Voltage Divider Calculator

How to use this tool

Choose the job in the first box. "Output voltage" takes the input, R1 (input to output) and R2 (output to ground), and gives the output, ratio, current and power in each resistor. "Find a resistor" gives the missing resistor for a target output, with the nearest E12, E24 or E96 values, or the best pair in a range. "With a load" adds RL across R2, "Tolerance" gives the range at 0.1% to 10%, and "Common conversions" lists standard answers.

The input people most often get wrong is the unit box: 10 with kΩ selected is 10,000 Ω. Next is which resistor is which. R1 is on the supply side, so swapping them turns 5 V into 7 V when 12 V meets 14 kΩ and 10 kΩ. The load goes across R2, not in series. The tool models ideal resistors and an ideal source at DC.

What the calculation is doing

The output is Vout = Vin × R2 ÷ (R1 + R2). The same current, I = Vin ÷ (R1 + R2), flows through both resistors, and the power in each is I² × R.

To find a resistor, the tool rearranges this: R2 = R1 × Vout ÷ (Vin − Vout) and R1 = R2 × (Vin − Vout) ÷ Vout, then shows the E series values either side with their error. The pair search tries every R1 in your range with the standard values nearest the exact R2, and keeps the smallest error, preferring the larger total resistance on a tie.

A load RL across R2 replaces R2 with R2 and RL in parallel. The Thevenin equivalent of the divider is a source Vth = Vin × R2 ÷ (R1 + R2) behind Rth = R1 × R2 ÷ (R1 + R2), and the loaded output is Vth × RL ÷ (Rth + RL).

For tolerance, the output falls as R1 rises and rises as R2 rises, so the true extremes are R1 high with R2 low and R1 low with R2 high, and the tool lists all four corners. The estimate 2 × tolerance × (1 − Vout ÷ Vin), used in TI note SLVA450B, is close: for 14 kΩ and 10 kΩ at 1% it gives 1.17% against exact corners of −1.16% and +1.17%.

The ratio is the design, the load is the catch

A divider sets a ratio only while nothing draws current from the output. Once something does, it acts as a source with output resistance Rth, and the output droops by Rth ÷ (Rth + RL).

The usual rule of thumb is a load at least 10 times Rth. Wikipedia's article on impedance bridging quotes Holman and Tomlinson, Sound for Film and Television (2012), for a minimum of some 10 times the source impedance, and lecture notes from the Edinburgh physics department put it at two orders of magnitude for a voltmeter reading a divider. Ten times accepts about 9% error, and 100 times about 1%.

It is also why a divider is a poor power supply. To hold 5 V within about 1% for 20 mA from 12 V, Rth must be near 2.5 Ω, which means R1 of 6 Ω, R2 of about 4.29 Ω and 14 W wasted for about 0.1 W delivered. A linear regulator would dissipate 0.14 W (7 V × 20 mA). Scaling a voltage into a high impedance input, such as an ADC, is what a divider does well.

For an ADC, the converter sees Rth as its source resistance. Fiore's embedded controllers textbook gives 10 kΩ or less for the ATmega ADC, and the 30 V battery monitor below has an Rth of 19.1 kΩ because it uses large values to save current. Analog Devices notes that source resistance lengthens settling, and leakage current through Rth adds leakage × Rth to the error. TI note SLVA450B suggests divider current 100 times the leakage.

Tolerance, noise and what the tool leaves out

On the 12 V divider of 14 kΩ and 10 kΩ, 1% resistors give 4.942 V to 5.058 V (−1.16% and +1.17%), 5% parts 4.711 V to 5.294 V, and 0.1% parts 4.994 V to 5.006 V. The error is about 2 × tolerance × (1 − ratio), so it is nearly double the tolerance for a small output ratio. The corners are a design limit, not a forecast. SLVA450B stacks a 2% reference, 1% resistors and 1% leakage error to reach 4.73%.

Thermal noise is V = √(4kTRΔf), about 13 nV per root hertz for 10 kΩ at room temperature according to Wikipedia. A divider's output noise is that of R1 and R2 in parallel: 9.83 nV per root hertz, or 0.98 µV rms over 10 kHz, against a 12 bit, 3.3 V ADC step of 0.81 mV.

The model is DC with ideal resistors and an ideal source. It leaves out temperature coefficient, self heating, the ADC sampling capacitor, stray capacitance and supply resistance.

Common conversions with standard values

Each row is the best pair of standard values in its range, found by testing every pair. Ranges are 1 kΩ to 100 kΩ, except the battery monitor (10 kΩ to 1 MΩ) and the ADC row (1 kΩ to 10 kΩ).

Some are exact: 9 V to 5 V in E24, and 12 V to 5 V in E96 at 93.1 kΩ and 66.5 kΩ. E12, E24 and E96 come from IEC 60063, with usual tolerances of 10%, 5% and 1%.

A worked example you can check on paper

The default view is 12 V across R1 = 14 kΩ and R2 = 10 kΩ. The total is 24 kΩ, the ratio is 10 ÷ 24 = 0.4167, and Vout = 12 × 0.4167 = 5.000 V. The current is 12 ÷ 24,000 = 0.5 mA, R1 drops 7 V for 3.5 mW, R2 drops 5 V for 2.5 mW, and the supply delivers 6 mW.

Add a 10 kΩ load: R2 and RL make 5 kΩ, so Vout = 12 × 5 ÷ (14 + 5) = 3.158 V, a loading error of (3.158 − 5) ÷ 5 = −36.8%. Rth = 14 × 10 ÷ 24 = 5.833 kΩ, so the load is 1.71 times Rth, and 1% needs at least 99 × 5.833 = 577.5 kΩ.

For standard values, 14 kΩ is in neither E12 nor E24, so the nearest E24 value is 15 kΩ, giving 4.800 V (−4%). E96 has 14 kΩ. Changing both resistors, 18 kΩ and 13 kΩ in E24 give 5.032 V (+0.645%). With 1% resistors the default gives 4.942 V to 5.058 V.

Mistakes that catch people out

Ignoring the load. A divider that gives 5 V on paper reads low once the next stage is connected. Compare the load with Rth, not with R2.

Swapping R1 and R2. R1 is on the supply side. Swapped, the default divider gives 7 V, not 5 V, which can damage a 5 V input.

Choosing huge values to save current. A 5 V divider of 1 MΩ and 500 kΩ draws 3.3 µA, but its Rth of 333.3 kΩ turns the 25 nA leakage in the TI example into 8.3 mV.

Assuming 1% parts give 1% out. The default divider can be 1.17% off before any reference or ADC error. Use 0.1% parts where the ratio matters.

Rounding one resistor. 15 kΩ over 10 kΩ gives 4.8 V, 4% low, when a pair of E24 values gets within 0.645%. Run the pair search.

Popular calculations

Each one opens the calculator with those figures filled in.

Everyday divider outputs

The resistor for a target output

Best pair of standard values

What a load does

Tolerance range

Common conversions in standard values

How this was checked

The maths reproduces published examples from seven sources: Simon Fraser University's Thevenin notes, the Edinburgh physics lecture notes, UMass Applied Electrical Engineering Fundamentals (which prints 4.08 V where rounding gives 4.09 V), Kuphaldt's Lessons in Electric Circuits, TI SPRAD89, TI SLVA450B (4.73%) and Wikipedia (13 nV per root hertz).

A separately coded nodal solution agreed on 200,000 random circuits to within 4.4e−16 of the input voltage. The pair search equals an exhaustive scan of 1,863,640 pairs over 160 cases, and the nearest standard value matches an exhaustive search on 3,000 cases. The tolerance corners equal the extremes of a 25 by 25 grid, and 2,700,000 Monte Carlo draws never left them. The E12, E24 and E96 lists equal IEC 60063 as printed on Wikipedia and agree with Vishay document 28372. Every address above was opened and checked.

Common questions

How do you calculate divider output voltage?

Multiply the input voltage by R2 and divide by R1 plus R2. For 12 V across 14 kΩ and 10 kΩ that is 12 × 10 ÷ 24 = 5 V. R1 is on the supply side. The formula assumes nothing draws current from the output.

How do you choose resistors for a target output?

Fix one resistor, then use R2 = R1 × Vout ÷ (Vin − Vout) or R1 = R2 × (Vin − Vout) ÷ Vout. For 3.3 V from 5 V with R2 = 10 kΩ, R1 must be 5.152 kΩ, and the nearest E24 value, 5.1 kΩ, gives 3.311 V (+0.341%).

How much does a load change the output?

By the fraction Rth ÷ (Rth + RL), where Rth is R1 and R2 in parallel. A load 10 times Rth costs 9.09%, and 100 times costs 0.99%. Use at least 10 times Rth, and 100 times when the output feeds a measurement.

Can a voltage divider power a circuit?

Only one that draws almost nothing. Holding 5 V within about 1% for 20 mA from 12 V would take resistors of 6 Ω and 4.29 Ω and waste 14 W. SparkFun's tutorial says not to use a divider to supply power to a load.

How much error do 1% resistors add?

About 2 × 1% × (1 − Vout ÷ Vin) at each end. For 12 V across 14 kΩ and 10 kΩ the exact corners are −1.16% and +1.17%, so 4.942 V to 5.058 V.

What values should I use to scale a voltage for an ADC?

Set the ratio so full scale lands just under the converter's limit. Keep Rth within the source resistance allowed, 10 kΩ or less for the ATmega ADC according to Fiore's textbook, and divider current at least 100 times the input leakage, as TI suggests. For 10 V to 3.3 V the table gives 6.8 kΩ and 3.3 kΩ in E24.

Related Calculators and Reference

• Ohm Law Calculator: the voltage, current, resistance and power behind every row above.
• Resistor Color Code Calculator: read the bands on the E24 and E96 parts you pick.
• LED Resistor Calculator: a series resistor that drops voltage the way R1 does.
• Capacitor Calculator: for a capacitor added across R2 to steady an ADC input.
• Voltage Drop Calculator: a different thing, the loss along a cable run, not two resistors.
• Texas Instruments application note SLVA450B: worst case accuracy of a resistor divider.