 to the session on multistage amplifier. Learning outcomes are at the end of session students will be able to explain the requirement and the need for the multistage amplifier as well as they can analyze the various parameters of the cascaded amplifiers. These are the contents. Physically as voltage gain or the current gain depends on the circuit parameters and the device parameters, but again there is a physical limitation on the gain of the amplifier and that can be fulfilled with the help of multistage amplifier. So, to drive the low device with the large current we need this multistage amplifier as well as for the composite electronic circuit we need this multistage amplifier. It will also provide the large amount of power to boost the signal. To achieve this what can be the requirements? To pass the AC without distortion I need to block the DC through the coupling network. So, we need to use various coupling networks as well as we need to find the arrangement to pass the signal applied at the input without any distortion. For that we need to use capacitor over there. Again there should be low loss for the signal. So, coupling network may cause the loss. So, we must keep that loss at the minimum value. While passing the signal impedance matching is very important factor. Therefore, equal impedances at the various frequency must be obtained. So, figure 1 shows the block diagram for multistage amplifier where we are connecting the output of first stage to the input of next stage. So, this will be carried forward till the number of stages that is here state n is shown. So, now observe this carefully. So, this is what VIN that is input to the stage 1 and this is VS which is applied input. So, we have the VS equal to VI1. So, we have to find various parameters with the help of this analysis like input to the stage 1 is VS equal to VI1 and the output of stage 1 is VO1. Now, we can find the gain voltage gain of the first stage just by taking the ratio of V01 upon V0 VI1 nothing but output voltage by input voltage. So, while doing so you will get the input to the next stage also. So, we can say input to the next stage is nothing but output of previous stage as we have shown here. So, this is what this is V01. So, we can find V01 equal to AV1 into VI1 and that will be seen that is input to the second stage AV1 into VI1. In this way we can find the gain of the amplifier in the different stages. But keep in mind we have to calculate the gain which is serving some load. Therefore, see this output of nth stage is Vout and therefore, we can say AVn equal to Vn upon Vn minus 1 where Vn is the output voltage of nth stage and Vn minus 1 is the input to the nth stage nothing but this is Vn minus 1 which is output of previous stage. Therefore, we can say Vout equal to Vn equal to AVn into Vn minus 1. So, once you know the value of load you can calculate the voltage at that particular point and you will get the overall gain as an amplifier. So, overall gain of amplifier can be calculated as Vout which is the output of nth stage of amplifier and VS which is the input to the first stage. So, I can say this as V01 upon VI1 which is the gain of the first stage and that of the complete single amplifier I can say it is just Vout upon VS. Vout is the output of nth stage, VS is the input to the first stage. So, in this way if I get the ratio of V0 upon VS what is this? V01 is output of first stage, VS is input to the first stage. In this way V02 is the output of second stage and V01 is the output of first stage which is nothing but VI2 nothing but what input to the second stage. So, in this way we can say here multiplication of the gain is the overall gain of amplifier. If I express that in decibels I will just take log on both sides and multiply it by 20 I will get the equation which provides the gain in decibels. So, this is 20 log A1 plus 20 log A2 up to AN. So, you can add that gain. So, this is AB in dB. Now, from this discussion just calculate the overall voltage gain in decibels. So, this is the voltage gain of the three stages that is 6300 and 160 respectively. Now, you can convert this gain which is a ratio in decibels and then you add these values. So, your answer is 119.65 decibels. Now, see this is my n stage amplifier. I need to find the various parameters of this like input impedance, output impedance and voltage gain and the current gain. So, what I will do I will just find this is as the kth stage that is one particular stage. So, out of n stages I have just taken an example of kth stage and carefully observed I have considered the log diagram for the common emitter cascaded amplifier. Therefore, I have the input as base and emitter output as collector and emitter terminals. Therefore, I have this rck minus 1 which is the load for the previous stage and this is vk minus 1 that is input to the kth stage, rck is the load for the kth stage and vk is the output of the kth stage. Therefore, I will calculate the voltage gain with the formula which we have derived in the single stage amplifier that is ai into rl upon ri. What is av? Av is the voltage gain. So, for the kth stage I will say avk equal to ai k upon rin k into rlk. So, rlk is the effective load at the kth stage and the current gain can be calculated as the same formula which is for single stage common emitter amplifier minus hf by 1 plus hrl that is minus hfe by 1 plus ho into rln for the common emitter configuration. Rln is rlk which is my effective load for the collector. Now, r in n equal to, r in n is what? Input impedance of the nth stage. Always keep in mind for any particular stage you have to go for the calculation from the last stage to the first stage. Therefore, I have taken ai n formula and r in n formula, formulas are same because my amplifier configurations are identical. So, for n minus 1 stage I can calculate rln minus 1 as rcn minus 1 into r in 1 r in n upon rcn minus 1 plus r in nn. That is so, because always there is a parallel combination of the resistances that is this is load impedance of the previous stage, this is the input impedance of the nth stage. So, this is my effective load rln minus 1, same way I can calculate for rlk. So, you can calculate the current gain as i out by ibi, ibi is for the previous stage. So, when I consider the theory of the two port network, I will consider this current is coming inside the port or to this particular point that is output point. Therefore, I will write it as minus icn because the direction of the current representation is opposite. So, I will write minus icn upon ibi. So, icn is the collector current of the nth stage. So, same here I will calculate the current gain just by making the ratio of these different values that is these different stages. Now, keep in mind I am taking the ratio of the two things. When I say i1 upon ib, I am taking the ratio of collector current and the base current, but further I am taking the ratio of collector current to the collector current. Therefore, I am writing here dash sign. So, I will write the ai1 equal to collector to base current of the first stage whereas, this ai2 dash ai3 dash as a collector current gain of the second third stages. In this way, I will get the equation for my overall current gain as the combination of both the things that is collector to collector to collector current gain and the collector to base current gain. In this way, I will get this ai k dash equal to ick upon ick minus 1 and the overall kth stage value is aik into rck minus 1 upon rck minus 1 plus r in k. So, this is my overall gain for the kth stage that is current gain. Now, for the impedance I have input impedance and output impedance. So, i out equal to h o minus h r h f upon r h r s this h i and in this way I will get the output impedance equal to 1 by y out 1 and this is my r in n. So, r out k dash equal to parallel combination of r out and rck. So, source this will be the source impedance for the next stage. So, in this way I can also calculate the power gain that is av into ai and putting the value of av as ai into rl upon rl, rl is rcn, ri is r in n types of coupling like register capacitance, impedance coupling, transformer coupling and the direct coupling. These are the references. Thank you.