Voltage divider calculator

How this is calculated

With nothing connected to the tap, the same current flows through both resistors and the output is the fraction of the total that R2 occupies:

V_out = V_in · R2 / (R1 + R2)

That is where every other calculator stops, and it is only true while nothing is connected. The moment you attach a load, the divider stops behaving like a voltage source, because it never was one. Thévenin's theorem says exactly what it is instead: a perfect source of the voltage above, in series with a resistance of

R_th = R1 ∥ R2 = R1 · R2 / (R1 + R2)

and your load then forms a second divider with that resistance. The output you actually get is

V_load = V_out · R_L / (R_L + R_th)

so the fraction you lose is R_th / (R_L + R_th) — and notice what is not in that expression. The input voltage is not in it. The individual resistor values are not in it. Only the ratio of the load to the source resistance. A 1 kΩ divider and a 1 MΩ divider with the same ratio give the same answer with nothing connected and completely different answers in a real circuit, and the source resistance is the only thing that distinguishes them.

The source resistance is the whole story

Three separate things people get wrong about dividers are the same thing wearing different clothes.

The output sags. Covered above: droop is R_th/(R_L+R_th). Nothing else matters.

The node is slow. Every load has capacitance — an oscilloscope probe is 10 to 15 pF, a length of hookup wire is a few pF, a microcontroller pin is a few more. That capacitance charges through R_th, so the tap has a time constant of R_th · C, a −3 dB bandwidth of 1/(2π R_th C) and a 10–90 % rise time of 2.2 R_th C. A 1 MΩ divider into 20 pF has a 20 µs time constant. It is a low-pass filter you did not ask for.

It picks up noise. A high source resistance is a high-impedance node, which is an antenna for capacitively coupled interference, and it generates its own thermal noise of √(4kTR) — about 4 nV/√Hz at 1 kΩ, 13 nV/√Hz at 10 kΩ, 128 nV/√Hz at 1 MΩ. It also turns any input bias or leakage current into an offset: 1 µA of leakage across 100 kΩ is 100 mV of error, which is a great deal more than the resistor tolerance you agonised over.

So the scale of a divider is a genuine trade-off with an optimum in the middle, and it is not the same optimum for every circuit. Low resistance means low droop, fast settling and low noise, but it wastes current continuously. High resistance saves current but sags, slows down and picks up. Most signal-level dividers land between 1 kΩ and 100 kΩ; a battery monitor that must sip current lives at 1 MΩ and pays for it by needing a buffer.

Why the “ten times the load current” rule sometimes works

The traditional rule of thumb is to size the divider so it carries ten times the current the load draws. It is a proxy for the real rule, and it is a leaky one: how much error it actually leaves depends on how hard you are dividing.

• A gentle divider — output at 90 % of the input — following that rule lands at about 1 % error.
• A halving divider lands at about 4.8 %.
• A ÷10 divider lands at about 8.3 %.

That eight-to-one spread is why the same rule of thumb feels dependable to one engineer and unreliable to another: they were dividing by different amounts. Size against the source resistance instead and the answer is the same every time — R_L ≥ 100 · R_th for 1 % accuracy, ≥ 20 · for 5 %, ≥ 10 · for 9 %. The calculator reports both numbers so you can see them disagree.

How accurate the ratio can actually be

Only the ratio of the two resistors matters, never their absolute values, and that has a pleasant consequence: the worst-case ratio error is not twice the resistor tolerance, as you might fear, but

ΔV_out/V_out ≈ 2 · tolerance · (1 − V_out/V_in)

The harder you divide, the closer the error creeps to twice the tolerance; the gentler the division, the more the two resistors' errors cancel. Two 1 % resistors halving a voltage give exactly ±1 %, not ±2 %. The same pair dividing by ten gives ±1.8 %. A pair producing 90 % of the input gives ±0.2 %, better than either resistor on its own.

Two further effects work in your favour and are worth knowing about, because they are why precision dividers are built the way they are. Resistors from the same reel are correlated — they were trimmed on the same machine from the same material — so their errors partly cancel rather than adding independently. And temperature coefficient cancels almost entirely in a ratio, provided both resistors are the same type and at the same temperature. This is exactly what a resistor network buys you: two or four resistors on one substrate, specified not for absolute accuracy but for ratio accuracy and tracking tempco, often 0.05 % and 5 ppm/°C when each individual element is only a 1 % part. If a divider has to be accurate, that is the component to reach for, not two tighter individual resistors.

The calculator's worst-case figure deliberately assumes no correlation and pushes both resistors and the supply to opposite extremes at once, which is the honest bound for parts bought loose from different reels.

Feeding an ADC through a divider

This is the most common use of a divider and the one with a hard, checkable constraint. A successive-approximation ADC — which is what is inside nearly every microcontroller — does not present a steady load. It presents a small capacitor, typically 5 to 30 pF, which it briefly connects to the input pin at the start of each conversion and which must charge to within half a least-significant bit before the converter starts deciding bits.

Charging through your divider's source resistance takes

t = R_th · C_s · (N + 1) · ln 2

for an N-bit result, which is 7.6 time constants at 10 bits and 11.8 at 16. If that is longer than the converter's acquisition time, the sample is taken before the capacitor has finished charging and every reading is low — not noisy, low, and consistently so, which is a nasty bug because it looks like a calibration problem rather than a timing one. The calculator computes both numbers and reports the largest source resistance that would fit in the time available.

This is the origin of the “keep the source impedance under 10 kΩ” line in AVR and PIC datasheets. It is not a mystical rule; it is that inequality solved for R with those parts' defaults. Lengthen the acquisition time in your ADC configuration and the limit moves. Three ways out, in order of preference: lower the divider's resistance, lengthen the acquisition time, or put a capacitor of about 10 nF from the tap to ground so the sampling capacitor steals its charge from that instead of through the resistors — the last of these costs you bandwidth, so it suits a battery monitor and not a signal.

Two more ADC-specific traps. Input leakage current flows through R_th and appears as an offset; at the 1 µA some datasheets permit, a 100 kΩ source is 100 mV out. And on a part that multiplexes one converter across many pins, the sampling capacitor arrives still charged from the previous channel, so a high-impedance channel next to a very different voltage reads wrong on the first conversion — the usual fix is to sample twice and discard the first.

A divider is not a level shifter

Using a divider to bring a 5 V logic output down to 3.3 V works, is done constantly, and is worth understanding the limits of.

The static part is fine: 5 V through the right ratio gives a valid 3.3 V-family high level with margin. The dynamic part is where it fails. Your source resistance and the receiver's input capacitance plus the wiring form a low-pass filter, and the rise time is 2.2 R_th C. A 10 kΩ/20 kΩ divider — R_th = 6.7 kΩ — into 15 pF has a 220 ns rise time. That is fine for a UART at 115 kbaud and hopeless for SPI at 8 MHz, where a bit is 125 ns. Keep the rise time under about a fifth of a bit period, which usually means keeping R_th in the low hundreds of ohms, at which point you are burning milliamps continuously on every line.

Two situations where a divider is not merely slow but wrong. It is unidirectional, so it cannot be used on I²C or any other bidirectional bus — the standard answer there is the single-MOSFET bidirectional level shifter. And it does not translate upward, so a 3.3 V output driving a 5 V input that needs a true CMOS high level needs a real translator or a buffer, not a resistor.

When a divider is the wrong answer entirely

As a power supply. A divider cannot regulate. Its output moves with the input, with the load, and with temperature, and any current the load takes is current the divider must waste in order to be stiff. Powering anything through one is a mistake that ends in a hot resistor and a circuit that misbehaves the moment it does something. Use a regulator.

As a precision reference. The output tracks the input exactly, so a divider from a supply rail is only as good as the rail — which is typically ±2 % to ±5 %, dwarfing the resistors. If you need a reference, use a reference; if you need a fraction of a reference, divide the reference and buffer it.

Where the load is unknown or variable. Any circuit whose input impedance changes with state or temperature turns a fixed divider into a wandering one. Buffer the tap with an op-amp follower and the problem disappears: the follower presents megohms to the divider and ohms to the load, and it costs a few cents. This is the single most useful thing to do with an accurate divider.

FAQ

What resistors do I need to get 3.3 V from 5 V?

The ratio is 0.66, so any pair where R2 is roughly twice R1: 10 kΩ over 20 kΩ, or 1 kΩ over 2 kΩ. But if you are doing this to power something at 3.3 V, stop — see above. If you are doing it to bring a logic signal down, mind the rise time.

Does it matter which resistor is bigger?

R2 — the one from the tap to ground — sets the output as a fraction of the input, so a bigger R2 relative to R1 gives a higher output. Swapping them inverts the ratio.

Why is my measured voltage lower than the calculation?

Almost always loading. Enter your load in the calculator and the two numbers will agree. If you are measuring with a cheap multimeter on a high-impedance divider, the meter itself is the load: 10 MΩ is standard, but 1 MΩ meters exist and a 1 MΩ meter across a 1 MΩ divider reads badly wrong. An oscilloscope probe is 1 MΩ on ×1 and 10 MΩ on ×10, which is one of the real reasons to use ×10.

What power rating do the resistors need?

Take the dissipation this page reports and fit parts rated for at least twice it. Signal-level dividers are almost always inside a 1/16 W part; it is only dividers straight across a high-voltage rail that need thinking about, and there the voltage rating of the part matters too — a standard 0603 resistor is rated for 50 V and a divider across 400 V needs several in series regardless of the power.

How much current does a divider waste?

V_in divided by the sum of the two resistors, continuously, whether or not anything is connected. The calculator reports it, and also what it adds up to over a year — which is the number that matters for a battery-powered product, and the reason such products either use megohm dividers with a buffer or switch the divider off with a MOSFET between R2 and ground when they are not measuring.

Can I use a potentiometer instead?

Yes, and its source resistance is worst in the middle: a pot of total resistance R has R_th = R/4 at the halfway point. A 10 kΩ pot is therefore a 2.5 kΩ source at mid-travel — which is why a pot feeding a high-impedance input is fine and a pot feeding anything else usually wants a buffer after it.

Does the divider change if I put a capacitor across R2?

It adds a low-pass filter at 1/(2π R_th C), using the same source resistance as everything else here. Putting a matched capacitor across R1 as well makes it frequency-independent — the capacitive divider ratio matches the resistive one and the response is flat. That is a compensated divider, and it is exactly what the trimmer in an oscilloscope probe adjusts.

How does this relate to the other pages here?

The buffer this page keeps recommending is sized on the op-amp gain calculator, whose noise-gain reasoning is the same idea from the other side. The time constant and the −3 dB corner are the RC filter page's material, the E12/E24 rounding is the same code as the LED resistor page, and if you are building an attenuator for audio rather than a sensing divider, the decibel converter turns the ratio here into the dB figure you actually want.