Peter Wheeler: Let's just repeat what we did really quick. We have a source voltage, sorry
we actually have a power supply voltage collector. We have a resistor in the collector circuit.
We come down to our bipolar transistor and in this case here let me just take this to
ground and we bring the signal in.
So we said that for a signal coming in, we had the voltage gain here equal to the collector
resistor over the r e prime resistance, which is that resistance in the emitter. Okay, that's
a gain in voltage, how about current?
Actually, we do have current gain. The current gain is equal to your beta. It's the multiplier
of basically the current you get through the collector over the current that you get through
the base. That is your beta. So we have current gain, we have voltage gain, this is a common
emitter transistor configuration, and therefore, because voltage times current is equal to
power, our power gain is equal to our voltage gain times our current gain (Ap=Av*Ai).
Substituting in, we can do that quickly, we would have beta r c over r e prime for our
power gain.
That is fine, but that's not the only configuration we have. We have other configurations. You
can bring the signal in on the base, of course, and take it off the emitter. You can bring
the signal in on the emitter and take it off the collector. Let's look at those two other
configurations.
The one we'll look at right now is a common collector. Again, similar to the common emitter,
but in this case here, we will not have a signal inversion because we are going to take
the signal off the emitter here and we are going to bring the signal in on the base.
Small c and small e, the AC equivalent of the emitter resistor.
Again, if I look at this particular model, this is our current source which is again
I b Beta. This is our emitter resistance, r e prime, this is our base. What I see here
looking in here is I see r e prime plus, this is all the transistor, plus this resistance
here, which is the AC equivalent of the emitter resistance. We will talk about that in a second,
I'll talk to you showing you why you have an AC equivalent versus a DC equivalent on
resistance. It depends on what extra devices you put on in parallel with it, the load devices.
Right now, we don't have a load, but we do have an emitter resistance. When I'm coming
in here at the base, the voltage that I see here at the base is the result of this emitter
current due to AC flowing. So my base voltage is going to be, let's look at this. V-in is
going to be
the current flowing through the sum of those two resistors.
Now the signal that we are going to take here is from this point, so the output voltage
is going to be the emitter current that we have flowing here times the AC equivalent
of emitter resistance.
Given that R e sub prime is kind of small compared to R e, this may be on the order
of 10K, this may be on the order of 10 Ohms, greatly out-ratioed resistance, what we effectively
have is Re/Re, which is 1.
Hmm. How is that? We had a gain of Rc/re prime here, but the gain for the common collector
amplifier is one for voltage. But we still, for what we put in to what we get out, we
still have a current gain, so we still overall have a power gain.
Now, that's all good. Again, just from a simple sketch, we had a transistor that looked like
this and so we realized that with the in versus the out, the emitter is common. Here we've
got a transistor that looks like this, because we've taken the signal off the emitter, okay?
What we've got is a common collector in this case here. This is the input, this is the
output.
Now I can change this up again and I can basically bring a signal in on the emitter, and then
take it off the collector. Let's see if I can just fit that in. Let me just do it here.
I'm going to take a signal off there, and I am going to bring it in here. This is my,
and this is my AC equivalent collector resistance and AC equivalent emitter resistance. Now
the thing is in order for this circuit to work, what I need to do, first of all we bias
this up of course in order to get a bias in there, but what I need to do is put a capacitor
to ground.
What that does is, again remembering that the impedance of a capacitor is equal to 1/2(pi)fc,
that essentially replacing all of the capacitors in an AC analysis with a short circuit is
something you should do. What we end up looking at here is essentially, let me short this
point, you've got an emitter resistor, this one here, in parallel with my emitter sub
r sub e prime resistance.
You know that because this is a short circuit, so from this model here I'm bringing my emitter
resistance r sub e prime to ground and putting it in parallel with Re, looking at the voltage
at this point which is Vb, which is V, as you can see two resistors are in parallel,
the rule as you know is it will be much less than the lowest resistor.
In this case here, a very low resistance in. So a low resistance in, approximately Re Prime,
and if that' is the case, then when we have our AC emitter resistance times Re prime in
order to give me V, and I look at my output from here, which is going to be effectively
IeRc=Vout.
Let me come down here for a common base. For a common base, my gain for voltage is going
to be, if we say Rc or Re is canceling out here, it is going to be Rc/Re prime. So slightly
different than the common emitter, okay? But in this case here, there is no current gained
from the base to the emitter.
It's like in our water configuration when we had a faucet, we were putting our finger
over the bottom of the spout. That's where we're modulating the current. In this case
here for a common base, the current gain is one. You get a voltage gain with a common
base, but no current gain. You get a current gain with a common collector, but no voltage
gain, and you get a voltage gain and a current gain with a common emitter.
Why these different configurations? The reason why is depending on the signal that you're
bringing in, what is its source? If it is a high impedance source, let's say from a
PH sensor, then you don't want to load that signal down. You want to offer a high resistance.
If it is a low impedance load like a speaker, an 8 Ohm speaker, again you want to match
the output from your amplifier to the load to maximize the power transfer.
So different configurations using a model like this and looking into a model like this
is, depending on where you are, if I come at Re, this is the impedance I look in. If
I'm coming on the base, again, these two combined is what I'm coming in. Again, looking at the
output load, seeing what that looks like to my supply, the load I'm supplying, again we
can play with those particular numbers. Thanks.
Let's just summarize what we've picked up today. Of course, just as a recap, you do
know from Ohm's law that voltage equals current times resistance, so that's good. I need you
to try to keep in mind that we are looking at, this is a current source, that we're looking
at this current source that feeds this r e prime which is your transistor now for the
AC model, and that r e prime is equal to 25 millivolts over the DC current that you established
by setting up your bias.
I also would like you to keep in mind that the impedance of a capacitor is 1/2(pi)fc.
Thus, if you have something, for example in the collector like this, and you have a capacively
coupled load resistor like this, what's actually happening is that ... if you short the capacitor
like I've asked you to, then you've got essentially, the voltage at this point, my output, is two
resistors in parallel.
What we say is is that we use a small c to reflect the fact that in an AC analysis, if
you short the capacitors through, you're actually putting resistors in parallel. That resistance,
therefore, due to AC, will be less than the Rc that is seen from a DC analysis.
We also have another situation that you'll come across as well, is we may put the capacitor
here. What happens is for your DC bias setup, you will see RE, and that will establish your
quiescent point, your normal operating currents.
But, from an AC perspective, we end up shorting this out, and so your equivalent circuit from
an AC perspective is this one. Okay? So, short the capacitors, use this particular model
for AC, recall that:
Voltage equals current times resistance from Ohm's Law - V=IR,
Collector current is equal to Beta, I b : (Ic=Bib)
Voltage gain is equal to Vout over Vin : (Av=Vo/Vi),
Current gain, more often than not just the beta : (Ai=B),
Power Gain, : Av*Ai,
and I think you will put a good dent into your AC analysis of transistor circuits.