Diode graphical solution

Now I want to use a diode in a circuit and we'll see how we, uh, 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 going to be a special kind; it's going to be an LED diode. So it's going to 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 going to figure out how to use an LED in a real simple little circuit. In our circuit here, we're going to have a resistor of 330 ohms, and we'll make this battery 3 volts, so it's like two AA cells. What we want to find out is how much current is going to flow around this circuit. Let me label this here; let me, let me label the voltages across our components. We'll have (V_D) which is this voltage here, and we'll have (V_R) which is the resistor voltage, and that's that voltage right there.

Now I'm going to show you a plot of the IV curve of our diode. You can see right here, around 7 volts, the current rises rapidly when there's 7 volts across the diode. 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 (I_S \times e^{\frac{qV_D}{kT}} - 1). So that's the IV characteristic for the diode where this is the voltage across the diode right there. The corresponding equation for the resistor is (I = \frac{V_R}{330 , \text{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 somehow solve for (V_D) and (V_R). But what we're going to do instead is we're going to solve this by graphing, by a graphical method. Here's a graph of the diode, and this is the (V_D) scale; this is the voltage across the diode, and this is (I) up here.

So what I want to do is plot the resistor curve on here as well. I want to plot the resistor IV curve on this same plot. Now, in this expression, I have (V_R) instead of (V_D). So let me see if I can work on (V_R) here. Let me try to figure out (V_R) in terms of (V_D).

So I can derive (V_R) as (3 , \text{volts} - V_D), and let's put this into this I expression here. The Ohm's law expression now becomes (I = \frac{3 - V_D}{330 , \text{ohms}}). Let's work on this a little more:

[
I = \frac{3}{330} - \frac{V_D}{330}
]

And this is starting to look like the equation of a line. Let me write this to recognize it as an equation of a line:

[
I = -\frac{V_D}{330} + \frac{3}{330}
]

So ( \frac{3}{330} ) is 9 milliamps. So this is the equation of a line and the slope is right here, which is (-\frac{1}{330}), and the (I) intercept is 9 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 (V_D) equal to 0, then (I = 9 , \text{milliamps}). So here's (V_D) equal to 0 and it'll go through 9 milliamps. So that's one point on the line.

What else can I say? Set (I) to 0. I can set (I) equal to 0, which means I'm on the voltage axis. What I'm actually going to do is I'm just going to look at my circuit and figure this out in my head. Setting (I = 0) means there's no current in this resistor, which means there's no voltage drop across that resistor.

That means that this voltage here is the same as this voltage here, and I know the voltage here—the voltage here is 3 volts. So that means the voltage here is 3 volts, because I know the current is zero. So let's go over and put that point on our line. When (I) is zero, (V) is 3 volts, so there's another point on the line.

Now we have two points, and we can draw a line between them 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 (I)s equal to each other, and that's basically where do these two lines intersect.

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: it's 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, here, let me erase this a bit to clean it up. Let me take out these two things here that was the resistor load line that we were talking about, and now for our solution, we have (I = 6.8 , \text{milliamps}) and (V_{D} = 0.7 , \text{volts}). So that's how you do a graphical solution with a diode.