Voltage divider | Circuit analysis | Electrical engineering | Khan Academy

Now I'm going to show you a circuit that's called a voltage divider. This is a name we give to a simple circuit of two series resistors. So I'm just going to draw two series resistors here, and it's a nickname in the sense of it's just a pattern that we see when we look at circuits.

I'll show you what the pattern is. The pattern is we have two resistors in series. It's no more than that. We assume that there's a voltage over here; we hook up a voltage over here like this. So that's called an input voltage. We'll call it (V_i) for (V_{in}). Then the midpoint of the two resistors, and typically the bottom, that's called (V_{out}).

So we basically just have a pattern here with a series resistor driven by some voltage from the ends of the two resistors, and we're curious about the voltage across one of them. So now we're going to develop an expression for this. Let's also label our resistors; this will be (R_1) and this will be (R_2). That's how we tell our resistors apart, and we're going to develop an expression for this.

Let's first put a current through here; we'll call that current (I). We'll make an assumption that this current here is zero; there's no current going out of our little circuit here, and that means, of course, that this current here is also (I). So it's continuous all the way down.

Now we want to develop an expression that tells us what (V_{out}) is in terms of these two resistors and the input voltage. So let's go over here and do that. The first thing we're going to write is, we know that using Ohm's law, we can write an expression for these series resistors.

On this side here, Ohm's law, we'll put over here (V = I \cdot R). In the specific case here, (V_N = I \cdot \text{what} \cdot \text{times the series combination of} , R_1 \text{ and } R_2), and the series combination is the sum (R_1 + R_2). I'm going to solve this for (I):

[ I = \frac{V_N}{R_1 + R_2} ]

All right, next step is going to be let's solve for let's write an expression that's related to (V_{out}), and (V_{out}) only depends on (R_2) and this current here. So we can write:

[ V_{out} = I \cdot R_2 ]

I'll solve this equation for (I) the same way:

[ I = \frac{V_{out}}{R_2} ]

Now we have two expressions for (I) in our circuit because we made this assumption of zero current going out. Those two (I)s are the same. So let's set those equal to each other and see what we get:

[ \frac{V_{out}}{R_2} = \frac{V_N}{R_1 + R_2} ]

So now I'm going to take (R_2) and move it over to the other side of the equation, and we get:

[ V_{out} = V_N \cdot \frac{R_2}{R_1 + R_2} ]

And this is called the voltage divider expression right here. It gives us an expression for (V_{out}) in terms of (V_{in}) and the ratio of resistors. Resistors are always positive numbers, and so this fraction is always less than one, which means that (V_{out}) is always somewhat less than (V_{in}) and it's adjustable by adjusting the resistor values.

It's a really handy circuit to have. Let's do some examples; we'll put that up in the corner so we can see it. I'm going to real quick build a voltage divider that we can practice on. Let's make this 2k ohms, 2000 ohms; we'll make this 6k ohms, or 6000 ohms, and we'll hook it up to an input source that looks like, let's say it's 6 volts.

Like that, and we'll take an output off of this right here is where the output of our voltage divider is, and we'll say that that is (V_{out}).

So let's solve this using the voltage divider expression:

[ V_{out} = V_N \cdot \frac{R_2}{R_1 + R_2} ]

Which is (6) volts times the ratio of resistors, and (R_2) is (6k) ohms divided by (2k) ohms plus (6k) ohms.

Notice this always happens; the k's all cancel out. That's nice, and that equals (6 \times \frac{6}{2 + 6}) is (8). And if I do my calculations right, (V_{out}) is (4.5) volts. So that's what a voltage divider is.

And if you remember at the beginning, we made an assumption that this current going out here was about zero. If that current is really small, you can use this voltage divider expression, which as we see up here is the ratio of the bottom resistor to both resistors.

That's how I remember it; it's the bottom resistor over the two resistors added together. If you think the current is not very small, what you do is you go back, and you do this analysis; you do the same analysis again but you account for the current that's in here.

So that's the story on voltage dividers.