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.
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.