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Current time:0:00Total duration:8:07

- [Voiceover] Now I wanna
use a diode in a circuit, and we'll see how we solve
circuits that include these nonlinear diodes in them. So I have a circuit here
with a battery and a resistor and a diode here, and it's
gonna be a special kind, it's gonna be an LED diode, so it's gonna give off light. This is a kind of diode that's
manufactured to generate photons of light when it has
a current flowing through it in the forward direction. They're pretty cool and
you see them all the time in electronic components. So we're gonna figure
out how to use an LED in a real simple little circuit. And in our circuit here,
we're gonna have a resistor of 330 ohms, and we'll make this battery three volts, so it's like two double-A cells. And what we wanna find out is how much current's gonna
flow around this circuit. And then we label this here, let me label the voltages
across our components. We'll have VD, which is this voltage here, and we'll have VR, which
is the resistor voltage, and that's that voltage right there. So now I'm gonna show you
a plot of the iv curve of our diode. You can see right here, around 0.7 volts, the current rises rapidly
when there's 0.7 volts across the diode. And let's start out by analyzing this, let's start with the same
tools that we always have, which is, let's try to
write some current laws for these two things. So for the diode, we write a current law that looks like this. The current is equal to IS times e to the qv on kT minus one. So that's the iv
characteristic for the diode, where this is V diode right there, and the corresponding
equation for the resistor is i equals V resistor over 330 ohms. That's just Ohm's Law for the resistor, and i in both cases is this i right here. Now if I wanted to, I could
set these two expressions equal to each other and
solve somehow for VD and VR, but what we're gonna do instead
is we're gonna solve this by graphing, by a graphical method. Here's a graph for the diode, and this is the VD scale, this is V diode, and this is i up here. And so what I wanna do is I
wanna plot the resistor curve around here as well. I wanna plot the resistor iv curve on this same plot. Now, in this expression,
I have VR instead of VD, so let me see if I can work on VR here. Let me try to figure
out VR in terms of VD. So I can derive VR as three volts minus VD. And let's put this into
this i expression here. The Ohm's Law expression now becomes i equals, VR is replaced with three minus VD, all over 330 ohms. And let's work on this a little more. i equals three over 330 minus VD over 330, and this is starting to look
like the equation of a line. Let me write this to recognize
it as a equation of a line. Minus VD over 330 plus three over 330 is nine milliamps. So this is the equation
of a line, and the slope is right here, is minus one over 330, and the i-intercept is nine milliamps. So let's see if we can plot this line. This is actually called a Load Line. That's just a nickname for
this kind of expression that you get when you
have a resistor connected to a fixed power supply
above and the resistor is hanging down from it. You get this characteristic
equation of a line that has a negative slope,
which is really distinctive. Let's see if we can plot this line. Now, it's a line, so all we
have to do is find two points that solve the line and then
we'll be able to draw the line. So if I set VD equal to zero, then i equals nine milliamps. So here's VD equal to
zero, and it'll go through nine milliamps. So that's one point on the line. But what else can I set to zero? I can set i equal to zero, which means I'm on the voltage axis. And what I'm actually gonna do is I'm just gonna look at my
circuit and figure this out in my head. So I'm setting i equal to zero, and i equal zero means there's
no current in this resistor, which means there's no voltage
drop across that resistor. And that means that this voltage here is the same as this voltage here. And I know the voltage here, the voltage here is three volts, so that means the voltage
here is three volts, and that's because I
know the current is zero. So let's go over and put
that point on a line. So when i is zero, v is three volts. So there's another point on the line. And now we have two points
and we can draw a line between 'em, like that. And what we've drawn is the Load Line for this 330-ohm resistor. You remember, back over
here we said we could solve these two equations by setting the two is equal to each other, and that's basically where
do these two lines intersect? And they intersect right here. That's the solution to our problem. So this intersection
point is the solution. It's where the resistor
current and the diode current are the same, and that's that point there. Now I can just read off my answer. Right there is about 0.7 volts, and the current over here,
if I read off the current, just straight across there, it's about 6.8 milliamps of current. So now we actually just solved our circuit using a graphical technique. And what that says is, let me erase this a bit to clean it up. Let me take out these two things here. That was just the resistor Load Line that we were talking about. And now for our solution, we have i equals 6.8 milliamps, and V, V diode equals about 0.7 volts. So that's how you do a
graphical solution with a diode.