Okay, on this slide here, I kind of up. We're going to go over what I told you before real quickly, and I've got a circuit right there. Notice I've got an amp meter and the amp meter in a volt meter. Okay? So just take a look at that. All right, we've gone over.

We've gone over this in the previous slide, IB is the current flow through the base element, I see is the current flow through the collector element, ie is the current flow through the emitter element now equals IC plus IB. And I show you electron flow. So we go this way, in the database junction and we go this way The emitter to collector or I think they I think the correct way is VCE voltage collector emitter okay. And again we know that the beta is the amplification factor. So, I B which is the current to the base times the beta which is the applet fake vacation factor will equal the current flow through the collector and alpha is the quality of the transistor where it would be IC over IE and I gave you an approximate breakup of the current where 98% of the current flow through the transistor element flows through the collector which we call IC for collector current 100% flows through the emitter, I and then i B which is the base current we've got approximately 2% is my audience Here's my equation.

So what we're stating here is if I see was 10 milliamp hours and i B was point one milliamp hours, what would be the beta of this transistor? That would be 100. Okay? Because 100 times point one equals 10 in this case 10 milliamp hours, then we have the L for and I just did this, I see is 10 milliamp hours, ie is 10.1. When I do my math, I get an L for 99%. Okay.

All right, so So the point I want to make here is this is a real stat and it's used to adjust the current flow through the base and if I adjust the current flow through the base, I then adjust the current flow through my collector, which will give me a varying voltage drop across the transistor. Okay. And it's not a complete circuit. I don't know why but they usually put a resistor in there. And they didn't do it here. But yeah, you'll see in the next slide, we put a resistor in there.

So that's it. So this is a little bit of a review that we talked about the previous slide. All right, so let's look at the next slide. Maybe it'll be a little bit more clearer. All right. Okay, here's my circuit.

By the way. This is called a common emitter. I did I put that in there. By better, let me put it here. Okay, that is called a common jmeter circuit. And the reason it's called a common emitted circuit, because here's my input here.

And here's my output. I take my out here, what's common to both my end and my out? My emitter right there, okay? Remember E, B, C, all right? The emitter is common to both the input and the output. That's why it's called a common emitter circuit.

All right, let me stop clear the slide off. Okay. So, what we have here what what what is this? This is a curve for this transistor. Okay, so let's let's let's look at that. This curve and skinned see what we've got.

Okay. Now here's my base current right here see where it says IB alright. And here's my collected to emit a voltage and we go from zero to in this case 18 volts okay. So when I set my base current, depending upon my collector voltage which is called Vcc that's the supply voltage that will supply my transistor with with voltage your power I and the amount of current That I select and we're going to go through this. Okay, we'll give me this load line, this red line that goes from 14 amps. All right.

All right, this is my current flow through my collector. And this is my voltage from collected to a meter right here on the bottom. Yeah, let me stop and clear the slide. So you can see that collected to a metal voltage VCE. All right, that's the voltage right there. And then we have collected current IC and milliamps is right here on this axis here.

And that's this current flow there. Okay. All right. So, what I want to do because this is a linear amplifier, I want to set my VC voltage between collector and emitter right there. Because if I have my if I have VCE in this particular circuit, what's common to both my input my output, it's, it's my emitter. Alright, let's say I put a ground there.

I see now I'm inputting something between my base circuit and ground and I'm taking an output between my collector and ground. Okay. But before I go through that, I've got a set what they say, set up my DC parameters, which is my steady state parameter on my transistor. And this could be an amplifier. Okay, that way when I input a signal, I'm, I'm gonna get what I want. Now think about it.

If I got a note some Note Let's 1000 hertz note. All right, now I want to amplify that note, I want to hear 1000 hertz note. And so I want I want to get, I want my input. Let me I'd love to take that back. I want my output to reflect my input, but maybe with some amplification, because that's what an amplifier does. Okay.

So what we're going to do here is we're going to set this red, which is called the load line. All right, and we're going to find these two corners or these two endpoints. All right. So what do I need to do first? Well, I want to limit my collected current to 14 milliamp hours. All right, if you remember from Holmes law, how do I do That well I selected my Vcc to be what 18 volts.

So, the most current I want to flow have my the most current I want to flow at my DC steady state. point on this circuit would be my power supply voltage divided by my maximum current. All right. That will give me the value of RL because when all the voltage is across RL and I shot my collected to emitter, all right, all my voltage is going to be across RL. That's where I'm going to get my maximum current flow. So, here I have my 18 volts.

I'm looking for 14 milliamp hours right here. And when I do my math, my IRL, which stands for resistor load is going to be 1285 ohms which I placed there, okay. Okay, now I'm gonna, I'm going to set I'm going to go into the next step but I'm going to erase the slide. Alright, so now I found 14 milliamp hours and that's when I got a shot between my transistor what happens now that I found this in what how do I find this in? Well, I find that in if I assume that that transistor is either open or out of the circuit and I measure here, so if that transistor is out of the circuit Where's all my supply voltage going to be? It's going to be across here to ground because that's going to be an open circuit.

So now I put the other end of my load line on my voltage. And then what I do is I draw a nice smooth line, which is I show you what read from my current to my maximum voltage. Alright, so now I've gotten two points of my load line. And do you remember what I said originally, okay, I'm going to amplify so I want my, my output to look like my input only bigger. Alright, what's the sense if I put a sine wave in on my input and I get a triangular wave on my output it that's probably not going to work, especially if, if it's music or or someone's voice is something or I'm I'm trying to amplify a signal. For a specific reason, and I want the same type of characteristic in out as I get in, right, so that that's not gonna work.

So, I want to find now we're working in if I haven't stated it, we are working in the linear the linear region. Okay. All right. So what I want to do is I want to find a nice point in the middle of my load line. All right. And if I measure this way, there it is.

If I look at that point, what do I have? I have approximately seven milliamp hours flowing through my collector. And I have approximately I'm gonna say about approximately nine volts between my collector emitter and Where's the rest gonna be? It's gonna be across my resistor guy. Alright, it's gonna be across my resistor. Okay, so how do I do that?

Well, if you look here, these curves are my current flow through my base. So if this is my center point, I've got 60. This curve represents 60 micro rappers. And this curve here represents 40 miles. grandpa's I mean, I want to go in the middle. So I'm gonna pick 50 micron hampers.

So I'm going to set ib 250 micro hampers and again, I just use ohms law, I've got a supply voltage at 2.5 volts. As you know my voltage drop between my bass in a metre is, is a point six volts. So now I just go through my nap, I do my math. And guess what? This guy here is gonna be 38 kale. Right 38 kale.

All right. So let's, let's, let's look at this. Okay, so what does this mean right now? Well, I've got 18 volts, that means I've got nine over here and nine over there. Because nine nine equals what? Nine plus nine equals 18.

That's my supply voltage. So, when I go up, this is gonna move down this way isn't it this curve or that line is going to go this way when I increase my current flow through my base All right. So what we can say here is the voltage from here to here is V RL which is the voltage across my load resistor. And the voltage from here to here is V C, E. Alright, because look at it if I go down this way, I might decrease my base count What's gonna happen this point is going to swing this way. And we're going to get more voltage across my voltage from my collected to emitter, and then less voltage across my resistor because we have less current flow through it. Okay, some people will use this terminology.

As I increase my bass current through my transistor, I turn it on more. And as I decrease the current through the base junction on my transistor, I turn it off more so you can use what you want. But the point I'm trying to make here is from here to here is the voltage across the transistor VCE. And the voltage from here to here is the voltage across the resistor vrl. And as this shifts and you'll see in a minute, is this shifts, we get a swing across my collector. All right, that should represent the same type of swing, I get in my base.

And we'll see this down here. All right down here in these two steps, which I'm going to do. All right. I'm going to clear the slide off, and we're going to talk about those two right now. Okay, so now we've set our DC point or our steady state point, and I'm just going to emphasize that right there. All right.

And now let's say I put a two volt peak to peak signal in there. All right. All right. Now, what does that mean? Well, when I go positive, with my two volt peak to peak signal, remember, I've got two volts peak peak, this is one volt peak and one volt peak. This is my zero.

All right, my zero line. So now my supply voltage is 2.5 volts. So what am I going to get? I'm going to get to point five volts plus one minus point six is going to be 2.9 volts. We found RB which is that resistor there. It's called resistor, the base.

And again, I just use ohms law. So when I'm at my point here, I get 76 micro ampere. So let's, let's find 76 micro amperes on here, and I would say that is right there, maybe. All right, so I'm gonna draw a straight line as best as I can. Look as best as I can. And we'll say that that's 5.5 volts, right?

All right. So now let's find the other point. So now, again, we've got a two volt peak to peak signal. So now I'm looking at this and it would be 2.5 minus one, minus point six, I do my math, and I get 0.9 volts. All right, I solve for my current. My current is 23.

Micro hampers. Alright, so I'm gonna find 23. It's right about there. So now look at what we got. All right, we've got 5.5 What would you say that is? I'm gonna say like, 13.5 maybe So if I take the difference between 5.5 and 13.5, what do I get?

Okay, if I do my math, and I say 13.5 minus 5.5, I do my math and I've got eight volts peak to peak. What did I put in? I put in two volts peak to peak, I got out eight volts peak to peak, I have a gain of four. All right, gain of four and amplification, which is not very good. But we're just I just was trying to make a point. So we got an amplification factor of four.

Okay. Um, the only other thing that I want to mention is it's 180 degrees out of phase meaning that when I my input goes Hi my Put goes low. That's what I mean. So, but the character Other than that, the characteristics of my input waveform should be very, very slim color to my output, except for that phase reversal. And that's a common characteristic with common, a common emitter circuit. Okay, we have amplification, but the input is out of phase with the output by 180 degrees, meaning it's opposite.

So when one goes high, the other goes low. When one goes low, the other goes high. All right, okay, that's a good intro. We're going to go on to the next one, which again, we're going to do and we're going to go into this in a little bit more detail when we do when we do transistor amplifier amplifiers. Alright, so let's stop here and go on.