Voltage Divider Calculator
Voltage Divider Calculator
Output voltage of a two-resistor divider, how much a load pulls it down, the current and power in each resistor — and a design mode that picks a standard E24 or E96 pair for the voltage you want.
Voltage divider
Vin = 12 V, R1 = 30 kΩ, R2 = 10 kΩ, no load
Voltage divider formulas
- R1, R2
- top and bottom resistors; Vout is taken across R2
- RL
- the load’s input resistance, in parallel with R2
- R1 ∥ R2
- the divider’s Thévenin (source) resistance: what the load or an ADC input sees
Worked example
Vin = 12 V, R1 = 30 kΩ, R2 = 10 kΩ, no load
Vout = 12 × 10 ÷ (30 + 10) = 3 V
Current = 12 V ÷ 40 kΩ = 300 µA
Power: R1 = 9² ÷ 30 kΩ = 2.7 mW, R2 = 3² ÷ 10 kΩ = 900 µW
Source resistance = 30 kΩ ∥ 10 kΩ = 7.5 kΩ
With a 100 kΩ load: R2 ∥ RL = 9.091 kΩ, Vout = 2.791 V (−6.98%)
Scaling a voltage into a 3.3 V ADC
| Maximum input | Exact R1 | Nearest E96 R1 | Vout at maximum | Error | Source resistance |
|---|---|---|---|---|---|
| 5 V | 6.667 kΩ | 6.65 kΩ | 3.003 V | +0.10% | 3.99 kΩ |
| 12 V | 30 kΩ | 30.1 kΩ | 2.993 V | −0.25% | 7.51 kΩ |
| 24 V | 70 kΩ | 69.8 kΩ | 3.008 V | +0.25% | 8.75 kΩ |
| 48 V | 150 kΩ | 150 kΩ | 3.000 V | +0.00% | 9.38 kΩ |
How a voltage divider works
Two resistors in series across a voltage share it in proportion to their values. The same current flows through both, so the voltage across the bottom resistor is Vin × R2 ÷ (R1 + R2). With 12 V, 30 kΩ and 10 kΩ, the output is 3 V and 300 µA flows. The ratio is all that matters for the output, but the absolute values set the current wasted in the divider and how stiff the output is.
Loading. Anything connected to the output draws current and appears in parallel with R2, pulling the voltage down. A 100 kΩ load on the example drops the output to 2.791 V, 6.98% low. A rule of thumb is to keep the load resistance at least ten times R1 ∥ R2 (about a 9% error or less), or a hundred times for about 1%. Enter the load’s input resistance to see the exact figure. A divider is not a power supply: if the load current varies, the output varies with it, and a regulator is the right part.
ADC inputs. The commonest use is scaling a battery or supply voltage down to an analog-to-digital converter’s range. The ADC’s sampling capacitor has to charge through the divider’s source resistance R1 ∥ R2, so high values give low readings or crosstalk between channels. Microchip’s ATmega328P datasheet, for example, says its ADC is optimised for sources of about 10 kΩ or less; the page warns above that. A 100 nF capacitor from the ADC pin to ground is the usual fix when you need large resistors to save battery current, at the cost of a slower response.
Transistor bias. A divider also sets the base voltage of a bipolar transistor stage. The base current is a load on the divider, so the divider current is usually made around ten times the base current to keep the bias stable as the transistor’s gain varies.
Design mode works backwards. Choose R2 and the output you want; the page computes the exact R1 = R2 × (Vin ÷ Vout − 1), picks the nearest standard value from IEC 60063’s E24 (5%) or E96 (1%) series, and shows the voltage the real pair gives. For 12 V to 3 V with R2 = 10 kΩ the exact R1 is 30 kΩ; the nearest E96 value is 30.1 kΩ, giving 2.9925 V (−0.25%). To go the other way — a known current and voltage to a resistance — use the Ohm’s law calculator.
Frequently asked questions
What is the voltage divider formula?
Vout = Vin × R2 ÷ (R1 + R2), where R2 is the resistor from the output to ground. 12 V with R1 = 30 kΩ and R2 = 10 kΩ gives 3 V.
How do I choose R1 for a given output voltage?
R1 = R2 × (Vin ÷ Vout − 1). For 12 V down to 3 V with R2 = 10 kΩ, R1 = 30 kΩ; the nearest 1% (E96) value is 30.1 kΩ.
Why is my divider’s output lower than calculated?
The load is in parallel with R2. A 100 kΩ load on a 30 kΩ / 10 kΩ divider pulls 3 V down to 2.791 V. Enter the load resistance to include it, or use smaller resistors or a buffer amplifier.
Can I use a voltage divider to power something?
Only something that draws a tiny, steady current, such as an ADC or op-amp input or a transistor base. For anything else the output sags as the current changes; use a linear regulator or a DC-DC converter.
What resistor values should I use?
Low enough that the load (or ADC input) is at least ten times the source resistance R1 ∥ R2, high enough that the current through the divider is acceptable. For battery sensing, 10 kΩ to 100 kΩ is common, often with a capacitor at the ADC pin.
Related calculators
References
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. Chapter 1: voltage, current and resistance; Ohm’s law; power in resistors; voltage dividers and Thévenin equivalents.
- IEC 60063:2015. Preferred number series for resistors and capacitors (the E6, E12, E24, E48, E96 and E192 series). International Electrotechnical Commission.
- Microchip Technology. ATmega328P datasheet, ADC section, analog input circuitry: the ADC is “optimized for analog signals with an output impedance of approximately 10 kΩ or less”.