 for gain, gain bandwidth product, power dissipations and stability is one major issue. We have done most of the analysis and also evaluated last time from the given data for a phase margin of 60 that C L, C C should be related to C 2 and we said that C 2 is essentially encompasses the load capacitance and total load given to assist and puff. So the minimum value of C C which you can use is 2.2 puff or it can be greater than 2.2 puff. So based on this we continue now. So if I have a choice of making C C I can either choose 2.5, 3, 3.5 or even 2.2 itself because that satisfies the condition of A. As I kept saying earlier larger the C C value I choose, greater stability I will get but then there is a problem of bandwidth. So I do not want to go too far from 2.2. At the same time I do not want to be very close to 2.2 either okay. So I, there are 2 possibilities I have looked into but maybe finally I decide I will choose 1. Either I will choose 3 puff or 2.5 puff. So the further calculations will perform with 2.5 puff as C C. You can choose 3 and do same analysis again or you can choose 3.2 and start doing everything okay. The values will keep on changing, the method may not. The only problem you must remember any value I choose here will have implication of all W by L's and also the gain and at that time you should see that gain should not below 4000. So the issues are little tricky. So I believe that if you close to 2.2 you choose things will happen okay. Because all this phase margin we have taken care of AB 4000 or something. So some way this value is good enough in my opinion. But this my opinion does not count we should check it if necessary. So you know the slew rate is given to us is dv0 by dt greater than 10 volt per microsecond and since our last time I said the current which C C can receive is the maximum I ds5. So based on this I can I have 2 possibilities of I choose C C or 3 puff. This will be 30 microns. If I choose 2.5 puff this will be 25 microns. So I am going to use I ds5 or what is our I ss for defam as 25 microns okay is that okay. So choice of I ss or what we now call I ds5 is now chosen from choice of C C. So it will be different if you have a different C C for example 3 it shows 30 2.4 25. So it depends it is proportional. So 10 so you can even choose higher than this because you are asking higher than dv0 by dt less than volt but I am choosing 10 okay. Why these numbers are given so that let us say in the first round you do not meet specs you are looking for. So you come back and replay right from here okay you say look come back and redo again okay. So I have since I have met the specs I am not going to do second time but in case I would not have probably I will have to do second time. So the an individual arm of a defam the I ds1 and I ds2 will be the half of this current which is 12.5 microns. So why these values are needed because GN's are calculated for these currents okay. So we will like that we also please remember that whenever I write Vgs I will be leaving 1 by lambda Vds term lambda being smaller and then I can always write Vgs is equal to Vt plus under root of 2 I ds upon beta dash w by L. Why this relation because if I know this or I know something here I am able to hit W by L through this relation. So one way of reaching W by L is this expression okay. So I like to use this expression once a while when I think I can I have other things known so I may evaluate W by L for given transistor and in this case obviously the over voltage or excess voltages under root 2 I ds upon beta dash w by L. The second third parameter given to us third or fourth order number is input common mode range it is not really values and form isomer evaluation I should say we have been given Vn minimum and Vn maximum. Please remember this word minimum and maximum are slightly what are you would say they are not very correct what I essentially means the main word is taken from Vss and max word is taken from Vdd okay whenever we calculate but the actual value this Vn min is the max value of that which is required and this max should be the minimum what I want is that clear to you. So anything greater than 2 is fair enough anything less than this also is good enough because swing is larger fair enough so please this slight statement making is very some so please what I am saying the value given to us is the this is the minimum value of maximum which is 2 words so if I get 2.2 long fair enough this is the at least I should have minus or if I am below that I am still fair enough I have larger swings is that clear. So this maximum words are not really the actual max they have taken max word was given from calculation from the uppermost values Vdd value down and Vss value up. So this confusion should not happen that then you see you are not meeting space actually you are improving the specs so what is common mode range is we are saying we are trying to say that in this input swing the transistor remains in saturations is that clear that is the range I have given you so that is if at even real value it will be still besides if you go below Vgs minus Vt may not be smaller than or equal to Vds then it will come out of saturation so these are the values so something going beyond is fair enough it is still if I increase Vds fair enough now how does it hurts me is that clear so this values are essentially taken as mean max but actually they are max means so there is some catch word this which I use so I thought maybe in the end I may show you something different values say oh it has not met in fact it might have done better than what were expected from me. So from this mean mean if you see the figure I may have to check the figure I should keep it separate now so that every now and then I need that we have done this expression earlier in when we derived those expressions you can see from here the Vn essentially is if I calculate from here it is Vss plus Vds at 5 this voltage plus this voltage plus Vgs of 1 is that clear to do this point I start from here this voltage plus this voltage plus this voltage is Vn is that clear so Vn mean is essentially Vss plus Vds at plus Vgs 1 so and this value has been given to us is minus 1.125 you can see that one was added by me because when I did some calculation 1.5 I went out of the way because device came out of saturation so I suddenly came out of okay I put one extra there okay so this expression and Vgs 1 can be written as Vt 1 plus under root of 2 Ids 1 upon Vt lw by l1 is that clear just now I wrote Vgs expression so I reused that expression in writing V minimum I may not use immediately this value which I am saying but these are the expressions I am going to tell you okay this I have finished we also see Vm Vn max max is taken from where from the Vd side so you come from here you come here then you come here then you come here and then you come here is that clear I repeat you come from here then you come here then you come here then you come here okay so this is the Vn max which has the minimum value I want really so that the device remains in saturation now if this function I repeat so it will be Vdd minus Vt 3 or Vt P3 you should write if you wish so that is a P channel device minus this is Vgs 3 I am writing what is this Vgs 3 okay minus Vds sat for 1 plus please remember Vds sat please just this is Vds sat plus Vgs 1 is that clear so Vds sat 1 plus Vgs 1 is given to be as 2 volt okay so if I write 2 is equal to Vdd minus Vt P3 and this is n channel device so please write Vt n1 so and what is Vgs 1 minus Vt so we actually rewrite in that expression so I get 2 is equal to Vdd minus Vt P3 minus this number but this essentially is Vgs minus Vt is Vds okay so that value I substitute as Pt n1 I keep doing this and finally from this because I now know this I know this I know this now the only catch I said that last time I did say that here I have used it 0.7 as the typical value of it is okay and I said you that normally in maximum values you should choose the worst cases or higher ones so that maximum values of base you should choose which will give you that bar better than the normal with this calculation I figure it out W by 3 is just 2 okay which essentially is what is the problem if W by L of P channel device is smaller the resistance is higher so nothing wrong but there the resistance is 1 upon gm3 which may not be then higher lower I mean I want that to be higher so I feel this 2 is slightly smaller value so I did some tricks you write down this yes yes actually if you see that science have been taken this is 0.7 minus no this that in writing we are already taken care of the sign okay Vgs minus Vt so Vgs has to be negative and Vt has to be negative then only so it is always magnitude subtraction because you cannot exceed the value of given to the supply okay so using this these are standard expressions how much W by L I got for the terms of 3 is 2 we also said that in all of the designs M1 is same as M2 by size as well as Vt and M3 is same as M4 by both Vt as well as sizes so once I know W by L 3 I as well know W by L 4 so okay I in my opinion since my choice the typical values of Vt I got into a small value of the have you written down I keep saying these are only 3 ways just write this expression and maybe this which I and this is also sometimes rounded by me so check it because this is I always and calculate at times not using calculators okay is it okay so if I use now Vt P max as minus 0.85 Vt P minus 0.55 that is I say bar of 0.15 is what I expect max mean values to shift from the typical same way between minimum 0.5 and this is 0.85 so now if I say to get the value of V in max I as that has to be 2 would so I must subtract the highest value of Vt P because then the then it should satisfy 2 okay then I should also satisfy the most value of ETN because then it will be the max worst case and therefore it will still should satisfy my two requirements if I now substitute with this I can recalculate W by L 3 and it turns out to be 12.5 which is 12 okay so just think of it if there is a bar on Vt the W by L's can change by 5 to 6 times this is an issue which you I wanted to bring to a notice that for me as a calculation I may say 2 is good enough what is that big about it but in real life the value may change as much as 5 to 6 times if your Vt bars are such high values is that clear. So this the designer should keep in mind that there will be a maximum values always associated with the data given to you even they will specify 0.7 plus minus okay 0.15 okay so they will specify that delta and therefore right now I have used it same Vt and same delta Vt but for P channel and N channel Vt will be different and the delta will also be different so maybe you have to do little more calculations correctly for the real life situation for simplicity I have made equal everywhere so that at least I get the numbers of my kind of numbers I am looking for okay coming back to your VIN what is the VIN max value you can see from here from here if I want to reach VIN one method is Vgs3 plus drop across M1 Vds1 plus Vgs1 now if this value has to be say minimum a maximum for the worst case whatever I subscribe that still should reach to that is the value given to me so when I say okay so I will Vgs1 I know is Vt plus under 2 under 2 current flowing through this so I choose the maximum of Vtp because that is the worst case available to me when I calculate this drop this drop is nothing but Vgs minus Vt okay so I actually now see to it that drop is the maximum available to you so I also you accordingly put Vtn so that the because it will be subtracted out of this please remember Vgs minus Vt is videos so if I have to subscribe out of it I should choose minimum value of Vt so that that is the highest drop across which it will come so that the still it will satisfy your two word requirement this is what I say V in minimum or maximum but what does that mean if something is lower than this fair enough sorry higher than this is fair enough because these values are chosen worst case as in fact they will not be that bad so this value will actually exceed to work but that is nothing wrong with us because devices are saturated even with these values is that clear so anything beyond is still satisfiable is that clear so the value we choose is called worst case situations if that satisfies the saturation condition then the other conditions are always going to satisfy so we say okay we are well within our limits to actually see transfer remains in whatever worry in all this no transfer should come out of saturation if this is the case that is videos must be equal to Vgs minus Vt or greater than Vgs minus as long as that condition I satisfy I am always so age is equal so if that is satisfied there anything higher will videos higher will be better now so I am safer when I say I take the worst cases and then I say okay it does not reach that fine thank you that is the way we actually assume the values point because I am asking more swings see my worries are that how much mean I am allowed in so larger input swing is better and with that also devices define worst fantastic if it becomes 2.2 I am fair enough it is still how much because try to satisfy as long as this conditions is satisfied I am least bothered about saturation condition should not be actually defeated by me that is that minus value become plus so V in max V mean cannot be plus it should be minus as such okay for the higher end of the V ID what is your looking for minus V ID by 2 to plus V ID to win this range device remains in saturation I am trying to see if it is it does not go to minus side remains positive and I do not have a swing is that clear so with the minus 1.5 if I put I find I went the other way subtraction so I said okay I will use that value okay so that still remains negative value okay you do calculations I did I did not bring those but I did and therefore I removed them actually okay so I do now is that I got W by 3 equal to W by 4 equal to 12 you can also for the sake of where you can say 2 it is okay I mean nothing wrong if 2 you put it and everything else is satisfied I have no objection I am using 2 better if it is 2 why because it will reduce capacitance it will reduce also it will increase resistances so everything is fine area will reduce so I should not really boost W okay but I am my worries are that I may I may lose something else and because I know typically it is around 10 so I thought okay I will see that what is the worst case situation I use and I will increase it that is the trick this you will not know because you are not seen actually so because we have seen it we know okay at least boost it little bit and how do I boost it I figure out this is the easiest way of looking at it and we can it will be higher than 2 anyway okay so once I declared so how many things I have now calculated I have calculated W by L 1 and W by L 2 W by L 3 and W by L 4 so different cases are now fixed by me okay sorry one we are not yet reached one we will calculate we will calculate from some GM 1 requirements when it which will come from the other area we will calculate yeah I am sorry only we have calculated 3 and 4 okay you are right very good now as last time I said before we proceed further our assumption is stability was that the pole at pole 0 at C because of C 3 in the different does not contribute to any stability issue because these poles and zeros are far far away from the the last year also okay the issue I assume Z 1 even it is higher than that and if that happens I say it is so far away and fortunately both are on the left half lanes so actually it is not going to affect you in any sense as far as you are this but this statement cannot be taken on a face value so I did evaluate and check it whether those values are available in real life if you do not calculate 99.9 nothing goes wrong but for the validity I must say I will calculate this is your define this is very equivalent circuit okay this is your input then C 3 is the capacitance G 1 GM G 1 G O 3 why this is GM 3 because this is diode connected so 1 GM 3 and this is my VGS on the other side from this arm GM to V in GM to V in plus GM 3 V now this current which I am current I am showing is the current flowing in this is actually getting mirrored is that correct it is getting mirrored so that mirror currents are shown here so this GM 3 V in G O 2 G O 4 in parallel and if I solve this circuit which you should be able to how do I solve these are resistances these are currents which I calculate please solve this this is a standard defined theory I do not ask me to redo the define again okay so I calculated the current transfer game for this define which is V 0 S by V in S and it turns out to be why I use GM sounds I do not suddenly because I say 1 upon RO 1 upon RO becomes difficult so if I write G is I just add G so sometimes I do G sometimes I do R you can make your choices as much this is if you know myself I have been teaching this courses law many industries this G business is with him he always like to write GVs is I so that is why sometimes when I am teaching with him I also follow his way so these expressions are picked up from my other old slides so okay so I got this big expression GM 1 upon 2 G O 2 G O 4 GM 3 plus G O 1 plus G O 3 upon GM 3 G O 1 G O 3 plus SC 3 that is the impedance due to this solve Christian law and get this expressions plus 1 now I can say GM 3 is much larger than G O 1 and G O 3 and then we start being some interesting calculations I get minus GM 1 upon this this can be expressed as 2 GM 3 plus this 1 and a GM 3 plus 1 mega so that becomes 2 GM 3 plus SC 3 upon GM 3 plus a 3 when we say GM are larger than yours okay for me if you keep doing the same this we can say the third pole you can say from here third pole is because of this which is minus GM 3 by C 3 this 0 means this is DC DC gain this is pole this is 0 so P 3 is minus GM 3 by C 3 which is GM 3 upon 2 CGS 3 by 2 CGS at that point CGS of 4 and CGS of 3 are in parallel so I just made 2 CGS now what is Z 3 2 GM 3 SC 3 is coming so it gets 2 GM 3 upon 2 CGS 3 which is GM 3 by CGS 3 so which is higher the 0 is still at higher frequency than the pole 3 yet we are not the jada and a come a come hand over jada high denominator is smaller so the value is larger denominator is higher means value is smaller is that clear is that okay therefore Z 3 is twice of P 3 that means there is still ahead of P 3 but both are always sign minus left of plane fantastic no issues okay they are far far away from P 2 they are far away from GBW so we say okay if values which I calculate for this comes out to be in giga I say thank you very much okay so now to do this have you ever noted down as I said this I am only showing you for the sake of this you should not say why I leave that so I actually taken a data and substituted this what is typical CGS should be 2 3rd C ox okay is in saturation typical value which CGS can be given is of course there is a overlap capacitance so you may add something additionally but typical value as a this we can choose as 2 3rd C ox but that C ox is per unit area so what is the area of this transistor W 3 into L 3 what is the technology I am using pointed microns so I choose lengths as pointed it is not compulsion I am choosing it you can choose some other values of length and still do everything see us from the data sheet is 2.47 femto farad per sending microns square 2.47 10 to power minus 15 femto farad are not 15 then 2.47 femto farad per microns square is the value of C ox I multiplied by 12 micron by 0.8 microns W 3 by L 3 into L 3 this is in micron square now is that clear so I do not have to write 10 to power something because this itself is now taken in micron square so I just multiply it so I get the capacitance CGS as 15.6 femto farad okay is that clear CGS I want to calculate G and 3 because the pole and 0 are related to G and 3 so I substitute G and 3 is 2 beta W 3 by L 3 ideas 3 what is ideas 3 half of ideas 5 which is 12.5 microns so if I substitute beta P as 50 into 10 to power minus 6 ideas 3 is 12.5 into minus 6 W by 3 W by 3 is 12 I get 12.24 into 10 to power minus 5 C months W if L is pointed it cannot be 12 okay so some small number will change you are right I am sorry I probably assume one and did that but you just calculate from there you are very good no no so it does not matter so that I say I am not changing you are right but he is right for a pointed micron W by 3 it will be more than 12 okay may be 10 by 0.8 so around 13 point something okay instead of 12 TK so if I got so what is the pole now GM 3 by 2 CGS so I got a pole of 3.9 gigahertz if I add CGS more it may slightly go down because then it will be slow it may come to little 1.8 gigahertz or 2 gigahertz even with additional capacitance if I put I mean I am still in the range of gigahertz okay please remember this is not gigahertz this is in variance per second so of course co-frequency level 2 power say divide karna chaiye so F3 jo hai pole ke li FP 3 is around 625 megahertz where FZ 3 is 1.25 megahertz gigahertz how much is your game bandwidth given to you 6 megahertz 600 10 times. So we can say that C3 the poles and 0s due to C3 for the case we are solving is of inconsequence as far as the stability issue go otherwise you may have to even take transfer function which is third order and then equations will become even more complicated is that clear in our calculation we thought C3 does not exist so I thought why I did not take it so I wish to verify myself okay I said okay I said you but maybe for other case because I had data which is from other technology 5 micron so I said let me check with the data available for 0.8 and then I figure out that even with this I am far far away from GBW and therefore P2 also so anywhere I am not close by okay so I say do not worry too much about C3 so that is why in all analysis in many books they not even they will not show this C3 okay the reason probably is for a given technology this will normally not affect any of your stability issues and therefore they do not consider but for our sake I want to now clarify that why designers sometimes do not tell you but use it anyway so this is what they use it they know by experience this will be too far away forget it okay forget it the reason why because CGS is in what value so that number ratio itself is telling you that you will be far away from this situation CGS is in the order of femto ferrats okay this is the reason why it will always be far away from it is that okay to everyone okay so now once we say that okay I am not going to utilize this transfer function for 3 though I did this evaluation for the heck of it now I said I will like to calculate GM at W by L1 so here is the calculations we know game bandwidth product in radials not in if it is in hertz multiplied by 2 2 pi so it is GM 1 by CC or which is same as GM 2 by CC and that is given to you as 2.26 mega that just writing 6 mega is not correct that is not the frequency should be written that is in actually radians okay so you should write in radian process you should always multiplied by 2 pi to make it hurts so if this is the value which is given then I calculate GM 1 GM 2 is 94.2 micro Siemens S is not second micro Siemens so what can I calculate the IDS 1 is known beta n dash is known so all that is missing is W by L okay so I substitute these values here okay and I get the value of W by L equal 1 is equal to W by L2 which is equal to only 3 which is correct the reason why the lower transistors are always because of the capacity effect they hold that should be smaller than the loads is that correct that is how the large current recipe channel device this is N channel device so matches to be there so N channel device should be smaller than P channel device for same currents to flow through okay so this value which I got is now you can see the value I got is 3.23 in our layout editors there is nothing 3.23 it is only integers are allowed okay so never use very random numbers okay always try to go to the nearest integer and in most cases I did not do it you should actually make it in binary numbers 248 kinds in layouts the simplest way of doing designs are 248 but the problem is if I use I have to reevaluate for all of them and check whether I am right so I am not doing it so I am keeping 3 but in real life simulation you do not mind is it not so you will put that value at higher value also and check if it works out fine okay why I am not doing it I am telling you as such this all values are never actually used by us in designs okay because the layout editor does not give accuracy of such second decimals okay so we normally do not play too much nearest value to that we use and we also check right now of course I do not know whether this is worst case we always see to it that the worst case is chosen so that this also if it is satisfied if I choose higher value it will always satisfy this way I do not know whether it is the worst situation but I just rounded to 3 what should be the what what values I got now W by 1 W by 2 W by 3 W by 4 so defam came main charo transistor I know now their values so which is the last value in this defam the 5 which is your source okay since that so that the next calculation should be now performed for W by L 5 is that clear all my procedures are based on individual how to reach each value of W by L yes length better if you increase the length the lambda change length is allowed to be changed it is not but again it should be integers let us say it is 0.8 the next value will be 1.6 no but lambdas will vary at different length lambdas are different so you have to from the spice you have to find which is lambda I agree with you I choose to let us say for that okay for that 12 value you must get your R0 correctly for the lambda to different now is that correct if R0 changes all gains will change all poles will shift so there is an issue which is not just by putting 2 okay you can always use it of course this value will never come less than 1 okay the least you will get is 1 okay but in case it comes this is doable okay is that okay so the next value I am interested in this so I use my mean minimum value we decide my this is 1.1 to 5 I rewrite that expression anyway so I got we decide 5 from this calculations ideas when I know beta and as I know W by L1 I know this I know this I know this I know so I calculated 5 as 0.246 a minus of minus is 2.5 plus so you are back is of you must have observed there you know I wish my side line like a bar one point one two card them in the school 1.5 say what bar gag now you just look at it if I put 1.5 so cover cover the data randomly choose Karthik Hena to get problem was like that you got the point what could I if I put 1.5 here this will have become negative okay so I just was met why come Karthi I mean so that it becomes at least positive this is taken other side now okay so once I know beauty side 5 so what can I calculate what is beauty side 5 is equal to by current I can calculate ideas sat is ideas 5 is beta n dash by 2 W by L5 into we decide square pjs minus pt so I did ideas 5 expression liquor and from there I evaluated W by L5 okay which is 7.58 which is rounded to 8 why I am saying I am not worried too much because these are the specs you will start with your design and verify is that clear but if you have no such value you may choose 16 you may choose 4 so where do you initial values of W by L be chosen this gives you very good guess for that okay so in 2 or 3 terms of spice if you will actually get the all specs correct punch away what is the next one now the gain stage okay you know I always say the defam defam is equal gains but normally defam is not still gain stage the next stage of amplifier is called gain stage okay now if I say W by 6 I go back to see my phase margin relations I figure out from there if you check our earlier GM 6 by GM 1 should be larger or equal to 10 I assume the equal value GM 6 10 GM GM 1 I have just calculated okay so GM 6 is 9 42 micro Siemens from there what do I what can I calculate what is the next thing I am interested in because there was a data hey we out max what is we out max from your expression output per VDD minus VDSAT 6 okay minus now calculate VDSAT 6 if I know GM I should be able to evaluate VDSAT 6 to do this I do little trick I write idea 6 expression which is half beta P W by L instead of this I write VGS 6 minus VT 6 square into this differentiate okay so I know GM 6 equal to half half cancels 2 cancels so beta P 6 2 of this so your VGS minus VT but that VGS minus VT is VDSAT so I got beta P into VDSAT 6 as my GM 6 is that okay so now I have W by L 6 is essentially equal to GM 6 by beta P does 6 into VDSAT 6 but VDSAT 6 okay that I will now calculate is that okay to all from here I got a relationship between GM 36 okay madam sorry well different shape may jay or a VG K cell and a lambda term is small so you can always start neglecting whenever this GM's calculation starts whenever GM calculate start so W by L 6 is GM 6 upon this what is V out maximum we are given to you VDD minus please look at the figure just a minute VDD minus this is the view is that clear so I say okay VDD minus VDSAT 6 is my V out max so VDSAT 6 is VDD minus V out max so 2.5 minus 2 which is 0.5 is that correct if I know this if I know this I already evaluated this so what do I calculate W by L 6 that is the trick in design once I get VDSAT 6 then I know V out max is VDD minus that so I actually evaluate VD out max is given to me as 2 volt so I get VDSAT 6 as 0.5 volt so W by L 6 is GM 6 upon beta P dash VDSAT 6 substitute this is 50 10 to the power minus 6 this is 950 to 10 to the power minus 6 so W by L 6 is 37.67 yeah abhijay and that is the most worrisome part in design okay that is what I am going to tell you why I this value I am not accepting I said fine W by L 6 if I calculate I get 40 value roughly and for this the ID 6 is 235 micro amps what is the maximum IDS 5 I am playing 25 micro amps this is roughly 10 times I am pushing in the last transistor okay size of W by L of 40 that is not very much good because this seems to be excessive power consuming device okay because I said there is something I must have done wrong or at least I am not getting the correct value which I should get okay so I said okay I have another way of calculating W by L let us check with that now you can come back to this circuit do you see this M4 M6 also forms a mirror your connection the other up go actually can't hold on circuit but actually connection the high it is like this current there is no everywhere gate so this mirror is mirroring in this in the ratio of M6 to M4 I just let you this VGS is equal to this VGS okay this VGS minus V2 is this VDSAT which is same as this VGS if this VGS this is very close to this value this current will be mirrored in M6 by some ratio of size M6 to M5 okay you try this this is what we did okay my expression okay so ideas 6 by ideas 4 is ratio of sizes however that means and that is the verification what does that mean VGSG6 equal to VGSG4 you are condition mangana now I am going to go is condition clear equal to Gm6 Nikala Gm4 Nikala inco yes so yes G6 is equal to VSG4 so I calculate Gm6 I calculate Gm4 the ratio of the 2 is W by L6 by W by L4 so as long as I prove you that these are equal my assumption of Gm6 by Gm4 is same as W by L6 by W by L4 Gm4 I can calculate why because I know this current I know the size so I calculate this as 12.24 10 to the power minus 5 micro Siemens or 122 micro Siemens Gm6 by Gm4 is ratio of W by L6 by W by this this I know this I know this I know is it okay everyone written down Gm4 is 122 micro Siemens either 0.4 micro Siemens so that Gm6 by Gm4 into W by L4 is W by L6 which is now 92 how much earlier 40 now I increase it to 90 okay that seems to be but if I now calculate VDSAT6 for this I get the value of 0.203 then I calculate the ID SAT6 current for this which I get Mohammad value 95 microns this is compared to 235 this is acceptable limits for me is that clear so I chose my tricks I say if I now choose larger size of W by L I can get different VDSAT value and for which then the ID SAT or ID 6 will be 90 odd microns. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . boost it to 90, okay. So that now at least our currents are now within my acceptable range. Also you can find from here how much is V out minimum you will get, can anyone tell? So 2.2 or something, so which is fair enough because we are expecting no, so it is single more, so fair enough, okay. What is the next transistor left now? 7. Please remember I kept telling you earlier also the current in is not decided by M5 but decided by M7. So this current is flowing here, this current is fixed by U, okay for GM sized. So the ratio of this should be such that it sustains the current of flowing from M6. Is that point clear? I repeat, the current in this arm is not decided by this. This is the low transistor there, okay. This is the driver, current is picked up from here which actually goes around through M7. However M5 and M7 are in mirror, okay. So for this current and for this current I will then make a size ratio, okay so that this current flows through also M7. So it looks as if I am bicing to that but actually I am not bicing randomly, same current I am not passing. I am bicing it correctly such that it flows that much current in M6. Is that point clear to you? I repeatedly saying do not push same current here because in that case this current will be different. So come from M6, figure out what is the current you are expecting to get that W by L for that W by L. Use that current for M7 and then find the ratio of this to this. This current I know, this size I know, this current I know, so I know the size of M7. So M5 and M7 are in current mirror configuration. So I6 is I7 and also I6 is I7. So I get W by L7 is 8 times IDS6 upon IDS5, 95.8 into this. So I get a value of 30.65 so nearest value is 30. Normally how much I should put? 32 but I am not sure whether it will change we decide how much I kept 30. Is it clear M5, M7 are in mirrors but M7 is now drawing 95 microns current. M5 is only drawing 25 microns current. So the ratio of M5 to M7 should be 30 so that it actually matches to plus into multiplied by value of W by L5. So it is 8, 8 into this ratio is roughly 30.65 which is 30. Now I verify that whether VDSAT is okay. So I substitute this value of in IDS7 and evaluate VDSAT for that. That comes out to be 0.23. If I take it from VSS it becomes minus 2.27. How much was expected? Minus 2. So I am not worse off I am still better off. So with this new calculation for W by L6 I am actually improving the output. So remember what is the best swing you should have? VSS to VDD. Any amplifier output if it can swing from VSS to VDD is the ideal situation. The minimum if you can further go down to as VSS it is still better. So nothing goes wrong if I go below minus 2 because we are actually expecting ideally full wave to wave which probably may not happen because saturation. What does this condition is trying to say? Device is still in saturation. Even with this this will still remain into saturation so good enough okay. And you are improving the lower margin of the V out. So that is larger swing is there we increase beyond 2 here we reduce minus of that. So very good okay. So in some sense we are getting better swings than what we are actually asked to do. Is it okay? The most value what is left now sizes we have done except the bicing side okay. Now what we do is to calculate power. What is the current flowing in this? How many arms please there are either the case or sir local loop there are either basically. How much currents are flowing? One current in this arm, one current in this arm and one current in this arm and one current in this arm. Into VDD is the so together I will call it 2.5 VDD because I am going to put equal this value so I say twice okay plus current due to I 5 into VDD plus current in this into VDD so that is the net power dissipation. Is that the each arm current multiplied by VDD some all of them that is the net power dissipations. So I said okay I buy a circuit plus I defamp plus I gain stage into VDD minus VSS because the swing is full to rail to rail so 5 volt. So I find out I buy a circuit IDS 5 plus IDS 6 into VDD minus VSS. I buy a circuit is how much? I buy a circuit is sum of this current plus sum of this current 9 and 12 okay. So if I write that I 12 plus IDS 9 and now I assume the mirror is 1 to 1 what do I assume M9 and M2 or mirror is 1 to 1 what is the current I want to so I said okay I cannot choose any current so I said since 5 is running 25 I have half it okay. So I have chosen size 4 here size 4 here and how much current each is drawing 12.5 microns. So if I do that by mirroring this as I said this is my choice you can choose any other value because mirror does not matter any 2 you can start with 2 1 that ratio will change is that clear to you. So any value I have for the simplicity I twice so if I do this W by 9 is same as W by 12 because I kept them that way so each is drawing each a size of 4 and each is drawing half of this current which is 12.5 so the bias current total is 12.5 plus 12.5 into 5 which is 0.125 milli watts how much 1 to 25 micro watts so it is now we also want to find the R value bias K lia with the R I will find out if I know VGS for this which is same as VGS for this then this value minus this voltage divided by 12.5 is R please remember this voltage minus this voltage divided by R is 12.5 so I did this and I calculated resistance to be VGS minus VT is VDSAT so for this value I calculate VGS 12 as 0.946 volt is that okay this is the 0.246 which I got saturation voltage plus 0.7 which is the VT here also you can choose the worst cases and figure out what should be the correct this and for this I calculate is that okay VGS please take it 5, 9, 12 are in mirror okay their VGS are same their VDS also are same okay one of the VDSAT 5 is known to me so I can evaluate VDSAT for this current I know so I know this I know this voltage if I know this voltage this voltage minus VSS minus this voltage divided by R is the current 12.5 okay so your expression Nikola R is equal to VDD minus VSS minus VDS 12 into this which gives me a value of 325 kilo ohms okay to get the currents of 12.5 micro ohms I need for this WBIL of 4 I need a size of a resistor of 325 kilo ohms. How much power do I have? I have 0.125 available how much was it? 2.5 so how much is available at 2.375 milli watts is still available for me for the limit for defam plus gain stage by Scatadiana. Now let us calculate the power for defam plus gain stage how much is the current in defam? 25 micro ohms how much is current in I6 or I7? 95.8 micro ohms multiplied by 5 you get this 0.6 milli watts okay this may be a 0.125 Jodh Vidya Jai it is 0.725 milli watts which is one-fourth of what was given to us. So we ask given WBILs and currents are chose and well within the power deception limits. So what is the next game you can play? Increase currents highest limit you can then gain recalculate all values of WBILs and see whether still remain in saturation but it satisfy conditions better is that correct? Because at the end of the day this may be sufficient because low power is one of the major criteria so we are anyway achieving it but let us say you suddenly say 20 micro ohm per micro second you want then probably you may boost the current but recalculate everything again with this no IDS 5 and do all the till you satisfy again your power and fluidity requirements keep doing as much as you wish okay and at the end you will find some better optimal in some case you may increase to 40 micro ohm volt per micro second that is very very high speed but you may do low power compared to what limits you are given. So it depends on the spec given to you you may actually still play with IDS values IDS 5 values and if you play that value then please remember every other WBIL will change okay and then you will see what sometime device means there is a minute that is over your max behind mix clearly so then I find it became the other side so verify that those limits are also then modified by so you can get some but what is the last thing left for me to evaluate is it okay which game we are interested in the mid-game game that is the DC game okay. GM 1 by this is a total R O 2 parallel R O 4 is essentially by denominator G O 2 plus G O 4 GM 6 is G O 6 plus G O 7 so if I multiply it becomes because current in each is IDS by 2 is that clear IDS 5 by 2 is that clear IDS 5 by 2 so this current is what is your specification currents IDS 5 now given to me is for N channel it is 0.04 for a P channel it is 0.8 length not equal to 1 micron what is the length we have chosen 0.8 so we believe that this is same this some you are asking you increase the length this lambda may go down is that clear to you then you have to recalculate other things so I know GM 1 I know GM 6 I know I know ideas 5 I know ideas 6 so I substitute 2 into 94.2 into 942 into 10 to power minus 6 into minus 6 minus 12 this is 0.09 so this is both are 0.09 so 0.09 square this ideas is 25 microns ideas 6 is 95.8 but just put 95 microns so AV 0 comes out to be 9,000 to 25 volt per volt. 9000 what was expected 4000 which is far away so now you are another game I do not want really this big game okay so I can release this okay so I will use the currents all it is the size so now keep there are parameters which were in hand keep playing but every time you change something face up there is no other way we can do this but that is how what we do is we go on one calculation and submit it to spies and believe that it knows better than me and therefore except whatever they say okay that is the way all of us actually believes is that clear you have to do because gain is one of the major parameters of open is that correct so no compromise can be made if this goes below the specified value unfortunately the way I did this simulation which is what easiest to think in my this and some books journals also are given this is not my original okay I am not claiming it I did it some many years ago but it is not that it is my original I am copying indirectly from somewhere okay right now I am not saying it but means I have done 8 times so I know this please remember this is the crucial factor we will have okay what is the problem if I use this value higher value the bandwidth will go down okay so some limit they say no no I want at least this much bandwidth then I cannot play with this game so I say okay so on each a curve the bandwidth okay this is how opams are actually get designed okay how many pages I spend you do not know almost 42 and there are 4 more to show okay so can imagine the design is not very casual approach design requires lot much thinking in lot much evaluations and then hope for the best and at the end of the day hope for the best is that clear if you could have calculated first you could have solved much of his worries but right now I did not know GM I did not know 6 I did not know these G's that time so I could not be I still I got all W by else I cannot evaluate these when I evaluated then I already coming to an end size then I started looking for gain then I was hoping you on each other. Now you think of it this the issue which I am saying is the most crucial let us say okay so this is to lambda this is to lambda that is lambda square terms are in so any change in lambda square move a brain go ahead you got it is that okay any change in lambda 0.04 point 0.02 Karthik the four times is it should be clear if I just half it I may figure it out that I know I do not get the gain even okay so therefore this evaluation of lambda from the spice fall is a very crucial factor at every length okay is that clear to you so that is something which any designer must first do these data has been chosen from a normal known value so they came correct but in case they are not actually you must evaluate is that right for given sizes of your various W by size killer ideas, videos, carrot, that is the actual design things go last but not the least part of this before we finally go and which I am not calculating right now I am just showing you this is also my old slide I am just reporting to you one of the technique which we use in this design was splitting up the poles what is the method I use I put a mirror capacitance and compensated so what did I compensate the non-dominant pole was shifted far away by so much factor whereas the dominant pole came closer towards the other side okay and then I am worried whether my bandwidth is okay so one other technique is that why not compensate the non-dominant pole itself okay since you have 0 so if I instead of just capacitance I put in series this circuit figure if I have maybe I can show you something here instead of this I should use something like this this is my R this is my CC please remember this R is created by bias something here called VA called low no one is telling you it should mean saturation and non saturation what is the choice you have to do is R from that if R is required smaller choose put such that it is in non saturation if R required is higher go for saturation is that clear so I want to specify what we are from where this we I can get that by a circuit may want diodes like I am there would be in charge but thank you don't give you use like never use case I may use cup voltage nickel once that voltage is known you fix that voltage here connect here okay once you connect here this R is known to me because the W by I calculate for this okay one that RC time constant is known I have now new Z Z1 which is minus one upon RZ CC minus CC GM 6 which is negative on the left half plane please remember if RZ is larger only then this will go to left half plane smaller the RZ this mode is dominant than this and in that case it will become positive 0 so is right half plane 0 can be shifted to left half by increasing RZ value is that clear to you just look at the 0 if I want to make this is the minus here anyway if this is smaller then this will become positive so I am on the right half plane if I start increasing this value RZ is that clear this is minus then this is minus so plus over right half mega if this is larger than this the pole is 0 is on left half plane so and then where that there should value I should put that Z1 equal to the P2 so I will cancel that P2 once for all because P2 are necessary with this CC so I say CC I don't want to increase too much is that clear now I say okay that split has to be then why a split was necessary because P1 P2 are very close so stability fire was an issue for me so I say go and nickel that on the bath if I have to remove that pole without splitting or small splitting then I will be able to actually keep CC smaller is that clear to you that is how I must remove the second pole this is called nulling this is called nulling so bring the 0 exactly on the pole and that's P2 value which you have so P2 is given to you as minus GM6 by 6 and if I equate this by this formula here I get RZ is equal to 1 upon GM6 CL plus CC C2 RZ GM6 minus 1 so that I can now I know GM6 where I evaluated so this is the expression so RZ can be realized once I get this RZ value then using this expression if it is in saturation and once actually get the correct expression and evaluate RZ for that that RZ value for that W value evaluate put that value such that Z1 is exactly at P2 is that clear once you modify this non-dominant pole the only pole with you is P1 and that P3 is still away so you are not caring so only one single pole you have which you can then think of what CC has used because so that it should not be too close to 0 axis okay that is the trick so RZ allows you to do that is that clear to you and then your bandwidth essentially is not directly heard by this method so to improve your bandwidth one way is to null the non-dominant pole one keeps CC lower is that clear so this method of nulling is very powerful the only problem is what is the only problem you will get all these values are varying with all kinds of variations so getting this relationship is not very easy that they null so some still will occur but that still will be by magnitude will be so small that it will not interfere P1 in any sense that is the trick of that so do not think exactly you can null okay it is very difficult okay. So they will not actually null it but very close to null so we said it does not affect you very much is that clear so at the end of the day your output I have made a choice of 2.5 for a 5 of 60 and Z1 of 10 GBW this was my choice as a designer your choice to choose this is that clear this 60 degree is my making you choose 70 you choose 50 55 your choice you choose 10 times 50 times I already given you table how much they what value it will shift the DC gain is 9,000 to 25 at 25 microns of current expected was only 4,000 okay above 4,000 so we already achieve that bandwidth because of this larger gain how much is the bandwidth came 4 kHz so very low bandwidth opium we have received okay not even megahertz the reason is our game bandwidth itself which was very small okay 6 megahertz okay is that clear if your game bandwidth point is here so the slope if you put it by so much gain P1 will be very small any way okay so then adjust that value so that your bandwidth are higher okay for the reduction is CC because you are improving P2 so you actually when more left side P1 became even lower so that CC can then help you by RZ so that you can still maintain little higher P1 value okay we have maximum is 2.3 expected by 2 we have minimum expected was minus 2 so you are minus 2.27 P dissipation 0.725 you are expecting less than 2.5 I am one third of it or one for almost one third of it channel length used by me is 0.8 and the W by L size I got is 1 and 2 or 3 3 and 4 or 12 5 is 8 6 is 92 7 is 30 8 we are not calculated because I do not know for those value of how much are there I need because for a given CC find P2 and then get that value of this I will write now put it question mark minus 4 12 is 4 I also did not talk about 10 and 11 what are those values how much you are that may decide their W bias drop across them okay and the bias resistors use in my analysis is 325 kilo ohms okay so do you agree that whatever specs given to me I have achieved almost every one of them is that correct almost as every one of them so that means I have designed an op-amp for given specifications finally this is what the circuit looks. You can decide these values corresponding to adjust this value so that the transistor may remain in saturation or in non saturation whichever R you are looking for and correspondingly create RZ there we are calculate W by L here W by L here W by L here W by L here W by L here W by L here W by L here here here here here here so I am able to evaluate every size of a given transistor for given specification of my choice is that is that clear to you okay CC is a variable CC is not given to me is that clear to you CC I will make a choice from the phase margin I choose okay so if that will not be equal to CGD values it is much higher. You got it so that value physically you have to put you have to create a mass capacitor right there how do you keep a mass capacitor. Call it okay. So you actually create CC by a mass capacitor okay please remember the only thing this is also is also very important because the cc will be function of this voltage, you know, is oxide thickness here, that will decide the, this should make an inversion and it should also in the size of cc you are looking for per unit area, so W by L you have to adjust for this value as well to get the cc value. And please I am telling you it is not small capacitor, it is very large capacitor and much of the area in chip is actually taken by cc's, transistor itna or capacitor itna, yes of circuit milkage such hote hote hote, that is the kind of values you get, is that clear to you? So please do not think, another capacitors you can make but that is very difficult because if you have a technology which is called poly oxide sorry, yeah poly oxide or AB poly it is called POP, if you have more than one layer of oxides available and double poly process then you can create poly oxide poly or at times metal oxide poly, okay. So do not use substrate, use the upper layer of poly value and use another capacitor on that. The advantage there is these are independent of biases, is that clear? These are like a standard capacitors, additional mask, paisa but if the technology has that then no additional mask because that process is anyway sitting there, okay. So generally in most capacitors in RF circuits are always of this kind because LC is so strictly see I need no variations I tolerate, okay. So I will use double poly process so that I can get this capacitors of my choice, okay. So please take it that it is not that this technology but RF technology is not same as analog technology, okay. So next time when you come we will start of ICMR again verify those issues today is already up. .