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How a Potential Divider works
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What is a Potential Divider?
A potential divider could be thought of as a voltage divider. A Volt is just a unit of ‘potential difference’ in electric
charge between two objects.
How does it work?
A potential divider is a simple circuit which takes advantage of the way voltages drop across resistors in series.
It is a very useful and common circuit and is widely used in our range of electronic kits.
The idea is that by using two resistors in series it is possible to divide a voltage and create a different voltage
between them. In the example below two identical resistors are in series with a power supply. The total voltage
across the circuit is ‘Vin’ however this total voltage is split between our two resistors meaning ‘Vout’ is at a different
voltage. The amount by which the voltage drops over across each resistor depends on the relative values of each
resistor and the total resistance.
The formula for working out the voltage drop across two resistors in series is:
Vout =
Vin × R2
R1 + R2
One of the most useful ways to use this circuit is to replace R2 with a variable resistor. If R2 can be controlled by
turning a dial then Vout can also be controlled. This is used for tuning or volume control in many circuits such as in
our FM Radio Kit V2.0 or 3W Stereo Amplifier Kit.
Another common use of the potential divider is to replace R2 with a sensor such as an LDR. Then as the resistance of
the sensor changes, Vout changes as well. This change can then be used to trigger a transistor or can be fed into the
input of a microcontroller. For more details of how to do this, please visit our How an LDR Works resource.
How a Potential Divider works
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Example
This is a worked example of using the formula above to calculate the missing resistor value for a circuit.
Look at the circuit below and take note of the values that are known. Vin is 5V, R1 is 1KΩ and R2 is 10KΩ
Next, substitute the known values into the formula
Vout =
Vout =
Vin × R2
R1 + R2
5V × 10KΩ
1KΩ + 10KΩ
Vout = 4.55V
How a Potential Divider works
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Example Questions
Now try finding the missing values in these three examples.
How a Potential Divider works
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Answers
1)
4𝑉 =
5𝑉 × 5𝐾Ω
𝑅1+5𝐾Ω
2)
1𝑉 =
5𝑉 × 𝑅2
1𝐾Ω+𝑅2
3)
𝑉𝑜𝑢𝑡 = 1𝐾Ω+1.5𝐾Ω
5𝑉 × 1.5𝐾
𝑅1 =
𝑅2 =
5𝑉
4𝑉
= 1.25𝐾Ω
1𝑉×1𝐾
5𝑉
𝑉𝑜𝑢𝑡 =
= 0.25𝐾Ω
7.5𝑉𝐾Ω
2.5𝐾Ω
= 3V