In today's video we're going to be looking at an AC model for a bipolar junction transistor. We've seen the DC model. That looks something like this. Here's the base, a mirror, and then a collector. Between the base and the mirror we model that is just a PN junction. So we know that's going to be 7-10th of a volt for a silicon device, they're abouts. And up here we have a controlled current source. The collector current is equal to beta times the base current. And with this model we were able to determine various sorts of biasing schemes, how well they worked, things of that nature. So now we return our attention to an AC model because we want to look at amplifiers eventually. That's where we're going for. So what's the AC equivalent of this? Well, the key here is to recall what that junction curve looks like. So when we look at VBE versus IC, and we get this standard kind of diode, quote unquote, curve, very abrupt sort of transition somewhere around 6 or 7th tenths of a volt. Now, if you have this in a DC case, in other words if we have a certain current flowing down through here, that current would be represented by a certain point. Let's see right here. And we could figure out an effective DC value. In other words, you could imagine, all right, this represents this current, this voltage. So if I simply took the ratio of this current here, and this voltage here, right, VBE over IC, we could get basically a resistance value. If you wanted to plot that, you get something that kind of goes like that. And as we would bias this higher and higher, you know, if we were up here, then that would be that much steeper. So that would indicate a smaller resistance. If we were down here somewhere, much flatter curve, much larger resistance. So that would be the DC case. You could think of that as just being, like I said, sort of a resistor that depends on what the actual current is. Well, when we move over to the AC case, things are a little bit different. We can't just sort of draw a straight line through this point. That doesn't quite cut it. So repeating this for clarity. Now when we operate at a certain point, we have a bias. So we'll just say it's right there. And now the AC signal is going to be writing on top of this DC signal. So it's going to be swinging back and forth. As the AC signal comes in, this is going to be some kind of sinusoidal kind of thing very basically. This could be through the base emitter voltage or the base current coming in. We find that this is going to swing up a little bit and then swing back down a little bit. So what we have in this case is not the same as what we had in the DC case where I could just kind of go from the origin out. In fact, the effective resistance here can be thought of as a line drawn tangent to the operating point. In other words, the Q point in the circuit. So you could imagine a line like this, which is tangent to this operating point. The slope of this line is what the AC resistance of that diode looks like. And you can see how dramatic a difference this makes if we were to look at a higher biasing current. Now if we were over here somewhere, now the AC current is going to swing back and forth like this, well, the line tangent to that is going to be really steep. Compared to as if we had a DC case, yeah, there is certainly an increase compared to the blue, but it's not nearly as radical as what we see over here. Or for that matter, if we were way down here, now this thing is virtually flat. All right, so how do we find the slope of a line tangent to that point? Well, what we would have to do is actually plot this equation, right? What is this equation that we have plotted? Well, that's the standard shockley equation, which basically says that your collector current is equal to the reverse saturation current times the quantity of E, the base of the natural log, raised to VBE, your base interval, times Q, which stands for the charge on one electron, 1.6 times the time of the minus 19th couloums. And that is divided by N, which is a quality factor. That's usually somewhere between 1 and 2, K, which is Boltzmann's constant, 1.38 times 10 of the minus 23rd joules for Kelvin degree, and then T, which is the temperature in Kelvin's. You take that minus 1, and that's essentially the equation for that curve right there, all right, this curve. So what we would have to do is take the first derivative of this equation after having plugged in all of the appropriate values for this thing. At room temperature, if we were to use typically a value of T of 300 K, which would be kind of a warm room temperature, 80-ish degrees Fahrenheit, plug that in with the other constants, you will wind up with an IC that's equal to IS times E, this is approximate of course, 38.6 times VBE. So we'll take the derivative of this, and we wind up, there's sort of the bottom line, we wind up with a resistance value, that's equal to 26 millivolts divided by the emitter current, which of course is the same as the collector current. You can use those interchangeably. So this value is appropriate for a silicon transistor at roughly room temperature. As we would change temperature, you know, this T value increases, decreases, and we will see a change in the value of R. This R value is what replaces the diode. So our model winds up looking like this. We have a current source, this is the AC current source, IC is equal to beta IB. Technically, this is the AC beta versus the DC beta, but it's very close. This is the value that we just computed, we call this our prime E. So we can just say, look, the value for our prime E is just 26 millivolts, a constant divided by whatever the collector current is. Now this kind of gives you a clue into why we were so interested in making sure that the collector current was stable, because without a stable collector current, this R prime E is going to change. And as we will see very shortly, as our prime E changes, this is going to impact things like the voltage gain of our amplifier. So we would like to have a very stable amplifier. In other words, I don't want to make a thousand amplifiers and have some of them with a gain of 100 and some of them with a gain of 150 and some of them have gain of 32. And we want them all to be very, very similar. So that's one reason why we want stability in terms of the DC. Without that, the R prime E won't be stable. And as we, as we shall see, without our prime E being stable, it'll be much more difficult to have these other circuit parameters be stable. All right, so that is the basic model that we have. This works well for low frequencies, audio frequencies. At very high frequencies, this is not sufficient. There are other characteristics that we would want to include, for example, there is a small resistance associated with the base of the transistor. Sometimes referred to as our prime B, we also have some capacitance to deal with. So here's our beta IB again. So there is a junction capacitance that we would be interested in. In other words, you could imagine sort of a small capacitance probably in the picophair ads, modest picophair ads across here. There is another capacitance associated with the collector to a meter from the collector to the base. There's also a resistance associated with the current source, just like there would be in the DC case. So the point being here is that these capacitance values, values are small enough to ignore. The internal resistance here is large enough to ignore, and the R prime B is small enough to ignore. So we can pair this down to just this. So we have a current source, we have an R prime E, and we'll pick it up next time.