SizingKitElectronics · resistor networks
Voltage divider calculator
Enter a supply and two resistors and this returns the output node’s voltage, the source resistance the next stage sees, the current the pair burns standing still, and how far a connected load pulls the output down. Give it a wanted output instead and it searches every E12, E24 or E96 pair within three decades for the ratio that lands closest. It is free, needs no account, and ranks every pair in the series while you type.
- 100% free
- No signup
- E12 / E24 / E96 stock
- Loading and tolerance
- Prints its sources
Two resistors you already have
R1 sits between the supply and the output node, R2 between that node and the return. Swapping them inverts the ratio, which is the mistake that produces an output larger than the input.
The rail the top of R1 connects to.
The upper leg. Raising it lowers the output.
The lower leg. The output is measured across it.
Input resistance of whatever reads the output. Leave blank for open circuit.
Default 1% — IEC 60063 pairs a series with a tolerance band: E12 with ±10%, E24 with ±5%, E96 with ±1%. Read the tolerance off the part or its reel label. Thin-film parts reach ±0.1% and ±0.05%; a matched divider network is specified on the ratio rather than on either resistor, and beats both.
OPEN CIRCUIT
3.8367 V
WITH THE LOAD CONNECTED
3.7179 V
| Division ratio, R2 ÷ (R1 + R2) | 0.319728 |
|---|---|
| Source resistance seen by the load, R1 ‖ R2 | 3.2 kΩ |
| Shift the load caused | -118.8705 mV (-3.10%) |
| Standing current through the pair | 816.3 µA |
| Burned in R1 / in R2 / in total | 6.66 mW / 3.13 mW / 9.8 mW |
| Output across the ±1% band | 3.7847 V to 3.8891 V (±1.36%) |
| Most power this node can deliver, at a 50% sag | 1.15 mW |
Two resistors at ±1% do not give a ratio at ±2%. To first order the ratio moves by 2t × (1 − k), where k is the division ratio: a divide-by-two is as good as one resistor, and only a divider that barely divides at all approaches the full doubled figure. The table above works both ends exactly rather than to first order, so the two numbers differ slightly at wide tolerances.
Or name the output and shop for the pair
Every combination of two stock values within three decades of each other, ranked by how close the ratio lands, then by how near the sum sits to the total you asked for.
Must be below the supply — this circuit only divides.
24 values per decade.
Lower stiffens the node and burns more; higher does the reverse.
| R1 | R2 | Output | Off by | Standing current |
|---|---|---|---|---|
| 24 kΩ | 9.1 kΩ | 3.2991 V | 0.027% | 362.5 µA |
| 18 kΩ | 6.8 kΩ | 3.2903 V | 0.293% | 483.9 µA |
| 4.7 kΩ | 1.8 kΩ | 3.3231 V | 0.699% | 1.846 mA |
| 20 kΩ | 7.5 kΩ | 3.2727 V | 0.826% | 436.4 µA |
| 3.9 kΩ | 1.5 kΩ | 3.3333 V | 1.010% | 2.222 mA |
| 15 kΩ | 5.6 kΩ | 3.2621 V | 1.147% | 582.5 µA |
| 22 kΩ | 8.2 kΩ | 3.2583 V | 1.264% | 397.4 µA |
| 4.3 kΩ | 1.6 kΩ | 3.2542 V | 1.387% | 2.034 mA |
Ranked against IEC 60063, series E24 — the twenty-four two-digit values a 5% resistor is made in. Two stock values reach almost any ratio because the pair has two decades of freedom, so the limit on a divider is rarely the ratio — it is the tolerance band above and the load below.
How to size a divider that still reads right once something is connected to it
The identity takes ten seconds. The two corrections around it are what decide whether the number on the meter matches the number on the schematic.
Fix the scale before the ratio
Type the total you want R1 and R2 to add up to. That single choice sets the standing current, the heat and the source resistance all at once — 1 kΩ total on 12 V burns 144 mW and holds the node stiff, while 1 MΩ burns 144 µW and lets a multimeter probe move it. Nothing about the ratio changes when you slide the total up or down a decade.
Enter what will be reading the output
Put the input resistance of the next stage in the load field. A CMOS ADC pin is gigaohms and moves nothing; a 10 MΩ multimeter across two 1 MΩ resistors costs you 4.8%; a 1 MΩ input on that same pair costs 33%. The shift is printed in millivolts as well as percent, because a bias of 40 mV matters to a 12-bit converter and is invisible as a percentage.
Read the tolerance band, not just the nominal
Set the tolerance to the one printed on the parts you will actually solder. The plate then shows the pair of outputs the two extremes produce, and that band — not the six-decimal nominal — is what your firmware has to tolerate or calibrate out.
Technical specifications
| Search space | E12 (12 values), E24 (24) and E96 (96) per decade, paired across ±3 decades of relative scale: 4,032 combinations for E24 and 64,512 for E96, re-ranked on every keystroke |
|---|---|
| E24 is transcribed, E96 is generated | Eight of the 24 E24 values sit off round(10^(n/24)) — 26 became 27, 29 became 30, 83 became 82 — so E24 ships as a table. E96 rounds exactly, so it is built from its definition in one line of code |
| Ratio tolerance rule | Worst case ≈ 2t × (1 − k) to first order; the tool works both extremes exactly. A halving divider from ±1% parts is ±1.0%, a 10:1 divider ±1.8% |
| Loading model | Thevenin: source resistance R1 ‖ R2, output falls by R_L ÷ (R_L + R_th). Exact for a resistive load at DC |
| Maximum deliverable power | V_th² ÷ 4R_th, reached when the load equals the source resistance — at which point the output has already fallen by half |
| Input range | 0.1 Ω to 1 GΩ per leg, 1 µV to 1 MV supply, loads to 1 TΩ. Units accepted as typed: 4700, 4.7k, 2M2 as 2.2M |
| What it does not model | Capacitance at the node, the source impedance of the supply, resistor self-heating and the temperature coefficient of the parts — a divider drifts with all four and this page is DC and isothermal |
| Privacy | No network request is made after the page loads; resistor values and supply rails stay in the tab |
Frequently asked questions
Why is the output lower than the calculator said once I connected my circuit?
Because your circuit is a resistor too, in parallel with R2. The divider behaves as a source of R1 ‖ R2 — 10 kΩ over 4.7 kΩ presents 3.2 kΩ — and anything drawing current through that resistance loses voltage across it. Put the next stage's input resistance in the load field and the tool prints the exact shift. The fix is either a lower total resistance or a unity-gain buffer between the node and the load.
Can I run an LED, a relay or a 3.3 V regulator from a divider?
No. A divider is a reference, not a supply: the most power it can hand to a load is V_th² ÷ 4R_th, and it can only do that with the output already sagged to half. A 12 V rail through two 10 kΩ resistors offers 1.8 mW at 3 V instead of 6 V. Anything that draws real current wants a regulator, and an LED wants its own series resistor sized on the current rather than on the voltage.
Do two 1% resistors give a 2% output?
Only in the limit where the divider barely divides. The ratio error is about 2t × (1 − k), so a divide-by-two from ±1% parts is ±1%, a 3:1 divider is ±1.33%, and a 10:1 divider is ±1.8%. Both resistors moving the same way partially cancels in the ratio, which is why matched divider networks are specified on ratio tolerance and beat two discrete parts of the same grade.
What total resistance should I pick?
Low enough that the load's effect is acceptable, high enough that the heat and the standing current are. The old shop rule — make the divider current at least ten times the load current — is just the loading formula in disguise, and it is worth replacing with the number itself: this page prints the actual millivolt shift, so pick a total, read the error, and move a decade if you do not like it.
Which of the two resistors is R1?
R1 is the upper leg, between the supply and the output node; R2 is the lower leg, between that node and the return, and the output is the voltage across R2. Swapping them gives 1 − k rather than k, which is the source of most reports that a divider produced more than its input. The tool refuses a target above the supply for the same reason.
How do I divide an AC signal rather than a DC rail?
The same ratio holds for a sine wave as long as the stray capacitance at the node is negligible at that frequency, which stops being true above roughly 10 kHz on a megohm divider. Beyond that the node has a pole and the divider becomes frequency-dependent — put the R and the node capacitance into the impedance page to see where the corner lands, or compensate the divider with a capacitor across R1 the way a scope probe does.
The pair search returned resistors I cannot buy. What happened?
It returned values from the series you selected, scaled to sit near your total — so an E96 result may be a 1% part your local supplier stocks only on reel. Switch the series to E24 or E12 and the list falls back to values that appear in every assortment box, at the cost of a slightly larger ratio error, which is printed for each row so the trade is visible.
About resistive dividers, Thevenin equivalents and the E-series
The divider is the first circuit anyone learns and the one most often used past its competence. Two resistors in series across a supply put the fraction R2 ÷ (R1 + R2) of that supply at the node between them; the algebra is exact and never the problem. The problem is that the same two resistors also form a source with an internal resistance of R1 ‖ R2, and the moment anything conducts at the node, that internal resistance starts eating the answer. Thevenin’s theorem is the tidy way to say it: to everything downstream, the divider is a battery of k·V_in behind a resistor of R1 ‖ R2, and it can never be anything better.
That equivalence is what turns the design into a trade-off with no free direction. Dropping the total resistance stiffens the node and shrinks the loading error, and it raises the standing current and the dissipation in proportion; raising the total does the reverse, and above a few hundred kilohms leakage, humidity across the board and probe capacitance start to matter. Neither end is wrong — a 100 Ω divider feeding a comparator and a 10 MΩ divider on a battery monitor are both correct answers to different questions. What is wrong is choosing the total without looking at either consequence, which is why this page prints the current and the dissipation next to the voltage rather than behind a link. If the current itself is the thing you are chasing, the Ohm’s law calculator works the plain V, I, R and P identities in every direction.
The stock values come from IEC 60063, which fixes how many values a decade of resistors is divided into and pairs each count with a tolerance: twelve values for ±10%, twenty-four for ±5%, ninety-six for ±1%. Every series from E48 up rounds exactly to 10^(n/N), so this page generates E96 rather than typing it. E24 is the exception — eight of its values were nudged off the geometric rounding to numbers the trade already used, which is why 27 kΩ exists and 26 kΩ does not, and it is stored as a sourced table for that reason. The resistor color code calculator sets out the series and the bands that encode them, and tells you when a value you read off a part is not a member of any series — usually a sign that a band was read in the wrong order. When the divider is feeding something reactive rather than resistive, the node stops being a plain ratio: the impedance calculator gives it magnitude and phase at a frequency, and the RC time constant calculator gives the settling time R1 ‖ R2 produces against the capacitance hanging on the node.
Where these resistor values are worked out
Every number on this page is worked out by JavaScript running in the tab you are reading it in. Nothing you type — loads, lengths, nameplate ratings, the rates your utility charges you — is uploaded, logged or kept, which is also why the calculators carry on working in a mechanical room with no signal.
The pair search runs 64,512 combinations for E96 in the tab, on every keystroke, with no round trip. That is also why it can afford to rank every pair rather than offering one resistor and rounding the other.