 Hello everyone. Today we are going to talk about microwave amplifiers. In the previous lecture you had heard about different types of transistors. So, today we will see the application of transistors for designing microwave amplifiers. However, we are going to start with an inverting amplifier op-amp. The reason for that is you are already familiar with how op-amps work. So, let us see a simple example of an inverting amplifier and we have a design statement here that design an inverting amplifier for a gain of minus 1000. So, an inverting amplifier configuration is shown over here, where we can see that there is a feedback resistor R 2. In some books they write R f and here the resistor is R 1 and we can say that the gain of this particular amplifier is nothing, but v out divided by v in which is given by the expression minus R 2 by R 1. The problem statement is we have to design this for a gain of minus 1000. So, we have to choose the values of R 1 and R 2. So, there are several possibilities of choosing R 1 and R 2. So, we can choose R 1 as 1 ohm, R 2 as 1 kilo ohm or 10 ohm, 10 kilo ohm, 1 kilo ohm, 1 mega ohm. So, now which one is the best choice among these different values of resistors? All of these values will give us gain equal to minus 1000, but are these practical values. So, this is where the difference comes when you are designing an amplifier. You must know what are the parameters we should choose, so that it fulfills different requirements. So, for example, over here if we choose R 1 equal to 1 ohm, then what will happen? Input impedance looking at this point is equal to R 1 and that would be equal to 1 ohm. Now, for an amplifier generally desirable characteristic is that it should have a very high input impedance. So, that means, if we choose a value of 1 kilo ohm, then R 2 comes out to be 1 mega ohm. Now, of course, 1 mega ohm resistors are available, but if you think about a practical 741 op-amp, then the practical 741 op-amp has an input impedance between these two ports is equal to 2 mega ohm. So, this 2 mega ohm somehow comes in parallel with this particular resistor. So, if you choose this value of 1 mega ohm, I can tell you you will not get a gain of 1000. In fact, gain will be slightly less than that. Now, there is another problem associated with the op-amp and this is the typical gain variation with frequency. So, let us see what we have over here. So, you can see that there is a peak gain of around 10 to the power 5 over here and then the gain is reducing. Now, typically for a 741 op-amp gain bandwidth product which is what is written over here is 1 mega hertz. So, if we design an amplifier for a gain of 1000, then we draw a line here corresponding to gain of 1000 and we can see that the corresponding frequency will be 1 kilo hertz. So, that means, this particular amplifier will be only useful to amplify the signal up to 1 kilo hertz. So, that means, it cannot be used for audio signal because audio signal bandwidth can be up to 20 kilo hertz. So, if you try to use up to 20 kilo hertz, you would not get this particular characteristic of very high gain. So, does that mean that we cannot use 741 op-amp for a very high gain? In fact, I would generally recommend do not try to design these op-amps for very high gain of 1000. I generally recommend that you design this particular value of gain in 2 or 3 stages. So, for example, if you use 3 stages, then each stage can give us a gain of 10, 10, 10. The another thing is if you want to use only 2 stages, then you can use something like maybe a gain of 10 and 100 or maybe gain of 31.6 and 31.6. So, that you can get overall gain of 1000. Another thing which I would generally recommend that for most of these application is better to design non-inverting amplifier. We would be giving input at this particular point and this would be grounded in that particular case gain will be given by 1 plus R 2 by R 1. So, here the advantages if we give input here input impedance is very high. So, one can realize a high input impedance amplifier. There are many applications where we have to amplify a very low signal and that signal may be at a very low frequency it can be even a DC voltage. So, for example, if let us say a sensor is connected at this particular port over here and this sensor can be a temperature sensor, thermocouple and so on and these sensors generally give very small voltage that could be of the order of 1 micro volt or 1 millivolt. So, we have to amplify that particular signal with a very high gain. So, you can use sometimes these kind of a configuration, but be careful you should know what is the application for which application you have to design the amplifier. Now, from here we will shift to the very high frequency and that is microwave amplifier, but I just want to bring to your attention one more time the gain decreases as the frequency increases. Now, this is 1 megahertz if we go to 1 gigahertz gain response will be extremely poor for 741 op-amp. In fact, it will not even work, but I want to mention there are several op-amps they have a gain bandwidth product of 1 gigahertz to even a few gigahertz. So, if you want to use op-amp at very high frequency it is better to use those op-amp than to use 741 op-amp. So, let us go to the next part how to design microwave amplifier using transistors. So, here we have taken an example of a BFP 520 transistor this is actually known as low noise silicon bipolar transistor. It is available for Infineon company and you can actually click on this particular thing and see the data sheet of this particular transistor. So, I just want to mention a few things over here. So, this transistor is to be biased for VCE equal to 2 volt IC equal to 10 milliampere the transistor looks like this over here. So, we have a terminal 1 which is base terminal 2 and terminal 4 both are emitter 2 and 4 should be connected with each other. So, that we have a emitter which is common and this is collector and you you must be familiar with let us say common emitter amplifier common base amplifier common collector amplifier. So, some of those can be used, but with little modification. Now generally speaking a manufacturer at microwave frequency does not give H parameters, but it actually gives S parameters. So, these S parameters are measured using network analyzer. So, one of the port of the network analyzer which is actually generating different frequencies is connected to the input port of the transistor and the output port of the transistor is connected to the other port of the network analyzer which does the measurement. Of course, the network analyzer can do measurement in both the directions also. So, by using this network analyzer one can actually measure S 1 1 S 2 1 S 1 2 as well as S 2 2. So, here the plots of S 2 1 and S 1 1 are given. So, let us see what we have here. So, this is the frequency from 0 to 6 gigahertz and this is the gain value you can see that it varies from 0 to up to 25 these are numeric values. So, let us see how magnitude of S 2 1 varies. You can see that as frequency increases S 2 1 value is decreasing. So, let us just look at some typical numbers here at 2 gigahertz frequency S 2 1 is equal to 10. So, gain is equal to S 2 1 squared. So, that will be equal to 100 and if we now take 10 log of gain that comes out to be 20 dB. So, gain at this frequency is 20 dB. So, at this particular frequency you can say which is approximately 0.6 gigahertz gain is going to be equal to S 2 1 square which is 20 square which is equal to 400 and 400 is equivalent to 26 dB. But you can see that as frequency increases gain keeps on decreasing and one can see that somewhere at 4 gigahertz now gain is just about 5 square which is equal to 25 and as frequency increases further gain is reducing further. Now corresponding to this now let us see we have S 1 1 plot one can actually see here that at lower frequencies S 1 1 is very very poor. If you look at a value of 0.8 for S 1 1 and recall reflection coefficient is equal to S 1 1 which is equal to 0.8. So, reflected power will be square of this which is 0.64. So, that means, 64 percent power will reflect back. So, one should really do something at this particular frequency. So, if you want to use this particular transistor at lower frequencies we must try to optimize the input impedance of this particular transistor. You can see that at around 3 to 4 gigahertz reflection coefficient is relatively less than 0.3 which corresponds to reflected power of about 10 percent which may be acceptable. But if we know that we are going to operate at a frequency of 3 gigahertz or let us say Wi-Fi frequency of 2.5 gigahertz then we know what is the value of S 1 1. So, this can be improved by providing impedance matching network. Now the similar thing needs to be done at the output port also S 2 2 also again may not be equal to 0 at all these frequencies it may have a finite value. So, we need to provide output impedance matching network. So, for a given device generally speaking S parameters are specified for given biasing condition by the manufacturer. So, now what we are going to do we are going to look at the expression for how to find out gamma input of a device. Now this concept is common to any general device, but since this topic is microwave amplifier. So, I have written here in the bracket amplifier. So, let us see what we want to do over here and why we want to do it. So, first thing is we want to find out gamma input. So, let us look at how to find out gamma input. So, let us see over here. So, here a given device is specified by its S parameters at the desired frequency and given biasing conditions. Now you can see here this is the incoming wave A 1, here is a incoming wave A 2, this is a reflected wave B 1, this is a reflected wave B 2, this device is now connected with the source which has a source impedance and there is a load over here. So, the objective here is to find out gamma input ok. So, how do we start? Let us start with S parameters we have already discussed about S parameters in our previous lecture. So, S parameters are defined in terms of B 1, B 2 and these are S 1 1, S 1 2, A 1, A 2, S 2 1, A 1 plus S 2 2, A 2. Now let us just look at the another thing here which is gamma L. So, how do we define reflection coefficient? Generally speaking reflection coefficient is defined by you can say reflected wave divided by incident wave. So, now if you look at this side here what is the reflected wave looking in this side that will be A 2 and what is the incident wave that will be B 2. So, gamma L is given by reflected wave which is A 2 incident wave which is B 2. So, from here we can find out A 2 is nothing, but equal to gamma L times B 2. So, we substitute this particular value over we substitute this value of A 2 in this particular equation here. So, B 2 is equal to now S 2 1, A 1 plus S 2 2, A 2 is gamma L B 2. So, now we simplify this particular thing for B 2. So, this term will come to this side and then we divide it. So, B 2 expression is given by this equation number 4. So, now what we do? We actually put this value of B 2 in this particular equation over here, but after we write A 2 equal to gamma L B 2. Why we are doing that because our objective is to find gamma in which is equal to B 1 by A 1. So, we want to get B 1 by A 1 we have already removed A 2 from here as you can see from this equation. Now we will remove A 2 from here. So, let us see how we proceed. So, this is how we proceed B 1 is S 1 1 A 1 plus S 1 2 A 2. A 2 is gamma L B 2 and B 2 is given by this particular expression. So, now we solve that equation and we get this particular expression over here. So, what is this expression? Gamma input is equal to S 1 1 plus S 1 2 S 2 1 gamma L divided by 1 minus S 2 2 gamma L. Now generally for a device we are familiar with S 1 1 now some additional term has come over here. Why this additional term has come over here? The reason for that is the S parameters of a device are defined when you put match load at the input or output side. Over here it is not terminated with the match load, but it is actually terminated with some unknown load impedance Z L. So, if Z L is unknown if it is not equal to 50 ohm then gamma L will not be equal to 0. But suppose if it is terminated with 50 ohm in that case gamma L will be equal to 0 and if gamma L is equal to 0 we put 0 over here gamma input will be equal to S 1 1. So, that is how S parameters are defined. Let us look at another thing also here and that is S 2 1 we know is a forward gain from here to here and amplifier should actually send the signal from the input side to the output side. What is S 1 2? S 1 2 is if we give input at the output side what is the output? Generally speaking we would like S 1 2 be close to 0. So, in this case also if we put S 1 2 equal to 0 then what we will get? Gamma input will be equal to S 1 1 and in that particular case it does not matter what is gamma L. So, what it really means is that if amplifier is only I am sending signal in this direction and not sending any signal over here this is also known as a unilateral case. In fact, we would like an amplifier to be perfectly unilateral case that means it only send signal in this direction it does not send signal in this particular direction. If it does send some signal in this particular direction it is known as bilateral case. Now, this equation can be further simplified. So, let us see what we have here. So, from here you can see that denominator is 1 minus S 2 2 gamma L which remains same here this particular thing is multiplied. So, you will get S 1 1 minus S 1 1 S 2 2 gamma L and plus this term over here that can be simplified in this particular form over here. So, where delta is given by this expression here what is delta over here actually speaking it is nothing, but determinant of the S matrix determinant of S matrix will be S 1 1 S 2 2 minus S 1 2 S 2 1 will use this particular equation later on. So, now what is the purpose of finding gamma in actually speaking the purpose of finding gamma in is that we find what is the value of gamma in after considering all these things here then we design impedance matching network over here. So, that maximum power transfer can take place from the source to this particular device when maximum power transfer will take place from the source to this particular device that will be the case when gamma S looking from this side is equal to gamma in conjugate. So, you all know that for maximum power transfer load impedance should be equal to source impedance if it is real impedance, but for complex impedance load impedance should be equal to complex conjugate of the source impedance. So, that is what we need to do. So, first we find out the value of gamma in by using this expression and then we design a impedance matching network. So, that gamma S of that particular thing will be equal to gamma in conjugate. Now, by using the same concept we can find out gamma out of the device also, but before we look into that I will just tell you intuitively what you should expect. So, if instead of in if it is out this will be B 2 divided by A 2, S 1 1 will become S 2 2, S 2 2 will become S 1 1, S 1 2, S 2 1 will swap each other. So, it will remain same and gamma L will become gamma S. So, let us see. So, this is a derivation of gamma out of the device. So, again what we want to find out? We want to find out gamma out is equal to B 2 by A 2. Now, in order to find out the output impedance or gamma out we must make source equal to 0. So, you can see that source is now made 0 since we had taken a voltage source. So, 0 voltage source would be short circuit. Had there been a current source then we would have made current source equal to 0 that would imply open circuit. So, now objective is to find gamma out equal to B 2 by A 2. If you look at this expression here I had just mentioned to you that this can be derived just by looking at the gamma in expression, but over here let us quickly see how we can do the derivation. Again we start with the S parameters which are given over here. In this case now let us define gamma S. What is gamma S? Looking at this particular side over here and for gamma S what we have to again say it is a reflected wave divided by incident wave. So, if you are looking from this side what is reflected wave A 1? What is the incident wave B 1? So, from here we can write A 1 is gamma S times B 1. Substitute this value of A 1 and simplify you will get this particular expression and this expression can be written in this particular form where delta is nothing, but S 1 1 S 2 2 minus S 1 2 S 2 1. So, now we will find the expression for the gain and here we are going to use Mason signal flow rule. Generally speaking this Mason signal flow rule is taught in the control theory, but we can use the same concept over here also. So, let us see we have S parameters of the device. We have a load here and there is a source impedance. Remember this is not 50 ohm. This ZS will be chosen properly so that we can provide impedance matching network later on. So, here this source is supplying BS waveform. So, now let us see how we can actually speaking built this signal flow graph. So, we start again with the S parameters over here. We also know what is A 2? A 2 is gamma L B 2. You can just quickly check here what is gamma L? Gamma L is reflected wave which is A 2 divided by incident wave. So, that gives us A 2 equal to gamma L B 2. What is gamma S? So, gamma S we had seen previously it was A 1 by B 1, but that was the case when BS was equal to 0. If BS is present in that particular case you can see BS is also coming. So, A 1 will be now equal to you can see from here gamma S B 1 plus BS. So, now these 4 equations are represented in this graphical form over here. So, let us see how we can build this particular signal flow. So, let us start with let us say B 2. So, B 2 is given by S 2 1 A 1. So, B 2 we locate A 1. So, A 1 multiplied by S 2 1 plus S 2 to A 2. So, we locate here A 2 this is S 2 2. So, these are known as path gain. Now, let us look at the other equation which is B 1. So, B 1 is S 1 1 A 1. So, B 1 is S 1 1 A 1 plus S 1 2 A 2. So, plus S 1 2 A 2. Now, we need to look at this equation here. So, A 2 is gamma L B 2. So, A 2 is gamma L times B 2 and A 1 is given by this expression here. So, let us see this is A 1. So, B 1 times gamma S and plus BS is coming from here. So, now let us see what is the expression for Mason signal flow. So, this is the transfer function which can be used to find out the gain of the amplifier. So, what it shows over here? This transfer function can be used to find out the transfer function from let us say point 1 to point 2 ok. So, what this expression has here? Let us start with the denominator. You can see here it is 1 minus summation L 1 plus summation of L 2 minus summation of L 3. So, let us see what are these things. So, you can see here L 1 L 2 L 3 these are nothing, but sum of all first order second order and third order loops. I will define in the next slide what are these things here. Now, here we have in the numerator you can see here there is a P 1 this is the gain of path number 1 this is the gain of the path number 2 and what are these terms here? These are the terms these are the different loops which do not touch the path 1 ok and these are the loops which do not touch the path P 2. Now, let us see how we can find out the various parameters for this particular case here. We want to find out the gain which is given by B 2 divided by B s B 2 is the output B s is the input. So, let us see what is the path we have there is a only one path from B s to B 2 and we can write that path gain is nothing, but equal to 1 multiplied by 1 into S 2 1. So, this is the path gain there is no other path to go from this point to this particular point. So, that means, B 2 is equal to 0. So, these are the different steps now. So, step 1 path. So, we can actually see that the there is only one path which is equal to S 2 1. I just want to mention here that in to define the path no note should be touched more than once ok. So, now let us see what are the first order loops ok. So, there are three first order loops in this particular case and the loop gain for these three loops given by these expressions here. So, this is the first one second and third one. Now, what is the second order loop? Second order loop is product of any two non-touching loops. These two loops are not touching each other. So, hence this term comes over here you can see that is a product of these two terms over here. There is no third order loop because there are no three non-touching loops and hence this particular thing will be equal to 0. So, now we can write B 2 divided by B S. So, let us look at the numerator it is nothing, but P 1. There is no loop which is not touching this particular path hence other terms are 0. So, we are left with only S 2 1. Let us see now the denominator denominator is 1 minus summation of the first order loop. So, you can see here S 1 1 gamma S then S 2 2 gamma L then this term over here. So, that is the summation of the first order loop and then plus summation of second order loop ok and there is only one term. So, that comes over here. Now, this expression can be simplified in this particular form. We will see later on how this particular form will be useful to design the gain of this particular amplifier. Now, we are going to define three different types of gain. Now, till now you might be more familiar with only one gain you know output voltage divided by input voltage or output power divided by input power. So, why we are defining three different types of gain over here. In fact, I just want to tell you actually all these three gains are related to each other in a very simple manner. We are going to first start with operating power gain which is given by P L divided by P in. What is P L? Power delivered to the load. What is P in? That is input power. So, from here let us go to this expression here which is transducer power gain. Transducer power gain if you see the difference is only that P in is equal to P available from source. So, what will be the maximum value of the power available from the source? When P input will be maximum. So, P input will be maximum when gamma in is equal to gamma as conjugate. So, what we have here that input impedance of the device should be complex conjugate of the source impedance. Now, let us see what is the maximum available power from a given device that is defined by this expression over here. So, compared to G T what is the difference here? Now, P L is equal to P available from the network. So, this is the maximum value of the power which can be delivered to the load and maximum power will be available when gamma L is equal to gamma out conjugate. So, that means maximum available power from a device would be when input as well as output impedances are complex conjugate of the source and load impedances. In that particular case we actually say that this is the maximum available power from an amplifier. So, now let us just quickly look at the expressions ok. We will start with the transducer power gain and transducer power gain expression is given by this term over here. So, why are we starting with this particular expression? The reason for that is the load may not be in our hand the load may change depending upon the requirement, but power available from the source can be optimized what we can do here that we can actually find out what is the gamma input of the device and then we design impedance matching network such a way that maximum power transfer takes place from the source to the input of the active device. So, that is why we are starting with the G T ok. So, what is G T now? P L divided by P available from the source. So, this is given by half b 2 square minus a 2 square y b 2 square is the wave which is going to the load and a 2 is the reflected back signal from the load. So, hence we deliver to the load will be given by this expression here. We take b 2 square outside then this will be 1 and this will be a 2 divided by b 2. So, that will be gamma L. So, that is the expression for P L. Let us see how we can find out expression for P available from the source and that is equal to half of b s square. So, this part is kind of obvious, but we have another term over here. Now, this expression will be equivalent to this particular expression if gamma s is equal to 0. So, if gamma s is equal to 0 this will be 1 minus 0. So, this is kind of obvious that P available from the source is half of b s square. Now, why this term comes into picture? The reason for that is if it is not properly matched then what will happen part of the wave will get reflected back. So, let us just take the worst case assuming that gamma s is equal to 1. So, if it is 1 what it would mean 1 minus 1 will be equal to 0 and that would imply that power available from the source is infinity. What does that really mean? Well, you can think differently. See if everything is reflected back then we can conceptually say well infinite power is available even though actually speaking it is not available you are not supplying anything. So, in general we always try that gamma s should be done properly. So, that power available from the source can be maximized. Now, let us look at the expression for g t. So, P L divided by P available from the source. So, half and half will get cancelled. So, b 2 divided by b s that expression we have derived in the previous slide. So, that will have several terms. Let us just look at other terms also in the numerator we have 1 minus gamma L square that is written right over here. This term was in the denominator since it is divided. So, this will go up over here. Now, all these terms over here they are coming because of this expression of b 2 divided by b s. So, you can actually speaking see here this term is related to everything in the source side this term is related everything to the load side. So, basically what you can see over here this is the gain term corresponding to the source side and that can be optimized by properly choosing impedance matching network at the input side. Now, this is the term which corresponds to the load side this can be optimized properly by designing a proper output impedance matching network. But I still want to mention over here you can see here this is gamma in this is not the way it was written as b 2 by b s. I just want to mention here you have to do little bit of a simplification gamma in is given by s 1 1 plus s 1 2 s 2 1 gamma L divided by 1 minus s 2 2 gamma L substitute that expression you will get what we had derived in the previous slide we can see over here gamma out is there. So, here gamma out whereas, this is s 2 2 over here. So, again for gamma out you have to write the expression of gamma out which is s 2 2 plus the other term simplified you will get the same expression as in the previous slide. In fact, these two expressions are exactly identical just that the representation is slightly different. So, here you can see these are the input side these are the output side. So, in the next lecture we will start from here we will start with the expression g t and then we will derive other expressions. So, just to summarize today we started with a very simple inverting operational amplifier where we did look at the design problem of gain equal to minus 1000 and how resistors should be chosen appropriately. Then we looked at s parameters of a transistor and we noticed that s 2 1 of the transistor keeps on decreasing as frequency increases. So, at higher frequency really speaking we have to optimize the gain properly. So, that we can get a higher gain and to obtain the higher gain what we need to do we have to find out gamma in we have to find gamma out and then we have to design impedance matching networks at both input and output sides. So, that maximum transfer of the power takes place from input to the output side. So, thank you very much we will see you next time.