 So, we will take one numerical, now here it refers to a figure, so I will first, technology is great. Alright, so the question is, you can see a figure in front of your screen, you need to calculate the value of beta dc and beta ac, okay, when VCE is 10 volt and I see, Sir, beta dc is 133.33, beta dc is 133.33, okay, close to that. The beta ac is 200, right? Beta ac, approximate. No, although you might have done it the correct way, but you are not getting the correct answer. So, VCE is what, 10, 10 volt, this line correspond to 10 volt and I see as 4, this one, okay. So, first of all, let's try to find out beta dc, okay, because that is easier. So, now tell me when you have VCE as 10 volt and I see as 4 milliampere, you can, you know, you see that this point is not on the graph, are you getting it? This point does not belong to the graph. What you do, you take two points that belong to the graph, this point and this point and then take the average of these two. Getting a point? Sir, what is closer to the graph on top, right, you mean just assume it to be on the graph? No, you can't do that. You don't know what are the values in between, you can't linearly extrapolate, okay. You have to exactly take it on the graph, okay, and then you assume that beta dc is uniform for a particular VCE, okay, and just take the average of two values. That's how we conduct the experiment, right, whenever we conduct experiment, we have multiple values, you take the average and you say that is the closest to the actual one. Nobody knows what is the true value. Yeah. Okay, so this point, you can say that this is let's say point number one at point one. We have collector voltage, how much it is around three? Three milliampere, right, and how much is the base? IB is what? Twenty, right, and at point two, collector is what? Collector current, roughly, can I say 4.5? Yes, sir. And base current is simple, 30 microampere, okay. You take the ratio of these two, beta dc, take the ratio of these two, beta dc, and then take the average of these two. You'll get 150, right? See, nothing is wrong, whatever you guys did. It's just that, I mean, this is the experimental finding. So this is how you deal with experimental things. You find out multiple points on the graph, then you take the average, alright. Now, how will you find beta ac? Can you go by the same logic and try to find beta ac? Should I do it? That's 100, sir. No. Okay. Now, at VCE, which is 10 volts, so this is 10 volts, can you tell me? Sir, is it 150 again? Yes, it is 150 again. See, I want to find out beta dc for this point, right? So I'll find out across this point, what are the two points that are on the graph, okay? Those two points are point number one and point number two, okay? So delta ic between point one and point two is what? Point at this point has 4.5 milliampere current and this point has 3 milliampere current. So delta ic is 4.5 minus 3, which is 1.5 milliampere, okay? Then on the graph, delta ib corresponding to delta ic is what? Simply 30 minus 20, microampere. So this is 10 microampere, fine? Just take the ratio of these two, you'll get 150. Do you guys understand what is done here? Yes, sir. This chapter is little counter-intuitive. I mean, you will not be able to intuitively think and do it very smoothly. But then look at the solution, you'll be like, okay, this is no-brainer, okay? But then this requires practice. So you need to solve questions from HCV or whatever books you may have, all right? And definitely there will be one or two questions in J-Mains, 100%. There will be a few questions and they will be very, very easy to solve if you have done some practice. See, nobody is going to ask you which chapters, questions have you solved in J-Mains. They'll just count the marks. Whether you get the marks by solving a question on semiconductors or rigid body motion, nobody cares. Just the number of marks. Okay, so let's... Now that we have discussed the basic, you know, the construction of the transistor, we can now talk about the usage of transistor, okay? So let's see how we can use a transistor as a device, okay? Write down the heading transistor as a device. Okay, so I'm going to simply plot common emitter characteristic... Sorry, not characteristic, I'm going to plot common emitter configuration. It will have a voltage, of course, as VCC, okay? Which is connected between base and emitter. There is a resistance like this. And when we have measured the characteristic, the voltmeter was connected here in this zone. But now we are connecting it outside. Okay, so don't get confused. It's the same figure, all right? So why are we earthing it? Why are we earthing it? I mean, it's just a reference point. It's like whenever you solve a mechanics question, you draw a horizontal line and you say, that is my zero potential energy, zero gravitational potential energy, but nobody knows where the zero potential energy exists. Similarly, whenever you have electrical circuit, you assume something as zero potential. But that need not be zero potential. You're just assuming it to be zero and every other potential is measured corresponding to that. So if I say that, you know, this point, if I say this point has a voltage of 5 volt, I am saying this point has 5 volt more than this point. So this is like a reference point for all the voltages. Are you earthing it? Okay, yes. So it looks like this. And then this is RB, this is BBB. And then you'll again have a voltmeter here to measure the input voltage. This is input voltage. So this is essentially the same circuit, okay? Just that the emitters are not there. The voltmeters are still there. All the voltmeters are there. But emitters are removed. Everything else remains same. This is IC. This is IE. E and this is IB. Okay. Now this is, okay. The first device out of transistor is switch. So write down transistor as a switch. Okay. Now can you apply Kirchhoff's loop rule in the input side as in this loop and in that loop? So I want you to apply Kirchhoff's loop rule here and there and write down the equations quickly. So you have IE equal to IB plus IC? No, but that is not the loop equation, right? That's the junction rule. Okay, sorry, sorry. VCE equal to VCC minus ICLR, RL, RL. Others, where you see RL? There is RC and RB. Sorry, RC, RC. I thought that was RL, sorry. I meant RC. So VBB minus IB, RB equal to 0. Okay. So let me first write down Kirchhoff's loop rule equation for input side. Okay. Anyways, YouTube also there are four or five. Okay, input side. Sir, I wanted to tell you that he couldn't come because it was his birthday. Happy birthday. Okay. So let's talk about input side. Input side, we have VBB then minus IBRB and then what will be there? Then there is a junction potential between base and emitter. So this minus VBE is equal to 0. Knitting it. So VBB, why I'm separating VBB as in why I'm keeping VBB on the left-hand side? Because VBB is nothing but input voltage. Okay. So input voltage is VBB for VBE plus IBRB. Okay. Now let's talk about output side. Output side, you will have VCC minus what? What is the current over here? So IC. IC is the current. Voltmeter will not draw any current. No. So ICRC, okay. Minus what? VCE. VCE is equal to 0. Now what is VCE? VCE is the output voltage. Okay. So I can say that V out is VCC minus ICRC. Fine. So this is equation 1 and this is equation 2. Knitting it. Now if I increase, suppose let me first talk about if input voltage is close to 0, then what will happen? Any guesses? So then even output voltage will be very low. Why? Because it's near 0, right? IB will be very small. When IB is very small, that is below 0.7, there will be no IC. So if VI is let's say 0.4, what will happen then? No current. Oh, no, sir, no current, sir. No current anywhere. So entire transistor will be shut down. Yeah. So if entire transistor shuts down, then even IB, IC, everything should be, all the currents will be 0. Are you getting it? So if all the currents are 0, output voltage will be maximum because this is negative. This is getting subtracted. Are you getting it? If so VI is very close to 0, V out is max. Getting it? Any doubts? Sir, but isn't VBE the input voltage as we discussed earlier? VBE, V Bombay as a friend. Isn't that the input voltage? VBE as input, no input is, see this is where you are creating input voltage. Or you are probably getting confused with the earlier slide, is it? Yeah, yeah. Let me open that. What is this one? Here? Yes, sir. So that's what it is? Yeah, sir, but then we consider VBE as... Let me now open this one. Yeah, correct, correct. See here, see you can, what is your input is, what is your device first of all? My device is, when I've considered the characteristic of transistor, I have just considered the transistor. I have ignored everything else. But when I'm using it as a device, then this entire thing is a device. This RB is a part of the device. RC is also part of the device. Now I am going to supply input voltage from this side. And I'll get output from this side. That's a very good doubt, actually. Okay, so... Sir, one more doubt, sir. I understand how when input voltage is very, very small, how output voltage is max? You understood that if input voltage is very small, all the currents will go down to zero? Correct. And look at the formula for output voltage. VCC minus ICRC. If IC is there, it will be less than VCC. Sir, there won't be IC because VI is very small. That's what I'm saying. That is why it is maximum value. It's maximum value is VCC. Okay, okay. Got it, got it. Okay. And suppose your VI is large. Your... See, don't get confused by naming. I may say input, I can say that... I can name it something else. I'll say that it is V Siddhant, okay? It doesn't matter. Okay, I can name it anything. So it's just a convention. So if input voltage is large, now what will happen to output voltage? Any guesses? It comes minimum. How much minimum as in? It will tend to what? Zero. It will tend to zero because after you increase the input voltage beyond 0.77, 0.7 volt, current will suddenly rise. It will behave like a piece of metal. So if IC increases, okay, it will decrease the output in such a way that it will tend to zero. Okay. So you can see that just by changing input voltage, you are able to change the output voltage. Okay. So this can be used as a switch. Your output voltage could be connected to the fan and input voltage could be a button in your hand. Okay. So from your hand, if you just change the input voltage, the output voltage will get changed. Suppose you want to switch off the fan, you increase the input voltage, output voltage will go down to zero. And if you want to increase its velocity, you just make input voltage zero, the fan will run in the full speed. Okay. Sir, I have a lot of doubts, sir. Generally, we have VCC at a higher potential than VBB, no? Yeah. Okay, no, that's all I want to know. See, you have VCC at higher potential than VBB just to be doubly sure that that is reverse biased. Okay. Okay. Just as a caution, you are making it higher potential. All right. So if you plot a graph between input and output voltage, let's say this is output voltage on y-axis and input voltage on the x-axis, the graph will look something like this. All right. So you can divide this graph clearly into three zones. Okay. This zone is cut off region. This zone is active, active region. And this one is referred as saturation region. Okay. So this is how you can use transistor as a switch. Right. Now let's talk about how we can use transistor as an amplifier. Okay. When I say transistor as an amplifier, I do not mean current amplification. What I mean to say here, that I'm going to use it as a voltage amplifier. Okay. Let's say, let's see how it can be used as a voltage amplifier. So did we just see it as a voltage amplifier as well because we just found a relation between Vi and Vo. So? Like, okay, never mind. Never mind. Never mind. See, voltage amplification, I don't mean to find out the ratio of input and output voltage. I want to find out what is the ratio between change in output voltage and change in input voltage. Okay, delta Vo by delta Vi I am interested, not Vo by Vi.