I mean this class as well as evening class of 5, I think we will be doing the actual design of an opamp. So those who feel that is not relevant mean not come in the extra class but today is the day when I will actually show you how to evaluate for given specifications. How do you make a choice of different doubly bias? Okay, so coming back to the issue which again will be used in the opamp design is the stability issue. Please remember the as I keep saying by joke that when you design an amplifier, you tend to find that it does not work an amplifier, it acts like a growing system, instability grows and vice versa occurs when you want oscillator to work, it does not oscillate at a frequency, it dance down. So the issue of probability is related to something feedbacks and we have been looking into feedback for quite some time, I may quickly say the closed loop gain that is there is a feedback available which has a beta network and in this case the beta is also a function of frequency, like a capacitive will always be a function of frequency. So beta is also a function of S, in the amplifiers which we designed earlier in our earlier second year classes, there we always show that you know series resistance, series feedback through our source resistance or so you always took a very simpler situations. But in internal circuits like opamps, the capacitance will play dominant rule and therefore beta is not constant like R2 by R1 plus R2 coin, but it will be something like a function of frequency. We also defined that the loop gain is called the AS into BS, call it A0 if you wish. I think that 0 stands for open loop gain, not the DC gain, it is an open loop gain maybe you should write well, then at any frequency the loop gain can be written as A0 j omega into BJ omega and in a phasor, I suppose all of you know phasors, the magnitude e to the power j phi omega is the phasor representation of this function and where phi omega is the phase angle. Now if I really see clearly that at a given frequency omega 0 and I figure out this phase omega at that frequency is 180 degree, then we figure out that this loop gain at that frequency is essentially minus but real because there is no cos 180, sin 180 cos 180, so it is minus 1 and therefore this quantity is real and negative. Now if you can see what is happening, if this quantity is negative and real which is what this phasor is telling me at a given frequency, I am just trying to recapitulate what we did or what we did not do correctly or otherwise or what you learnt in second year. The reason why I am interested in doing this kind because ultimately there are number of ways stability criteria can be tested for any second or third order transfer functions control theory course if you are really doing it or have done it, you are well aware of all other techniques, root locus techniques or the liquid criteria or T methods but we are interested to show only one method which is the Bode's plot method because it gives you some direct relationship with the components which we are going to use. No one to say that those techniques will not result in the same but this is only to take case that we are only looking for Bode's plots. So for the Bode's plot all this analysis is done only for the sake of finding out when stability has or when does not have is decided by looking at the Bode's plots. So if you see this quantity is negative and real, obviously denominator is going down because that quantity is real and if this quantity real obviously closed loop gain is higher than open loop gain okay which essentially means that growth is starting okay that means closed loop gain normally should be less than open loop gain but in this case if that happens at that degree 180 degree above we will figure out that this quantity will then push closed loop gain higher and we say instability is set in simply because positive feedback has started. So obviously if I do not want positive feedback to occur so the loop gain maximum the degrees up to which I should go is 180 degree or preferably lower so that this positive feedback does not come into that is the stability issue okay. Typically if you plot the loop gains for magnitude and phase one can show that this is the standard Bode's plot per loop gain versus frequency in log scale same way omega versus phi j omega goes from 0 to 180 degree and this 180 may come from the network one may come from the phase of the amplifier itself. So in phase component may start increasing as feedback increases and therefore more and more growth can be actually seen instability occurs. So now how do I get rid of this or how do I figure out that I am safe okay. So that is what the condition which we are going to show if the beta network is constant that is resistive that is much easier to see from the Bode's plot how much is that stability issues are but if it is not then probably you should plot loop gains okay I will show you the other figure also just note down I think these are very trivial okay just to show you typically there is a pole and right now I am assuming only single pole system you can have multiple poles and we can have two slows down okay 20 dB per decade may become minus 40 dB and third may get minus 60 dB this is just to get a single pole and this is the wave phase associated with this. So if I really plot for a given case for a say two pole system this is called second order system which has a feedback and if I plot loop gain in dB versus log in frequencies the first pole occurs somewhere at P1 okay and that is our bandwidth okay and then the way it is chosen right now is this pole all the way a single pole is occurring till the gain bandwidth point is reached that is 0 dB gain is achieved. Beyond that let us say second pole occurs at GBW okay at it may occur earlier or it may occur later but this is the mid case where the second pole starts at GBW itself then the fall will be minus 40 dB per decade 20 from this 20 from the other and if you now see if you plot this into the in the case of phase values at every pole we know 45 degree it should show from 0 to 45. So it is shown since it is shown from minus of this I am starting from 180 so somewhere at the pole it must go 45 degree down from this maximum value 180 so at 135 this is the pole and till the second pole occurs there is no another fall should occur so it should keep constant but at this point you want another 45 degree from here that means at 45 degree this point must get hit which means the slope should be then 90 degree like 45 degree per decade so in two decades it will reach phase of 0 degrees. Now this essentially is trying to tell that the when the gain bandwidth product point or GBW point that is the loop gain is 0 at that time the phase is not 0 phase is not 0 but it is some positive quantity shown here what does this essentially means it is essentially trying to tell you that before the this reaches 180 degree when the positive feedback starts the gain has already become negative so the numerator will start increasing because gain itself has become negative because phase has appeared already from there. In this case gain was positive and if 180 point occurs then then there is the issue of positive feedback to occur so one can see from here the I can have the second pole start from here or I can have second pole start from somewhere down of course it may still pass through this point the point is if the second pole occurs somewhere earlier to this that means 45 point will come somewhere here and then it is likely that GBW by that time it would have crossed 180 because if it is occurring earlier at 135 next 45 down it may reach 180 even before gain bandwidth point is reached which means you are guaranteed in positive feedback system because gain is still positive. So this means the minimum the I should not say minimum the safest point because 45 degree is from here that means the safest possibility is that 5 phase margin what we declared is this value should be at least 45 somewhere here also it is safe I am not trying to save but you do not know how much pole is then away from this when I decide whether you will come very close to GBW or you may be even reaching 180 before GBW is attained. So for the safety margins that 40 we know is 45 per decade so even if one decade shift is there you are just at 180 okay if the pole is for one decade ahead earlier then you are actually hitting 180 point and any problem starts in paracetics because in the circuits we believe that we have assumed all the paracetics properly but there are metal line in actual chip there are metal lines there are polylines there are metal poly metal oxide metal lines there are many lines at some nodes they may contribute to some paracetics okay and those each capacitance will give out another pole and let us say it happens to be earlier than this then your stability criteria is in hewires okay so at least 45 will save you up to 180 okay another pole will come so okay 180 will hit. So you are still in the margin of safety you may say 30 degree is also safe 10 is also safe yeah I am not saying it is not safe but for the circuit performance to be guaranteed the minimum phase margin you should keep is 45 degree please take it 0 above everything is safe even if this point is reached just here it does not matter okay because it is fair enough that is still not positive feedback but that may change because of the other paracetics for example even the gains which you thought on a single chip it may have some devices may have different W by or different parameters and the adjoining transistor or adjoining other chip may have not the same value variability which may again push you to other values which may actually have that means some chips will be stable some may not be at that risk we cannot take. So the minimum phase margins you should hold for any amplifier is 45 degree so please do not think that 0 is not safe 0 is safe 0 plus safe but safer is 45 it is higher it is even safer okay now to make it higher obviously you can see if this pole occurs even far not from here but somewhere here that is even safer because the second pole 135 value will now reach even closer this and by then gain bandwidth has already more than 135 degree it is showing okay that means more than 45 it is showing so if the pole second pole occurs beyond gain bandwidth point safer for you always safer but minimum it should occur is at the point itself because the pole should not start or second pole should not start earlier than gain bandwidth because otherwise that 45 degree in risk situation I am not saying still it is safe if it is close to that still it is safe but not very safe to be a 100% safe I would prefer the second pole should be outside GBW point or at least at GBW point is that clear so why we always look for that GBW point is at this point we are looking for phase because phase margin is measured at that point is that correct so at that point we like to see whether sufficient margin is provided for me so that the loop gain does not allow you positive feedbacks is that clear so that is the way all stability people start looking is that clear now this case is shown only for two poles if there are multiple poles multiple problems but in general there are techniques which you have learnt splitting poles or nulling the poles or nulling zeros we can do tricks to actually get rid of some of these poles problems and still can attain what we call stable amplifier is that clear the two techniques which will use later in my design one is using the pole splits the other may be nulling the pole itself okay by 0 so this issue why this 135 or 45 should be very clear that why keep everyone says 45 45 45 is not sank or sank otherwise but one pole another pole may give you another 45 degree per dB decay that may create huge problem for you in stability and to avoid that always go beyond 45 so the minimum phase margins which amplifier use is 50 degree minimum then you may say can I use 90 yes you can but then the what else it will happen will have to see if you keep increasing phase margin does it affect something else okay and if that affects then we will have to go back okay reduce the phase margin so it should preferably I may show you later somewhere that between 55 to 70 is typical phase margins are used in almost all analog amplifiers is that correct 55 to 70 72 some designs are but around that value and not beyond 75 any time okay is that clear to you so these are not figured out by just by we do some evaluations we do some experiments actually on chip and we form by knowing parasitic now we figured out okay this is the range at least you should hold so that stability is guaranteed please remember we are always thinking of two pole system if there is a third pole more values more problems become to see to it that the third pole actually never occurs before GPW at any cost okay at any cost okay that should be moved away or removed moved away or removed whichever you can that is the best for you okay the technique which I do not know my second year then second year students remember just to give you that interesting figure which we use you can always find from the open loop need not plot loop gains of course the assumption in all this decision is beta is independent of frequency easier to do it you can see from here log a beta essentially can be written as log a minus log of maths nothing great nothing simple now if that happens now I plot log a I plot a as the normal open loop gain so I let us say quickly let us say right now two pole system this is my a open loop then I assuming beta is constant which in our last case is not but in case it is resistive feedbacks I can put log 1 beta is the one value because a beta is constant so I draw a line somewhere here which is log of 1 upon beta so what is this value loop gain this is the loop gain so what is this value therefore loop gain is 0 is that point clear this is the point where loop gain is 0 now if you take the phase for this it will reach 45 degree this will reach 135 degree and as long as you are you are within this range your feedback 1 by beta point you are always more than 45 degrees phase margin you can create is that clear to you so the what is the thumb rule if you have an open loop gain see to it that your feedback does not go beyond second pole it should be within first pole and second pole if you can fix your this value there you are 100% stable is that clear so that is how actually from the border plot by just looking you can say oh it is stable okay so this is the technique which normally many books show but I thought that you should know why they show they instead of showing T or a beta they show simply this they say okay this is this at this point so anyone between this if you are 1 beta 1 log 1 beta lies then you are always safe because you are now not going beyond 135 anyway so 45 below you will never reach and therefore you will always be safe in this so a thumb rule is adjust your feedback such that you remain between first pole and second pole as long as you maintain that you will be safer okay so these are some tricks of the feedback networks those who are done control theory that may look trivial but that is what we say there are different way of looking things Bode's plot actually gives you some idea of numbers and that is why I follow Bode's plot now if you say I have I can have phase margin of 45 above okay the issue is not still trivial there is another problem seen because Bode plot is what kind of response you are looking frequency domain you are always in frequency domain but if you see much of the amplifier properties are actually found by what response step response that is how we actually evaluate how the outputs will properly go with the inputs okay so here is a time domain analysis if you do for the loop gain you say you are sorry it is V out by AV 0 so if you vary FIM from 45 to 90 which is what you say by stability is the range up to which you should work and if you plot the output this is called what is it called normalized output versus normalized frequency or time now if I plot this I have figured out for different FIM I have plotted V out okay normalized V out 45 degree is this so at 45 I figure out there is a ringing going on you can see this rises and comes lower and it may take more than I only showed in one but it may take two three rings to finally settle settle okay so this will be settling time okay this will be a huge settling time okay if I increase phase margin which I did of course these figures are not up to the scale these are just a representative scale if I go for 55 I figure out that the peak goes down and it also reaches the constant point faster I have increased to 60 I find very little overshoot and it dies down very fast this dies down come from what is the damping factor which in the second-order function you have 1 upon 2 Q as we say is actually a damping coefficient so we figure out that if phase margin is around 60 the damn the damping is so that it reaches and almost settles to be 0 if you increase further 70 it takes even longer time now to that may not ring but it may take longer time to settle and if you go further it may not reach or it may take long enough to settle is that clear so if you see this time response and you look your stability issue where would you like to pay your FIM somewhere between 45 55 to 70 at best is that clear so in this range phase margins are always chosen is that clear because then I am assuring you that output will not ring even if it does little bit it dies down very fast this is our second-order any nonlinear system you study any transfer function and this can be proved otherwise is that clear so why FIM 45 to 70 as 55 to 70 because I figured out that other than the other requirement which may occur even the time response may create a issue which at time please remember here I am only showing you say 1.4 times sometimes if you lower you can see this may become twice and then it will take hell of time to settle and within your time frame it will never settle is that correct so lower phase margins are also risky for even settling times is that clear and therefore we always prefer from the 2.1 is of course the stability issue 45 is of markers so I say okay 10 degree above or at least 5 degree above so 50 to 70 is all that I should look for in my designs is that clear so you must be finding that every time I stick to 50 to 70 this is the reason why I am sticking to 50 to 70 not that I cannot design at above 70 not that I cannot design below 45 but the risk of stability and transient is very obvious and I must look into both before I decide what margins I have is that clear so this issue which is not some books do some many books do not even consider that why so I thought from the second-order transfer function theory you can see you can evaluate how much is the range really available to you is that clear to you so please take it that these values which I choose yes yes this is omega t is time response omega is radian per second so omega t is radians okay do you regard sky that is theta theta is in time frame okay so having having decided that in all my designs I will keep phase margins between 55 or 50 to 70 and not go beyond is that correct or no not go below these two values so bound I have already fixed this is my range up to which so any design now I will do in amplifiers or opens or anyone this factor will tell me whether I am okay or I am not okay is that clear so the first thing I will what is the design thing will be in design analysis what we did we actually are given function and plotted and figured out what is fine in real life what will be this they say okay this is the game this is this then I will have to make my choice okay this fine I choose okay and then I saw if I meet all my specs and still stable thank you very much otherwise change my fine do it again till I satisfy everyone okay that is what design is otherwise is that clear in phase margins are never specified that is your design spec that is what you will control okay okay so before going to that today we are going to hopefully solve the op-amp issue but before op-amp actual numbers we come let us see the analysis part as well this is a typical two-stage single ended op-amp with parasitics available I think this C1 C2 everything last time I have already explained which values of capacitance as to C1 and C2 only thing now I added is C3 okay and we will see whether that has some influence if not we will just forget about it okay but we will figure it out we will not like to say C3 has no influence okay when it will not have an influence when the time constant associated with this becomes extremely small RC time constant is very very small then what we will say that the frequency at which it will come it will be far away from GBW so damn clear okay but if it does then we will have to actually figure out this pole as well 99.99 this value is because of the W bias parasitics available to you always safe enough to think that C3 does not affect but we will solve that also okay so this is a M please remember this is a defam you have VN input here and the current in M5 which is your ISS current for this is decided from the mirror side okay will actually show final circuit for op-amp there we will put the bias circuit here so what M5 decides the mirror current coming from the mirror because there will be a whatever I parallel it will come here and that is which value I will choose here I will choose the ISS or what we call I5 current is the major current making all decisions for us why it is making most decisions because we will see later this will not only decide the GM's this will also decide the slew rates okay therefore and also power dissipation so everything is going with what current I choose at M5 so another parameter for my design is I5 or ISS okay that decides my everything so I like to see what sizing M5 I should have to create as much current I may have mirror any size but I should have to mirror here with that much current which is my design parameter is that clear so the issues are now getting clear okay that I must look very carefully what M5 size comes I also should see that C3 R C3 is not very important and then as move ahead now I say okay this is a D-fam the output of D-fam which is single ended because this is diode connected loads is fed to the P channel driver which is M6 and right now you you may forget CC or you may right now maybe you can put dotted lines on that if you wish this is then which is what is the purpose of CC it is the feedback capacitor which we are putting between output and the input is that correct now this is the input this is the output so we are actually giving a feedback for the gain stage this C1 we already said is the bunch of all capacitance which last time I had right same way I have said C2 also has all the capacitor plus any external capacitor this there is the issue we will see that if this C2 is a function of external load you have a problem what is the problem that means something else I connect I will have to redesign and chip nothing can be designed as of now we will think yeah C2 is known and does not vary how do we get rid of this we will see whether is there is a possibility is that point clear my worries are chip is once designed there is no something add on there that is it so you do not know okay then you have to design over design for something and over design is over sizing over power dissipations which means you have paid lot of penalty for doing nothing which may never actually come into picture but it comes then you should be ready so you actually thought and put it so if you know it it is independent it does not influence so much then you say kuch bhi aage load of Mireoke so let us see how we do that that is also another problem which opium designers have to look for this C2 influence though looks very strong should be actually not as strong how much I can do is our game because I remember C2 will contain capacitance external to the chip itself which it will drive next okay so this is typically another thing you should other day someone asked so I may clarify for those for others 99 probably new and therefore did not ask or did not know and did not want to ask either the someone was saying is this current and this current is same it will not be it need not be the current is decided by M6 is that correct because gm of this is going to decide the games outputs so what current this is going to pass has to pass through this so if whatever current is passing here if they are same jolly well good you put one to one ratio otherwise adjust ratio here to here to get this current M7 is not deciding the current here M6 is deciding the current M7 gets it and then it figures out if I get connection in mirror what sizes I should keep so that this current translates into this current so please remember though biasing looks to me from this side current source side it is not decided from this it is decided from the gm of this M6 is that point clear gm of M6 is going to decide the gains bandwidth so whatever I you should want wish to pass here will pass here and that must be adjusted by W by L to suit mirror requirements is that clear so never think it is going to the current which will be decided by M6 is that clear issue so this last time after the class some two gentlemen came and they were asking so I thought may be clear today that M7 is not mirrored biasing though it is mirrored it is decided by M6 current requirements is that clear so it may happen in some cases they may be equal but in some cases they may not be equal for gms you are requiring different here so what can you do is that clear so please remember these are design issues we should keep in mind because this designers should be the best analysis people because they know what hurts when that is why designers should also be able to solve problem correctly because otherwise they do not know which is hurting us most we have done this again I repeat for you this the equivalent circuit of two-stage op-amp this is my gm1 evane this is RO2 parallel RO4 C1 CC gm we will also see two cases CC removed and CC put gm6 v1 RO6 RO7 C2 this is the AC equivalent circuit of two-stage amplifier and I think this last time if I am not written please write down these are the C1 C2 values which means in real life once you decide the W by else you will have to verify whether this is reaching C1 and C2 values so there is an issue so this gain function is going to decide C1 C2 values so there is a loop kind here as well adjust your C1 C2 okay please note down because these are of course these are given in books so it is not that I have invented or something just see the transistor and that node which you are looking which parasitic capacitance for each transistor comes there please remember one more thing circuit wise you must know unless the second terminal of the capacitor is AC grounded that node does not receive any contribution from that capacitor is that point clear at this node a capacitor is sitting here and if this is not grounded this does not play any role AC or DC DC means AC for AC it is ground is that correct so unless there is other end is going to be grounded at that node they do not add to so only those capacitance like if this substrate is not grounded let us say floating then the CDBs and this will not be there because substrate is not actually grounded so those capacitance does not exist is that clear so please verify that all those nodes at all the capacitor that node the other end of them should be AC grounded even if it DC means AC grounded down okay and this is a compulsory requirement to add any node any capacitance okay everyone has written down C1 C2 but what are C1 C2 I thought maybe I will explain normally books do not give C1 C2 you I only thought that you should know what C1 C2 actually are coming into our picture of course some books may be giving I say Menage K statement Galatasar this is CGD 6 or 5 6 may confusion that this is 6 7 word huh B1 is across that is the input to the gain stage okay that is the input this is all AC huh though it may look capital non-capital all our ACs no confusion I have once told you now onwards unless said otherwise there is no DC we are only solving equivalent circuit all our AC components okay is that okay do not have two equations. Using the Kirchhoff law nodal analysis G V plus I plus I plus G whatever it is I wrote the equations for the two nodes at V out and V1 only two nodes V out and V1 so I wrote this equation from the equation one I evaluated the value of V1 because I am interested in V in and V0 relationship is that correct so I replace V1 from this relation from here okay then V1 is essentially equal to V0 SCC GM1 V in bracket R1 upon 1 plus R1 SC1 plus SCC do you get them and these are just corrected the terms and figured out what is V1 V1 occurs here V1 occurs here V1 occurs here this does not have this does not have so only V0 SCC minus G1 V1 this R1 is multiplied everywhere and then you get this expression if I substitute this into second equation what do I why I want to eliminate V1 and only get relationship between V0 in V0 and V in so remove V1 you just someone asked me so where is V1 is the output stage first stage gain I am first stage output that we figured out now substitute into second and then write the same GM6 into V1 plus V0 by 1 upon R2 plus SC2 plus SCC minus SCC again V1 is appearing here so minus SCC R0 V0 SCC minus GM1 V1 upon this this now again there is an issue because there are V0 terms here there are V0 terms here so collect V0 terms and then on the right get V in terms divide V0 by V in to get your gain is that clear a V0 term and a V0 term okay if I do that is it okay I want to write why are we doing all this I want to design an open so I am trying to correlate the gain function the slew rate function power everything with whatever spec I have okay so I am trying to get the relations and what is the major relation I will get from these the poles and the 0 because they are going to decide some way my GM okay because pole is essentially a GM by C now that means GM is now decided not just by a factor but also decided by the poles available to you and that is why we are interested to get those expressions that correct is that okay so we get the transfer function of gain V0 by V in of course to be precise you should write it is AVS so it is a transfer function and as most books this has taken just to convert it to what books give so I figure I made some constant as they have given in the expression so that it looks identical to second order system which many books actually tell okay so we wrote M is equal to GM 1 GM 6 R 1 R 2 N is equal to R 2 C 2 plus C C R 1 C N plus GM 6 is there is no S here Q is R 1 R 2 C 1 C 2 C 1 C C C 2 C C this may say term collect Karli him and so I have taken GM 1 GM 6 R 1 R 2 outside so I get 1 minus GM 1 this GM 6 sorry so then we had 1 minus SCC by GM 6 so how the transfer function looks now it is very familiar transfer function something on the top is trying to give you 0 and something from the denominator since it is a second order term I am going to get two poles please remember each capacitor contributes a pole so there are two capacitance C 1 C 2 use so there should be two poles second order system so AVS is M times one my M in the bracket 1 minus SCC by GM 6 totally divided by 1 plus NS plus QSS square a GM R 1 R 2 a bar do no sides and become the only cut guy is that okay this transfer function is very familiar so I thought I should represent a second order stand if you see this Q is essentially the term which will get there the only difference is Q is used with N okay and 1 upon 2 Q they write which is the 2 Zeta Zeta is the damping factor so I did not want to correlate there but I just made some second order system please remember whatever goes with S is 1 upon 2 Q and that is the Zeta Zeta is damping if you see ringing is essentially for the Zeta factor that fine can be therefore represented in terms of Zeta that is exactly what we are saying so that choice of fine is nearer to your Zeta is that okay everyone has the formulae we are still not done design we are still working on analysis but some issues which will want to clear so if you see the denominator ABS it is 1 plus NS Q is a square which can be written as 1 minus S by P 1 into 1 minus S by P 2 partial fractions I expand it 1 minus S by P 1 S by P 2 plus this technique sub books media every control books media but I just thought I will do it for my own sake my will get so if I expand this I get the only thing I made I mean of course analog people know this I said in general the first pole which is my dominant pole is much higher I mean much away from if I take such decisions then it is easier because the second term then I can do so I said is 1 minus by P 1 S square P 1 P 2 then P 1 is because this is N so it is minus 1 upon N so and then that is 1 upon P 1 P 2 is Q so P 2 is minus N by Q or 1 upon P 1 Q so now I have both poles available to me P 1 and P 2 since I know N and Q I can now substitute N and Q in terms of P 1 for P 1 and P 2 N and Q in what terms R 1 R 2 GMC all so now my actual circuit components will actually appear I need not have done this I could have do if you are smart enough you know which term from where you can see those terms from 0 time as constant value you can actually know which terms are coming alternatively solve Christia of law and get the same relationship what is the problem why I said I do not want to do the 0 value time constant on this because I will miss the 0 in the 0 time constant situation you always miss the 0 in our case 0 has a very dominant influence therefore I did not show you the other technique but pole still can be figured out by using 0 time constant value both dominant by open circuit and short time short circuit you can evaluate so using this I have the pole 1 upon R 2 C 2 C plus R 1 C 1 C C plus GM 6 R 1 R 2 which is my denominator now for this P 1 for typical amplifier GM 6 is quite large so therefore GM 6 R 1 R 2 is larger than R 1 R 2 so it is only this and this is always larger than R 1 C 1 R 2 C 2 so one may say GM R 1 R 2 is greater than R 1 plus R 2 all to say P 1 can be just written as GM 6 R 1 R 2 C 3 you can verify this is what the value will get from 0 time constant method okay so that you could have directly figured out but this is what it is and therefore the please remember pole has to be negative is that magnitude wise how magnitude we say positive but actually it is the negative value which you are getting pole we what does that negative value means on sigma j omega axis pole lie on the left half plane okay that is one the criteria will do Niquiz criteria says as long as your minus 1 0 you circle and the pole is within you are stable otherwise not stable so that is exactly what we are achieving by now figure it out P 2 okay I also should say C C is normally larger than C 1 and C 2 okay C 2 is larger than everything but this is larger so we say the dominant pole will come from GM 6 R 1 R 2 R C C so this is the dominant pole the second pole which is minus N by Q or plus 1 by P 1 Q okay the first pole and therefore substituting that I get the second pole we assume that C 2 is larger than C 1 and C C is also larger than C 1 then P 2 pole can be roughly written as GM 6 C 2 C C or GM 6 by C 2 which also you will get from 0 time constant technique so you need two poles from the second order function so you got two poles could you now see this is an issue that output capacitance because P 2 is decided by GM 6 and C 2 if load is higher this pole shifts left side yes okay and we also have said that is 0 exist which is the 0 part of the non-numerator which is GM 6 everywhere the gains please remember where the feedback is going on at the gain stage the poles are hit by the gain stage and the first one which you are saying should come from C 3 but that I am I will show you that I am not looking into them okay okay so rewriting whatever I wrote here in nutshell is the following will look I will say final version may be around if you are not written you can write down here the gain function is M 1 minus SCC by GM 6 1 plus N N P N N Q I already M N Q I already defined the first pole is minus 1 upon GM 6 R 1 R 2 the second pole is GM 6 C C upon this and since C 2 is greater than C 1 and CC is also greater than C 1 this is GM 6 by C 2 and finally this we already derived I am just trying to summarize it and there is a 0 which is GM 6 by CC is that correct so you are two poles and a 0 yes you want last sheet now they are just restatement in nutshell so how do you calculate phase margin for such this I have figured out Phi M is equal to 180 minus phase of each of these P 1 P 2 and Z okay but what will be the phase of Z plus 45 degree that is minus 45 degree so what we do is tan inverse minus is tan minus of tan inverse the sky advantage like us goes flip curtain W by Z is Z by W let us see it now is that okay everyone has written down I repeat this third last one is Z 1 is GM 6 by CC P 2 is GM 6 by C 2 this is the relation I am going to use please check it the 0 is at GM 6 by CC and pole second pole is at GM please remember Phi M is decided not by P 1 so much but by P 2 position where is it so this is C 2 and this is CC and they are going to decide how much stable I will be is that okay is that final because of me so if I have two poles and a 0 last time I said the phase margin I wrote there is tan inverse omega by P 1 of course it should not be called omega normal phase phase margin should be written at the omega at what point where you measure this should not be M this should be only Phi but phase margin is chosen where when the frequency is at GBW is that correct so we write so this need not be called M Phi M generally because of the GM and CC values Z 1 will be far away from GB okay so by the time if the second pole has already gone to minus 80 dB or minus 100 dB so another 20 dB plus will make it minus 80 dB but it already you are well within your margin side so you are not worried because the gain has fallen down sufficiently below and then 0 please remember 0 will push the gain plus 20 dB per decade but if it is already minus 100 okay 80 dB is also okay fair enough okay so that is something you have to understand that this 0 should be away from this we will check this for different value of Z how much phase margin we can evaluate this is one problem we will solve okay so please remember this I changed tan inverse GB by Z 1 because I said it is all or minus because otherwise should have been plus Z by W so I inverted it so I have to now calculate what do I calculate from here no no I will assume phase margin I will check it what is this I will assume this and then what I will get P2 so I am only interested in P2 I will say this is always known this I will make it known then I say okay I have all that I need a relationship between gain bandwidth and P2 because that is the one which is deciding major this for me stable or not stable is that clear so the tricks of the trade is fix this somewhere better this will come nicely you know how if you know this for this given value I get a relationship between P2 and GB that is what essentially now P2 has GM6 by C2 is that correct gain bandwidth gain is known to you P1 is known to you so the bandwidth is known to you so you know GBW so from this I can know if I can get C2 or relationship between C2 and others and therefore I can say W by GM6 that is all the design is about that okay yes I told you know madam it should be head by W tan inverse minus of minus of tan inverse is same so it is called okay now before we quit this area before we start open design there is an issue this was all done with CC available I should have done the other isn't it without CC care so let us do without CC if I remove CC I can see my input poles and out poles or without any Calvary zero time constant technique I know the first pole is one upon R1 C1 the second pole is one upon R2 C2 is that correct since C2 is larger than C1 what will happen R2 R1 may be same is that correct C2 is larger than C1 we don't know right now why because that feedback was adding to C1 now this may be something different than that C2 is that correct O1 plus AV0 dekhar dana wo term nahi hai toh abhi koan thota hai koan bada we will have to check really by the values now if I have this new poles with CC removed is that clear no feedback that is open loop system AOL same equivalent circuit the CC here has been taken away okay now thota interesting case Hora as CC laghanika wo will what is CC is also called Miller capacitor no compa compa compa is doing but it is essentially Miller capacitance so the compensation is also called Miller compensations because a Miller capacitance which is doing the job is called Miller compensations so if we have a CC the two poles are P1 dash is GM 6 R2 and just to make equivalence correct I put R1 C1 upon C2 CC by C1 and P2 dash is GM 6 by C2 so you can see from here in the earlier case it is P1 is only one upon R1 C1 is that correct without CC GM 6 R2 is gain so obviously and this CC is larger than C1 so obviously these quantities P1 dash is reducing so P1 is larger than P1 dash is that correct however if you look at P2 P2 dash which is one upon R2 C2 but we know R2 is R01 R06 parallel R07 GM 6 is always larger than one upon R0 because there are mega ohm this is in tens of kilo ohms okay one upon GM okay so what does that mean so I can always say P2 dash magnitude wise is much larger than P2 is that correct so what has happened P1 has reduced P1 dash by CC but P2 dash has magnitude wise enhanced is that point clear to you P1 let us say it was somewhere here it moved to okay I will show the figure and then it will be much more easier you figure just this word which you often use in Miller compensations or Miller capacitance is split the poles okay it is called splitting poles yeah but that is what I said you also C2 is normally higher simply because the load is sitting there if unloaded case may issue but normal mean external question of course who goes for me say that is the reason but I do not want that so strong dependence which is I will remove that I do not want I should know a priori what is the load okay so I will try to see what you are asking is that okay so if I show you on the figures very now this gives you everything in nutshell if I see now loop gain sigma j omega on this scale complex scale or complex graph my 0 is positive sitting here my dominant pole was P1 R1 C1 which was here and my next pole is R2 C2 which is here okay now these are like this as before CC applied when I applied CC I figured out P1 reduced or other is the other side so P1 dash reached here but P2 actually enhance the other side so P2 dash reached here P1 dash came closer to 0 so even a P1 P2 were closer by putting the Miller capacitor I separated them so one of the case we were all the time saying that there should be a dominant pole is guaranteed if I put a CC capacitance in natural system CC exists from where CGD exists so partially up to split here because CGD is always sitting here but that value is very small this CC may not be that small actually may actually decide to put extra capacitance to suit your split requirements is that correct so do not believe that CGD is sufficient in some cases maybe I am not saying it will never be but generally it is not okay if you use that CGD your phase margin will be close to 45 and then you will have a problem so you want phase margin to be at least 55 or something and that is my worry so let us look at the loop gain Bode plots this is my original pole C1 this is my second pole okay if I put the plots for phase you can see at the first P1 how much should be the phase 45 degree down so this is 180 so I must reach 135 at P1 is that correct and second pole is away so it goes down and reach till 90 it remains till second pole it sees and when it sees second pole at that time it should become another 45 degree which is 135 that is 45 here okay now this 45 is at the second pole is that correct and if I extend it down you reach at this value at GBW you have a very marginal phase margin okay still I would not say it is unstable but very close to unstable instability is that okay so if I do not put Miller I may have a situation of course this still depends on C1 C2 values but typically if you see it may have worst case maybe that you may actually get zero phase margins kind of thing okay or close to that which is unsafe to a great extent is that clear now I say since CC causes old splitting so P1 dash the come over so we first pole now with CC is reached here the P2 which is away now we can have a choice we can initially at least say that this P2 dash should be at the gain bandwidth where at the gain bandwidth point why did I say so because at least at that point then 135 degree is what I am expecting so 45 degree at least I am getting that is that clear I repeat this P2 at least should not before it should cross GBW point at best at least okay because at this point then if I plot the base I at least have 45 degree phase margin is that correct because at that second pole I should get 45 degree so the minimum margin I now reached is 45 so they correct so what should I do here this should continue to have minus 20 dB further and then the second pole should start in which case this 135 point will be somewhere here so the margin will be larger than 45 and then you say sir we are safe okay so this pole how much to split is also the problem starts somewhere else if you keep splitting too much this P1 may actually go into the right half plane so you cannot put CC as much you like and you suddenly find instability created because of the P1 itself is that clear so we took all this top may have me cabula as if this does not exist 0 does not exist both do we have to do it we can play with Z then I have another parameter for 5 okay now this part is what we say controlling right half plane poles control right half plane zeros in a such a way that this CC requirement is slightly reduced is that otherwise what is the base this you increase CC feedback increases this will further go down ahead because you want this second pole should cross GBW so you will prefer what larger CC but then this P1 should not hit the imaginary axis okay if it is imaginary it is still okay very low bandwidth but at least stable but if it crosses then you have a problem you have a 0 equivalent 0 on that okay now this is an issue which makes something else you must think is that clear so it is not just see must be something else there which may change the poles okay so I may add another equivalent pole or 0 whatever I will call so that CC is not increased very much because then I cannot I do not have to play too much with this but still my cutting point is much away from it so that I say I am safe now okay so this is what we will do the method I already explained at the pole from the top value the slope 45 per decay means 45 degree it should hit it should continue until 90 because J vector says 90 degree but if the poles come before that then even there itself another 45 should start because that much slope should change 90 dB then and start decreasing right there but normally we assume they are away so we are allowing this to settle up 90 and when the second poles appear at this point another 45 must show so that is how you graph with without this we see that phase margin is close to 0 or plus but with CC you are shifted because of the splits we are shifted the P2 points such that we otherwise just drawing that becomes difficult so you always draw lines so that you know where to draw things huh if you are really calculating value from this don't because then you actually take a graph and then put on because this is not of course these are called log scales these are linear scales so lock semi log paper you have to learn so phase margin improvement is that this issue is clear last slide and then we will stop we will start with this this is that clear to you as anyone has a doubt now that why CC CC is one which is controlling now next slide everyone so nutshell what do we say the bandwidth gain bandwidth point gm1c1 plus ccg1 plus cc and P2gm6 the minimum phase margin 45 degree forces you gm by gmccc by c2 is that correct this is equal that is at the point where phase margin is 45 so at least you should not go beyond this preferably what you should do gm6 by c2 should be greater than gm1 by cc is that correct and if that happens you have larger phase margin to play with is that correct right now what is the minimum the corner cases gm1 by cc must be equal to gm6 by please remember the gain bandwidth is something to do with the defam's drivers gm1 or gm2 they are equals so gm1 by cc is the gain bandwidth to be actually it is c1 plus cc cc is larger than c1 so gm by cc so the ratio of cc to c2 is ratio of gm1 to this is that clear to you so we must be now from the all that 5m value you choose you probably get some ratio which will be able to get ratio of gm's and gm's are proportional to currents so you will get the ratio of currents that and currents are proportional sizes so you get the ratio of sizes okay okay we can increase cc but we lose bandwidth hence to avoid reduction in bandwidth but to improve 5m for stability we can control flight half plane 0 okay which we are right now thought cushioning here to bring it towards left half plane if I somehow shift that to left half plane that can help a lot to nullify a pole okay and that if happens then we say cc requirement will be minimum just sufficient for you and you are to do this we will see later cc need not be alone it should be r series to cc this will create another time constant r cc okay that will give you the 0 shifts out and if 0 shifts right half plane left half plane you are much of the worries could be that is you sit that pole on something so null cargo that is called nulling okay so this is the thing which will do today evening it last line so now you show you see that okay theory has been discussed importance of every issue has been thought at least if not discussed fully and now actual values will see what do we give That will give you some idea how one gets sizes.


Hello, welcome back to Exotic Wine Travel. I am your host, Matthew Horkey. So we're gonna do tasting of a very small producer today. Actually, I've become a friend of ours. Her name's Joe A. Herd. She's a master wine, a Brit, actually producing small batches of wine on the island of Guar in Dalmatia, in Southern Dalmatia in Croatia. She only produces four labels. We're gonna take a look at these days. Very small batch. She is, we've considered a friend. She did send us these samples, but the video is not sponsored in any way. We're just tasting and we're gonna see how these wines hold up. It's really cool because we have four wines here. We've already tasted one. We're gonna taste three. We have tasted these vintages. No, sorry, we have tasted these wines before, but in the past, like I tasted this one when it was a barrel sample. So I'm really excited to see how it was. We started off tasting last night. Already the Ahern Poship from Guar 2016. Beautiful Poship. You know what? It makes me excited, actually, because she's not doing Poship on Guar anymore. She's gonna start doing Poship from the island of Cortula, where Poship is actually from. And I'm really excited to see how it is. This Poship's beautiful. You're gonna have the right tropical fruit flavors I associate with Poship, the flintiness, the mineralness, the salinity, big body that does need some air.  I still think it was 4.0 out of 5 on the Vevino. I thought it was excellent, but let's move on to the real tasting, shall we? We have the Ahern Wild Skins from Guar 2016. This is bog, these are these crazy varieties, bog denouche, kooch, and Poship on the skins. It is an orange wine. Joey Herd mixed with spontaneous yeasts, all these, as little salt fights as possible. I guess they could be in these hipster natural wine types bars. You can see the beautiful color here. Let's give this a go here. You're gonna love the shrooms behind the camera. Has the yeastiness. You're gonna get with orange wine. Apricots, wow. I had this like six months ago, and this is a lot better than I remember. Orange peel, lemon, apricot, beautiful nose. Little bit of ginger, spice, ginger cake spice, let me give it a little bit of a go. Nice texture, weight. Sorry, I was doing an Airbnb start, I always went with that, but just went fine. Little bit of spiciness. This is so much better than I remember. Do you like the nose on this? It's funky and funky in a good way. Very juicy.  It's a lot of finesse, citrus. I usually take a walk to this side, but yeah, I enjoy it. I actually really like this better than I remember. I'm gonna say, I'm gonna say 4.2, 4.3 out of five on Vivino. And it's a beautiful wine. 4.2, 4.2 out of five. So easily a 90-point type wine, beautiful type wine. It's so funny, because there's a dog running around here and she's trying to keep it out of the house. So, we're moving on here. The next wine we have, we have the Ashern Rosina made of this weird odd grape called Darnakusha from high elevations in Hoare. This was the first wine she released. I tasted this two years ago, not this vintage, but the first vintage she released two years ago when it first came out. I love this wine. Sherene loves this wine, more of a serious rosé. This is more kind of orangey in color, orange-saminy in color. Still exactly what you remember. Got a lot of strawberry, rose, mineral-tight notes. That is textured and big as I remember. Camer, I'm a... Oh, yeah, syringes in. So, you like a lot? What would you describe this as? What kind of flavors? It almost has this cherry syrupy finish to it as well. A little bit peaty. I like that the finish has texture, a little bit of astringency. I wouldn't call it tannin, because it's like tannin-plus acidity that gives this nice fresh astringency on the finish. It's a structured, rosé, beautiful nose. This is a weird grape that you don't see very much in Croatia, made in a variety of wine, Darnakusha. I kind of describe this between... For me, it tastes almost like halfway orange, halfway rosé, with texture. Yeah? Yeah, yeah, yeah, yeah. I'm thinking, yeah, you're right, man. I might be confused and wonder, huh, is this... Wow, I wrote about the first vintage. This is the newest vintage I taste. Wine that's very easily, for me, 4.3 out of 5 on Vivino. Like I said, these two wines above, I think, honestly, gonna score 90 points and above in any time of critic tasting. Good stuff. This is the kind of... Very, very juicy. Okay, the last wine we have, we have the Ahern Plovetsmalli Southside, 2014 from the Southern Slopes of Hvar. Now, I tasted this wine when it was a barrel sample two years ago. Sirene was a little bit scared to taste this wine. We're gonna see how it shows here. First of all, beautiful, true Plovetsmalli color. It's not super dark. Kind of like a little bit... Like kind of a Barbera-y Nebbiolo color. It's got some browning on the edges. Comes from the South Slopes of Hvar. Very steep slopes, right down tumbling down into the sea. Old bush vines. Let's give this a go. You're gonna love it. A lot of strawberry notes, a lot of Mediterranean herb. Rose petal, very floral. Plovetsmalli can be almost Nebbiolo kind of ask when it's done, right? Wow, really good wine. Medium body. You get some tar notes, small grippy tannins. Sirene, taste this. I think you're gonna be more impressed with this than you think. What do you think? I like the texture a lot. It's really smooth and harmonious. What's in the tissue? 2014. Yeah, I like the palette a lot. Very cleaned up. For me, another wine. For me, it's 4.3 out of 5. I think this is an excellent rendition of Plovetsmalli. For me, I really like the producers on Hvar when it comes to Plovetsmalli. The Plovets tends to be softer. So, you guys check these out. That wine, you have to drink it over a couple of hours. It's so new and that it's... Yeah. I got a lot of fine new ones. 4.3, 4.3, 4.2, 4.0. All some excellent wines. I think these wines are available in Viva and Croatia. I think they're also gonna be available in Wine and More. I'll put a discount code. We have discount codes for Wine and More for our viewers. Just know if you do order from Wine and More, we do get a small commission for whatever that's worth. I don't know if these are available on Wine and More yet, but just check it out. I'll put it up on the screen. Let me take one more sip of this before I end. Beautiful wine, beautiful wine. So guys, if you like this video, follow us on Facebook. And please subscribe to our YouTube channel, Exotic Wine Travel. I will see you in the next episode.