Voltage Divider

In one sentence: A voltage divider uses two resistors in series to give a smaller share of a supply voltage. The share is set by the ratio of the two resistances.

$V_{\text{out}} = V_{\text{in}} \frac{R_2}{R_1 + R_2}$

Read it out loud: “V out equals V in times R 2 over R 1 plus R 2. V out is the output voltage, measured in volts. V in is the input voltage. R 1 is the top resistor and R 2 is the bottom resistor, both in ohms. So the output voltage is the input voltage times R 2 divided by the total of R 1 and R 2.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Powers of ten, and the prefix kilo: A power of ten is 10 multiplied by itself a set number of times, so $10^{3} = 1000$ . The prefix kilo (k) means $10^{3}$ , one thousand. So $10\ \text{k}\Omega$ is $10 \times 10^{3}\ \Omega$ , which is 10 000 ohms. Converting kilohms to ohms matters only if the two resistances use different units. If both are in kilohms, the units cancel in the ratio.
Subscripts: in, out, 1 and 2: A subscript is a small label written under a letter, and it tells similar quantities apart. $V_{\text{in}}$ is read ‘V in’, the voltage going into the divider, and $V_{\text{out}}$ is ‘V out’, the voltage coming out. $R_1$ and $R_2$ are read ‘R one’ and ‘R two’, the first and second resistor in the chain. The labels name the parts, and they do not change the numbers.
Ground, the 0 V reference: Ground is the reference point in a circuit that is called 0 V. Every voltage is measured relative to it, the way heights are measured from sea level. The bottom end of the divider is usually connected to ground, so the output voltage is the voltage between the middle point and ground.
Series and the shared current: Two parts are in series when they sit one after another on a single path, so the same current passes through both. A current entering one part must leave it and enter the next. Because the current is shared, the voltage across each resistor depends on its resistance, and the supply voltage is shared between them.
Load, and why it matters: A load is any device connected to a circuit that draws current from it, such as a sensor, a display or another resistor. When a load is connected to the output of a divider, it draws current from the middle point, and that changes the output voltage. The divider formula assumes nothing is connected to the output, which is why a load changes the answer.
Buffer and voltage follower: A buffer is a circuit that copies a voltage to its output while drawing almost no current from the source. A voltage follower is a common buffer built with an amplifier, where the output voltage equals the input voltage. Putting a buffer between a divider and a load means the load cannot pull the divider’s output down, so the formula stays correct.

What each symbol means

Symbol	Meaning
$V_{\text{out}}$	The output voltage, the voltage across $R_2$ , measured in volts (V). It is the voltage that the divider delivers to the next part of the circuit. In the example it is 3 V before any load is connected.
$V_{\text{in}}$	The input voltage, the supply voltage across both resistors together, measured in volts (V). A battery of 9 V gives $V_{\text{in}} = 9$ V, which is the whole push that the two resistors share.
$R_1$	The resistance of the top resistor, between the input and the output, measured in ohms ( $\Omega$ ). A bigger $R_1$ takes a bigger share of the supply voltage and leaves less for the output. The example uses $R_1 = 20\ \text{k}\Omega$ .
$R_2$	The resistance of the bottom resistor, between the output and ground, measured in ohms ( $\Omega$ ). The output voltage is measured across this resistor, so a bigger $R_2$ gives a bigger output. The example uses $R_2 = 10\ \text{k}\Omega$ .

The formula piece by piece

$R_2$: Start with the bottom resistor. $R_2$ is its resistance in ohms, and it is the part whose voltage we want. The output voltage is the voltage across $R_2$ , so its resistance sets the share.
$R_1 + R_2$: The total resistance of the whole chain. Because the two resistors are in series, their resistances add, and the total is the opposition the supply has to push the current through.
$\frac{R_2}{R_1 + R_2}$: The fraction of the total resistance that sits in the bottom resistor. Because the same current flows through both, the voltage across $R_2$ is this fraction of the supply voltage.
$V_{\text{in}} \frac{R_2}{R_1 + R_2}$: Multiply the supply voltage by that fraction. The result is the share of the supply that appears across the bottom resistor, which is the output voltage.
$V_{\text{out}} = V_{\text{in}} \frac{R_2}{R_1 + R_2}$: The whole formula in one sentence: the output voltage equals the input voltage times the bottom resistance divided by the total of the two resistances. A bottom resistor that is half the total takes half the supply voltage.
$I = \frac{V_{\text{in}}}{R_1 + R_2}$: A standard second form, the current through the chain. The same current flows through both resistors, so it equals the supply voltage divided by the total resistance. It is the step that gives the voltage across $R_2$ by Ohm’s law.
$V_{\text{out}} = I R_2$: The output voltage from Ohm’s law, with the current from the previous form. The voltage across $R_2$ is the current through it times its resistance, and putting the two forms together gives the divider formula again.

What the formula is doing, and why

Two resistors connected in series form a chain. The same current flows through both, so the voltage across each one depends on its resistance. The supply voltage $V_{\text{in}}$ is shared between $R_1$ and $R_2$ , and the output is the voltage across $R_2$ , the bottom resistor. What goes in is the input voltage and the two resistances. What comes out is a fraction of the input.

Why is the output a fraction? The same current $I$ flows through both resistors. The voltage across $R_2$ is $V_{\text{out}} = I R_2$ , and the whole supply is $V_{\text{in}} = I (R_1 + R_2)$ . Dividing the first equation by the second cancels the current, which leaves $\frac{V_{\text{out}}}{V_{\text{in}}} = \frac{R_2}{R_1 + R_2}$ . That is why the bottom resistance goes on top: the output is the part of the supply that falls across $R_2$ . If $R_1$ and $R_2$ are equal, the output is exactly half the input. If $R_2$ is small compared with $R_1$ , the output is a small fraction of the input.

The divider only gives a clean output when nothing else draws current from the middle point. If you connect a load, such as a sensor or another resistor, from the output to ground, it sits in parallel with $R_2$ . That lowers the effective resistance of the bottom half, and the output drops below the value the formula predicts. The formula assumes a load of infinite resistance, which draws no current.

As a rule of thumb, keep the load resistance much larger than $R_2$ , for example at least ten times larger. Another option is to put a buffer, such as a voltage follower, between the divider and the load. The rule of thumb is a practical guide, and the exact effect depends on the circuit. A divider sets a voltage, but it wastes power and cannot supply much current without the output dropping.

An everyday picture

Think of a tall pole with a clip that can slide along it. The total length of the pole is the input voltage. The clip sits at a point along the pole, and the length of the lower part, as a share of the whole, is the output voltage. Move the clip and the share changes. Pulling some of the pole’s height away at the clip, which is what a load does, lowers the output.

Worked example: A divider that gives one third of the voltage, then gets loaded

A 9 V supply feeds a divider with $R_1 = 20\ \text{k}\Omega$ and $R_2 = 10\ \text{k}\Omega$ . First find the output with nothing connected. Then connect a sensor that draws current as if it were a $10\ \text{k}\Omega$ resistor from the output to ground, and find the new output. The k means kilo, which is one thousand, so the resistors are 20 000 ohms and 10 000 ohms.

Find the unloaded output. Use the divider formula: $V_{\text{out}} = V_{\text{in}} \frac{R_2}{R_1 + R_2} = 9\ \text{V} \times \frac{10}{20 + 10} = 9\ \text{V} \times \frac{1}{3} = 3\ \text{V}$ . The $\text{k}$ cancels in the ratio, because the top and bottom are both in kilohms. The bottom resistor takes one third of the total, so it gets one third of the 9 V.
Find the bottom resistance when the sensor is connected. The sensor sits in parallel with $R_2$ , between the output and ground, so the two resistances combine as a parallel pair. Using the two-resistor shortcut, $\frac{10 \times 10}{10 + 10} = \frac{100}{20} = 5\ \text{k}\Omega$ . Two equal resistors in parallel give half the value of one of them, so the bottom half drops from 10 k $\Omega$ to 5 k $\Omega$ .
Use the same formula with the new bottom resistance. The top resistor is still 20 k $\Omega$ , and the effective bottom resistance is now 5 k $\Omega$ . So $V_{\text{out}} = 9\ \text{V} \times \frac{5}{20 + 5} = 9\ \text{V} \times \frac{5}{25} = 9\ \text{V} \times 0.2 = 1.8\ \text{V}$ . Since $5 \div 25 = 0.2$ , the output is one fifth of the supply.
Find the drop caused by the load. The output fell from 3 V to 1.8 V, which is $3 - 1.8 = 1.2$ V lower. As a share of the unloaded output, that is $\frac{1.2}{3} = 0.4$ , so the load pulled the output down by 40%. The sensor did not change the supply, but it changed the divider.
Check the loaded result. The loaded output must be less than the unloaded output, because the load adds a parallel path that lowers the bottom resistance. Here 1.8 V is less than 3 V, as expected, so the result is consistent with the physics of the circuit.

Answer: With nothing connected the output is 3 V. With the 10 k $\Omega$ sensor attached, the output falls to 1.8 V, so the load pulled the output down by 1.2 V. To avoid this, a buffer or a much larger load resistance is needed.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Putting the resistors in the wrong place in the formula. The output is across $R_2$ , so $R_2$ goes on top of the fraction.
Forgetting the load. Any device connected to the output draws current and changes the result, often by a large amount.
Using the divider to supply a large current. A divider sets a voltage, but it wastes power and cannot supply much current without the output dropping.
Using the formula for resistors that are not in series. The formula assumes the same current flows through both resistors.

Where you will meet it

Reading a sensor, such as a thermistor or a light-dependent resistor, with a microcontroller that measures voltage.
Setting the bias voltage at the base of a transistor in an amplifier, so that the transistor works in the right range.
A potentiometer, which is a variable divider, used as a volume control or a dial.
Scaling a high voltage down to a level that a measuring instrument can read safely.

Questions people ask

Why does the output go down when I connect something?

The load draws current from the middle point. That current flows through $R_1$ as well, so the voltage across $R_1$ rises and the output falls.

Can a voltage divider power a motor or an LED?

Usually not well. The divider sets a voltage, but the current it can supply is small and the output moves when the load changes. A resistor in series with the LED, or a proper driver circuit, is the better choice.

Does the divider need a battery?

It needs a voltage across the two resistors. The supply can be a battery, a power supply or a signal source.

How do I pick the resistor values?

Choose the ratio for the output you want, then pick values that are large enough to keep the wasted current small and small enough that the load does not change the output much.

Current Divider: A current divider splits a total current between two parallel resistors, and the smaller resistor always takes the bigger share.
Ohm's Law: Ohm’s law says the push on a wire, called voltage, equals the flow through it, called current, times how hard the wire resists that flow, called resistance.
Kirchhoff's Voltage Law: Kirchhoff’s voltage law says that if you walk all the way around a closed circuit loop, the voltage you gain equals the voltage you lose, so the net change is zero.
Resistors in Series and Parallel: Resistors in a line add up, and resistors side by side always give a total smaller than the smallest one.