 CMOS RF integrated circuits today is lecture 16. As part of lecture 16, we are going to finish with the fifth module that is bandwidth estimation techniques. So, as part of that I am going to discuss the method of short circuit time constants. And then hopefully time will permit and if time does permit, then we are going to start the new module called on a wide band amplifier design. So, that is the plan for today. First I am going to try to finish the old module and then we shall start the new one. So, earlier in the previous lecture we were discussing the method of open circuit time constants. So, the method was basically you have a network full of resistors and capacitors you have got an input and an output. So, step one the question is what is the bandwidth of this network this is what you want to find out. So, to do this what you first do is you pull out one capacitor out of the network. So, you have pulled out one capacitor out of the entire network and then the next step is that you kill all the other capacitors inside the network. When I say kill all the other capacitors this is the method of open circuit time constants. So, you open circuit all the other capacitors inside the network. Then you replace the capacitor that you have pulled out by a voltage source and find out the current that it generates. So, you replace this capacitor by a voltage source and find out the current that it draws. So, that voltage source divided by the current will give you the resistance. In other words you do not have to do any of these replacements. In other words what you need to do is you find out the impedance looking into the network from these two terminal points between these two terminals. What is the impedance that you see once you have that got figured out then R times C that resistance times C will give you the time constant corresponding to that particular open circuit time constant. This was the method. Now, we discussed that this works when our circuit does not have any 0's. It works in the absence of complex conjugate pairs of poles. So, this is the general technique. So, we did quite a few examples of this and we figured out how we can use this particular technique to our advantage and how we can work wonders. Now, what the method of short circuit time constants addresses is this fact that open circuit time constants do not address. This method does not address any 0's. So, the method of short circuit time constants addresses just this. It helps you figure out what are the 0's of the circuit. Now, to do this you have to be very careful. You have to first know which are the capacitors responsible for poles, which are the capacitors responsible for 0's. Failing to do this is going to land you into trouble. So, we first need to know in advance which capacitors give us the poles, which capacitors give us the 0's. If we do know them then the method of short circuit time constants is as follows. All of the capacitors that are responsible for 0's have to be short circuited when you kill them. All of the capacitors that are responsible for poles have to be open circuited when you kill them. So, let us relook. I have got a network. I have got a lot of resistors, you know g m's etcetera, and I have got some capacitors which give me poles. I am writing p for them, and I have got some capacitors which give me 0's. I am writing z for them. Now, the method is as follows. You pull out a capacitor responsible for a 0. You pull that capacitor out and you kill the other capacitors inside the circuit. When I say kill the other capacitors, what you do is you open circuit all the pole capacitors. You short circuit all the 0 capacitors. Now, what you need to find out is the time constant that you see looking in from these two terminals or rather the resistance that you see looking in from those two terminals. Resistance times the capacitance that will give you the time constant. You do this for all the 0's that you have all the 0 contributing capacitors that you have. That will give you 1 by that total time constant is going to give you the lower cut off frequency of your circuit. So, the method of open circuit time constants is going to give me the higher cut off frequency. The method of short circuit time constants is going to give me the lower cut off frequency. In other words, if I pull out the capacitors responsible for the poles, kill everything else. When you kill everything else, you have to follow the same strategy. You short out all the capacitors responsible for 0's. You open out all the capacitors responsible for poles. So, you pull out all the pole contributing capacitors. Find out the sum of the time constants. 1 by the total time constant will give you F 2. You pull out all the 0 contributing capacitors. Find out the total time constant. 1 by that time constant will give you F 1. The proof of this is very similar to the proof of open circuit time constant, the earlier strategy. So, the proof is very similar. Here, the assumption is that my transfer function has only 0's and no poles. It is a very strange assumption and that is how it is. So, we are going to assume that we have got everything only 0's in our transfer function. So, these are all the 0 contributing capacitors. This is what it gives. You do not include the pole contributing capacitors in the short circuit time constant calculation. Similarly, you do not include the 0 contributing capacitors in the open circuit time constant. So, this should solve some problems. What problems is it going to solve? It is going to solve the decoupling capacitor problem. I mean you are all familiar with this kind of a circuit. Now, typically what we do is these two capacitors, actually these three capacitors are supposed to be large capacitors. Remember, these are not parasitic ones. These are capacitors you intentionally put there and you make them large. So, that at A C the signal goes through, at D C the signal is blocked. That is the thinking behind putting those capacitors. These are capacitors that contribute 0's and not poles in the system. If you say that these are large capacitors, I am going to treat them with the O C time constant method, then you will get ridiculous answers. As in you are killing your bandwidth every time you make the capacitor larger, that is not the case. These capacitors affect the lower cut off frequency. So, the next obvious question is how do I know which capacitors contribute poles and which capacitors are responsible for 0's? I mean there are so many capacitors here. There is got to be some capacitance here, here, I do not know where else. All kinds of capacitors are there everywhere and some of these you are saying are contributing poles. Some others are contributing 0's. How do I know which is which? How do I know which capacitors to include for my O C time constants which to include for short circuit time constants? The answer to that is you have got to decide and the way you decide is this. If you open the capacitor and see that your performance improves, then it is got to be contributing a pole. So, let us look at it. Let us call it 1, 2, 3, 4, 5, 6. If I open out capacitor number 1, the signal does not reach the base of the transistor. So, clearly performance is not going to improve. So, it cannot be a pole contributing capacitor. If I open capacitor number 2, then the configuration becomes like a emitter follower gain reduces. So, therefore, performance decreases. So, therefore, it cannot be a pole contributing capacitor. If I open capacitor number 3, then once again signal does not reach the load and as a result it cannot be a pole contributing capacitor. If I open up capacitor number 5, then life is better. I have got less capacitance looking into the base of the device. So, therefore, performance is going to increase. So, therefore, capacitance number 5 is going to create a pole. If I open up capacitor number 4, Miller effect is dramatically reduced. So, therefore, it is a pole contributing capacitor. If I open up capacitor number 6, capacitor number 6 is basically loading the output. You reduce the load, performance is going to improve. So, therefore, capacitor number 6 is also creating a pole. So, this is how you decide which is contributing a pole which is contributing a 0. Alternately, you can also do the following. You can short the capacitor and if performance degrades, then it is a pole. If performance improves, it has got to be a 0. You can do either of these two experiments to figure out if that particular capacitor is responsible for a pole or for a 0. Now, your homework problem is to figure out capacitor number 4. That is going to be your homework problem. I would not give you the answer right away. Capacitor number 4 creates both a pole and a 0. So, how do you figure out what needs to be done? I would not give you the answer and think about it, try to work on it and then we shall see. So, with this I am going to close the topic of open and short circuit time constants, bandwidth estimation. So, we have done enough. We know how to work with a circuit. So, given a circuit, we can compute quickly and estimate its 3 dB frequency, both the upper cutoff frequency and the lower cutoff frequency. So, we know how to do that right now. Next we are going to move on to the next module that is wide band amplifier design. So, we have learnt how to estimate bandwidth etcetera. But of course, these techniques have a couple of very serious flaws. Flaw number 1 is that you cannot handle poles and 0s at the same time right and flaw number 2 is you cannot handle a pair of complex conjugate poles or 0s for that matter. Strange things are going to happen if you try. So, consequence of this is that you cannot handle circuits which have both inductors and capacitors. Unfortunately, that is how our life is going to be. So, we wanted to design a wide band amplifier and this is where we had started from right. This was our first cut circuit. We did the method of O C time constants. We figured out that it has got some bandwidth what was it 75 megahertz with some rough numbers I gave you. Then in the next step we improved it we added one more device and it was improved. Then in the next step we added another device it improved even further and the final step we added even one more device and I got a bandwidth of what was it 250 megahertz or 280 megahertz or something like that reasonably good bandwidth. So, I got something like 300 megahertz I have forgotten the numbers right. How did we do this? We kept on adding devices and solving problems. So, wherever we had a problem we recognize the problem we added another device over there to fix the problem. So, that was our design strategy alright. Now, this works when we talk about integrated circuits. You can keep adding devices because devices are free or an integrated circuit can use as many devices as you want right might cost you a little bit of area, but really I mean believe me it is nothing compared to what you are going to burn in terms of other accessory circuits. So, ESD protection and this and that is going to cost you much larger area penalties then this extra 1 or 2 devices. So, 1 or 2 devices is no big deal digital circuits routinely use millions of transistors right. So, why cannot we we can use many as many devices as you want to use. Now, earlier before integrated circuits made their head way when products used to be built out of discrete components lot of engineering had taken place people sat and figured out how to make wide band amplifiers even though they were not on an integrated circuit right. So, there in those times the philosophy was that every device I add is a penalty I have to pay for that device I have to buy that device I have to put that device over there every active device is a penalty it is costly passive devices I still have to buy it, but it is not as expensive as the active device active devices are costly each individual active transistor is a price penalty can I do something with passives instead can I improve the bandwidth of this amplifier with passives. So, this question was explored has been had been explored long back and people came up with some rather nice solutions that are still interesting even though we are on an integrated circuit even though we have been you are allowed to use as many transistors as you possible transistor is free even with that philosophy even then there are some beautiful things about the older philosophy where transistors were expensive passives were not as expensive there are some nice things about that and we should explore those strategies as well. So, coming back what can I do to this circuit without using more transistors more MOSFETs I am not allowed to use more MOSFETs what else can I do to this circuit to make it have a wider bandwidth. So, this is the question what can you do to this circuit to make it have a wider bandwidth now the answer that came up was this you know R L and C L were anyway split in my older topology as well add with extra devices I had split up R L and C L. So, I am giving myself that freedom. So, the first suggestion is that why do not I add an inductor over here what does this do. So, let us take a look we are going to compute the impedance looking into the load network earlier the load network was R L in shunt with C L and Z L earlier was basically R L in shunt with 1 by j omega C L in other words this is equal to R L by 1 plus j omega R L C L. So, this was my load earlier what is it now. So, now it is R L plus j omega L in shunt with 1 by j omega C L which is equal to R L plus j omega L divided by something like this and you can simplify this a little bit it can be simplified a little bit in this fashion. So, what have we got we have now got we have now arrived at a system which has a 0 and 2 poles. So, earlier our system had just 1 pole now Z L has a 0 and j omega in 2 poles where is the 0 what is omega such that the result is 0 omega should be equal to minus R L by L j omega what is j omega such that not omega I am sorry this is fine. So, if you replace j omega with minus R L by L then you get the numerator to be equal to 0 and you get nothing at the output. So, if you plot on the S plane if you plot the 0 the 0 is over here at minus R L by L as you increase L the 0 moves closer and closer to the j omega axis next where are the poles. So, I just rewrite this 1 plus S times R L C L plus S square times L C L that is my denominator polynomial. So, what should S be such that the denominator is equal to 0 please let me know if I am making too many mistakes over here I doubt there are too many mistakes. So, this is my this is these are the 2 locations of the poles. Now, if R L squared is more than 4 times L by C L then the locations of the 2 poles are on the real axis if R L squared is less than 4 times L by C L then they are going to become a pair of complex conjugate poles. So, I am going to rewrite this condition in a fashion that suits me how I am going to write it is as follows I am going to say R L times C L has got to be 4 L by R L for it to be real and similarly R L times C L has got to be less than 4 L by R L for it to be complex conjugate pair. So, this is my rebranding I am going to further rebrand this and say R L times C L is really the time constant from R L and C L and L by R L is the time constant from L and R L. So, we are going to rebrand this it is the same result I am going to rebrand it and call this R L times C L as tau C and I am going to call L by R L as tau L L for inductor and C for capacitor. Further what this means is that if I write tau C by tau L if I call this something what you want to call it let us call it eta all right let us call let us define eta as tau C by tau L and this will basically give me that condition number 1 basically means that eta should be greater than equal to 4 condition 2 is eta has got to be less than 4. So, let us rewrite our so I am going to rewrite this in this fashion instead of talking about L and R L and so on and so forth let me just take R L common outside numerator is going to be 1 plus j omega times L by R L L by R L is really tau L and in the denominator I have got 1 minus omega squared times C L times L C L times L is also equal to C L times R R L times L by R L. So, this can be recast as tau C times tau L and then the third term is plus j omega tau C. So, I have neatened it up a little bit and then further what I am going to do is I do not like too many time constants I am just going to use this eta variable and brand everything in terms of eta. So, everything is in terms of tau C and eta I tau L is equal to tau C by eta. So, this is my load network now if you examine the load network earlier it used to look like this. So, at D C the value was equal to R L and then at the frequency of 1 by tau C it starts dropping at 20 dB per decade and of course, the gain is g m times or rather I have forgotten the output impedance over here does not really matter the gain is g m times the output impedance with this load. So, that is also going to exhibit similar behavior the net gain is also going to exhibit similar behavior fine. Now, let us examine what is the new load impedance the new load impedance has got a 0 where is the 0 the 0 is at eta by tau C. So, this is 1 by tau C and if eta is more than 1 then the 0 comes further down if eta is less than 1 then it comes earlier. So, let us say it comes earlier in that case you will be off to a flying start will start off from 0 at 0 frequency your impedance is equal to R L and then at eta by tau C suppose eta is less than 0 less than 1 I am you fly off at plus 20 dB per decade then you wait for your poles where are the poles. If eta is less than 1 then what have we got here if eta is less than half then what have we got we have got a whole squared no. If eta is equal to 1 then we have got a pair of complex conjugate roots at 1 minus it is one of those cube roots of 3 you remember it will be if eta is equal to 1 then you have got something it is one of those cube roots of 1 I am sorry. So, you will get these two points you will get all three coefficients 1 1 and 1. If you draw the root locus plot then what you are going to find is that as you change eta the locations of the poles change in this fashion they traverse an arc with center right over there that is what you are going to find as for example, as eta tends to 0 if eta is equal to 0 then you do not get I mean your 0 is really becoming very important you have got to multiply everything out by the by eta the term in the middle the term with s. So, you have got to rewrite this as if eta approaches 0 then what you are going to get is basically two poles on the j omega axis that is what you will end up with is that right might not be correct if eta is equal to 0 then I just get 1 pole at 1 by tau c I will have to check on that what you are basically doing is as you tweak this value of eta the location of the poles changes right what you are going to find out is that at eta equal to 4 at eta equal to 4 you have got the two poles to be on top of each other on the real axis exactly at eta equal to 4 you are going to get two poles on top of each other two real poles on top of each other. So, R L so basically you will get minus R L by 2 L as the two poles the entire quantity beneath the square root is going to become equal to 0 at eta equal to 4 eta less than 4 we are going to get complex conjugate pair of poles eta more than 4 we are going to have real poles. So, if you remember your control theory what this leads to are what happens when you change the locations of the poles and what happens when you have got two poles the poles move and then from one point they start arching out in the form of a complex conjugate pair what happens do you remember first of all the root locus plot I think I have made a mistake in the root locus plot it does not go in this fashion I think this is what it does correct me if I am wrong the two poles are at where are the locations of the two poles there has got to be mistakes here is this correct. So, these are the locations of the two poles and they do move seem to move in the form of an arc towards the j omega axis all right it does not matter what matters really to me is the value of eta itself if I make eta more than 4 then you see that the quantity under the square root is positive and as a result you get two real roots if I make eta less than 4 then the quantity under the square root is no longer positive and I get a pair of complex conjugate roots as I make eta lesser and lesser this quantity becomes larger and larger in the imaginary value now once we understand this more or less let us plug in some numbers let us say eta equal to root 2 let us start with eta equal to root 2 when eta is equal to root 2 what I get for the pair of complex conjugate poles is minus square root of 2 plus minus square root of 2 minus 4 times root 2. So, what is square root of 2 minus 4 times root 2 this is definitely going to be less than 0 if it is less than 0 then you are going to get a pair of complex poles all right one can show that at this particular value of eta you get the maximum bandwidth what is the bandwidth the bandwidth is the frequency at which this impedance becomes 1 by root 2 times the magnitude of Z L now becomes equal to 1 by root 2 times R L all right in other words the frequency at which this particular quantity 1 plus j omega tau c by eta divided by 1 plus j omega tau c minus omega squared tau c squared by eta. So, the frequency at which the modulus of this is equal to 1 by root 2 can we work this out it is possible to work it out modulus of numerator divided by denominator is the modulus of numerator divided by modulus of denominator modulus of the numerator is 1 plus omega squared tau c squared by eta squared this is modulus squared and modulus of the denominator is 1 minus omega squared tau c squared by eta the whole squared plus omega squared tau c squared and I am saying that this is got to be equal to half to figure out what my bandwidth is. So, let me do a little bit more of reshuffling and I am going to get 2 plus twice omega squared tau c squared by eta squared is equal to 1 minus twice omega squared tau c squared by eta plus omega to the 4 tau c to the 4 by eta squared plus omega squared tau c squared these 2 quantities have got to be equal and then you solve for omega to get the bandwidth. So, let us remove and then let us simplify. So, I have got 1 plus omega squared tau c squared times 2 by eta squared plus 2 by eta minus 1 minus omega power 4 tau c power 4 divided by eta squared equal to 0 and a little bit more of mathematics and that will give me my bandwidth I now have to solve for omega. So, I have got a quadratic. So, let me recast my quadratic to the appropriate signs. So, I am going to put plus over here minus over here and minus over here and actually does not matter does not matter. Let us solve this for omega and that should give me my bandwidth can I solve for omega or omega squared. So, quadratic I am it is a quadratic in omega squared. So, in the first step I will just have to solve for omega squared it is going to give me 2 roots for omega squared which one are you going to take it is a good question let us hang on a little bit. So, omega squared is going to be is going to be given by how do you solve the quadratic equation your famous formula. So, this is going to give me 2 roots now thankfully both of these roots are going to be real quantities no complex quantities involved looks like no complex quantities are involved unless you really goof up no you just cannot goof up over there no complex quantities at all both are going to be real quantities. And little bit more simplification is warranted let us cancel out tau c squared tau c power 4 and let us get rid of this minus sign right. And so, this is basically how we are going to compute omega squared and from this you are going to compute the bandwidth. So, we will use this result in the next class and proceed from here you are going to see nice things when eta is equal to 1 eta equal to root 2 and so on and so forth. So, let us stop here and we will proceed from here in the next class. Thank you.