 Okay, this is the 23rd lecture and this is our problem session 6th, I hope the number is correct, on frequency response of small signal amplifiers we will work out one problem and then we shall work out a couple of problems on differential amplifiers. The first problem that I consider is question number 3 in tutorial sheet 6 and the problem is this, we have a Vs, 10k is the source resistance, 10 microfarad is the coupling capacitor, then the base is biased in a peculiar manner, there is a resistance of 100k, another resistance of 100k and the middle point is connected to ground through the capacitor whose value is 10 microfarad, the collector is connected to this point and this point is connected to plus 10 volt and the resistance of 3.3k, 10 meter is grounded, as you see this is the DC feedback biasing that is the current, the current that flows through 3.3k is I c plus I b and that I b, I c flows here, I c flows here and I b flows here, cannot flow this path therefore it comes to the base, we have done this kind of biasing, the only thing that was not done was this capacitor which is for obvious reasons to prevent AC feedback that is input signal cannot go to the collector directly because of this capacitor, output signal cannot go to the base because of this capacitor, okay, this capacitor is a bypass and then the output is taken from the collector through a capacitor again 10 microfarad, all 3 capacitors are 10 microfarad and then a 4.7k, this is the actual load and this voltage is V 0, the transistor parameters are beta equal to 100, I do not want that marble from there, beta equals to 100, V a equals 150 volt, C mu is 2 parts, C pi is 5 parts, C 0 is 0, no stray and R x equal to 100 ohms, even the R x is given in most of the problems, we conveniently ignore it because it increases our problem of analysis, okay, the first thing we have to do is to find the problem is to find out A V S J omega that means we have to find out all the frequencies, all the critical frequencies and we shall have 3 critical frequencies corresponding to the 3 capacitors here, okay, this capacitor is place the role of C E bypass, you see the usual biasing is if the resistance comes here, then we require a bypass at C E, C E bypass was avoided but one capacitor have to be introduced here, so the low frequency response shall be affected by this capacitor, this capacitor and this capacitor therefore I expect that there shall be 3 critical frequencies omega L 1, omega L 2 and omega L 3, let us call these capacitors as C 1, C 2, C 3, we therefore have 3 low frequency critical frequencies omega L 1, omega L 2, omega L 3 corresponding to the 3 capacitances and there shall be 2 high frequency critical points corresponding to C pi plus C m, similar capacitance C pi plus C m, we will combine them, we do not want to analyse 3 node circuit and the other critical frequency at high frequency shall be due to C mu plus C naught at the output, C naught is fortunately 0 therefore we expect that omega H 2 corresponding to C mu plus C 0 shall be much larger than omega H 1 due to C pi plus C m, these are the qualitative observations from the circuit, therefore what we expect is because the capacitance is 2 puffs, it is much smaller capacitance as compared to C pi plus C m, as you will see C m will be very large, it will be about 200, I am sorry about 50 times C pi, okay, we will see this but in general qualitative observation of the circuit reveals that A V S J omega would be of the form A V S O divided by 1 minus J omega L 1 by omega, 1 minus J omega L 2 by omega, this subscript corresponding to the corresponding subscripted capacitors, 1 minus J omega L 3 by omega multiplied by 1 plus J, you must distinguish between the low frequency and high frequency, it would be omega divided by omega H 1 and 1 plus J omega divided by omega H 2, is it clear? Any more explanation needed? You must be able to write this expression just by looking at the circuit, just by looking at the circuit. How are we sure that there are no zeros? Yes, we are not sure unless we draw the equivalent circuit but we have experience with the equivalent circuits of this kind, we have experience already, if C mu was not absorbed in C m, the Miller capacitance and the output capacitance that is C mu was not absorbed in C 0 at the output, then we know there is a 0 at minus G m divided by C mu plus G m divided by C mu, we have already done that but since right from the beginning, we look at the complication of the circuit and we say sorry we have to use Miller, we cannot use a 3 node circuit and therefore that 0 shall be avoided in the analysis, anyway this is much further from any omega H that you can imagine and therefore this 0 will not come, the first thing that one should do is to find the parameters and therefore we require a DC analysis, let us look at the DC analysis this is 10 volt and this current is I sub C plus I sub B therefore I can write 10 volt as equal to this drop, drop across 3.3 plus the drop across 200 K, this 2 can be combined as far as DC is concerned plus 0.7 and that shall give me I sub B agreed, so my equation becomes 10 equals to beta is given as 100 and therefore I C plus I B would be 101 times I sub B, this has to be multiplied by 3.3 K, this gives the drop in 3.3 K, then the drop in 200 K is due to I sub B and then we have V B E which is 0.7, this gives me I sub B as equal to 17.4 micro amperes, that means I sub C which is 100 times this would be 1.74 therefore I sub B, I sub C is 1.74 milli ampere, you must now check because of many hurdles that we have passed, we have seen circuits which behave very peculiarly, we go ahead blindly calculating but ultimately we find that one is cut off and one is saturated as in question number 2 of minor 1, therefore V C E we must check this although it is not required in any of the parameter calculations, we must check this routinely and we must say that this is not neither negative nor comparable to 0.2, okay, V C E is 10 minus the drop in 3.3 K which is 3.3 K multiplied by 1.74 milli ampere, actually we should have made 1.754 okay because it is I C plus I B but well I did not do it, you can do it, I found this as 4.25 volt, this is not critical to include 1.754 that is I B also because the voltage is much above the saturation voltage, okay. Once we have found this, once we have found I C, we can find G M as I C by V T, you take V T as 26 millivolt, then it becomes 0.067 mole okay, then you can calculate R Pi as beta divided by G M, beta is given as 100 and the result is 1.494 K and R 0 can be calculated as V A divided by I C, 150 divided by 1.74 milli ampere and this comes out as 86 K, the 3 parameters of importance have been obtained, these all these 3 depend on I C therefore we had to calculate the DC conditions, okay. Now let us look at the mid-band equivalent circuit, to find out A V S 0 okay, the mid-band equivalent circuit would be like this, V S then at N K what would be this resistance, it is not the parallel combination now, at mid-band 100 K goes to ground, is that clear? Because of that capacitor C 3 it is 100 K, not 100 K parallel 100 K okay, alright, this point must be recognized, then we have R Pi which we had found as 1.494 K and the voltage across this is V Pi with this polarity, we have the output current generator G M V Pi, G M is already found at 0.067, 0.067 more, then we have, well we will combine everything into 1 resistance, R L prime is R C which is 3.3 K parallel 100 K shall come in parallel from the collector to ground through via C 3, then we can also include R 0 which is 86 K, this is 3.3 K R C and this calculates out to 1.86 K, oh we must take that into account, yes parallel 4.7 K, that is how I get 1.86 K okay, therefore the, I hope I did calculate, yes, now therefore A V S 0, A V S 0 would be minus G M R L prime okay, minus G M R L prime multiplied by V Pi divided by V S, agreed? So this is, this is calculated by inspection, by observation, no equations indeed, let me write this, minus G M R L prime, I am writing this separately because of, I am not writing numerical values right away and I am not calculating because of a specific reason, what is the reason? I need this value, why? For calculating the miller capacitance, so I keep it separately, these are tricks of the trade, otherwise you will have to go back and calculate again, so I do it in one step, then my V Pi over V S is 100 K parallel 1.494 K divided by R S which is 10 K plus this resistance, 100 K parallel 1.494 K and this comes out as, this is minus 125 multiplied by this value and I calculated this to be equal to minus 16 okay, I do require this value 125 because I have to calculate C T, let us calculate in one step, C T the total input capacitance is C Pi plus C Mu 1 plus G M R L prime, I know everything C Pi, I know C Mu, I know G M R L prime, this becomes 257 path, this is the equivalent capacitor and therefore the Omega H 1 due to the input capacitance would be given by, Omega H 1 would be given by 1 over C T 257 path multiplied by the equivalent equivalent resistance across it which is 10 K parallel 100 K parallel 1.494 K and my calculation gives this value as approximately 3 times 10 to the 6 radians per second. The other high frequency critical point due to R L prime and C 0 plus C Mu which is at the output is equal to, is this clear or I have to show the circuit again? It is clear, so this should be C Mu which is 2 path multiplied by, I had calculated the equivalent resistance as 1.86 K and this is greater than as you can see very easily 250 multiplied by 10 to the 6 RPS, so Omega H 2 can be ignored, if you so desire and Omega H the high frequency 3 dB point would be approximately 3 times 10 to the 6 radians per second but since the question asked is to find AVS J Omega, you must include this okay, you must include the factor 1 plus J Omega divided by 250 multiplied by 10 to the 6, is the question clear? The question I did not ask find the high frequency 3 dB point, no, I asked find an expression for AVS J Omega and therefore this must be included, if the question was to find the high frequency 3 dB point this is perfectly alright okay, now let us go to the low frequency. Sir in recent analysis you are applying both the methods of time constants and middle approximation on the time type? No, no, the time constants method okay, that is a good question, let me answer this question, you see the only way that this equivalent circuit differs from the high frequency, if I want to convert it to high frequency, I apply this approximation of C T and this is C 0 plus C 0, now it is no longer the method of time constants and the analysis are exactly the same, they are exactly the same, if you write V Pi over V S this critical frequency income, if you write minus G N times this, this is the other critical frequency Omega H 2. So it is not, after the approximation the method of time constants and the method of analysis are the same, as I said the method of time constants has to be used when there are many capacitors, this is as good as analysis, it gives the same results as method of time constants and this is what was used to validate or give confidence regarding the method of time constants, that in simple cases the results are identical, in complicated cases no, sorry they are not identical, they differ, okay. Now the low frequency equivalent circuit, as I said in low frequency equivalent circuit there are 3 capacitors now and they are coupled to each other and therefore there is no other way but to consider the effect of each at a time, fortunately here you can consider the effect of C 1 and C 2 in one go, fortunately. So first you assume that C 3 is short circuit, let us see what the equivalent circuit is, equivalent circuit as long as C 1 and C 2 are decoupled from each other while the method of time constants and method of analysis are the same and fortunately we can do that here, you draw the equivalent circuit V s, it is always good to draw the equivalent circuit, 10 k then you cannot miss anything, 10 k, 10 micro farad, this is C 1 then we have the parallel combination of 100 k and 1.494 k and this voltage is V pi then we have the g m V pi, we do not require the exact values of this because g m does not determine the time constant then we have, now we cannot combine 4.7 k here because we are considering the effect of C 2, so 10 micro farad, this is equal to C 2 and we have a 4.7 k, so this resistance is the parallel combination of R 0 which is 86 k parallel R C which is 3.3 k parallel 100 k, anything else? No, that is it and this calculates out to 3.08 k, now here is a point to note, we have to calculate this separately, whereas we also calculated the parallel combination of this and 4.7 k, we could have done it in one step, anticipating that this resistance will be needed, we first calculate this, then we calculate 3.08 parallel 4.7, rather than putting 3 resistors in parallel and 4 resistors in parallel, we could have done that, just like we calculated A V S 0 alright, these are tricks of the trade with experience, you will know that I should not calculate the 4 together, I calculate 3 of them because I require this value separately and then the 4 is fine alright. Now you see that C 1 and C 2 are decoupled and therefore the critical frequencies due to C 1 and C 2 can be calculated from this circuit alright, omega C 1, omega L 1 for example due to C 1 would be 1 over, the capacitor is 10 micro farad and the given in resistance not the method of time constants or otherwise, it is the same, given in resistance would be 10 k plus 100 k parallel 1.494 k and this calculates out to 8.7 radians per second, omega L 2 the second due to the second capacitor comes out as 10 micro farad once again multiplied by 4.7 k plus 3.08 k okay, the current source is opened and this calculates out to 12.85 radians per second alright, these 2 are comparable, they are comparable, even omega L 1 squared and omega L 2 squared, one cannot ignore one with respect to other because the ratio is simply 1.4 okay, however these 2 are not the only only culprits, there is a third culprit C 3 which creates a complication in the calculation of omega L 3 and we have to draw that circuit separately and it is instructed to consider how this circuit is drawn you see, as far as C 3 is concerned if C 1 and C 2 are shorted then by equivalent circuit draw this carefully, we have this 100 k like a dumbbell 200 k on 2 sides and this is my C 3, this is the position occupied by C 3 okay, let us draw C 3, C 3 which is 10 micro farad then on the left, on the left comes R pi which is 1.494 k, this is R pi I am sorry this is V pi and comes at 10 k in series with V s, I am drawing the equivalent circuit considering C 1 and C 2 are short and C 3 effect of C 3 and on this side on the other side we have the G and V pi, G and V pi then in parallel width, how much is this, this now will contain 4.7 k also, so 1.86, 1.86, 1.86 it will not contain 100 k then, it will not contain 100 k, wonderful, it cannot contain 100 k now, very good, so we have to calculate the value, this gives us another lesson that we should not have calculated those 4 resistances at one time, we require these 3 values separately okay, who is this boy who pointed out this main, okay, that is very good, I appreciate it, we have made a great blunder if you have included 100 k, why, why great blunder because R 0 is comparable to 100 k, is it not 9, 86 k, suppose R 0 was not there, if it was simply R C parallel R L then the error would not have been much by ignoring that okay, so we have to calculate this and now the problem is that looking from here, looking from C 3, what is the resistance that it sees and therefore what we have to do is to find out the equivalent thing in it, we have to short this and then we have to replace this by lengths of a voltage generator V and calculate the current I, then V by I will be equivalent resistance, it cannot be obtained by inspection unfortunately, so let us draw a clean circuit on a clean slate and then proceed, we have a V and this current is I, on the 2 sides there are 200 k's, 200 k's then we have a G and V pie, we cannot open it because this is a control source, V pie is not 0 okay, what is V pie, V pie is the potential division between 100 k and the parallel combination of 1.494 k which is R pie and R S which is 10 k, so this is V pie, this is V pie and V pie is not 0 therefore we cannot make this open and in addition we have the parallel combination of R 0, parallel R C, parallel R L, what we have to do now is to write okay, I can write V pie in terms of V by inspection correct, this would be 10 k parallel 1.494 k divided by 10 k parallel 1.494 k plus 100 k multiplied by V, so V pie I know in terms of V then I have to write a node equation here okay, I have to write I as equal to what is this current, this kind is V pie divided by this parallel resistance, so V pie divided by 1.494 k parallel 10 k, it is this current okay then this current is if I call this voltage as V 1 is V minus V 1 divided by 100 k, I now want to know what is V 1 okay, we also see that this current is equal to G m V pie plus the current through R 0 parallel R C parallel R L, so my third equation would be V minus V 1 divided by 100 k equals to G m V pie plus V 1 divided by R 0 parallel R C parallel R L, there are 3 equations now and the way to proceed and the systematic way to proceed is to replace V pie, so I have one equation to second equation and this is the third equation, from 1 I replace V pie, wherever V pie occurs I replace V pie alright and then from the third from the third I get V 1 in terms of V, V 1 in terms of V and then I substitute in this equation, I divide both sides by V, that is the input conductance, the reciprocal of which is the input resistance, it sounds complicated but it is not complicated because you will work in terms of numbers, I have not simplified this but I have I must assure you I have calculated and my calculation after some calculator approximations and slipiness approximation, my calculation gives omega L 3 as approximately equal to 2 R p's, so 2 8, what was the, what was the other case, 12.85, so who controls? No, this is the answer I was expecting and I wanted to correct, no it is the highest but even that does not control 12.85 because there is a frequency close to it, so this is a fit case, these 2 radians per second can perhaps be ignored compared to the other 2, 8 and 12 because 1 is 4 times 1 is 6 times, where we have calculated R 0 plus R c parallel R L, yes, 100 k comes here, 100 k comes here, this you have to calculate independently, this is what I said that you learn 2 lessons that you should not go ahead calculating the equivalent of 4 resistances because equivalent of 3 resistances are also required, you have to do it in 2 steps, okay, so who controls, who controls the low frequency response, omega L is controlled by omega L 1 and omega L 2 because omega L 1 by omega L 3 is approximately 4 and the square is 16 and omega L 2 divided by omega L 3 is approximately 6, the square is 36, okay but the total a v s j omega now has to be written in terms of what we found out for a v s 0, then 1 minus j 2 divided by omega that is omega L 3, 1 minus j is 8.7 divided by omega, 1 minus j 12.85 divided by omega multiplied by 1 plus j omega divided by what was omega h 1, 3 into 10 to the 6 and 1 plus j omega divided by 200, I did not calculate that, I showed which is greater, so you put that value, whatever value, this is the complete answer, yes, that is right, what do I do What do you do, you ignore, okay good question, how do you calculate omega L, what you do is you ignore this, you ignore this and you ignore this, is that the high frequency, then you have 2 frequencies, 2 cut off frequencies and therefore you have to write the equation a v s square of this, square of magnitude of this is equal to 2 and solve for omega L, okay, that is it, exactly like the question number 3 in the minor. Now in the rest of the time we work out a couple of problems on differential amplifiers because I find that we will come to the tutorial on differential amplifiers only in the next week, so let us do some differential amplifiers and you will see that this is a different brand of calculation, even the DC calculations are somewhat different from what we have done so far, the first question, draw the circuit with me, a simple problem R sub C, R sub C, this is 10 volt and R sub C is given as 100 K, not the large value of R sub C, this is a micro circuit, it is an integrated circuit, this R sub C is not a lumped resistance, it is a, what is it, it is another transistor, it is a current source, this output input is 100 K but that we will see later, suppose it is discrete, even in discrete differential amplifiers what we are interested in is voltage gain, not power gain, not voltage level or current level and therefore your signals can be micro volt order and your I sub C1, I sub C2 can also be in the range of micro amplifiers, alright because it is voltage which is of importance, okay, so R sub C is large, you have 2 transistors, it is a discrete circuit let us say, okay, we will calculate this is a discrete circuit, there is an R sub EE and the current through this is 100 micro amperes, this is equal to I sub EE and this is taken to minus 15 volt which is minus V EE, this is plus V C C and of course because it is an, because it is a differential amplifier, the 2 inputs B1 and B2, they are left as terminals to be connected to sources, the differential voltage is to be connected between these 2, as far as DC is concerned unless the source resistance is specified, this would be considered as grounded, okay, so that forms the heart of DC calculation, now the question is capital T is given as 25 degree C that is the normal room temperature which means that you can use V T equal to 26 millivolts, VA is given as 100 volt which means you know what is R0, R0 has to be combined with R sub C and they are now comparable, okay, so VA can no longer be known, beta as I said you must require very high betas for this transistor beta is 200 and the question is to calculate Q points of each transistor, naturally the Q point of any 1 transistor will suffice because the 2 are identically biased, Q points of each transistor you require the value of REE to be able to support the 100 micro ampere current, what value of REE is needed with these 2 supplies, 10 volt and minus 15 volt, note that they are not identical, in ICs usually they are identical but to bring variety into experience we assume them to be non identical and you also want the value of R ID, the differential input resistance, we require the value of A sub D, the differential mode voltage gain and fifth we require the value of the common mode voltage gain A sub C, okay, this will completely analyse the amplifier, so the first thing that we should do is to check the Q points that is find out I sub C and VCE, now to find I sub C our procedure would be slightly different from what we have been using earlier, you see what we do is we write, first we write 10 volt equal to, do you know this, this current, how much? AC is given, AC is 100 K, this must be half of this, 50 micro ampere because beta is large, okay, so I know RC, I know 50, I know the current and therefore I know the drop here, then what I want to find out is this, VCE, okay, so 10 volt is equal to the drop in RC plus VCE, then we could have gone through this but we do not, we go via this because I know as far as DC is concerned this point is grounded, so this voltage is 0.7, with minus and plus, agreed, that makes a shortcut, I do not have to go through this, so my VCE equation becomes 10 equals to 100 K multiplied by 50 micro ampere plus VCE then minus 0.7 where I go from the emitter to the base, which gives me VCE as equal to 5.7 volt, wonderful, so 5.7 volt and 50 micro ampere is the key point of either transistor, now to find out REE, yes, it is not clear, okay, 10 volt, how is this connected, plus 10 and negative ground, so 10 volt is equal to the drop in RC plus VCE, we have come here, when I go to ground back via the base of transistor 1, that is why minus 0.7, this voltage is 0.7 okay and I found a VCE, how do we get 50 micro ampere, how do I get 50 micro ampere, this current 100 micro ampere is approximately twice I C, I C 1, actually it is IE 1 plus IE 2 but since beta is large, this is approximately I C 1 plus I C 2 are identical and therefore it is half, any other question, now to calculate REE, to calculate REE what we do is, we come from here and go to the negative supply, we write a KVL, so what should I write, 0.7 plus REE multiplied by 100 micro ampere, then I have to go to ground via this supply, minus 15, plus minus, plus minus, then minus plus therefore minus 15 which gives me REE as equal to 143 K, is this point clear, okay, I come from the base of transistor 1 from ground 0.7 plus the drop in REE plus whatever potential this point is at, this is the minus 15, so that is what I write, you have to be careful in this calculations okay, you have to find short circuit, short short cuts, short circuit, short cuts okay, that is what distinguishes an engineer from a non-engineer, I will not say scientist, a non-engineer okay, so I have found REE and once I know these parameters then other parameters calculate very easily, GN is I C that is 15 micro ampere divided by 26 millivolt and this calculates out as 1.92 millimole, once I know GN then I know AD, AD is minus GN, the differential mod gain minus GN RC parallel worth R naught, small R naught, I cannot ignore it, RC is 100 K and what is small R naught, GA was given as 100 volt okay, 100 volt divided by 15 micro ampere which is equal to 2 megs, 2 megs is 2000 K, 20 times this but even then because of high gain calculations it includes this, be kind to R 0 and do not ignore it, this in my calculation this comes out as minus 183 okay, then I require an R ID, R ID which is equal to twice R pi, now what is R pi that is 2 times beta is 200 divided by 1.92 millimole, you see how large R pi is approximately 104 K, why does it happen because of the low values of collector current okay, so no longer any of these can be ignored, R X fortunately is an ohmic resistance, it still remains 100 sub ohms, it does not depend on current but R pi cannot be ignored, R 0 cannot be ignored and this is 208 K, this is a high input impedance amplifier, finally I have to calculate the common mode gain which as you know is minus GN RC divided by 1 plus twice GN times REE, this is also an approximate expression but fairly accurate, minus GN RC already calculated minus 183, 1 plus 2 into 1.92 into 10 to the minus 3, 1.92 millimoles multiplied by 143 multiplied by 10 to the 3, yes, yes that is correct, thank you, that is why I used 183 here, I missed that term, 1 plus 2 GN RA and how much is this, 143 approximately 0.3, it is a fraction, you see this 10 to the 3, 10 to the 3 cancels, 3.8 take this as 4, 4 times 14 516, so approximately one third, approximately 0.3, minus 0.3, what is the CMRR then? 183 divided by 0.3 that is more than 500, CMRR is more than 500, okay, the third question, I will indicate the question and leave it to you to work it out. Third question is design a differential amplifier, no longer analysis, what I want is given you have to give me the circuit, design a differential amplifier that meets the following specs, I shall briefly discuss how to proceed, the specs are RID equal to 2 meg, differential input resistance must be 2 meg, A sub D differential gain I require minus 500, CMRR I require 54 decibels, these are the ways people specify practical specifications and V C E Q that is the transistors Q point must be a very safe value 5 volt is specified, assume that the transistor beta is 200 that is you are given transistors a lot of transistors, a lot not many I mean a lot, maybe a lot of 10 on that of 12 identical transistors in the beta is 200 and VA is 150 volt, this is what is given you are required to design, now what does design mean, design means you will have to specify V C C, you will have to specify V E E, you will have to find out R E E, you will have to find out R sub C what else, that is all, that is all there is to it, okay how do you proceed, first, first from this specification you know what is R pi alright and if you know R pi and you know beta therefore you know GM, if you know GM then you know I sub C the collector current okay, agreed? From R ID okay let me draw the chart, from R ID find out R pi then combine the information on beta to find GM, combine the information of R pi and beta to find GM, from GM you find out I sub C okay alright, now V C E Q is given right, therefore from V C E Q and I sub C, no, no, can R C be found out? No, V C is a V C condition, V C C is equal to I sub C multiplied by R C plus V C E Q minus 0.7 and therefore from this you can find out R sub C okay, then wonderful, so you cannot do that, no, V C C has to be found out, from the gain A sub D minus GM R C parallel R0, R0 is known because you know I sub C and V A has to be given and therefore and GM is known therefore from AD you can find out R sub C and if you know R sub C then combine this with the given information on V C E Q and I sub C to find out V C C, next how do you find V E E? From this now wait a second, from the CMRR it is given in DD, we first have to convert it to a fraction, a number okay, first have to convert that is 54 equal to 20 log 10 of CMRR, from which you have to find out CMRR as a number and that CMRR is equal to how much twice this have to be here at the end of your names, it is scratch it and the number and say AD upon AC. No, we do not know what is AD, it is CMRR, CMRR is twice GM REE, so if you do the CMRR and you know GM, you know REE okay, CMRR was AD upon AC minus GM R C divided by okay, AC was minus GM R C divided by 1 plus twice GM REE, 1 is known and therefore the ratio is twice GM REE, since you know GM, since you know CMRR, you can find out REE and if you know REE, finally you find out V E E and the design is complete, would you care to stop a minute and note down what my calculations gave okay, my calculations say that V R sub C is equal to 2.7 meg, then V C C is equal to 17.3 volt, REE is equal to 1.25 meg and V C V E E is 13.7 volt, not very nice numbers but this is unfortunately what the design is. Do you require anything else? Okay, that is all for today.