1 R2 but how much is I1 because of the virtual ground Vn-0 divided by R1 is the I1 so Vn by R1 into R2 is my Vx is that okay Vx is minus Vn by R1 into R2 now we say at this node Vx I2 is I2 plus I4 is I3 a current a current must be a current is node per sum Kia subcode is that okay I am summing current as node Vx entering current is I2 entering current is I4 leaving current is I3 now you say why opposite it does not matter I will put minus sign correspondingly anyway okay so I say I2 plus I4 is I3 but how much is I2 I just now calculated how much is I2 per I2 Vx by R2 is that correct how much is I4 this is the number we have four yeah sorry yeah as a sign we have a minus lately I mean it is that okay I am opposite current Anna so Vx by R4 is Ix if I R4 for the sign minus like I say a minus lick them in a minus Vx by R4 is equal to Vx minus V0 by R3 is the I3 current sum of the two currents at the node is equal to the third current okay a minus sign they say now up to ultrasonic other than you have a minus lately I mean it corresponding a Vx by I as a current I say minus lately I mean it so this is equal to this I collected the Vx terms from all of that so I get Vx is equal to 1 upon R2 plus 1 upon R3 plus 1 upon R4 is V0 by R3 but how much is Vx Abhik it Na Nikala them in a Vx minus V in by R1 times R2 Nikala in a Vx post was substituted this is my Vx so minus V in by R1 in to R2 into 1 upon R2 plus R3 plus R4 is V0 by R3 is that okay equation message substitution Kihayi finally is can us are many V0 by Vn Nikala V0 by Vn Nikala so that is minus R2 by R1 into 1 plus R3 by R4 R3 by R2 now can you see if I did not have T network to gain concept a man R2 say what I have R3 or R4 turns all I have so may in key value adjust cut the home or additional twice thrice ten times the initial hay you have R2 by R1 now is key value Kobe may adjust is that point clear what did I say if this term would not have been there this is a normal amplifier R2 by R1 and if I want larger gain R2 I would have increased now I said don't increase R2 okay but adjust R3 R4 R3 R2 ratios which may not be very high okay resistance may not be very high and still boost the voltage gain is that clear so now is that point clear many times the limitation may come on your resistor values is that correct then do not say sir up may gain kise bhanu the tricks a a trick the case key all the tricks up idea lagao kya I say network kyo banaya gaita network theory man when we are teaching network theory T network pie network everything you are not very keen but we are only solving there so many I have put the khaya kidney what T network calm or a fight a mass I just showed you is that clear this is how you actually play the games okay so this essentially does whatever most of you have already done but not in this format I am sure this was not the format before we quit let us quickly see by same logic a inverting amplifier pala deka I mean non in what is inverting means minus R2 by N1 is a phase shift of 180 degree so we say inverting if I apply a input at the V plus okay and still we assume our is infinity and a well is high V minus is still V plus so yeah the a V in a then I calculate current I1 I calculate current I2 so I2 is V in minus V0 by R2 I1 is I2 is equal to V0 by I1 so V in by R1 please remember this is V in the a V in ha ha ha ha ha ha ha ha ha ha ha ha ha ha I1 0 minus V in by R1 is the I1 minus V in by R1 is V in minus V0 by R2 they are equal currents same current flowing both sides so all we are to be 0 by V in is 1 plus R2 by R1 plus sign so it is called non inverting gain amplifier non inverting amplifier V0 V in R plus now 1 plus R2 by R1 if I want a non inverting amplifier with a gain of 100 what should be ratio of R2 by R1 99 do not say 100 okay is 99 is that clear so correspondingly you must adjust the values but if it is a inverting amplifier then what should be the ratio or what should be the value 100 because this minus R2 by R1 is that clear so there is slight difference in ratios when we do inverting and non inverting amplifier a last circuit there is a non inverting amplifier with R1 not present and R2 also not present okay I am connecting V minus to the output okay there is no R1 there is no R2 I am just connecting V minus to the output and then apply input at the V plus terminal okay now you can see V0 is V minus same terminal I but V minus is V plus so V0 is V in because V plus is V in so V0 by V in is 1 is that correct so V0 is V in what is the circuit I say called voltage follower or a buffer voltage follower or a buffer now the second word why I wrote or buffer buffer cake quality what is the quality of a buffer that the input resistance and output resistance can be matched to any input side to output that is the buffer part is that here what is buffer means it is a resistance both jadaya or both come here both jadaya or both come here out there method don't know go beach may buffer cut this question as cut this question as the buffer requirement now let us take a case why buffers are required I have a sensor like a mechanical card means of a strain gauge it is a resistant but jadaya 100k if you have a strain gauge which creates a voltage of say V in with a series resistance of 100k and it has to be sent to a load of 1k a normal circuit a sensor car be zero V in 100k a series may score output per 1k car load could drive. So V0 by V in 1k upon 100 1k to keep them alive actually point 0 1 V in mila to pay levy inch hota retai sensor ka output bhoj hota retai okay up the output the KITNA Mila 100 time less Mila is that clear so the up go kuch phyline you know I have a small resistance here which has a large series of strength though actually it is getting a lot of voltage though input say become okay at least it should have crossed the input if nothing else. So I do not like such connections so I said okay it can make circuit does it on buffer car yeah it is a match Karjaya or yeah it is a match Karjaya it is car input is KSR match Kare is car output is KSR match Kare is circuit car as I do many Karjaya to this will not load the other circuit input will not be loaded by the output that is why you need a word buffer is that correct so pack is he be instrumentation amplifiers ke sath mein 1st stage kya hoga buffer stage hoga ki sensor unknown rathain ke resistance bhoj jala variable rathain ECG lia the meter hack jhansi source of problem has ka one mega ohm thak resistance as I think it is a shock my rathain bhoj resistance higher act now such outputs of ECG dikha gai nahi aapko bhoj you have to understand why we do all these games because in reality we do not connect otherwise so here is something I said he ke a voltage follower laga let me a 100 ke yeh hai yahaane wala output same hona chahiye at least utna to bhi main na chahiye aap condition kya hona chahiye is ke liye is ka input resistance hai kitna hona chahiye atleast 100 ke se jaya hona chahiye to load nahi karega ho jadee e resistance one mega ohm se jada hai to yeh is ka karan is ke sath match nothing will enter in that so only voltage will appear here that is followed at the output is that correct if this resistance is very very high which is what opamp will provide you this will only create a voltage which is proportion no current passing so no drop here is that correct no current passing in 100 ke so this voltage appears here it is followed as it is at the output and it does not worry what resistance you are putting at the output is that correct this is buffered it so why buffers is that clear buffers are essential part in any unknown sources okay otherwise like antenna of a mobile okay it has a 75 ohm balloons as we call okay internally antennas a baki circuit pachas ohm per etai square transform kar kya hona pata us ko malte balloon so is this question match kar deta everywhere you need some kind of an amplifier whose input should get protected from the output is that clear one K does not affect now 100 k side circuit because this has very large resistance allow all of its to appear here and as follower all that will come out is that here so abhi up is k across full V V in million is that clear now if you wish you can further boost it by R2 by R1 if you need the first stage must be a voltage follower is that okay thank you for the day.