Peter Wheeler: Okay, so let's get back to this.
Before I leave it, one last thing I wanted to do then is that this is a multistage and
we only talked about two separate stages.
This one with load.
This one seeing that Q2 and then Q2 with nothing loaded on it.
In that case, the truth is, what is the overall gain or let's call it overall?
What is the overall gain and it is equal to, in this case here Av1 + Av2 as long as these
are both in db.
If they were just the number that we had, the hundreds where we had the individual gains
without taking the log of it, then you would multiply the two together, but because logs
are exponents, as you know, when you multiply with common bases, you just add exponents
together.
The total gain then, in this case here would be 45.8 + 34.4 to carry 1, 35, 45, 70, 80.
80.2. 81.
That type of number, you just add the two together to get the overall gain.
What'd we say?
We said .2.
Is that right?
Carry one.
80db.
All right?
Okay.
All right.
Now let's just see of we can change this circuit a little bit, so what I want to do is I'm
going to capasively couple now.
I'm going to come down here with a so called stiff voltage divider and what do I want to
give it?
Let's say, how about 33k?
I'm going to make this 8.2k, so in this case here, what I'm going to point out is, and
we can go through the DC set up again, but I'm not going to do that.
I'm going to leave that for you to do.
We have basically .2k over 33k plus 8.2k times 12 volts to come up with a V at this point,
we drop .7, and we get V at the emitter.
This voltage over this resistor gets us a current.
This current through this resistor is a voltage drop.
Subtract that voltage drop from 12 volts.
You got the voltage here, which you know you voltage out.
Take this current, the emitter current that you've calculated.
Then divide it into 25 millivolts that come up with your RE2 prime.
It's going to be a new one.
I'm going to call it second second.
It's going to give you another new number.
Now, the point I'm trying to make here is that this total gain of 80.2 was direct coupled.
What this circuit is now seeing is this resistor, RC1, in parallel with R3, in parallel with
R4, in parallel with beta Re sub 2 sub 2.
All right?
All of that divided by Re prime 1, so let's look at that.
I said Rc here 1 in parallel with R3, in parallel with R4, in parallel with beta RE prime.
All of that over Re 1 prime and that will now be the gain.
You take this for 1 with load.
All right?
Here and that will change.
That number will change.
If you actually do the math and I would like you to do the math, you will see now that
as a result of being capacitably coupled, this gain now is equal to 33.9 dB.
So the extra load has dropped it almost .5 dB.
Not a lot, but it's there.
What you need to do is you going to go through the DC setup to come up with the R sub E prime
set up and then use that and plug it in to a formula like this would probably be given
to you on an exam simply because you can get lost in, "Oh I'm shorting this to that.
This gets shorted to ground.
That's the ground.
This goes to ground.
Okay.
Oh, they're all in parallel," so we tend to simplify that with the formulas that we can
give to you.
So I want to leave that with you.
I don't think there's a lot to say other than if we actually did this calculation again,
just so that when you're doing it yourself, you'll find the number is 148.1 ohms with
the new one.
You'll find that Ie2 as a result if you do the calculation will be 168 micro amps and
the voltage here, well, it's basically going to be this minus .7 is going to be 1.69 volts.
The base voltage, of course, this was Vb2 is 2.39 volts, so go through that.
That's your homework is to go and make sure that your calculations come up the same as
mine.
All right?
You got your DC parts.
I'm pretty comfortable with that by that this time in the video series, you're familiar
with the DC parts.
The next thing is to try and keep track of the AC parts so that you got the proper loads
in place so that you can complete the gain equations appropriately.
Okay, I'm going to move on to a different type of device now.
These are bipolar junction transistors.
So these are multistage BJT transistors.
This happened to be common ... Common what?
Common collector?
Common emitter circuits, okay?
The emitter is common to the input and the output.
The signal's taken off the collective for a common emitter circuit.
Be careful when you're looking at your formulas that you're not looking at, say for example,
if I took the output off the emitter resistor, this would be a common collector.
Make sure that you're looking at gain formula as appropriately to the circuit configuration.
But this is a good overview as to how this amplifiers work.
If you need to know what is the current gain because you want to know the power gain, well
basically, there's your current gain right there.
BJTs are current amplifiers.
Base in multiplied by beta gives you collected current out.
If you multiply the current gain times the voltage
gain, you'll get the power gain.
Okay?
If you do that at logs, you will add them together.
Okay, I'm going to leave that alone.
I think that's a very good comprehensive understanding of bipolar junction transistor multistage
amplifiers.
Let's move on to the final topic of the course, which is the enhancement MetalOxide-Semiconductor
Field-Effect Transistors.