Peter Wheeler: Welcome back.
Now we're looking at a different amplifier set because previously I showed you where
you could calculate the gate to source voltage, because ground was right here.
Now we've got a wrinkle in this.
Used for feedback purposes, you may find a resistor in the source, so a source resistor
in this case here.
Which means that you're going to, with the current that flows, you're going to create
a voltage here.
The voltage at this point here, VG-1, the gate, has to be, and the voltage that is created
across the source resistor, that has to be greater than the threshold voltage, in this
case to make sure that the transistor turns on.
The math for doing this is, we're going to go back and we're going to look at our friend,
Kirchhoff and see what he says.
He says, "Lets just take the voltages around the loop and they'll add up to zero."
For example, if we take the voltage that we're going to determine in a second, V gate 1,
and if we subtract the voltage for the gate source for the first transistor, and then
if we subtract the current that we're going to determine through this 500 ohm resistor,
the voltage that's going to be created by the current through that 500 ohm resistor,
that will equal zero.
We are then going to have an expression, so we can calculate this, and we're going to
do that in a second.
This one we don't know.
This one is to be determined as well, but we'll have an expression where through substitution
we can solve for our drain current.
Once we solve for our drain current then we can go back and solve for our day current.
Unfortunately, again, as I mentioned before, the transfer characteristics of this device
are not linear.
As a result we'll be using this.
I hope that's a familiar formula to you, how you saw the quadratic equation, something
with a square in it.
We're going to have an expression with the current squared, times current, plus current,
plus a third component, and then we're going to solve it by just substituting those coefficients
into this equation.
We will come up with a plus and a minus expression.
With that plus and minus expression we have to determine which one is real and which on
is not, and we'll see how that goes.
Let's do the simple stuff first.
The first thing is to figure out what is the voltage at this particular point.
We know that we have again a voltage divider.
We have 2.2 meg here, 4.7 meg here, 18 volts there, so it's 2.2 divided by the total of
this, 6.9 times 18.
If you do that math, and I'd like you to try that, 5.74 volts.
We know what the voltage at gate 1 is, 5.74.
We got VGS-1, which we're going to try and calculate in a second.
We have to try and calculate, so this is what we don't know, we've got to try and calculate
ID because we've calculated K from the data sheet 75 milliamps divided by, so V gate source
from data sheet was 4.5, V threshold is 3.
That's basically 1.5 squared, and that comes out to 0.0333.
That's the math for that.
What we're going to do is let's use that formula for ID.
We're going to have 0.0333 times V gate source, which we don't know.
We do know the threshold voltage, which is 3 for this device.
It was 1 for the previous device, sorry.
Minus 3, and that is all squared.
I just jumped ahead just to write the numbers out because it's a little messy.
We plugged in our values from our Kirchhoff's voltage laws.
We had a voltage here.
We have a voltage here and we have a voltage generated by the current resistor there.
We plugged all those into come up with an expression for VGS based on ID, RD.
Now we are trying to use the initial saturation formula to calculate what is that drain current
after that.
It is a mess, as you can see, and that's why I needed to break.
It comes down to what is called a A is the coefficient here.
B is the coefficient here, and C is the coefficient here.
We're going to substitute this coefficient in order to solve this particular equation
into this formula here, which is a solution for a quadratic equation.
The thing is you're going to get two numbers.
When you do it one way, and you can do the math, you're going to get something like 6.249
milliamps, and then you do it the other way.
One is plus and when you do it the minus, so let's put this one as the plus, and when
you do the minus one you're going to have a different number.
You're going to have on the order of 4.67 milliamps.
You're going to have two numbers.
Which one is the correct ID?
What you'll have to do is you take this current and multiply by that resistor and look at
the voltage that generated.
When you do it with the 6.249 milliamps you get a voltage of 2.575 volts, which is less
than the threshold voltage.
You need to have at least three volts to turn this device on.
When you do it with the 4.6 milliamp volts you'll get a different expression when you
plus it back into the Kirchhoff's.
By the way, yes, you're plugging this back into Kirchhoff's voltage loop equation.
You'll get another number which is going to be 3.365 volts, so this is volts, which is
greater than the V threshold.
That being the case, this is the correct current that we're going to use in our Kirchhoff's
loop equation.
That being the case, we've now got a VGS of 3.365.
We know what VG was.
We know what the voltage across here was, so that's like this minus 3.365 gets you the
voltage here.
The voltage here gets you the current through there, or actually we had the current already
so that loop is complete.
With this current now that we have known as to be correct, 4.67 milliamps, we can also
then get the voltage dropped at this point, so you can determine what VD is.
Basically going to be ID times R drain number one.
That's fine.
That then will be the voltage on this gate.
Now we go through the same process of a Kirchhoff's voltage laws.
You've got a voltage here.
You've got a voltage drop across here and then you've got a current through a resistor
here in order to figure out what is that particular gate source voltage.
When you've got that particular loop solved then you have a current that will drop through
this 560 ohms which gets you a voltage out here.
That is straight math, and I'm going to leave this part of it to you.
The trick again was understanding there's a Kirchhoff's voltage law, which is VG1 minus
VGS1 minus IDRD equals zero.
That then becomes the one that you substitute into your ID equals K, VGS minus V threshold
squared equation in order to solve it.
Again, because you have a square you will end up using this particular quadratic equation
to complete it.
Now, this is great.
This essential DC stuff.
Let's try and figure the gain after I clean up this screen a little bit more.
We're going to go back to this particular equation trying to determine what RD is for
both these loads.
Thanks.