 Okay, so we have already discussed the OTA part as a concept and as I also discussed the circuit. Today we shall start with some applications, these applications which a few of them I will show you. As I said there are many but the major ones I will actually like to tell you. And the first thing and foremost for which OTA is very popular is realization of active filters. What is the difference between active filter and passive filter that if you do not have any device only RC or I will see circuit we call it passive filter. Of course please do not think that passive filters are bad or something, passive filters are many times very good actually but there are limitations many times because of the realization on silicon and therefore it is better to go for active filters. We have already done in our second year active filters using opams, salamki kind of blocks. We say that we can always create using solid shape this or maximally flat filters, choosing the whether you wish to have specs. You can have second order, third order, fourth order, nth order filter designs, more accuracy can be built, cut-off can be very sharp, can be slow, 3 dB down point can be shifted all possibility exist. And if you are aware about the metallabs I think this is a very trivial job in the case of using metallabs. So you do work in some case, the first attempt of any filter design should go on metallabs so that you see the response and know what is going on. But we are interested more so in the realization of filters with using OTAs and one of the problems with most of the filters requires higher frequency cut-offs that is their bandwidths are required higher and that means if you look at normal passive filter it is time constants is 1 upon r c, these frequencies are 1, omega is 1 upon r c, either you reduce c which at best you can create out of a mass transistor with all parasitics there, everything there that is not very small value as we think. So you can reduce r but if you reduce r silicon has a problem that it is a limited on both sides, you cannot have very high resistance because the sheet resistance is not very high so area it takes is very, if you take a very low r then the exactness of that r because of the variation in parameters like mobility is very difficult to get 10 ohms, 1 ohms it is very difficult. That means 1 ohm may become 2 ohms also, on chip there may be many which are varying drastically it may be 50% variations and therefore very low resistance or very high resistances are not greatly realizable on silicon, in circuit we can always say 1 ohm and fair enough but when I do in silicon I find on 100 chips only 1 word, so that that is something which I am worried about, so active filters try to alleviate some of those problems and therefore the OTA based filters have become very popular. Essentially what we are doing we are replacing r by 1 upon gm and we know gm can be well because of feedback system we can make much more accurate gm values compared to passive r's okay and they are much more controllable by bias currents, so we can have a large value of variation on gms for which or even tunable we can actually tune it back okay by feedbacks, so something which is great in the case of active filters we would like to see, is that okay, why OTA where and see gm is coming from an OTA output is essentially I out, it will say trans conductance amplifier, let us start with the actual filters, typical OTA as I shown here is it has a trans conductant gm and in the case of gm we saw in actual circuit which I showed you if the k the multiplying factor k is used 1 then small gm and capital gm are same, otherwise k times small gm is essentially the capital gm and the way we said I out current is proportional to gm v plus minus v minus, so if we put v minus 0 obviously which we can do any time, so this is I out is gm times vn and I out by vn is therefore the trans conductance, so all that we are now trying to see, can I use this block to create a gm circuit which then can be used to make gmc filters, so that is the basic idea behind all these choices, so let us see is anyone anything there is nothing much written, there is nothing great in figure, all that I am saying I am trying to use I out gvn is it okay, the kind of thing right now shown here is open loop, is right now open loop, but we like to use most of them in closed loop configurations, so let us say this is the this I grant for say and this is the OTA which you have started with okay and then we always said I out is gm, let us say right now gm is equal to gm whatever if it is 1 you wish to put 1 okay times vn, now if I say if this is the system I load it by a capacitor C okay, is that okay, I can load this I out OTA, of course this I still grant it, of course I may remove the ground soon but right now I kept it in real life what I am going to do is this, so now what will happen the output voltage here is I out into j omega c, oh sorry I sorry sorry, I out upon j omega c okay impedance is 1 upon j omega c, so I now figure it out that I out has something to do with gm, there is some v in there sitting, so v out by v in has a function of gm by c kind of function, is that correct that is something I am seeing it here, so I said okay to control it better of course I did not show you but you can always say this is my I bias which is controlling my gm and now I do this is my v out I actually feed it back like this and here I actually put a capacitance this is loaded and with negative feedback, why we normally do not do positive feedback because it will become unstable because growth will start, so this is an interesting simple circuit, I can solve this now, I see something few things you just write, this is only step by step I am showing you I could have directly written this and solve it but I thought how do we realize this is how we started thinking, so my transfer function for this is v out by v in and if I look now for any transfer function of a low pass filter is it drawn everyone, if you see a low pass filter typically I am looking for let us say transfer function what can be this into omega, so I want ideally this kind of transfer function, sharp fall but if I say that okay it falls something like this, my cutoff is still the same 3 dB point of a normal gain amplifier kind, though whether you can tolerate this much is your requirement, we can decide how many 20 dB or 40 dB down you want, so additional poles may be required, additional some gains required but otherwise in normal A S omega function or A O J omega function if it has a single dominant pole it looks like a low pass filter okay, so that is the idea behind actually finding that v0 by v in if we find is a function of omega and we get omega 0 as its known frequency which is one of related to GNNC and I know what is the cutoff of this low pass filter, so if I use this last expression you can see from here this v out is connected to v minus that is v out is v minus, so for this case v out or rather v minus is v out okay, we also see one thing for this figure, what is v out essentially, i out essentially is, i out is gm times v plus minus v minus or gm one times v in minus v minus which is is that okay, I just replace v minus by v out because that is the connection I did, how if you look at this point, this point and this point are same, so what is essentially v out otherwise I can figure out, there is a i out current flowing in capacitor, so v out is i out into 1 upon j omega c, is that okay, so if I write now this i out here I get gm 1 v in minus v out by j omega c is v out, collect the term for v out, so I get v out 1 plus okay or rather if I would have done the other way, does not matter, 1 plus gm 1 by j omega c is equal to gm 1 upon j omega c into v in, if I find the transfer function gain function which is v out s, if they are all omega terms j omega times v in which is also a sinusoid then it is equal to gm 1 by j omega c divided by 1 plus gm 1 upon j omega c, I divide numerator denoted by gm 1 by omega c, so I get 1 upon 1 plus, please remember I am dividing this, is that okay, so it is j omega c by, if I define gm by c I define omega 0 as gm 1 by c, so this function can be written as 1 upon 1 plus j omega by omega 0, which is like a dominant pole sitting at omega equal to omega 0 because omega equal to omega 0 the gain is 1 upon root 2 okay magnitude, so obviously so it looks very trivial and it is, I just thought that people think I just copy somewhere and one can always do any time, so there is nothing great things happen, so a transfer function shows me that if I do this I can always get a low pass filter this, if I want this omega 0 to be larger what should I do, see of course it will be normally decided by availability of c's, so I must boost gms, so what do I boost bias current, k of course if it is fixed once I cannot change, may be 4 at best you may put, that means I should change the bias current, so the low pass filter with variable cutoff is now possible from v control okay, so I change the v control bias and I have different cutoff for the low pass, so it is a tunable GNC filter, is that correct, it is tunable GNC filter, that is the fun of OTA, that OTA allows you to do such tunability, otherwise what will happen for every case you will have to RMC you have to adjust and you cannot change inside a chip, you have only a pin which is v control outside, so you have to only play with that and still you are able to actually design a filter of your cutoff valance, is that clear, stability is somewhere better because there is a single pole and there is no question and it is always on the left half plane, so system is inherently stable Yes but when I tell you that some kind of a pseudo active device because j omega cc is not under root of b it comes one upon under root of b, now whenever there is some kind of such relationship the cv is parabolic, that means certain range you cannot separate the seeds, you understood what I said, if the curves are sharper like this, somewhere at very small values the c values are different, cannot be figured out how much, where the curves are normally used when you are actually looking for lot of difference that is in vc I will use it actually, maximum values, so I will use one value here on the curve and one value on the other side, then I have range of this I can use which is what we see wants, voltage control, so in voltage control oscillators we will use diodes or detectors, is that okay, but as I say here that voltage which range is very difficult to control, so this is much easier to control just, is that okay to you, so this is where I will say that this is a low pass filter which I can create out of a simple OTA by putting, if I want to put a high pass what should I do, where should I put a capacitor in the input side okay and still keep it negative feedback, please remember system will become unstable as soon as I remove the negative feedback okay, of course this is not the course I should talk, please remember that whenever opans are used, we always use this equation in all our circuit that v plus is equal to v minus, this is very suspect function okay, some other day we will use it that this is not true every time, you must verify this statement okay and that is why the value starts okay, so is that okay, low pass filter, so I also did for the sake of, of course this I got it, I mean did home, so that I just showed the technique, so this is the technique which I used, I want a high pass filter, so beyond the frequency of omega 0, I want everything to pass below that, I would not want anything to pass okay, this is not omega 0, maybe this is okay, so I used same OTA, now I put v plus as ground, I put capacitor in series to my input and took a feedback from v minus v out okay and then same equations I can use, I out is a function of v in v minus times this much, I out into this j and v out, I just write down this expression, just write Kishav last okay, one equation is coming from v in minus v by upon 1 into j omega c is the input current which is same as the IF current which is opposite of actually I out currents okay, so you calculate this and adjust the two currents and figure it out what is v0 by v in, you note down if you feel you cannot but otherwise the function I got therefore is j omega by omega 0 upon 1 plus j omega by omega 0 which is the same which is the trademark for a high pass filter with cutoff at omega 0, is that okay? So now I got two circuits, this is a high pass filter with a capacitance here, a low pass filter where was the capacitance at the output, oh so now I got trick, I say okay I have a low pass and I have a high pass and if I properly make the two combinations, I can make a band fast or band reject because depending on the cutoff of which I can, let us say if low pass filter cutoff is higher than cutoff of a high pass it will become a band pass but if it is the other way it may reject that band between the low pass and the high pass, so it is only a matter of adjusting cutoff for low pass and cutoff for high pass. So what essentially the chip available to you is a bioquad filter with OTA which is general purpose in the sense all four filters can be obtained by using the same block by just putting three input terminals different voltages either 0s or 1 or V1 0 and you can now create out of those blocks any filter, a low pass, a high pass, a band reject or a band pass. Another filter which is required very often in the case of communication is specifically the notch filter, so you can now think how to the notch, you bring omega 0 very close and you can create really really good notch filters, so there are the same bioquad can actually be implemented for low pass, high pass, notch, band reject, band pass and that is the fun of this bioquad filter designs or bioquad filter chip available, this is available chip so it is nothing great notch, it is not really band pass, you can say band pass with 0 here but since the 3D viewpoint any way slopes you down, you can adjust the two slopes such that the omega 0 is common, is that clear? It is a band pass at one frequency, it is a band pass at one frequency, but there is a noise associated ahead because of the slope down, so some part of the frequencies are passing but mean is input directly at that frequency is maximally output, okay, so you understood this is omega 0, so low pass, high pass 8 juggles up, it falls down either one, this is the low pass requirement, this is the high pass requirement, so you can understand what I am saying, this is like this, this is high pass, this is low pass, fix it, you get a notch, okay, of course this is not very easy because to get exactly this value for the 2 GM's which cannot be easily adjusted, so it depends how much notch you have, this notch has very high queues, many circuits required very high queues circuits, so these are required for high queues circuits, okay, so think of it, so I am just trying to tell you where simple things which we have can be implemented and realize whatever you are really trying to achieve in your actual hardware, okay, so where is the circuit which is the quad one, bi-quad, there are 2 OTA's available, right now I showed them that their GM's are the same, on a silicon chip this is not a very wrong assumption, as close I can make their properties can be almost identical, however on chip there are many chips, so one can 100% guarantee but one believes at least at that area the 2 GM's will be equal, this GM may not be same as the other chip GM exactly, but on chip they will be same property can actually, so this capacitor is here, this line has come so please do not draw that line also because I have, I did not want to re-draw again so I just interpose capacitance there, okay and this is a typical circuit which can implement everything, that is hello pass, high pass, band pass, band reject and even notch if omega 0 are adjusted, of course if I only want LP I can do one GM itself, only one HP, only one GM circuit but if I have to make 2, 3 such separate chips that much costlier, okay, so I will give you one chip which makes all of it, okay, so I repeat I can certainly achieve only for low pass, only for high pass but then I will have 3 chips to do or 4 chips to do each function, so which I will not like, so please I repeat again this is a C1 in series from this node to the V2 and there is nothing in between, there is no shorting in the capacitor, if a capacitor is shorted like this what is the output impedance? No, nothing wrong, I think you are justified in writing that, I just feel since it goes out faster so I thought we should not allow people to see something, we are as if intentionally we do not know things and putting things too, okay, now in this case we start with the case let us say first, the first one was low pass, so okay let us say this creates a low pass filter, so I said okay V2 and V3 voltages are grounded, okay, V2 and V3 voltages are grounded, so only V1 is present, V2 and V3 are grounded and now we start looking at these identities here, the V out 1 is essentially V plus 2, is it for the second transistor input V plus 2 as I call is same, V minus 1 is same as V minus 2 and is same as V out 2, connection there is nothing very great I am talking about and V1 of course I say V is my input voltage, so start writing now for each voltage, okay this small gm sometimes I replaced it by capital Gm, so just think right now K is 1, so you can you know in Haroo sometimes I write small gm, so take it K is 1 so capital Gm is small gm, okay, in case something happens ahead, so V out 1 I want to find the output of this, this it is nothing but the trans-conductor, V plus minus V minus, so V1 minus V minus into J omega C is the output, this is the current, this is the impedance at this node, now please take it this capacitance should be larger than any other capacitance at the input end, otherwise that will get modified, but still they are in parallel, so C may still be little different than the normal, it is okay, we have C2, okay, which is V out 1 minus V out 2, please remember V plus 2 is V out 1, so V out 1 minus V out 2, V minus 2 is V out 2, so V out 1 minus V out 2 into V out 1 and V out 2 and I will just pick up this V out 1 here and substitute that in 2, so that why I am interested, I am only interested in V out 2 by V in, that is my interest, so I just probably place V out 1 from this equation 1 and substitute into 2, okay, so substituting V out 1 and V out 2, so I get V out Gm by J omega C2, I substitute V out 1 from the earlier equation and correct the terms, so I got a V out 2 with Kitter may get V out 2 either we had, this hole I have shifted on the other side, so Gm omega C2 by Gm, okay that is what has happened, I started using small ones, okay, into V out 2 is Gm 1 V in upon J omega C1 minus of V out 2 which is minus Gm 1 J upon C1 plus into V out 2, correct the terms, 1 A term or 1 A term, all V out 2 terms were collected by me which gives me 1 plus Gm by J omega C1 plus J omega C2 by Gm 1 is equal to Gm 1 by J omega C1 into V in, so the trans function of the gain A J omega is V out 2 by V in which gives me this function, Gm by J omega C1 upon 1 plus J omega 1 J omega C1 plus J omega C2, if I replace for your standard thinking that J omega is over the S which is what if I go into the S domain or if I am in the frequency domain, then kind of function I am going to get, so if I replace it by AS this is Gm by C1, Gm by C1 SC2 by Gm 1, correct the terms now, so I get Gm 1 square upon SS square C1 C2 plus SG1 C1 plus Gm 1 square, if I put some constant looking things then this is C upon AS square plus BS plus C, C is constant, Gm is a constant value, so 2 poles, one of them is dominant which is the cutoff frequency for this low pass filter, so what which was the, which are the voltages I made 0, V2 and V3 I made it 0 and V1 I put it as Vn and I got the, from the same this a trans function which represents a low pass filter, is that okay, less to the next one, so this is a, so what is obviously our 3 voltages, you make in other case other 2 zeros and one of them and in the 1 4th case what can you do, 2 of them you make availability and one of them you make 0, so it will be the next 4th one, okay, so we will do one, we will not do all of it, I will just show you the 2 of them then you can repeat the other ones, so is it okay, so if I use the same circuit now V2 is Vn, okay, but V1, V3 are 0, is that okay and I believe it should give me high pass, I believe it, so let us see, okay, by same method which I followed just there, I figured out Vout 1, I figured out Vout 2, substituted Vout 1 into the equation for Vout 2 which I did just now, only thing now that a slight change has occurred, this voltage plus Vn is essentially the output, is that clear to you, whatever i out times this plus because that is at above, that will be additional term which is appearing there, if this would have been rounded this term would have been 0, okay, this is series, 2 battery in series, so I collected Vout term is minus gm1 Vout 2 by g omega c1 plus Vn and Vout 2 is same as gm1 Vout 1 minus Vout 2 upon g omega c2 is 2, substitute this Vout 1 here and solve for Vout 2 by Vn, is that clear, I repeat substitute Vout 1 here and then collect the term for Vout 2 and others had we leave the term for Vn, Vout 2 by Vn is the transfer function, you know many times book do not solve, so I solve and check things are okay and luckily they are always right, once I wrote to Rosalie some mistakes which in the newer versions they have corrected, you know every state does help to know how it is done but it does not give any physics ahead of it, it only tells you how to substitute, arithmetic is made here, I do not know how it says but that is that all that, so if I substitute 1 in 2 and collect term for Vout and Vn in the other side and I get these terms again and I get the transfer function as 1 upon 1 plus sc2 by gm1 plus gm1 by c1 okay and if you see if I take this sc1 and multiply numerator by gm1 sc1 so that goes up, this becomes gm1 sc1 this becomes sc1 sc2 into s that is s square c1 c2 and this becomes gm1 square okay, just multiply numerator denominator g1 into sc1 then other of multiply or up or down, if you get a term which is g1 sc1 upon s square c1 c2 plus sg1 c1 plus gm1 square, if this is g1 msc, gmc1 is called constant d, so d into s upon a s square plus bs, bs plus cr, d co-divide kardo to a0 for new other constant as upon a0 s square plus b0 s plus c0 is the transfer function and you know whenever the transfer function s upon s square kinds this is a representing a high pass filter okay and its dominant pole will be the cut off point okay, sorry this is it is a band pass okay, if what would have been a high pass s square sorry it is a band pass okay, so let us see the two frequencies are omega 1 omega 2 is essentially the band pass filter in which outputs are 1 transmission as a magnitude of 1 okay, so this is a band pass filter as I said you are right, by same argument if I make v2 equal to vn and then make v1 equal to v3 equal to 0, the 2 OTA I mean this is what we did right now, 2 OTA architecture behaves like a band pass filter, we can do something interesting now if we do v3 is vm and v1 v2 is 0 by same method which I did I can get a function of s square c1 c2 upon s square c1 c2 s c1 gm1 gm1 square which is the transfer function for a high pass filter, the only filter we have not yet done is band reject, so band reject means what is the constant case or band pass or band okay, so this if you have seen it denominator for all of them is same s square c1 c2 plus s c1 what is gm plus gm square, this is denominator for all 4 of them, the filter is only on the top, I mean whether which kind so if the final filter which is V1 is equal to V3 equal to Vm and V2 equal to 0 then I get this transfer function SS square C1 C2 plus GM1 square upon the same denominator and this represents a band reject filter. I repeated I think someone should not be worried about this essentially is being told all that I am saying sorry omega 0 1, omega 0 2 I adjust 2 blocks and then in between there is nothing and therefore it is rejecting those frequencies in between okay but equivalently saying it is. The problem is that one notes the other which in this case we have taken care if you are independent and we do not feel the leading of the other. The reason why I showed you this was that it takes care of the when you connect it loads the next stage. So we know what is happening because of the second state. If you are independently this and when you want to connect independent design will not take care of load unless you know what load I have to use there. If that happens it is same as what I did is that okay you have I mean what you are saying is I can use 2 blocks but then I may not get this just that when you connect the connection makes the difference and that is the circuit I have shown which connects the 2 blocks is that okay. So this type I understood that if I have a 2 stage amplifier I may design a first stage I must generally know what is the input stage or input of the next stage because that will load my first stage. So if you do not take care of that then your design will never be correct because the actual leading will deteriorate every performance. Either first evaluate also you must know this also loads the one and this also loads the other. So both sides the output and input and impedance have to be matched and then only power is maximally transferred. So game is to be played when you connect the things okay. And I said to you that okay this OTA can also be used as an amplifier. So let us say if I have a normal amplifier is that okay this is here done. So this is something the chip which is available from TI it is available from analog devices available national chip is available. Of course non-national is again TI but chips are still national numbered so these chips are available. But I must tell you that there is no chip which takes you whole of a band what they will do there will be still 8 to 10 chips which will be because otherwise the GM value will be too bad for them. So they actually adjust band wise filters okay that means from this megahertz or this kilohertz this much then this megahertz this this so each has a band wise this in which this operate they theoretically 0 to anything is possible from this okay. So that is why they can make lot much money you want this you buy this okay. Okay so here is something last part of my theory okay we have been doing this standard opamp based amplifier. So I actually connect the same OTA in an opamp feedback kind of thing instead of you asking me how I solve you are you have not done I will not say more than that. I have solved this as I could okay the reason why I put it because this is V minus minus it is used so I change the sign there but otherwise it is okay. So I figure it out I 0 is minus GM time this much V in minus this by R 1 is this current but this current is so in this criteria I can the criteria which all of you should know and that is the one which again I repeat and verify yourself. In all this analysis we say no current enters the terminal input of the terminal of OTA opamp the reason is to be that means the input impedance should be extremely high okay and the negative feedback dominates this is the second condition which allows you to repeat this okay. So if that is not so real circuit you solve the field out if both conditions are met only then these statements are valid statements okay. So then I this minus this minus this is something I 0 is minus I 2 collect the terms find I 1 and I 2 yes the reason is an opamp if you see it the currents are always shown in this direction that is essentially it is minus of the actual current but in opamp if you see currents are always shown in if you see opamp equal this is for that thing which is equivalent of an opamp if you see any opamp circuit we show I current in okay no this is different case I am only saying the difference of this into GM is the output current yeah it is GM times V plus minus V minus no you know this current which is coming here has no other terminal to go so the current here in this arm is same this is only a matter of node at that node the current is same no no no no no but please remember the impedance seen from here to here and here to here is not same when I said this okay well at best you can say here okay but that there is no other there is no other nodes of this current should be equal actually opposite sign that's why I added to add it I said if they are opposite sign that's why I say that in all opamps we put inside this so I just equated it okay so think of it yeah second day yes no no because this current essentially is moving this and I assume the direction opposite I put it minus this is only a question of putting minus sign for the equivalence you are doing okay so I calculate I1 I2 and then equated I1 I2 at least that you agree I1 is equal to I2 because there is no current enters the OTA and I substitute all this expression here and I get a very interesting relation V in by R1 minus V out by R1 upon 1 minus gm R2 is equal to gm by 1 minus gm R2 into V out collect the term for V out and collect the term for V in and get a ratio of V out by V in note down because in negative feedback system the output will be V0 is out of phase of that and therefore if it doesn't come minus R2 by R1 you will start telling me where is minus sign k now I want to bring this and that's why I assume the same direction as opamp uses so yesterday I was telling that under certain conditions only OTA is here like a good voltage control voltage sources okay so the condition will now bring it soon what could be the condition anyone quickly gm R should be very very high compared to 1 which many times you can always attain in OTA that's the purpose of OTA gm is higher okay so if you use don't use very small hours then gm R will be very large compared to 1 and then equations will be same as an opamp requirements okay is that okay everyone expressions okay so I figured out the gain V out by V is 1 minus gm R2 upon 1 plus gm R1 and yes if gm R2 and gm R1 are much larger than 1 V out by V is minus R2 by R1 which is what an opamp would have given so OTA with these conditions V is like an opamp is that correct so it is not my condition this one not necessary to show only I was saying that many a time people believe only OTAs are on chip how do they make opamps things are possible even using OTAs okay I agree that to some extent the load has not been shown to you so if I put a load that will take care of this actual impedance going down your viewpoint is very correct if I open in there if I see yes there is an issue but as soon as I load it that will be the output resistance so in that case it will behave like an opamp output okay you have very good point saying that in real life people use this and say and they don't think but they get the correct answers so why is the reason I said this is how they get the picture the last part of opamp based base is now one of the last part which I thought today of course I will not be elephant I will only show you the basic but something which is very relevant for those who are per se interest in joining a company which has more interest in mixed signal design like Broadcom, Qualcomm okay who are more interested in analog plus digital on chip and RF as well these chips require fully differential amplifiers okay they require in many applications fully differential that is fully word where essentially inputs are always differential so what is a fully means even outputs are differential therefore it is called fully differential okay some books some journal they only say difference amplifier but then you know I don't know whether difference at the input or some right difference output amplifiers which is also equally good way of telling I have followed some books which says fully differential I use this terminology nothing very different all that I am saying to input to outputs okay this is okay so we start the last part of opamp as I said there are too many things in real life actual designers should learn to do themselves but where to do and how to do is whatever course is trying to help I cannot replace everything I can only tell you why I look this way for example now I will show you how to get some things which is wearing me in this and that is the idea that if there are variations what do I do so I will show you this later so this is a fully differential opamp which essentially says if our V minus V plus our inputs V out plus V out minus and they are of course our opposite terminal terminals phase change so this is V out plus this is V out minus such systems are called fully differential amplifiers or differential output opamps so I keep telling they are mostly used in mixed signal designs for example if you want to make an analog multiplier which is essentially a digital requirement so you can use such circuits to actually quad multiplications as they call quad multiplier we can use analog blocks to do digital functions real or in ADC some of the ADCs you require differential outputs so these are essentially for mixed signal blocks their advantages they are certain whenever things happen good obviously everything must not be good one other thing which is better for two ended outputs double ended output compared to single ended is since the output there is limited by the like in the cost code or above you were you know V minus that much okay because each is controlled by the same I5 currents okay so there is a difficulty there to manage swings this may have much easier way of because you can independently handle this and this and swing may be much higher it may also have better frequency response or bandwidth simply because there is no miller poles here in the normal stage you put CC you did stability but you lost the frequency bandwidth there is no miller pole here and therefore or miller feedbacks therefore it will have a larger bandwidth requirement so in some this maybe before we quit last slide also in the same thing the biggest advantage of fully differential system is the rejection of noise of course along with the noise something else we will talk a few minutes we will talk let us say this is my fully differential amplifier and there are two parasitics at the two inputs and there are two parasitics at the outputs if I now see some noise signal comes at the input is that okay and our belief cannot be very absurd if you say the noise at the two terminals there will be two different noises unlikely event I am not saying it cannot happen but unlikely event so what will be the actual V minus it will receive this plus noise which is charging this capacitor if these parasitics are same because of the symmetry you have then this noise sitting with this and this at this terminal noise sitting with the other since both gates noise added to their input the difference will be independent of noise by same argument if there is a noise coming at the output the difference between VO plus or VO plus or VO minus will cancel the noise so that is the biggest advantage of having a fully differential system it rejects all noises whether at the input or at the output we haven't actually I should have said but I think some word finally I may say there is another problem in defense or opens the word we never talk we talked about CMIRs but we did not talk about PSRs so this power supply rejection ratio is essentially because power supply is not constant all the time the output varies and we believe that that should not vary it should be even as so much power supply goes output should not vary with the change in power supply voltage this is equivalent of that so if there is a change the difference will cancel it okay so PSRR is very high 100 dB plus in the case of differential systems there is 65 to 80 dB PSRR in the case of normal opens is tough but in the case of differential systems they cancel any variation is taken care so indirectly not saying there but it also improves your PSRRs because it does not allow you any variations at the difference actually only assumption is the variation in VDD is same as variation in VSS if that doesn't occur there will be some PSRR value which may not be very large but still better than normal system so is that point clear to you that why differential systems are preferred in certain stages and what is the difficulty with them because the difficulty is now going to be shown to you a few minutes and then next time we will solve that difficulty okay is that okay so the fully differential systems reject noise as much as possible and it also does one more thing can you think another noise kind of thing is something related to what we call offsets offset is also what is an offset of an opend that is the inputs are made 0 output should go 0 and if it doesn't we actually some way adjust V minus or V plus so that the for that the output goes 0 here it cancels because any change in both differences also cancelled out so in every sense differential amplifier seems to be very good but if you see one base do you think it is actually two opends I am using to create a differential amplifier ASR ASR so I am essentially using equivalently saying two opends together equivalently saying okay and one trying to cancel the back part of the other that's the one part in this okay is that okay so we will see one what is the major very soon and next time that is the circuit I will solve for you there is an issue which we will say the common mode voltage as it called for CMRR to be how we want the two devices should be as identical as possible on the fan gm1, gm2 everything W bar everything should be and we figure that then the CMRR will be infinite everything is fine delta gm is 0 denominator but that doesn't occur gm1 minus gm2 occurs and how close they are that decides your CMRRs now in this case that common mode voltage is not a constant quantity at all we will show you next time and if that happens at different this it will show you different CMRRs operations and we like to hold that common mode voltage and that is called common mode feedback circuit so additional circuit is required for fully differential system which is called CMRB circuit which is very important in any real design the first thing we ask people to design is CMRB circuit okay because that will then hold this common mode to a value okay we will see that little later just to go back to the open circuit and closed loop gains of a normal differential system if I see the open loop gain of a differential fully differential circuit it is nothing but Vi plus minus Vi minus upon V plus minus V minus okay if you make a single ended make one of the output 0 and then consider it so Vi plus upon so it is nothing but I cannot make a single ended one output divided by the difference you still have a single ended output so it is not this that this cannot be used as a single ended it can always be a single ended but then the major advantage of noise everything is lost the value is double you can as well as single ended which is much more stable everything is so good actually so if you consider one of the output is fully differential case and use negative feedback similar to that of a single ended of time as shown here here we involve the rule that Rn is very very large or to say current do not enter a pass or negative or positive inputs then only we say V minus H V plus this is we are invoking proof we had done earlier some way so a closed loop gain then will be minus R2 by R1 open loop of course V0 by Vn whatever it is okay external this to a differential case then the VO plus is the opposite of U minus if I extend this to this it will be opposite of that once I say this is like this now if I am now telling that it has a 2 different there 2 open there you will note down this and I will show the equivalent of that please do not think I actually put 2 different open there I am trying to say equivalent of this is what is inside the key so it fully differential amplifier can be realized from 2 single ended differential systems differential amplifiers J input, A output, A input, A output so as if 2 systems are going on is that clear to you 2 inputs same inputs 1 output same input other output so as if those system has this is the last slide for the day and we will start next time with C M M P circuit and why it is needed more okay last slide for the day not last this is equivalently same there is a differential amplifier can be fully differential amplifier can be created by using 2 single ended normal amplifiers defam so the way it is you have one defam whose inputs are V minus and V plus the output is and the other one one input is grounded the other is taken as a part of this resistive network we will talk about this later next time and we will say how these 2 2 systems together actually gives you V O plus V I which are equal in but in opposite phase this is given in Bios and Baker's book so what is the problem with this system which is I wrote down which is why we will have to do something more about it please do not think that this is how I am implementing I am just trying to see the principle of fully differential system is essentially equivalent of this you can see from here the difference voltage is divided equally and that is being paid here which makes this opposite R upon R plus 2 R ratio is adjusted so that this actually gets equal and opposite in signs so this is the game which we are getting there but just wait next time the catch word here is essentially because the 2 defams do not have identical inputs so if you use all your stability criteria system there the phase margin associated with the 2 systems are not identical and if you connect them the rest will come okay and therefore essentially what will happen that the bandwidth will get maximum on this when we started with saying that fully differential system has higher bandwidth but if you realize it like this then you may find your bandwidth you are losing actually that means something I must hold greatly correct so that my bandwidth is not affected this is the catch what is that common mode feedback C M C M F B does every time okay essentially what is common mode voltage V out plus plus V out minus or this divided by 2 average value of that I want to hold it but if we have changes then how do I hold it that means adding substratum of F max must be same as adding to the other so that the average value remains constant that is the what common and common mode feedback it holds that value okay and if it holds then much of your problems are solved so next time is that point clear so this will be first 30 40 minutes I will finish this year and be which is as I say it is very very important all these in real system will be everywhere C M F B circuit will be there and they start with the noise next time