Electronics calculators

Voltage divider calculator

Find the output of a resistor divider, or the resistor values for a target voltage, with the effect of the load, the current, the power in each resistor and the nearest standard values.

Calculated with electrician.wiki · https://electrician.wiki/calculators/voltage-divider/

Result

Vout = 2.878 V

9 V across 10 kΩ (top) and 4.7 kΩ (bottom) gives Vout = 2.878 V; the divider draws 612 µA and has an output resistance of 3.2 kΩ.

Output voltage (Vout): 2.878 V
Ratio Vout / Vin: 0.3197
Divider current: 612.2 µA
Power in R1: 3.748 mW
Power in R2: 1.762 mW
Output resistance (R1 ∥ R2): 3.197 kΩ

Tolerance: with ±1 % resistors Vout can sit anywhere from 2.839 V to 2.917 V; with ±5 % parts, 2.685 V to 3.077 V.

Show the work

1. Output voltage with no load
$V_{out} = V_{in} \cdot \frac{R_{2}}{R_{1} + R_{2}} = 9 V \cdot \frac{4.7 kΩ}{10 kΩ + 4.7 kΩ} = 2.878 V$
2. Current through the divider
$I = \frac{V_{in}}{R_{1} + R_{2}} = \frac{9 V}{10 kΩ + 4.7 kΩ} = 612.2 µA$
3. Power in R1
$P_{R1} = I^{2} \cdot R_{1} = {(612.2 µA)}^{2} \cdot 10 kΩ = 3.748 mW$
4. Power in R2
$P_{R2} = I^{2} \cdot R_{2} = {(612.2 µA)}^{2} \cdot 4.7 kΩ = 1.762 mW$
5. Output resistance (Thevenin)
$R_{th} = \frac{R_{1} \cdot R_{2}}{R_{1} + R_{2}} = \frac{10 kΩ \cdot 4.7 kΩ}{10 kΩ + 4.7 kΩ} = 3.197 kΩ$
Seen from the output, the divider is a 2.878 V source with 3.197 kΩ in series. That is what a load or an ADC input sees.

Ideal resistors at their marked values, at DC or low frequency (stray capacitance ignored).
A load, when given, is a pure resistance from the output to ground.
The tolerance range puts both resistors at opposite ends of their band, the worst case; real spread is usually smaller.

Report a problem with this result

How to use the voltage divider calculator

Pick what you want to find. Find Vout takes the input voltage and both resistors and returns the output voltage, the current the divider draws, the power in each resistor and the divider’s output resistance. Find R2 and Find R1 work backwards: enter the input, the output you want and one resistor, and the calculator returns the exact value of the other, then picks the E12, E24 or E96 value that lands closest to your target and shows how far off it is. Switch on the load to model whatever the output feeds (an ADC pin, a meter, the next stage) as a resistance RL; every mode then designs for it. Values accept the shorthand printed on schematics: 4k7 is 4.7 kΩ, 1M is 1 MΩ and 470R is 470 Ω.

The voltage divider formula

Two resistors in series carry the same current, so the input voltage splits between them in proportion to their resistance. R1 runs from the input to the output and R2 from the output to ground, and the output is the voltage across R2:

V_{out} = V_{in} \cdot \frac{R_{2}}{R_{1} + R_{2}}

Solved for either resistor, the same equation sizes a divider from a target voltage:

R_{2} = R_{1} \cdot \frac{V_{out}}{V_{in} - V_{out}}

R_{1} = R_{2} \cdot \frac{V_{in} - V_{out}}{V_{out}}

Only the ratio sets the unloaded output: 1 kΩ over 470 Ω gives the same Vout as 100 kΩ over 47 kΩ. The absolute values set two other things. One is the current the divider wastes, $I = \frac{V_{in}}{R_{1} + R_{2}}$ . The other is how stiff the output is. Seen from the output, every divider behaves like a source equal to its unloaded Vout in series with R1 and R2 in parallel, its Thevenin resistance:

R_{th} = \frac{R_{1} \cdot R_{2}}{R_{1} + R_{2}}

Connect a load RL and it forms a second divider with that resistance:

V_{out,L} = V_{out} \cdot \frac{R_{L}}{R_{th} + R_{L}}

That second step is why a divider that measures perfectly with a 10 MΩ multimeter can read low once it drives a real input. For the current and power in any single resistor on its own, work it out with the Ohm’s law calculator.

Worked examples

A reference from a 9 V battery

A comparator needs a threshold near 3 V from a 9 V battery. Ten kilohms over 4.7 kΩ is a common first try: what does it give, what does it cost in current, and how stiff is it?

1. Output voltage with no load
$V_{out} = V_{in} \cdot \frac{R_{2}}{R_{1} + R_{2}} = 9 V \cdot \frac{4.7 kΩ}{10 kΩ + 4.7 kΩ} = 2.878 V$
2. Current through the divider
$I = \frac{V_{in}}{R_{1} + R_{2}} = \frac{9 V}{10 kΩ + 4.7 kΩ} = 612.2 µA$
3. Power in R1
$P_{R1} = I^{2} \cdot R_{1} = {(612.2 µA)}^{2} \cdot 10 kΩ = 3.748 mW$
4. Power in R2
$P_{R2} = I^{2} \cdot R_{2} = {(612.2 µA)}^{2} \cdot 4.7 kΩ = 1.762 mW$
5. Output resistance (Thevenin)
$R_{th} = \frac{R_{1} \cdot R_{2}}{R_{1} + R_{2}} = \frac{10 kΩ \cdot 4.7 kΩ}{10 kΩ + 4.7 kΩ} = 3.197 kΩ$

9 V across 10 kΩ (top) and 4.7 kΩ (bottom) gives Vout = 2.878 V; the divider draws 612 µA and has an output resistance of 3.2 kΩ.

Scaling 0–15 V into a 3.3 V ADC

A microcontroller monitors a supply that can reach 15 V. With R1 = 47 kΩ, which bottom resistor puts 15 V at the ADC’s 3.3 V full scale, and what does the nearest E24 part give?

1. Rearrange the divider formula for R2
$R_{2} = \frac{R_{1} \cdot V_{out}}{V_{in} - V_{out}} = \frac{47 kΩ \cdot 3.3 V}{15 V - 3.3 V} = 13.26 kΩ$
2. Nearest E24 value and the output it gives
$V_{out} = V_{in} \cdot \frac{R_{2}}{R_{1} + R_{2}} = 15 V \cdot \frac{13 kΩ}{47 kΩ + 13 kΩ} = 3.25 V$
3. Error against the target
$error = \frac{V_{out} - V_{target}}{V_{target}} = \frac{3.25 V - 3.3 V}{3.3 V} = −1.52 %$
4. Divider current
$I = \frac{V_{in}}{R_{1} + R_{2}} = \frac{15 V}{47 kΩ + 13 kΩ} = 250 µA$

For 3.3 V from 15 V with R1 = 47 kΩ, R2 must be 13.26 kΩ; the closest E24 value, 13 kΩ, gives 3.25 V (−1.52 %).

The same ratio, loaded by a 10 kΩ input

Two 10 kΩ resistors halve 12 V to 6 V on the bench. Then the output is connected to a stage with a 10 kΩ input resistance.

1. Open-circuit output (Thevenin voltage)
$V_{th} = V_{in} \cdot \frac{R_{2}}{R_{1} + R_{2}} = 12 V \cdot \frac{10 kΩ}{10 kΩ + 10 kΩ} = 6 V$
2. Output resistance (Thevenin)
$R_{th} = \frac{R_{1} \cdot R_{2}}{R_{1} + R_{2}} = \frac{10 kΩ \cdot 10 kΩ}{10 kΩ + 10 kΩ} = 5 kΩ$
3. Output with the load connected
$V_{out} = V_{th} \cdot \frac{R_{L}}{R_{th} + R_{L}} = 6 V \cdot \frac{10 kΩ}{5 kΩ + 10 kΩ} = 4 V$
4. Current through R1
$I_{R1} = \frac{V_{in} - V_{out}}{R_{1}} = \frac{12 V - 4 V}{10 kΩ} = 800 µA$
5. Power in each resistor
$P_{R1} = {I_{R1}}^{2} \cdot R_{1} = 6.4 mW P_{R2} = \frac{{V_{out}}^{2}}{R_{2}} = 1.6 mW$

12 V across 10 kΩ over 10 kΩ gives 6 V with nothing connected, but the 10 kΩ load pulls the output down to 4 V (−33.3 %).

Feeding a 1.8 V input from 3.3 V logic

A slow 3.3 V control line has to drive a 1.8 V input. With R2 = 10 kΩ, what top resistor is needed? (For fast edges, the divider’s resistance and the pin’s capacitance form a low-pass filter; use a level shifter instead.)

1. Rearrange the divider formula for R1
$R_{1} = \frac{R_{2} \cdot (V_{in} - V_{out})}{V_{out}} = \frac{10 kΩ \cdot (3.3 V - 1.8 V)}{1.8 V} = 8.333 kΩ$
2. Nearest E96 value and the output it gives
$V_{out} = V_{in} \cdot \frac{R_{2}}{R_{1} + R_{2}} = 3.3 V \cdot \frac{10 kΩ}{8.25 kΩ + 10 kΩ} = 1.808 V$
3. Error against the target
$error = \frac{V_{out} - V_{target}}{V_{target}} = \frac{1.808 V - 1.8 V}{1.8 V} = +0.46 %$
4. Divider current
$I = \frac{V_{in}}{R_{1} + R_{2}} = \frac{3.3 V}{8.25 kΩ + 10 kΩ} = 180.8 µA$

For 1.8 V from 3.3 V with R2 = 10 kΩ, R1 must be 8.333 kΩ; the closest E96 value, 8.25 kΩ, gives 1.808 V (+0.46 %).

Best resistor pairs for common conversions

When both resistors are free to choose, the job becomes a search: for each standard R2 there is an ideal R1, and the nearest standard R1 lands a little off. The table below comes from the same engine as the calculator. It tries every value in a decade for R2, keeps the pair with the smallest ratio error, then scales the pair by a power of ten so the output resistance sits between 1 kΩ and 10 kΩ, which most ADC inputs accept. Multiply both resistors by ten if you need a tenth of the current.

Best standard pairs (no load), scaled so R1 ∥ R2 sits between 1 kΩ and 10 kΩ. Error is the ratio error before tolerance.
Conversion	Ratio	E24 pair R1 / R2	Error	E96 pair R1 / R2	Error
5 V → 3.3 VLogic level for a 3.3 V input	0.66	4.7 kΩ / 9.1 kΩ	−0.09 %	13.7 kΩ / 26.7 kΩ	+0.135 %
3.3 V → 1.8 VLogic level for a 1.8 V input	0.5455	10 kΩ / 12 kΩ	0.00 %	13.7 kΩ / 16.5 kΩ	+0.166 %
4.2 V → 3.3 VSingle Li-ion cell, fully charged	0.7857	3 kΩ / 11 kΩ	0.00 %	10.2 kΩ / 37.4 kΩ	0.000 %
12 V → 5 V12 V rail on a 5 V ADC	0.4167	18 kΩ / 13 kΩ	+0.65 %	14 kΩ / 10 kΩ	0.000 %
15 V → 3.3 V12 V system with headroom	0.22	39 kΩ / 11 kΩ	0.00 %	11.5 kΩ / 3.24 kΩ	−0.086 %
24 V → 5 VIndustrial 24 V on a 5 V ADC	0.2083	9.1 kΩ / 2.4 kΩ	+0.17 %	5.23 kΩ / 1.37 kΩ	−0.364 %
24 V → 3.3 VIndustrial 24 V on a 3.3 V ADC	0.1375	47 kΩ / 7.5 kΩ	+0.08 %	10.2 kΩ / 1.62 kΩ	−0.323 %
48 V → 3.3 V48 V telecom or e-bike pack	0.06875	27 kΩ / 2 kΩ	+0.31 %	107 kΩ / 7.87 kΩ	−0.346 %

E24 values are widely stocked in 1 % tolerance, so an E24 pair with a ratio error well under 1 % is limited by tolerance, not by the value choice. Notice that E96 does not always win: it is not a superset of E24. Values such as 1.2, 3.3 and 4.7 are missing from it (E96 has 1.21, 3.32 and 4.75), so a ratio that E24 hits exactly can be out of reach. The full lists are in the E-series resistor value tables.

How much a load pulls the output down

The loaded output depends only on how many times larger the load is than the divider’s output resistance. For the 9 V reference divider from the first example:

Loading the 9 V, 10 kΩ / 4.7 kΩ divider (no-load output 2.878 V, output resistance 3.197 kΩ)
Load ÷ (R1 ∥ R2)	RL	Vout	Error
1×	3.2 kΩ	1.439 V	−50.00 %
2×	6.39 kΩ	1.918 V	−33.33 %
5×	16 kΩ	2.398 V	−16.67 %
10×	32 kΩ	2.616 V	−9.09 %
20×	63.9 kΩ	2.741 V	−4.76 %
50×	160 kΩ	2.821 V	−1.96 %
100×	320 kΩ	2.849 V	−0.99 %
1,000×	3.2 MΩ	2.875 V	−0.10 %

The common rule of thumb, a load at least ten times the output resistance, still costs about 9 %. Staying under 1 % needs a load about a hundred times larger, and a 12-bit ADC that should be accurate to its last bit needs several thousand times. When the load cannot be that light, buffer the output with an op-amp voltage follower, or design for the load: Find R2 and Find R1 both do that when the load is switched on.

Practical pitfalls

Using a divider as a power supply. The output moves whenever the load current changes, and both resistors burn power all the time. To run a 5 V device from 12 V, use a linear or switching regulator; keep dividers for signals, references and bias points. An LED is not a divider job either: size its series resistor instead.
Tolerance stacking. Both tolerances act on the ratio. For the 9 V reference divider, ±1 % resistors allow anywhere from 2.839 V to 2.917 V; ±5 % parts widen that to 2.685 V to 3.077 V. The worst case is roughly ±2t(1 − Vout/Vin) for a tolerance t, so a divider that cuts the voltage a lot suffers nearly twice the tolerance of a single part. Use 0.1 % parts or a matched network, or calibrate in software. If parts come from a mixed drawer, check each resistor’s colour bands first.
ADC source impedance. A microcontroller ADC samples by connecting a small capacitor to the pin for a short acquisition time, and the divider has to charge it through its output resistance. Many datasheets give a maximum source impedance (often around 10 kΩ at the default sampling time) or a formula linking it to the sampling time. Exceed it and readings come out low and carry over charge from the previous channel. Lower the resistor values, lengthen the sampling time, add 10–100 nF from the ADC pin to ground for slow signals, or buffer the output.
Battery drain. A divider across a battery never switches off. 10 kΩ + 10 kΩ across a 4.2 V cell draws 210 µA, about 1,840 mAh a year. 1 MΩ + 1 MΩ cuts that to 2.1 µA (18.4 mAh a year), but its 500 kΩ output then needs a capacitor at the ADC pin. The alternative is a MOSFET that connects the divider only while measuring.
Power and voltage ratings. The calculator flags any resistor dissipating more than 62.5 mW and suggests a rating with a 2× margin. On high-voltage dividers the resistor’s maximum working voltage matters as much as its wattage: small chip resistors are typically rated between about 50 V and 200 V depending on size, so mains-sensing dividers use several resistors in series, with creepage distances to suit.
Temperature drift. General-purpose thick-film resistors typically drift ±100 ppm/°C or more, thin-film types ±25 ppm/°C or better. When both resistors are the same type and at the same temperature, much of the drift cancels in the ratio, which is why precision references use matched resistor networks.

Dividers also hide inside other circuits: a 555 timer’s ⅓ and ⅔ thresholds come from three equal internal resistors, which is why its timing barely changes with the supply. Calculate 555 timer frequency and pulse width with the timing calculator.

Method and verification

The calculator evaluates the divider equations exactly in double precision and rounds only for display, to four significant figures. A load is treated as a resistance from the output to ground, in parallel with R2; the loaded result is also checked against the Thevenin form above. In Find R2 and Find R1 the exact value comes first. The calculator then takes the standard values either side of it from the IEC 60063 series (E12, E24 or E96) and keeps the one that brings Vout closer to the target. That is not always the nearest value in ohms: when R2 is much smaller than R1, the output is almost proportional to R2 and the arithmetic midpoint, not the logarithmic one, decides.

Test cases include the reference divider (2.878 V, 612.2 µA, output resistance 3.197 kΩ), the 12 V half-divider loaded to exactly 4 V, round trips through all three modes with and without a load, a brute-force check of the best-pair search over every E24 combination, and loads too heavy for any R2 to reach the target.

Assumes ideal resistors and a resistive load at DC or low frequency. Updated 2026-10-14. Found a problem? Report it and we log every fix publicly.

Questions people ask

What is the voltage divider formula?

Vout = Vin × R2 / (R1 + R2), where R1 connects the input to the output and R2 connects the output to ground. 9 V across 10 kΩ and 4.7 kΩ gives 2.878 V. The formula assumes nothing draws current from the output.

How do I choose resistor values for a voltage divider?

Set the ratio first: R2 / (R1 + R2) = Vout / Vin. Then choose the size. The total R1 + R2 sets the current wasted, and R1 ∥ R2 sets the output resistance, which should be at most a tenth of the load (a hundredth for 1 % accuracy) and within what an ADC input accepts, often about 10 kΩ.

Why is my voltage divider output lower than calculated?

Usually the load. Whatever the output feeds sits in parallel with R2 and lowers the ratio; a load ten times the divider’s output resistance already costs about 9 %. Resistor tolerance and an ADC input that needs a low source impedance are the next suspects.

Can I use a voltage divider to power a device?

Not reliably. The output voltage changes with the device’s current, and the divider wastes power continuously. Use a voltage regulator: linear for small currents, a switching converter for anything larger. A divider is for signals and references.

How do I get 3.3 V from 5 V with a voltage divider?

For a signal, R1 = 4.7 kΩ and R2 = 9.1 kΩ (both E24) give 3.297 V from 5 V, with an output resistance of 3.1 kΩ. For a 3.3 V supply rail, use a regulator instead.

What is the output impedance of a voltage divider?

R1 and R2 in parallel: R1 × R2 / (R1 + R2). For 10 kΩ and 4.7 kΩ that is 3.197 kΩ. This is the resistance a load or an ADC input sees, and the reason the output sags under load.