Lessons · Engineering · the voltage divider
The voltage divider: two resistors share the volts by their size
Two resistors in series across V_in give V_out = V_in × R_2 / (R_1 + R_2) across R_2, the one you read across.
Hone is a place to practise a career, one idea a day. This is one of its lessons, written out in full and free to read without an account.
In beta. This lesson was written for Hone and has not yet been checked by a licensed engineer. Practice material, not professional advice. What that means.
What it is for
A 5 V microcontroller pin has to read a 24 V signal. Two resistors, chosen by this one line, bring it down. Choose them backwards and the pin sees 19 V and the board is scrap.
How to think about it
Name the resistor the meter is across; that one is on top of the fraction. Both of them go on the bottom. It is only Ohm's law twice: the current through both, then the volts across one.
Worked example
V_out = V_in × R_2 / (R_1 + R_2), where R_2 is the one you read acrossThe formula, with the one thing people get wrong named.
12 V across 3 kΩ then 1 kΩ, reading across the 1 kΩThe setup.
V_out = 12 × 1 / (3 + 1) = 3 VOne quarter of the resistance, one quarter of the volts.
The long way: I = 12 / 4000 = 3 mA, then V = 3 mA × 1 kΩ = 3 VOhm's law twice gives the same 3 V. The divider is a shortcut, not a new law.
Your turn
20 V across 4 kΩ then 6 kΩ; read across the 6 kΩ. Write the output.
V_out = 20 × / (4 + 6) = 12 V
Solve one, graded on the server
The trap
Putting the wrong resistor on top. The numerator is the one the meter is on. Swap them and 3 V becomes 9 V, and a pin rated for 5 V sees it.