Okay, let's look at the common collector configuration, also called the emitter follower. So let's let's go on here. Okay, so now the first thing that we want to do is we want to find DC parameters. All right, which is a steady state. And what we do here is we're going to find the voltage at the base, which is right here. And if you look, it's just a voltage divider, okay?

All right. So the voltage at the base is actually the voltage across RB two right there. And I just use this formula, okay, rb two over RB one plus RB two times my supply voltage. And when I go through that Math, I get six volts DC. So my steady state voltage or my DC voltage at the base is six volts DC. All right?

What's my steady state at the emitter? Well, it's my voltage at the base minus my VB drop. All right, voltage base to emitter is VB. So we got six volts DC minus zero dot six volts. And again, when I do my math, it's 5.4 volts. So right here, okay, I'm getting 5.4 volts.

So right here, I get what? I get six volts DC. And right here I get 5.4 volts DC. Now, my input is two volts peak to peak. All right. So now if I'm taking my Voltage off out here at V my steady state is five dot four volts, right?

I go up one volt. If I have in other words if I have a two volt peak to peak signal, it's a sine wave my peak is one volt plus 5.4 which is 6.4 volts and at the other end, I subtract one volt and I get 4.4. So you can see all right, I still have a two volt peak to peak signal, because what 6.4 minus 4.4 volts, okay, is going to be two volts peak to peak. All right, so I still have a two volts peak to peak swing. The only difference is or I shouldn't say difference. The only thing that that point here right here is allowed me to do is since I met the emitter of the transistor, and do you remember I II equals IC plus IB?

Well guess why doesn't IE have a lot more current, or capable of delivering a lot more current than my input IB. So I don't have voltage amplification. But I do have the ability to increase my current and therefore my sine wave or the characteristics of my sine wave. Or let me say it another way, the care the characteristics of my input signal is is exactly like my output. They're both in favor Okay, because it's an emitter follower that Bolton face, I just allowed to have, or be able to deliver more current to my output. And that's it.

All right. Again, it's called the emitter follower. Yeah, common collected configuration. But if you get out there in the field or if you hear some engineers or senior techs or whatever talking, they're going to use the term emitter follower. With that said, we're going to end this slide and we'll see on the next one. All right.

The next configuration we're going to look at is the common base configuration right here. All right, and there it is right here below that, and we take our output at the collector, and we put our input in at the emitter, okay. And one of the things that I want to mention is That we do get voltage gain here. Okay? So we do get voltage gain. But we lose current gain.

All right? Because if you remember, we are putting our input at the emitter and we're taking it off at the collector. And if you remember, what do we have? We have iy equals IC plus IB, right? Well, I got a little bit more current than I see. So very, very, there's a very small difference, but there is no current gain.

That's the point I'm trying to make. We have voltage gain, but no current gain plus the input and output voltage out of phase and I show you that right here. All right. Alright, so meaning when the input goes high, the output goes low. When the input goes low, the output goes high so they're out of phase. But the sick if the circuit is designed correctly, it should be a reasonable duplications duplication of the input signal with the only criteria is that they are out of phase, that's all.

Alright, so let's, the first thing we're going to do is we're going to go through our DC points, and then we're going to put the input waveform in there and show you how we get this 4.8 volts peak to peak out. Alright, so I'm going to stop here. I'm going to clean off the slide and we're going to start on top and we're going to go down. Okay, the first thing we want to do is find a DC equivalent So we can do that by finding IE, as you know, ie is a current flow here. Okay, and how to do that? Well, we've got VB B, which is three volts.

And we have a, a voltage drop across the resists cross the transient resistor across the base emitter junction of the transistor. And we know that it's going to be six tenths of a volt. So we've got 2.4 volts divided by the value of, of the resistor in the emitter, which we call our V which is 1200 ohms. I do my math, and I get $2 million. Okay, so we know that iy is $2 million. Now what I've done and a lot of people will do, because the error is so small, it really doesn't matter.

Is I assume that I IE approximately equals IC. And if you can see it here i say i equals IC plus IB. Therefore I can say I II approximately equals IC because I see at the low end is 98% of IE. Alright, it's between 98 and 99%. So I could have a 1% error. Quite honestly, I've got more of an error in my circuit design if I use run of the mill components, because run of the mill components of plus or minus 5%.

Okay, so I can, I can do this without a problem, I can do this. So right here, I'm going to say that I approximate equals IC. So I'm going to find The voltage across that resistor. So I say I see times RL, well, it's two milliamp hours, times 3000 ohms. And when I do my voltage when I do my math, I get six volts. So now I know that I've got six volts across RL right here, which I call vrl.

Okay. So now, I want VCB. All right, which would be right here. The voltage from collected to base since the basis at ground VCB. So it's Vcc, which is my supply voltage minus v rL, which is the voltage across this resistor that we just calculated in the previous step. And again, I do my math 12 volts DC minus six volts vrl What do we get there?

We got six volts. So what do I know right here right now that I have a two volt peak to peak signal in my American parents to milliamp hours and my steady state voltage VCB is six volts. So let's go on and put a two volt peak to peak signal in here and see what we get. So now we're going to put a two volt peak to peak signal in right here. And then what I'm going to do is I'm going to solve for my emitter current at both peaks, so for instance, three volts which is my steady state My steady state right there plus one volt DC peak all right minus my voltage drop across my base emitter junction divided by my emitter resistance right there. Okay, I go through my math I get 3.4 divided by 1200 ohms I get 2.8 milliamp hours All right.

So when I when my two volt peak to peak signal over here and I go one, one volt peak positive, my emitted current is 2.8 milliamp hours, okay, now, when I go to the other excursion when I go to To my negative peak, okay, I have three volts DC minus one volt, I still subtract the voltage across my base emitter junction which is zero dot six volts DC. I do my math, I get 1.4 volts DC divided by 1200 ohms which is Ari, I do my math and I get 1.2 milliamp hours of current. So my emitter current varies with that two volt peak to peak signal and it varies between 2.8 and two milliamp hours, which is my steady state down to 1.2 milliamp hours. Okay, so now all I need to do is find the voltage or Cross vrl at those points. All right, so now I'm going to do it right here, where I've got 2.8 milliamp hours flowing.

So 2.8 milliamp hours times 3000 ohms is 8.4 volts across vrl. So what's the voltage at my output, which is the C B would be Vcc minus vrl. Okay, I go through my math, and I get 3.6 volts DC. go the other way. All right. vrl one with 1.2 million campers.

When we're at the low side over here. All right, it's 1.2 million campers times 3000 ohms. I get 3.7 volts Same deal. Okay. I come up with 8.4 volts right there all right. So now what do I have my output waveform right is going to be well I should have put it here a point four and 3.6 and if I subtract 8.4 minus 3.6 what do I get?

I should get 4.8 volts peak to peak. So I have a four volt, a 4.8 volt peak to peak wave form. I put into I get 4.8 what do I have? It looks like I got a 2.4 gain. So that that this, this configuration is going to give me a 2.4 gain or amplification factor. And that's what it is.

Alright, that's pretty much it. Rewind this, go through it again. Look at it. And again, it's, it's not hard. If you've not done it before you saying, Oh my god, but go through a three or four times. Basically it's its own slaw, taking into some considerations and some assumptions.

The assumptions are good people. The assumptions are good. All right, it's not that difficult. If you do a couple of them. I'm going to tell you something. It's very easy when you do Two Three, and I'm sure that I'm going to give you some in there.

I just finishing up the video portion on this course, I always give you some problems. So what I'm going to do now is stop, go to the last slide in this course. And I'm sure they'll they'll be some exercises in there for you. So with that said, I'm going to clear the slide we're going to go to the next one. Okay in in the this slide, I've just given you a synopsis of the three amplify pnp transistor amplification configurations, common base, common collector, common emitter, okay, we all have a power power gain. Okay, each each one has a power gain.

And if you remember power equals I times he All right, so we all have a power game. They all well, bail No, not all don't have a voltage gain the common collector which is what what did I call that? I call that the emitter follower right? That's going to have a voltage gain lat little bit less than one. All right, so there's going to be no voltage gain, okay, current game they they all have current gain except for the common base, which we just did and the reason being remember iy equals IC plus IB, okay, the input impedance is glow on the common base, highest on the common collector and medium on the common emitter. What does that mean?

What do we mean by input impedance? Okay? When we go into qualifiers, I will explain that in a little bit more detail. But when I'm trying to put input signal to an amplifier, we'll use that as an example. Okay? The higher the impedance, the less current or less power that preceding stage has to deliver.

All right. And that's one of the reasons why some of you audio files, you hear the term preamp, right? Because what I need to do is I need to beef up that input signal a little bit more to drive the amplifier, because the amplifier probably has a low input impedance. So I need to boost it up a little bit. So they say we'll use a preamp, I preamp is actually an amplifier. Alright, it just gooses up the signal just a little bit so I can deliver it to the final amplifier so I can really get some power to my speakers.

And I'm using that as an example. All right, output impedance is how much current I can drive. So if you look, the lower the output impedance, alright, the more current and power I can deliver night fasion version we've talked about that. The only one that has a fasion version is the common emitter. And here are the implications or applications I should say. common base used mainly as an RF amplifier, common collector use mainly as an isolation amplifier.

And that's what I, I said when I spoke to you earlier I use the term buffer buffer in an isolation amplifier is one of the same isolation amplifier is a little, little bit more of a fancy name. Okay, and that common emitter is, is universally works well you. There's a lot of common emitter transistor amplifier configurations out there. They're very, very common. Alright, so with that said, I really hope you like this introduction of transistors and diodes. And with that said, This ends what I call my core courses.

So if you look on the platform, I've got around seven or eight what I call core causes. Alright, and now what we're going to do is we're going to take these cores, and we're going to build on them. We're going to talk a little bit more about amplifiers. I'm going to do one on power supplies. We're going to put some circuits up there and put all these little electronic components and make things up. I'll have some kits for you.

And we're just going to build upon this foundation. And yes, at some point, I'm going to do one on vacuum tubes. All right, and what I have in the back of my mind, and I'm not exactly sure when we're going to do this, but if you stay with me, we're going to build a vacuum tube amplifier. I'm not sure how many tubes we're going to do, and I'm just kind of looking at it now. But we're going to build a vacuum tube amplifier and I'll show you the theory of operations and so forth. So with that said, I wish each and every one of you will Wonderful, wonderful time and electronics.

I hope you've learned something from me. Electronics is a great vocation. It's done a lot for me. And if I've taught you just just some, just one or two points, let me hear from you because this is really something that I really enjoy doing. And it's been like I said, it's been good to me. So, without any further ado, I wish you all the best.

We'll see you in the advanced courses. Talk to you later. This is Owl and best is over. They're laying down. We'll see you next time around. Take care everyone.