Peter Wheeler: Hello again. Today I'd like to talk about bipolar junction transistors
used as switches and amplifiers. Let's start with switches.
If we take a bipolar junction transistor, let's put it looking like an NPN here, grounded
emitter. What we're going to do is we are going to take our signal out from the collector
and we're going to put our signal in on the base.
This can be turned on fully or turned off completely in order to act as a switch, in
order to give us a high or a low signal; a 1 or a 0; a true, have you, or false. Also
used in digital circuits. Let's see what happens when we try to use it as this type of a switch.
Let's put an ammeter in our switch circuit here, and let's measure all of this with a
volt meter. Now, in this case here, if I have the input signal at zero, then that's the
same because I don't have, as you recall this 0.7 volt drop, because I'm not turning this
transistor on, that's the same as the switch being open.
If this switch is open, no current can flow. However, the pressure I feel at this point
here across the collector emitter will be equal to the voltage I have up there.
When I turn this transistor on by putting a signal in there, by putting a voltage greater
than 0.7 in this case here, then my switch configuration looks like this.
What I see with the switch closed, that is to say with the transistor turned on, is a
short circuit, so zero, no pressure here, and this will have some value of current in
the collector. So this has collected current through here as we've shown before. You'll
see the current flowing. You'd get a reading.
Open: No current flow, voltage pressure's felt across the collector emitter.
Closed: Because you've turned the transistor on, current flows, but there is no voltage
pressure across here because effectively, this point is grounded.
Let's take it to the next step. We talked about the digital use of the bipolar junction
transistors; the switch, high, low, true, false, on off, conducting, not conducting.
Great for digital circuits, what about analogue circuits where you have a small circuit coming
in and you have to amplify it either from a microphone, cell phone, GPS signals, sensors,
pH sensors, thermocouples, low-leveled voltages that need to be amplified to be used in a
productive manner.
What we can do is, let's have a quick look at our DC bias circuit again. Now we set up,
in the previous videos, our bias point which is turning this on half way so that we have
room to either turn it more on or less on. Towards the off or towards the fully on. We
set up our bias resistors. This is the emitter resister, this is the collector resistor.
We have a gain function of some number. Let's just leave it like that.
The relationship in this bias was that when we have a current going into the base, we
had a multiplied current coming through the transistor from VCC source down through, and
we had a formula. We know that we use this relationship.
We are taking some liberty here because in truth it's slightly different. It's actually
Ic=BetaI sub B, but I sub C, in our case, is approximately equal to I sub E. So we'll
use the function as Beta times the base current equals the emitter current. Notice that I
am using capital letters as sub-scripts to my current, for example. That is to reflect
the fact that we are talking about a direct current. Direct current flows in one direction.
If that is the case, let's look at what we've set up here because at this point here in
the base - this is voltage - we have some value of voltage coming in setting up here.
That value, as you know, is {R2/(R1+R2)} *VCC. That is a situation right there.
Over here, we have a similar situation if we take our signal out this way and look at
this point here. Again, we will have a voltage and just ignore the fact that it's inverted,
because when this is on, this is ... pulled down to ground, and when this is off, this
is up. Regardless, we have a voltage up here much higher. Much higher. This will be V ... at
the collector and this is V at the base.
The collected voltage will equal ... so if you take the voltage at the supply up here,
of the collector, and we have this current flowing down, again a direct current, and
we have this resistor, Ohm's Law, that will equal the voltage at this collector point
right here.
Now that's fine, that's set up. That's DC, but the signal that we need to bring in and
therefore amplify is going to be an alternating current signal. Alternating current signals
are denoted by small letters. So we could have "A, small base current," for example.
We will have a small lettered collector current denoting the alternating current. Again, the
same thing for the emitter current, AC. So let's look at that.
We are going to bring in a signal that looks something like this from the AC side. It's
going to be, let's just do this one for now. It's going to positive some and it's going
to go negative. Now, this is an AC signal. This is an alternating current signal, which
means that we're pushing you a charge and then we are pulling it back from you. It's
as if I had a battery connected this way and then I switch the battery around and you see
the signal that way, all right?
So push, pull, push, pull, and as you notice, it is swinging around a reference point. In
this case, zero. So looking at this circuit, what we are trying to do is bring this alternating
current signal in to this particular transistor and then amplify it out.
Connecting this signal to this device, which already has a bias set up here, without affecting
this bias point, I need to couple it. I am going to couple it in with what is called
a capacitor. Now this is the symbol for capacitor.
A capacitor is a device like a wall. Nobody is coming through that wall, but you know
people are on the other side. The more noise they make, the more you will be able to hear
it through the wall. DC is steady. DC has what is called a frequency that is equal to
zero. AC is a changing signal. AC has some frequency greater than zero.
The push-back, the impedance that a capacitor gives us, has a formula. It's like this. This
is the push-back from a capacitor: Xc=1/(2pi*f*C). So for a frequency of zero, zero into this
number is a very very large number. However, with a much higher number here, we can bring
the impedance of that capacitor down. With a suitable frequency and a suitable sized
capacitor, we can make this almost a short circuit for our AC signal. Again, recalling
that across our switch when it is shorted like that, you have no voltage.
So we can short out a signal, specifically an AC signal because capacitors need a changing
signal in order to be effective; you need a frequency, we can short out an AC signal
with a capacitor and block, and block, and only pass the AC and block the DC signal.
What we're ending up with coming in here is an AC signal that rides on top of a DC signal.
The expectation now is with amplification is to get this result. When I turn this transistor
on, I am pulling this point down to ground. When I turn this transistor off, I am raising
this point up to the rails.
So as you can see, turning the transistor off will bring the signal down, turning the
... sorry. Turning the transistor on brings a signal down, turning the transistor off
allows us a point to rise up. This is the result we are trying to get.
The aim of the amplifier is to increase the voltage level from the input signal to the
output signal. What we noted in our DC analysis of our transistor is we used the model that
looked like, in this case here, we have an NPN material transistor, we used a model that
looked like two diodes back to back.
That being the case, we understood that at this point here, we had a 0.7 volt drop in
analyzing, what we did was of course as you recall, we went around a closed loop analyzing
our base voltages, our drop, and our emitter voltages in order to complete the formula.
For an AC model, we don't use this. We don't need this. We don't need this particular diode
set up because the 0.7 has been established already. We are riding on top of it; we don't
care about a 0.7 volt drop.
That being the case, what we do is we use a slightly different model for our AC side.
We use something that looks like this. Again, we have a multiplier to the base current that
comes into this transistor, and we have another very small resistance. We label this small
resistance R e (sub) prime.
If you recall from your days maybe when you experimented with a diode, when we had a diode,
we had a curve that looked something like this. This is your 0.7 point here, this is
a voltage, this is current, and there is this delta in voltage over current equal to the
resistance of the diode.
Bulk semi-conductive material has resistance. Now, we have resistance in the collector,
the base, the emitter, we have what were called our parameters, but the one that really is
of any note is this emitter resistance. This is the entire transistor. This is outside,
this is outside. This is where we're going to bring our base.
That being the case, when we are looking at this particular device, what we want to analyze
is the gain, for example, what are we putting in and what are we getting out? In a transistor
circuit, just like putting money in the bank, you put something in, you expect to get something
out that is more than what you put in, so let's just put this one together here real
quick.
What's going to happen is we are going to put in a small base current, small b. And
we expect to get this collected current, small c, times this collector resistor, giving us
a larger voltage. The voltage that's put in is actually the voltage that's developed at
the base here and when we look at that, because this is in the emitter circuit, the voltage
that will be developed will be I, small e, times re'. That would be the voltage developed
across these two points. which would be the voltage that we'll see at the base.
So let's look at this. You're going put a voltage in, and you're going to get a voltage
out. That will become our gain. The voltage that we put in is, in this case here, I e
re', and the voltage out is I small c R C.
The question is, "What is this re' Wheeler's talking about?" Well, really, it's a generation
of electrons thermally inside the device. For our purposes, we are going to use a formula.
It's 25 Millivolts, very small, divided by the emitter current that was set up from your
DC analysis. That current there. That is a value you typically see if you measure it
or even do the calculations somewhere around 10-20 Ohms.
That fine, that's where the voltage is going to be built up on the base, but let's see
what happens when we combine these two.
If we have gain, which is used as an A symbol, particularly let's say the voltage gain of
this device, if we look at the voltage gain, what we've got is the output, so that's IcRc,
right? Over the input voltage, Iere prime. Again, the emitter current that is formed
that by the amplifications of beta through this particular circuit, if we look at that
and we say for example that our assumption is our collector current is approximately
equal to our emitter current, again this current created by our alternating current signal,
then what we've got is this will cancel out and we will have Rc over Re prime.
So for our first configuration, this configuration of transistor, which is also called a "Common
Emitter Transistor Configuration," because if we look at this, the emitter is common
to the input signal and the output signal. For a common emitter configuration, the voltage
gain is equal to the AC collector resistance over r e prime in this configuration.