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LessonThe voltage divider · 16 min

The voltage divider: how and why it works

Very often a circuit has the wrong voltage for the job. A 12 V battery, but the microcontroller's input only takes up to 5 V. A 5 V signal, but a 3.3 V chip. A sensor that changes its resistance, but an input that can only read a voltage. The simplest answer to all three is two resistors in series: the voltage divider. It is probably the most used circuit in electronics.

R1 and R2 in series across 9 V. The same current flows through both, so each takes a share of the voltage. The voltmeter reads Vout: the middle point against 0 V.
R1 and R2 in series across 9 V. The same current flows through both, so each takes a share of the voltage. The voltmeter reads Vout: the middle point against 0 V.

Why it works

It is just the series circuit you already know. One current flows through both resistors:

I = Vin / (R1 + R2)

The voltage across R2 - between the middle point and 0 V - is that current times R2 (Ohm's law). Put the two together and the current cancels out:

Vout = Vin × R2 / (R1 + R2)
Vout
the voltage across R2: between the middle point and 0 V
Vin
the supply across both resistors
R1
the top resistor (from Vin to the middle)
R2
the bottom resistor (from the middle to 0 V)
Worked example 1: what does it give?

R1 = 20 kΩ, R2 = 10 kΩ across 9 V. What is Vout?

  1. Vout = 9 × 10 / (20 + 10) = 9 × 1/3.
  2. Vout = 3 V.
  3. R1 takes the other 6 V: twice the resistance, twice the voltage.

Answer: 3 V.

Worked example 2: design one - 12 V down to under 5 V

A 12 V lead-acid battery must be read by a 5 V Arduino input. While charging it can reach 14.4 V. With R2 = 10 kΩ, choose R1 so the input never exceeds 5 V.

  1. Design for the highest voltage: Vout / Vin ≤ 5 / 14.4 = 0.347.
  2. 0.347 = 10 k / (R1 + 10 k) → R1 + 10 k ≥ 28.8 k → R1 ≥ 18.8 kΩ.
  3. Choose a standard value above it with some room: 22 kΩ.
  4. Check: 14.4 × 10 / 32 = 4.5 V at most; a normal 12.6 V gives 3.94 V.

Answer: R1 = 22 kΩ, R2 = 10 kΩ.

InteractiveTry it: move the sliders
Output Vout-V
Current through the divider-mA
Power wasted-mW

Make both resistors 10× smaller: Vout stays the same, the wasted power grows 10×.

Circuit for: Try it: move the sliders
The circuit at the starting values
Vout as R2 changes (R1 = 10 kΩ, Vin = 9 V)0102030405002468R2 (kΩ)Vout (V)R2 = R1: half
Vout

The curve is not a straight line: Vout climbs quickly while R2 is small compared with R1, then levels off towards Vin. That shape matters for sensors - the divider is most sensitive where R2 is close to R1.

More than two resistors

Three equal resistors make two taps: 2/3 and 1/3 of the supply (6 V and 3 V from 9 V). Any tap's voltage is Vin × (resistance below it) / (total resistance).
Three equal resistors make two taps: 2/3 and 1/3 of the supply (6 V and 3 V from 9 V). Any tap's voltage is Vin × (resistance below it) / (total resistance).

Practice 1R1 = 4.7 kΩ, R2 = 4.7 kΩ, Vin = 5 V. What is Vout?

Hint

Equal resistors share equally.

Show the answer

Answer: 2.5 V

5 × 4.7 / (4.7 + 4.7) = 2.5 V.

Practice 2R1 = 10 kΩ, R2 = 40 kΩ across 10 V. Vout?

Show the answer

Answer: 8 V

10 × 40 / 50 = 8 V.

Practice 3Which change makes Vout larger?

  1. A bigger R1
  2. A bigger R2
  3. Both resistors 10× smaller
  4. A smaller supply
Show the answer

Answer: B. A bigger R2

Vout is R2's share: a bigger R2 takes a bigger share.

Challenge 4You need exactly 1/4 of a 20 V supply and have R2 = 3.3 kΩ. What R1 do you need?

Hint

R2 / (R1 + R2) = 1/4, so R1 = 3 × R2.

Show the answer

Answer: 9900 Ω

R1 + R2 = 4 × R2 → R1 = 3 × 3.3 k = 9.9 kΩ (10 kΩ from the E12 series gives 4.96 V instead of 5 V).

Challenge 5A ladder of 1 kΩ (top), 2 kΩ and 3 kΩ (bottom) is across 12 V. What is the voltage at the junction between the 1 kΩ and the 2 kΩ?

Hint

The resistance below that point is 2 + 3 = 5 kΩ out of 6 kΩ.

Show the answer

Answer: 10 V

12 × 5 / 6 = 10 V. (The lower tap is at 12 × 3 / 6 = 6 V.)