 Last time we said we did something about MOS diffamps. We will quickly, we do not do all of it but at least show you that similar diffamp can be made out of bipolar transistors and a typical bipolar diffamp is shown here. You have two bipolar NPA transistors T12, you have loads of RCs here, two inputs VN1, VN2 then the current source which is IEE shunted by its current source resistance REE, okay. Let IC1 be the current in T1, IC2 is the current in T2 and for, we like to see that VB1 – VB2 is equal to the difference signal and as usual we will apply VN1 is VID by 2 and VN2 as – VID by 2. Now let us say T1 has T1 and T2 both have transistor gain current factor beta same for both which we know it is equal to GM times R of R5, okay. Assuming T1, T2 identical okay you may find IC1 will become IC2 if signals are same RCs are same and betas are same, okay. But the only difference is this DC currents may be same but if this is positive and this is negative one current will increase and the other current will decrease because that is what the diffamp is. If one current increases because that voltage increases then the other current will decrease because the voltage there decreases from its DC value. So if I use this simple analogy which I just now said then an AC equivalent circuit of a BJT diffamp can be VID by 2 – VID for VO1 and VO2. We only calculate right now by symmetry half of the circuit. So we write VID by 2 plus minus is this is our R pi is there are no capacitances used here. This is the DC analysis or zero frequency analysis. Small signal I would say zero low frequency or mid band frequency analysis. Let us say the voltage across R pi is V pi then at the output between collector and this emitter there is a GM V pi as the current source shunted by RC. We forget right now about all parasitic resistances RE or VRS. They can be added and more complication of circuits I mean more complicated expressions may start appearing. Since the difference input is VID by 2 and we are right now assume there is no source resistance then whatever is the voltage source you are applying is same as V pi that is V pi is V in by VID by 2 and then we define the AVDD is VO1 divided by VID input signal which is GM V pi RC upon VID by 2 which is essentially GM RC which is similar like GM RD in the case of MOS transistors. So it is not very difficult to replicate whatever we do in the case of MOS that we cannot do in the case of bipolar. This is for different signal gain we can also do common mode. What is the common mode signal will be both will have same voltage VID by 2 VID by 2 or VID VID total is V common and if I do that analysis which you can read in the book this is very I am just trying to show you what we are doing. If I apply VID by 2 VID by both sides common mode and I divide this RE you look at my defam this RE if I divide into two parts 2 RE here and 2 RE here parallel combination is still RE okay. So if I do that then I find this circuit can be half circuit with 2 RE here and half circuit 2 RE here okay and then we actually write equivalent circuit V common voltage this is your R pi this is your 2 RE GM V pi for one arm half circuit GM V pi and this is your RC going to the ground and output is this solving this circuit which is A common mode AV common mode is VO1 divided by VCM is minus beta RC upon R pi plus 2 RC this just solve this network okay. How much is VCM here in terms of V pi and something the drop across 2 RE plus V pi is the common mode voltage is that correct the circuit has to be understood this is one loop and this is another loop okay to solve for the 2 loop equations and you get VO1 by VCM is beta RC minus of course current is downwards R pi 2 RE beta plus 1 and let us say beta is large enough which will be around 100 plus please do not confuse this beta with beta in the MOS transistor there beta is essentially beta dash W by L it is not here here it is just common emitter current gain. So if I say beta is larger than 1 and then RE times B is larger than this is roughly equal to RC by 2 RE this is like emitter degenerated amplifier is that correct emitter is degenerated by resistance of 2 RE so the gain is normally equal to RC upon 2 RE is that clear we have done this in any transistor if there is a source or emitter is degenerated by resistance the typical common emitter gain is load divided by the source resistance or the emitter resistance is that correct equivalent yes for the change in this one change will be plus and the other change is minus so RE there is no current equivalently same for AC signals one changes plus IC delta IC other is minus delta so when they pass through RE they cancel it is like R5 which we say no current goes okay equivalently same is that okay in the case of common mode both currents are passing through RE is that correct in the case of difference opposite currents pass is that clear and therefore we assume that they are equal if they are simul the transistor identical then there is no current in RE is that okay same logic which I described for RO5 in the case of MOS transistor is valid here is that okay so if I write my common CMRR which is AV DD by AV CM and I write the expression typically this expression appears to be 1 plus 2 GM RE is that correct if RE is infinite assuming current source is ideal then how much is CMRR in finite but since RE is finite there is a some kind of finite value of MRR and this value may be typically as I said 100 DB 80 DB to 120 DB the RE value which a current source will give will be such that GM times RE to 2 times GM RE will be of order 20 log of that if I get then it will be around 80 DB 80 DB to 120 DB which will be higher when RE is higher or lower when RE is higher that is better current source you will get higher CMRR when you have a little bad current source you will get lower CMRR what is the advantage of CMRR why we are interested in CMRR because we want the signal actually we only calling it there is a common mode and difference mode signals in fact we will only apply V in so any noise if we supersede superpose on the signals and we say both end will get same noise and since common mode rejection is very high the noise will not get into the output and we will say we will only get different signal outputs that is the purpose of defam is that clear to you is that clear so we are always designing CM of course CMRR is not a fixed quantity in design I may choose CMRR 80 DB 95 DB 100 DB and typically get a current source value for which RE I will choose and then correspondingly I may design my current source is that correct so it is not the inverse way I put something I first actually decide how much I should have 100 DB for this I found RE for a given GM DB or this beta is given and then I will find okay I should use the transistor here which makes a current source which will give me typical RE of this value is that correct this is what we everyone does this is just a small this think of it normally of course here we have everything as an NPN transistor but a good current source can be made out of a PNP transistor or a P channel transistor in mass why compared to N channel or NPN think of it if I only want a good current source I will prefer a P channel device or a PNP device here of course I have one technology in which everything is NPN so I am using NPN everywhere but in case I have options just to create an current source I will prefer PNP and PMOS transistors over NPN and NMOS is that clear this is something as a designer we must know if I want a current source I look for P channel devices or PNP transistor okay this is just to I am not doing great detail of this is just to show the analysis which we did for mask is similarly can be done for any bipolar circuit and method is identical in the sense give a equivalent circuit of the two cases whichever cases you are looking for and then solve the circuit and you will get both all the parameters of your interest I promised you last time that I will solve a problem for same MOS current mirror based loads and this is of course a designed actually designed the fam in our lab but I have just modified some values as well as what I did is there since we were designing in the case of design of a MOS circuit what do we actually design as a word the size of transistor that is the design parameter in my case I have whatever values I have actually got there I use them here and say what is the gain and what is the inverse problem has been shown to you or simpler problem in real life you will be given some value of gains value of this and find W by I so let us see a problem which is simple and actually wants to tell you what how do we solve such problems here is standard CMOS opamp with a current mirror loads and let us say you this defam is driving a large capacity load please remember this 5 puff is a large load why typical value do you know how much is CDB CDB are what 0.01 point kind of pops they have okay typical technology this is 100 times now those values so this load is excessively higher compared to any other capacitances in the circuit what does that mean essentially if I say this capacitor at this node has a very large value compared to all other parasitic capacitances what does essentially in the response term I am talking about let us say there are three more capacitances available to you actually the more than three are there but at least three and this is the fourth one so what does essentially I am talking to you when I say this capacitance is larger how many poles it will give a four capacitance will give me four poles which will be dominating the one with the CL because one upon RC so C is the largest there so the dominant pole will come from CL itself is that correct. So my assumption right now to tell you is that CL is governing the dominant pole and what is dominant pole in terms of frequency we say what is dominant pole essentially tells if I response it falls at certain frequency what does it tell bandwidth excuse you bandwidth is that correct so the bandwidth of this defam loaded with heavy capacitance CL will be decided by the CL itself is that correct larger the CL bandwidth will be smaller or higher larger loads smaller so this is an important parameter in thinking that if the loads are higher the bandwidth drops is that correct if the loads are higher the bandwidth drops okay. That is why we keep saying the value should not be too high or too low because otherwise it will open or short wrongly or at least we should get the bandwidth of your choice okay there is another term last time I introduced to you the output depending on the input changing the capacitor at this will charge or discharge charge through M4 discharge to M1 M2 or essentially saying change at V0 or V0 going to a value of given inputs will take final time is that work here will take some finite time what is the time it is associated with the time constant associated with resistance of M2 and M4 is that correct the times will be associated with this ours into the sea is that correct so we now want to we are saying how fast the output response when the inputs applied is that word clear to you how fast the output comes after application of an input is that correct and this is defined by the word calls slew rate SLEWRAT slew rate and essentially slew rate is rate change of V0 DV0 by DT what is slew rate DV0 by DT why I am talking a DV0 by DT can you think what is the current in the capacitor will be I is equal to CL DV by DT is that correct so how fast this DV0 by DT appears is called slew rate okay so someone has given me a specs that my defam should have a slew rate greater than 10 volt 10 volt per microsecond per microsecond is that clear or 10 into 10 to power 6 volt per second I am given VDD 2.5 volt VSS minus 2.5 volts I am given beta n dash and beta p dash as 110 and 50 so what is the mobility ratio I assume 2.2 2.2 so that is the ratio I am assuming of course this is a actual data from a actual cadence tool so I am using their model file okay let us say VTN and VTPs are same which 0.7 volt okay I am also missing but we should you should not do everywhere units must appear properly let us say lambda in is 0.04 per volt and lambda p is 0.05 per volt can you think why I gave higher value to lambda p then generally lambda is very close to each other but p will have higher lambda than marginally higher than n channels okay now we also they said that the at least the bandwidth I am expecting is 100 kilohertz preferably more than 100 kilohertz so is that word design clear to you they never say 100 kilohertz they say it should be greater than 100 kilohertz that clear it should be also greater than 10 volt per microsecond the minimum value expected is 10 volt per microsecond you get 1215 fine with this also satisfying fine is that good so that is the word design in analysis we do not give the greater than or something say this is the value so I want to as I said I have chosen this value from my design of a defam and therefore something I am using them something I am actually specify now one can see the current DC current in M1 and M2 is IDS1 IDS2 is I5 divided by 2 is that correct for the biasing okay for the biasing of a defam half the current in the M1 half the current in M2 now this is what something given to us okay I have not been given the value of I5 and that is what first thing I must know how much is the I5 but they have given me some more specifications and let us see why I chose you this example of a design so you should appreciate that analysis helps you to design and therefore you must see design as yours ultimate and look your theory to adjust to that here is something you will appreciate when I say there are three ways or three parameters which will decide my I5 value here are the three they have specified that the dissipation of this defam should not exceed 1 milliwatt okay what is the constraint they are given the net dissipation in the where is the dissipation occurs in transistors resistors are equivalent they are resistors I square R loss is going on I into V is the power loss in the transistors is that clear to you so we will like to see if they say power dissipated is less than 1 milliwatt then we say I5 into V maximum current is I5 maximum voltage is VDD minus VSS which is actually 5 that should be less than equal to 1 milliwatt is that point clear I have they say one less than 1 milliwatt okay so I can I need not get 1 milliwatt if I get 600 micro watts fair enough 800 micro it is better or 600 is even better but whatever I make a choice they should also satisfy the other conditions of fluid and the bandwidth if they do not then I will say which one okay okay so let us say one value I got it from this that this I5 current should be less than 200 micro is that correct which is the condition is coming from this 200 that the power dissipation should not exceed 1 milliwatt okay if that is so I take the second parameter for me first you read understand and then you write first is I5 has given me it should be less than 200 from the power dissipation requirement the second this is you read should exceed 10 volt per micro slew rate I just now said slew rate is I5 by CL I5 is the maximum current available either the last transistor will drive or sink so I5 is the net current available to you so I5 by CL is DB0 by DT which is your slew rate from here we say I5 SR into CL and therefore I am now saying I5 should exceed CL is 520 power minus 12 SR is 10 volt per micro second so 1020 power 6 so I5 should be greater than 50 micro amps is that correct to meet a slew rate requirement of 10 volt per micro second I should at least have a current which will be of the order of 50 micro and above okay there is a third parameter which is also governing the what is it governing the bandwidth okay because pole will be also decided by GM please remember poles will be decided by GM's GM is a function of IDD by I5 by 2 so I will now like to see for a given bandwidth how much is I5 is that correct is that point clear first I say power dissipation I got some second I say slew rate I got some value I say okay I am seeing the third parameter even the GM term is function of I5 and they are saying me this is the bandwidth I should have let us say then I must get some value corresponding to that as well so I got the second value third value is that now point clear this is design this is what part I am doing only partial design rest I will now substitute value but this is how we start designing we are given specs and we must meet every one of them we must meet every one of them yes as I tell you from the power supplied to the ground the maximum current available to is I5 current source is that clear so when the capacitor is charging all of it going through CL when it is discharging all of it is coming to lower transistors essentially saying when the slew rate what is actually DV by it is charging so the full current is made available for the charging full current is available for the discharge okay having done this I have been given the bandwidth as one upon okay now I say I already assumed you can actually find by pole theory that this is the dominant pole why I assume this what is the why under what condition I assume this that the CLC is at least one or two orders higher than all other capacitances and so is our okay and if that is so then one upon RC of this will be the smallest value and that will decide the bandwidth okay what is R out in this case Defam please look at our circuit of Defam and what is R out here at the output what is R out seem RO of M2 parallel of RO of M4 okay so if I do that then I get one upon GO2 plus GO4 to understand why I suddenly shift to GO Gs because addition is much easier than doing one upon R plus one upon R so I this is what the tricks of the calculations nothing great in that okay how much is GS G output resistance of a transistor ? ID is that correct the G oh sorry this is I5 I am sorry but you can see the next line I have written I5 so this is the expression which we are derived and this is what we are substituting so this is ? N plus ? T by I5 by 2 if I calculate this is 22.22 by I5 now they say bandwidth should be greater than 100 kilohertz so we say this should be greater than if I calculate now I5 from this I get a value of 70 micro amps okay so I have now three values got how many values I got first I got 200 micro amps second I got 70 micro I am 50 micro and third I got 70 micro amps now what should be the choice I can make which will satisfy all of them it should be larger than 70 but smaller than 200 so for the heck of it I chose 100 you may choose 125 you may choose 150 is that clear your right to choose those values because then these three conditions still will be satisfied but there is something else will occur and I will show you why that choice is not necessarily always correct you may have to boost this current further also okay how much you can boost up to 200 okay so we will see okay we just calculate for 100 and see what what results we get and then we say okay we are now given that I5 by 2 there is 50 micro amp W by L1 is equal to W by this 18.4 value is not very good value in design okay we will make it 20 in real life but calculation wise whatever I actually have got it I substituted back here I am just to make clear for you other uses are up to division Galata I claim in a exact value Joe Mary I have a dick I have so W by L1 is equal to 18.4 and 3 and 4 is equal to 8 this data actually we will find in design is that clear in design what will you find these W buyers actually we will evaluate for given specs right now I gave you this value and then I only calculate gains there I will be given gain now I am asking gains what is gain of a different GM times are out is that correct GM 1 or GM 2 they are a same so GM 1 or GM 2 are out I substitute that GM is 2 beta n dash W by L into 1 I5 by 2 divided by R a lambda n plus lambda p I5 by 2 if I divide this correctly by substituting these values I roughly get difference gain of 100 okay so in real life what they would have specified that the gain should be at least 100 and then reverse back to find right now I give you that okay find the DC gain why I call DC difference gain sorry not different difference gain because for this expression we derived for differential amplifier with difference inputs okay common mode what is if the CMRR is very high we say we do not consider AB common mode right now in any assuming it is to be so close to 0 okay so I calculated these values gain what else I had to find now I do not know I have got everything I only want to know this transistor 5 which is giving the current source what is the condition I am going to apply on that it should be in saturation and it should have a corresponding RE which will make CMR are very high is that clear so that is some indirect condition in my mind okay so I just want to see whether that can be guaranteed by me okay so I am now going back to my little analysis okay one more specs which I forgot to give you they will also give you input common mode range what does that mean the maximum and minimum inputs which this defam can go into linear modes is called input common mode range one will be minus one will be plus why it will be minus from minus VSS you will calculate is that correct and the upper one from VDD down you will calculate is that correct so one will be maximum from the VDD side minimum from the VSS side so I am actually I do not use range there I am only saying okay VICM minimum given to you is minus 1.5 what is the expression I wrote there please look at this expression this figure common mode range is calculated from here is that correct so that we in same is VDS side drop across 5 okay plus VSS VSS plus this drop plus VGS is input this plus this is this this plus this plus this I repeat VGS plus drop across this plus potential here signs will take care is that correct this potential is this this and this is that clear so if I use this expression which I showed it will be VDS add plus VSS plus VGS one putting the correct signs VDS at 5 is one minus VGS one okay but VGS one is the current in what which transistor M1 is that clear VGS one is giving a current in M1 what is the current in M1 for a DC value of I5 by 2 I is equal to beta by 2 VGS minus VT square is that correct so VGS is equal to 2 I is that correct Pi beta my plus VT is that correct under root of that this is the expression I mean is that correct it is what expression 2 I upon beta or beta into W by L under root of that is that correct is the VGS so VGS one is I5 by 2 half of beta and W by L please remember this is current is half my one transistor will draw half the current plus VTN because they VGS minus VT plus VTN solve this and I get VGS minus 0.922 volts how much 0.922 so if I substitute back in VDSAT what is the VDSAT value for the 5th transistor 1 minus 0.922 which is 0.08 volt so VDS saturation across the current source is 0.08 volt if VDSAT is so small or this is essentially VGS minus VT so small and I is very large so what what can quantity will I increase I repeat if VGS minus VT is very small square term is very small even smaller current is larger than which term is increasing to my balance left and right okay beta and this is constant so what W by L is that point clear so now if I use this value of VDSAT for the transistor 5 this is for 5 if I substitute this value in this current equation for the transistor 5 and assume that is in saturation then I get W by L is equal to how much 300 in an integrated circuit such a large W by L will take huge area is that correct so how do I increase or decrease this W by L this is design this is what we will do so what do if I want to reduce this W by L 5 smaller value say 20 50 what quantity it is you should see from there what should I do now this content is fixed for us VDSAT should increase if you say VDSAT has to increase so if I increase VDSAT VDSAT what should be the quantity here I should decrease increase this VGS 1 should decrease sorry VGS 1 should decrease if VGS 1 is to decrease what should I change there in the VGS 1 expression I 5 is that correct so now is that point clear to you why I 5 is a design parameter because if I want to change my W by L for 5 I have a choice still I 5 with me is that clear so I will increase that I further from say 100 to 125 recalculated back if this quantity has to reduce what is the GM what is the current is in this okay whichever side so if I have still 70 to 200 range so I should now start moving my I 5 whichever side you wish such that this W by L comes to around 50 or so or lower but then it should not still come out of any of the constraint of power bandwidth or this means that is the design word is that word clear design design means given a spec so I did half designs for you given a specification I actually iterate okay I got some values and I start iterating okay this is going to be too high so let us reduce that or less increase that I keep varying also please remember that's new rate which I chose is the 10 but maybe 12 if I choose some other this value will come so we must adjust your values as long as your bars are met and get your I 5 which will satisfy all three of them as well as satisfy sizes they are all dependent on each other this is essentially called defam design okay this is essentially called a differential amplifier design is that clear to you now what design is clear that what why it is called design I am varying to get my specs so you got the point there is some kind of a interrelationship I do this change something else changes if I change is something third changes but I should remain within a bound which is given to me okay I am allowed to change but I should remain within the given bound of specs this is what we say that so this problem is not the universal problem in defam or something this was specifically when we did some years ago this one defam design for my student so I showed you may exactly on price how this vary very rather cadence tool so I told him that now think of it and how to adjust so when first time he comes I show something second time I don't go on system okay now you know what to do so is that correct why I need to know theory because theory gives me where to start is that correct spy you must have seen a spice you have two n parameters you can vary n outputs you will get is that correct so where do you start do you want to have convergence to come after 20 years after you all are already in a strange any job with a big money you want to see just as fast as possible so your solution example I may give you which is very relevant in maths and all of you have done this often if I have a solution fx versus x this is the solution if I start my initial guess here I am iterating iterating iterating and after some years I give up I am not reached here but if I start here I cross immediate okay so this initial guess has something to do with your thinking and that will come from where analytical because analytical values gives you where to start okay and the accurate values will come from actual Simaoka they like actual models all parasitic everything there is that correct but essentially do some analytical analysis before going to system this word has to be understood by all electrical engineers that if you do otherwise solution will never converge okay is that clear so choice of your initial guess come from where some equivalent analysis you do then fix your values around that and let it it read them and it will give fast results then you can design get it fab it will test in time and it will be marketed with that correct so please remember why I keep telling all this why analysis is so important in my opinion otherwise you will keep doing analysis this and it will never come with a design values 100 parameters to vary there is a temperature everything else will come and you keep doing oh you go or you go come so some initial values should be correct so that you get the correct time outputs is that clear so learn analog circuit digital circuit whichever circuits you do for this purpose because at the end of the application is very time scheduled this finishes all that defam I wanted to say there are few more things left please read this Smith Cedra and me okay now we start today with a very important chapter or important topic which is not necessarily to only for us but it is true for every system in the world okay what is happening in Japan why why those reactors are going blowing one after the other the feedback is not working is that correct the feedback is not that is the problem is that clear and that is why I say let us learn feedback a bit okay a typical feedback system is shown to you here typical in the sense I am not saying whether it is voltage or current or anything actually I need not as an a also it should have been transfer function something HS or something we have a signal which is excess how many kinds of signal in electrical we can have excess either in the voltage or it can be current X0 is the output of a system X0 is the output of a system whether what can it be either it can be voltage or it can be correct so how many combinations it gives 4 voltage input voltage output current input current output voltage current output so 4 possibilities are there so I make it generalized statement excess is the signal X0 is the output depending on in real life you can adjust either voltage source or current source or current input current out what are you let us say there is a gain function which is written AOL what is OL stand for open loop that means without feedback with the gain of the so-called amplifier or any transfer function is AOL and they are also are a functions of frequencies then what I did whatever I receive an output X0 I take it back and return it to input in some ratio this is called beta is essentially called feedback factor please remember beta is normally of what kind it will be the frequency dependent or non-frequency dependent non-frequency different should be passive network it should be a passive network typically such that it does not give too much of frequency responses. However this diagram does not insist on it may be j omega also beta j omega also for us this is a to anyway in respect all that is analytically then solving becomes little tougher but otherwise you have to keep worrying about phases but right now assume beta is something constant but please remember this is not a condition we are taking it for simply ACL is a gain which we call when the feedback is applied X0 by excess when the feedback is applied is called closed loop since it is a closed loop we say that gain as ACL is that correct with feedback the gain is ACL without feedback gain is your open loop and close it typical feedback system shown here if I look for this expression X0 is now I have not seen actual sign here if it is minus or plus corresponding signs will also come right now it is more generalized a can be plus a can be minus so I am just putting some values but actual signs will be taken care in actual decisions is that correct for a normal amplifier gain is positive or negative why if I use a source amplifier source follower so do not always jump on it depending on the amplifier gain may have plus value or minus value I am using plus but if it is minus put correct sign okay so X0 here is a some kind of a summation given to you the part of feedback is such that the output of this feedback network is such that the output except which is coming here is some part of X0 so what should be the value of beta should be should we have limit what should be the value of beta should be less than or greater than less than 1 because otherwise the return will be larger than the output then this will further either damped down completely or grow up completely depend on the signs it gets so right now assume beta is less than 1 but that also is not a condition for generalized system and oscillator we will see that it matches okay so in amplifier we say it should be less than but in other cases any why one it should be one or one greater than 1 possible is that clear so we are right now not saying whether beta is whether but general amplifiers will have less than 1 what is the signal Xi coming here XS and let us say the way it returns to you XS-XS negative feedback it reduces the input signal summation is such this is signs are such of course it should be plus it should be shown both plus then only it will subtract AB plus minus a AB plus minus okay sorry then you have minus plus you are right so it should subtract basic idea is the signal is reduced by a feedback okay it reduces now what is the advantage of this kind of feedback you can immediately understand for some reason some reason the gain is not constant okay for some reason gain increases suddenly temperature is for example one of the case then what will increase X0 so what will be XF will be larger or smaller than earlier it will be larger so when it subtracts from input what will be XI then reduce now with that larger gain with lower input X0 will come down let us say X0 goes lower down then this XF will be smaller so XI will increase and a0 whatever it is so this means after few iteration as if whatever happens X0 will become constant so is the greatest thing what this feedback is given to us an amplifier is gain can be made that is the otherwise other parameters will keep varying the gain and I do not want gain to vary gain I want constant is that correct why I am one gain constant because gain means the output voltage which is going to be input of the next stage or next whatever system if that keeps varying that will have also a problem okay so we want to see that the gain does not change okay by using a feedback we cannot negative feedback we can always set the gain to a constant value however this creates a problem what is at the cost of work if you see XI which is XS minus FF XF is beta X and if I substitute back here I believe J omega X0 is AOL XS minus AOL beta X0 so the closed loop gain is X0 by XS which is AOL upon 1 plus AOL beta if AOL is negative it will get minus sign correspondingly a beta is the gain beta is less than 1 but a beta still can be larger than 1 because a is very high beta may be 0.9 0.1 0.2 0.7 but still a beta can be larger than 1 so what will be ACL if a beta is larger AOL beta is larger than 1 what is ACL value will be 1 upon beta 1 is neglected AOL beta is greater than 1 means AOL beta AOL by AOL beta means 1 by beta but beta what did I say it is a passive constant value so what has happened to the gain it became independent of the transistor parameters is that correct but what is the problem with 8 and beta is less than 1 again has fallen down is that correct at the cost of lower downing of lowering the gain we could stabilize the gain that is what feedback is trying of course it is always less than 1 who will use an amplifier is that clear so you cannot have a beta so high than 1 that it will be only 1 upon beta so there will be a dependence of AOL also but that will be over weighed by this denominator and therefore we may say the gain will become closer to a constant value and lesser than normal every time this is essentially the crux of if on the contrary there is a positive feedback then what will happen to this term excess plus XI XI will be excess plus except positive it adds to the input signal then what will happen to V0 XI will further enhance X0 will further enhance so what is the system going it is called growing system okay it will start growing towards infinity or saturated so essentially then it will not remain in linear mode and you will lose the game factors is that correct so positive feedback is not needed or not desired what the feedback we are interested will be negative is that correct I am not saying we cannot have positive feedback in fact you may have a combination which is what the oscillator does we make both positive negative such that they compensate every time so for a given frequency nothing change that is what we will do so is that very clear to you what is the feedback advantage feedback actually stabilizes the gain to a smaller value but it makes lesser dependent on device parameters or environmental parameters because beta is a passive network this is what is achievement of a feedback system now you think of it this can be any transfer function this can be any mechanical input this can be any mechanical output the system will be control in the some system mechanical system you can have optical can have optical system so any energy system can be created and feedback can still do the same job is that clear so it is not that electrical feedback is only 11 we are looking at because we are electrical arrow people will be using everyone will be using feedback very strongly because control is only through how much feedback I do okay and that is why control theory is very very strongly depend on feedback who are the best people which country has the best control people you know now most of the control theory has come from not from Germany or India or anywhere where I see country Russia Russia has the strongest control system that is why with all this big very trash looking systems they could put putting earlier than us because their control was much far superior okay let us continue in a normal feedback system which we will use open loop gain beta close loop gain and 1 plus a beta minus we will accept them actually a beta is covolta return ratio is covoltae since you may read books somewhere some may write return ratio some may write loop gains 1 plus a beta is essentially deciding how much is feedback return okay that is the amount of feedback it is provided these are definitions open loop gain close loop gain since output is part of the output is going to be fed to the input the circuit which allows you to take part out of it or pass on something output to this beta network this is called sampling what is it called sampling is that what we have output is sampled whether to return current or voltage in series or in parallel C4 of them okay so it is called sampling okay at the input the actual input signal is mixed with this feedback signal so this area is called mixing so there will be on input side you have a mixed circuit mixing circuit at the output side you will have a sampling circuit is that word clear so on the output the sample at the input we mix now depending on what many possibilities if it is a voltage okay and return is current okay then is a one kind of mixing will happen current and current current and voltage voltage and voltage all four possibilities can happen in this mixing and sampling okay so let us see there are four possible circuits to a here and to here and very interestingly you will see what do they move this is true for any I just now said all of it I just repeat if the a beta is larger it is only function of beta therefore constant whatever I said is this okay why are we so much worried in the analog circuits or circuits per say negative feedback properties so the first term as just now I said the gain reduces as we said due to negative feedback and this word we say gain desensitivity is called word is desensitivity now what is this word desensitivity is shown here the closed loop gain is a 0 upon 1 plus please remember right now a 0 is not given any sign okay in real circuits I will start putting minus plus do we have is that clear right now I am putting generalized so you may say some last time plus we have any or be general so if I differentiate this with reference to a 0 so da cl is da 0 upon a term I why though term a numerator car a denominator differential so we say if I is that okay differential 1 upon a 0 beta due to a 0 and one of a minus a 0 beta on the square due to the other term if I subtract this value I get da 0 upon 1 plus a 0 beta square so if I define what is sensitivity analysis means the change in that value to its original is that what we are what is sensitivity change in X with or DX by X what is the change with reference to the original value that is sensitivity ideally what we see how much so the ACL by ACL is da 0 this upon I divide by ACL on both side so I get 1 upon a 0 beta da 0 by a 0 I just divided da CL by CL both sides ACL I divide on both sides this is the expression this is 1 upon ACL then this this cancels so I get da 0 by a 0 multiplied by 1 upon a 0 beta what do you think I mean I do not tell you the a 0 by a 0 is sensitivity of what function open loop gain function it is higher okay then what will happen because there is a denominator term which is higher what will happen to closed loop gain sensitivity it will be lower and we say ideally it should be 0. So by adjusting this values we actually have reduced variation in the gains closed loop gains so we actually desensitized it is that correct is that what desensitivity is clear that we actually it was very sensitive so if it was very sensitive here this got desensitized so first advantage of feedback is desensitization or desensitivity so clearly we say percentage change in ACL is always less than percentage change in a 0 and therefore 1 upon 1 plus a 0 beta is also called desensitivity factor that term will be always in denominator greater than 1 so obviously change in a 0 will be much less in ACL because that factor is in denominator okay this is something we achieved that either a close loop problem come that is the first advantage which negative feedback will provide for an amplifier. Let us say you have a bandwidth is decided by which pole dominant pole separated so what do you expect will the bandwidth increase or decrease gain come on otherwise gain bandwidth is gain come current the bandwidth one name on it so how does it here is one which is essentially saying bandwidth enhancement what is first thing we did desensitize the gains now second is let us say you have a dominant pole at omega 0 which is bandwidth related so a mid band upon 1 plus a omega is open loop gain we call this am upon 1 plus a 0 this is our standard a 0 term in the expression which use this is now please remember there are no frequency term now it is first time I introduced frequency term so am upon 1 plus s omega 0 is a 0 omega 0 is your dominant pole or bandwidth pole related to bandwidth so I mean a kya kya a 0 ke jago a function deka no function a 0 upon 1 plus a 0 beta a 0 ke jago am upon 1 plus a beta ke laga deka so what I got am upon divided by 1 plus sm r 0 1 plus m same expression all that I replace a 0 by am upon 1 plus s by omega 0 I collected the terms and I got ACL 0 which I defined as am upon 1 plus why this word 0 is coming am is fixed or not what is am mid band gain constant is in a mid band volata a mid band gain a open loop or feedback kya to 1 plus am beta but am is constant beta is constant so the mid band gain under closed loop is ACL 0 what is mid band value we have a 0 we call 0 no there it is ACL 0 so ACL 0 is am upon 1 plus a beta and the denominator I have one upon s times or division may I have 1 plus am beta times omega 0 is that point clear mathematics it is a avatar ACL is AOLS upon AOLS beta AOL is am upon 1 a substitute kya yamper terms collection shuru kya and then am upon 1 plus am beta divided by s upon 1 plus am beta omega 0 plus 1 if we define this numerator as ACL 0 then it is ACLS is ACL 0 1 plus s upon 1 plus a beta times omega 0 the pole kya gaya be a beta is higher than 1 is that clear us so band width kya yam beta times the original bandwidth became the new bandwidth is that correct so feedback say kya disadvantage dikha tha aapko gain reduce over that but uska advantage mila me uski band width hanko additionally mid way ok mila aisa kya hona kya ya tha jaroori gain bandwidth constip hai abhitha kham ne cascord ni kya kya so I repeat what I said the gain is open loop closed loop gain is am upon 1 plus at 0 is am beta and the new pole is omega 0 1 beta which is your new bandwidth frequency because this is larger than 1 so omega f becomes much larger than omega 0 so we say bandwidth extension is is that point clear to you what two things I said today one I said by negative feedback I will reduce the gain ACL is less than you but if I do that I will get an advantage that the bandwidth will go a beta times 1 plus a beta times a beta times omega 0 will be new bandwidth so gain bada so frequent bandwidth come again come kya bandwidth bada bode plot me kya kya aapko bada bada bada bada bada bada bada bada bada bada bada bada bada bada bada bada bada normal amplifier a feedback kya or let us say a beta is 10 for the sake of simplicity yada other but a value ke liye that is so bandwidth kaha on iti abhi a decade agai yara times so let us say this is 11 omega 0 yadi a line me ne upara si drop early abhi abhi mera second pole hai dominant pole abhi aap kya kya kya kya kya kya kya kya reduce one aap yaya utne i terms 11 times okay somewhere abhi a exactly to me kya batang but somewhere here so game 11 times which aaya bandwidth 11 times agai to be this is exactly what Bode will show you we will discuss that word what we call stability you mean new stability but in this case I am telling it will not so is that correct this board a plot correct. So, why do you want to do this but it is a advantage over camilla gain desensitized only a parameter dependency come correct is that correct which is case are you have a reduction currently say I was not looking for this I was only interested the gain became constant which I achieved but in a by a product by a product I got higher so a feedback it now interesting hey about why we are so worried about feedback is or why we study so much system is controllable word is controllable so what do we actually change for control the feedback how much feedback we should provide so that the system becomes controllable within the range is that here so any system you choose please remember this has nothing to do with electrical per se we will only look electrical every time but otherwise feedback systems are nothing to do with electrical input outputs if greater or less go here is essentially feedback is that clear so conditional jabbi kuch outputs let that must come through some feedbacks is that clear next time we will start with something more about see desensitization of linearization this is very important for us a once for all may be around or a word Barmar may use nahi Karuna what I am talking in future is this I have signal XS XI is the input to the amplifier and XF is feedback signal X0 is the output and then what is being mixed and what is being sampled if all three of them XS XI if XF is voltages and X0 is also a voltage then the amplifier must be mixing voltage in series and sampling must be shunt kind which will be current we will see later I repeat we will have voltage mixing in series and current being sampled or actually voltage being sampled converted to see this right now you only write down so this amplifier when the all three inputs are voltages and output is also voltage and mixing is what we call series shunt and sampling is shunt these amplifiers are called series shunt amplifiers first stage series shunt okay first yet you know voltages and a b voltage a mixing series their output sampling shunt a is for a it is called kind of series shunt amplifier yet the XS XI XF voltages and but output currents then we will say mixing since is going to be at the voltage then it is called series series is that correct series by same logic if the inputs are currents and currents but the outputs are current or voltage it will be called shunt series and shunt shunt corresponding to this other names. So, first amplifier is shunt shunt a series shunt what will be the output voltage inputs what is voltage so what will be the amplifier voltage amplifier input here is voltage output current so what will be the trans conductance amplifier I by V okay the third input current high output be current high the current amplifier and forth input current high output voltage is the trans resistance amplifiers okay so we will have four possible amplifiers not do all of them and I will do at least two of the four and we show you that similar thing can be done for the other is that correct that will finish our feedback for normal amplifier see you then.