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Voltage Divider

See the information below this calculator for a deeper dive into how a voltage divider works.
Circuit Values
Please enter any three of the following four values for your circuit. The fourth value will be calculated for you.
Resistance value of R1
Units
Resistance value of R2
Units
V
Voltage value for circuit input voltage
V
Voltage value for circuit output voltage

What Is a Voltage Divider?

A voltage divider is a common and useful circuit that can be found in many devices. In fact, most volume pots in guitars and amplifiers work as voltage dividers. They create an output voltage that is a fraction of the input voltage. This fraction of the input voltage is proportional to the ratio of the two resistors used in the circuit. The four variables used are the input voltage ($V_{\text{in}}$), the output voltage ($V_{\text{out}}$), and the two resistors $R_1$ and $R_2$. With three known variables we can use the voltage divider equation to find the fourth.

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

This equation is derived from Ohm’s Law. When applying Ohm’s Law to our input voltage we know that

$$V_{\text{in}} = I_1 \times R$$

In $V_{\text{in}}$’s case,

$$R = (R_1 + R_2)$$

so

$$V_{\text{in}} = I_1 \times (R_1 + R_2)$$

For $V_{\text{out}}$ we know that

$$V_{\text{out}} = I_2 \times R_2$$

$I_1$ is assumed to be equal to $I_2$.

Solving for $I_1$ we find that

$$I_1 = \frac{V_{\text{in}}}{(R_1 + R_2)}$$

Solving for $I_2$ we get

$$I_2 = \frac{V_{\text{out}}}{R_2}$$

Combining these equations we get

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

Solving for $V_{\text{out}}$ we get our voltage divider equation

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

Above we mentioned that $I_1$ is assumed to be equal to $I_2$. This is very important when considering using a voltage divider. Let’s try this example. We have an input voltage of 10V. We’ll use two 10k resistors as $R_1$ and $R_2$ giving us an output voltage of 5V. We want to apply the 5V to a 100k load which would look like this:

We can see that $R_2$ is running parallel to our 100k load, altering our voltage divider. $I_1$ is no longer equal to $I_2$ since the current is divided between $R_2$ and the load. Using the resistor values in the parallel resistance equation we find that

$$\frac{10{,}000Ω \times 100{,}000Ω}{(10{,}000Ω + 100{,}000Ω)} = 9{,}090.91Ω$$

When we use that result with our voltage divider formula we can see the change in output voltage from the divider.

$$10\text{V} × \frac{9{,}090.91Ω}{(10{,}000Ω + 9{,}090.91Ω)} = 4.76\text{V}$$

The 4.76V is a 4.8% difference from our expected 5V output. You will need to consider if this change is critical to the circuit you are designing. When using a higher load the divider will be more accurate and a lower load will result in an even less accurate output voltage than the one in our example. This is one of the reasons why voltage dividers should not be used for supplying power in place of components such as voltage regulators or other power supply circuits. A unity gain op amp can be used following the voltage divider in circuits that require a more precise output voltage.

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