The rather simple circuit shown below is known as a voltage...

The rather simple circuit shown below is known as a voltage divider. The symbol consisting of three horizontal lines is represents "ground" and can be defined as the point where the potential is zero. The voltage divider is widely used in circuits and a single voltage source can be used to provide reduced voltage to a load resistor as shown in the second part of the figure. (a) What is the output voltage $V_{\text {out }}$ of circuit (a) in terms of $R_{1}, R_{2},$ and $V_{\text {in }} ?$ (b) What is the output voltage $V_{\text {out }}$ of circuit (b) in terms of $R_{1}, R_{2}, R_{L},$ and $V_{\text {in }} ?$

Transcript

00:01 So we're going to look at the operation of a voltage divider, just as the name implies, because of the two resistors being in series in the circuit, the voltage from the battery gets split according to the resistance.

00:21 But the divider circuit is a simple way to provide a variable voltage output to some other circuit, but there is an issue that if you just use a dvm to measure the voltage output, the dvm has infinite input resistance and will not drain current, but if you put some other type of load with a small resistance in there, you may be surprised at what you get out.

00:53 You may not get the predicted voltage out.

00:56 So we're going to just take a look at these two scenarios.

01:01 But we first start off with r1 and r2 in series with the battery.

01:10 And our goal is to determine an expression for the voltage output in this case.

01:17 And then we'll look at what happens when you put any kind of load resistor on there.

01:22 So we'll start off there in series, which means it's the same current that runs through everything.

01:30 And nothing gets drained out the outlet.

01:37 So i goes all the way around.

01:43 And with those two resistors in series, we can predict the current.

01:49 The r equivalent is just r1 over r2.

01:53 Sorry, r1 plus r2.

01:56 And the current is vn over the sum of the two resistors.

02:02 Then the voltage output is simply the voltage drop over resistor 2, which is just i times r2.

02:22 And since we know what the current is, that says the voltage output is vn over r1 plus r2 times vr2, the r2.

02:39 And we can get the ratio then, v out over vn.

02:47 Is equal to r2 over r1 plus r2.

02:55 And of course, the remainder of the battery voltage drops across r1.

03:02 So a simple splitting of the voltage.

03:06 Now, the mistake that many people make is to just put the load right across those output terminals.

03:14 And that's the second case that we're going to take a look at of why you don't want to do that.

03:23 And there is a simple fix.

03:27 So you usually use a operational amplifier, which will kind of serve as a buffer between the r2 and the rl, but we won't get into that.

03:43 What about loading the output with rl? what is v -out? okay, so i will redraw the circuit, but basically what we now have is the load resistor is now in parallel with r2.

04:20 So there is a parallel resistance in our circuit that we did not have before.

04:41 And here's ground...

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