thank you welcome to analog plsi design this is lecture one as we had seen in the introductory video in this course we will be almost exclusively focusing on designing and amplifier now this is a course for electrical department or electronics department so when we when we talk about amplification amplifying something we are essentially talking about amplifying an electrical quantity right so what is the most familiar electrical quantity that we can amplify obviously we can say that let's start off with voltage right so when we are planning to design an amplifier the question is are we planning to design a voltage amplifier right what do I mean by that so let's assume that I have a voltage source let's assume I have a voltage source be in let's further assume that it has an internal resistance RS right because no voltage source comes with zero internal resistance it's always a real life voltage source is always accompanied by an internal resistance let's call it RS where s stands for the source and let's say there is a load resistance RL right what are we interested in are we interested to build a block and put it in between RS and RL such that after we connected this block the voltage that will develop across RL right let's call it V naught are we interested in building a block let me just put a question mark here so are we interested in building a block such that V naught is equal to AV times V in where AV is greater than 1 so are we interested to build this one might say that if we have such a block we might as well call it a voltage amplifier right can call this oh voltage Amplified so is the course about finding out what is this stuff inside this box which can give us voltage amplification now some of you might turn around and tell me that hey what is so only about voltage in electrical domain we have another quantity that we are equally interested in that is the current can we make a current amplifier right why are you partial about amplifying voltages right so you will also be right so let's say that I have a current source we call it I in time source is again accompanied by an internal resistance RS and I have a load resistance RL what should I so is the course about building this box inside such that the current I naught that will flow into RL right I naught is equal to sum proportionally constant a i times i n such that a i is greater than 1 and if this is if this is realizable if we have been successful in realizing this we can call this ah current amplifier and you will be right right this if we are able to realize the Box on the top we can say that that box is a voltage amplifier if we can realize the box in the bottom circuit then we can say that the Box acts like a current amplifier however in this course we are neither interested in designing a voltage amplifier not a current amplifier what we are interested in is the following we are interested in building a box such that if I put this box in between the load and the source if I put this blocks box between the load and the source then the power delivered to the load right so the power delivered to the load be out right should be greater than the power that I am extracting from The Source in other words if the power that is going in is p in and the power that is getting delivered to the load RL is p out we are interested in find finding out this box such that e out is equal to let me call some proportionally constant AP times p in where AP is greater than 1. this course is focused in designing this box where the output power is greater than the power that is that I am extracting from the source The Source might be voltage source The Source might be current Source it doesn't really matter all we are interested in is to deliver an output power which is greater than the power that we have extracted from the source and why is this relevant this is relevant because there are several applications which demands this for example the most common application that you might familiar with is the right of a microphone and a loudspeaker the volume that comes out of the loudspeaker is much much stronger than the decibel level of the volume of uh of the volume that is a decimal level of the signal that is going into the microphone right so clearly there is a power amplification that is rather than that has happened it's neither a voltage amplification not a current amplification let me try to convince you with an concrete example of what I mean let's take the example of a humble Transformer I am sure you have gone through or you have encountered an ideal Transformer in your basic electrical engineering courses right so for example you have let's say I have this again this non-ideal voltage source and I have foreign and I put this Transformer in between and let me Mark this Transformers turns ratio as 1 is to n right what is a Transformer an ideal Transformer is nothing but all we need to know what the purpose of this discussion is an ideal Transformer is the following if this is V1 and I have a trans ratio of 1 is to n the other terminal will be the voltage across other terminal will be n times V1 right and also and also we know that if if I have a load resistance RL connected to the secondary of the Transformer that is on the right hand side of the Transformer in this case it that that load load resistance gets reduced by a factor of N squared obviously if n is more than one okay so these are the two things that we need to keep in mind in order to solve this problem okay so what I what are we interested in I'm interested in figuring out what is V naught so what is P naught over we are interested in we are interested in figuring this out okay so how should I go about figuring this out so we know that the resistance looking in it is RN for an ideal Transformer in this case will be RL Over N Square right so if I remodel if I remodel this circuit right if I remodel retro this circuit like this so this becomes V in RS and a resistance of value or a Over N squared and if I Mark these voltages as V one this voltage becomes E1 okay what is V1 V1 is nothing but the current that is flowing into this Loop that is v in over RS plus r l Over N squared times the resistance RL Over N squared okay so if that is the case if this is V1 what will be V naught now if I switch our attention to the circuit on the top I know V1 so what will be V naught from the transform or action we know that V naught is n times V1 which is equal to p in times n times RL Over N squared divided by press r l N squared correct so if I multiply both the numerator and the denominator by N squared what do I get I get V naught over V in is equal to n times r l by N squared RS plus RL okay so if I rewrite this here so V naught over V in becomes n times n times what r l by RL plus N squared X okay so let me redraw this circuit redraw this circuit be in press okay so if if RL is much much greater than N squared time RS what happens to V 0 over V in in this case v 0 tends to n times V in right V 0 tends to n times v n so this acts as a voltage amplifier right from the definition that we started off with right if I take you back to the definition that we start started off with in few minutes back it seems like a simple Transformer is able to satisfy this condition right okay obviously one might turn around and say that hey this is limited because this is true only for certain ranges of oil the RL has to be much greater than N squared times RS if that is the case I will I might get voltage amplification but nonetheless we are getting voltage amplification correct however note that we are not interested in these type of amplifiers because you cannot get power amplification out of the Transformer right if you are not convinced what I would like you to do is pause the video here and try to find out what is e out over B in in other words what I am asking you to do is to find out what is the power delivered find be out or what be in where e out is equal to the power delivered to RL right so which is effectively you can say that V naught squared average over RN and what is p in is equal to the total voltage V1 and the total current i1 the multiplication of V1 times i1 the average of that right so you can assume be in to be equal to BB sine Omega naught T right you can assume this and try to figure out what will be P out what will be p in and what will be the ratio of P out over p and we'll continue from there [Music] [Music] oh [Music]