Peter Wheeler: All right, let's look at the AC part.
Now, we're going to look at, again, the only thing I'm really interested in is the gain
of the different stages.
What I'm going to do, just for now, is I'm going to break this circuit here, and make
them two independent circuits and we'll tie them together in a second.
This circuit as it is, has nothing hanging off it, which means it has no load.
We're going to calculate the gain for this amplifier without a load.
The gain for stage one is going to be the voltage that's created across here by the
signal that comes in versus the voltage that comes in, and the voltage that comes in is
the voltage that's developed across RE 1 prime.
But because we have a capacitor here, this resistor is no longer in circuit to the AC
signal.
The AC signal will come in, and follow this path, all right?
It's essentially riding on top of the DC signal that is there.
The 0.7 doesn't even come into play.
It's a DC concept, that 0.7 volt drop.
The AC is going to ride on top of that.
The voltage created here will create a voltage out here.
If we look at what's happening here, is we have IE, so-called I small-E, all small C
in this case here, current that is generated here.
And that current is the same current that's going to run across the RE 1 prime.
If I look at IE RE, which is the output, over IC RE prime, then these two currents will
cancel.
And so the gain of this amplifier is going to be equal to simply, no load, is simply
the value of the RC circuit 22k ... We have 22k ohms divided by ... And RE 1 prime was
80.5 ohms.
And so if you do that particular math, you come out with a number of 273.
Okay.
A fairly large gain.
If you do 20 log to the base 10 of 273.3, you'll get a number more like 40 ... Off the
top of my head, 48.7 decibels.
Okay, so we actually have our first gain stage without load, no load, at 48.7 decibels.
We can do the same thing for AV 2.
We would have basically a signal in and a signal out.
The current that we had listed here, we had an IC current listed there, and we would divide
that ... Sorry, we would take ... Well, we'd use the same formula.
We'd use RC over RE prime 2 to come up, because this current will cancel that particular collected
current.
Again, for AV 2, same kind of math, no load… should put here NL for no load, okay.
No load is, same idea.
We've got 10k ohms, 10 times 10 to the 3 ohms, and we're dividing that by 55.9 ohms.
Ohms will cancel, dimensionless, and you get a number like 178 point ... We could round
it up, but let's say 89 anyway, and that as far as we'd like to see it in dB, is 20 log
to base 10 of 178.89, is going to be equal to 45.8 decibels.
One-tenth of a bel.
That's fine, but the truth is we need the signal to go from here to here.
Really, what's going to happen is this value of RC is going to see a load over here, and
what it see is, it does not see this particular RC 2, RE 2 resistor.
It doesn't see that because we have a capacitor here shorting out the AC signal.
But what it does see is the RE 2 prime.
Now if you look at this circuit, you've got a battery up here that goes to ground, and
an AC signal, when it sees a battery, can go into the battery and also current can come
out of the battery.
But for all intents and purposes, the battery stays at 12 volts.
It's a storage facility.
From an AC signal perspective, a battery is also like a big capacitor, which is like a
short circuit.
That means that the RE that we have here which goes to ground, and the RC that we have here,
which hits this, which is also grounded, are in parallel.
We're going to be looking at the load that stage one will be seeing is R small-C, which
is the AC determination of the R big-C in parallel with the RE 2 prime.
Okay.
I hope that works out for you.
And so that being the case, now there's another small problem here.
The actual resistance in here is not just RE prime.
It's multiplied by the beta of the transistor, which happens to be 125 in this case here,
and why is that?
If you look at the input resistance ... input is V in over I in.
And if V in is basically, like we showed there in that circuit ...
And this is a current that's IV times beta ... If the voltage in is essentially I, in
this case here, let's say IE times RE prime, and the current in is I small-B, okay.
The current in is small IB, what we've got is basically beta I, small B, RE prime.
Okay, sorry, let me clean that up a bit.
Instead of IE here, we've got beta IB.
Remember, beta IB is equal to IE is equal to IC.
So here we have beta IB, and these two IBs will cancel out.
The resistance looking in will be, in this case here, beta times RE prime.
We are actually in parallel, 22k is in parallel with beta, 125 times the value of RE 2.
If we look at that, then ... Oops, sorry, this is for this part here.
Let me just take this and lower it down a bit.
One more time: with the load, with load, the gain for AV 1 is going to equal R small-C
which is RC 1 here, in parallel with beta, RE 2 prime, RE 2 prime.
That's going to equal ... 22k in parallel with 125 times 55.9 is, 22k in parallel with
7k is, let's see, how does that go, over 80.5 ... It's going to come out to 3.3k over 80.5,
and that, when you do the math ... these are ohms, they're going to cancel, is going to
be 66.
And that, if you do 20 log base 10 of 66, it's going to be 34 point, what do you think,
4?
Yeah, 4 makes sense, dB.
With load, we manage to change the gain from this number, 48.7, down to this, because we
are now connected and current is going through here but it's also taking an alternate path
through that part over there.
Pretty good story, and the story to understand, though, is that it's not just that RC.
When you look at the formula, when you see the formula for gain and you see RC over RE
prime, this is a small C, which means it's an AC C, which means it's this in parallel
with other loads that the AC signals sees.
Okay, so that's a good start.
Now I'm going to change the circuit slightly again, and I want to do one that is capacitively
coupled with a stiff divider, and we'll see how the numbers change with that particular
biasing.