Audited 29 Jul 2026·Last updated 28 Oct 2026·6 citations·Tier 1·0 uses

LED Resistor Calculator (Series and Parallel Arrays)

Find the series resistor for one LED or a string: exact ohms, the E24 value to buy, the real current, resistor wattage, efficiency and a headroom check.

LED Resistor Calculator

The rail feeding the LED and its resistor. Use the real voltage, not the nominal one — a 'nine volt' battery is 9.5 V fresh and under 7 V flat, and the LED current follows it.
V
From your LED's datasheet, at the current you intend to run. It is not a fixed number: it depends on the die chemistry, so it varies by colour, and it rises with current and falls with temperature. Kingbright specify it per die and quote a ±0.1 V tolerance on the figure.
V
In milliamps. 20 mA is the usual test and rating point for a 5 mm indicator LED, but check the absolute maximum on your part — exceeding it causes light output to degrade and the part to fail early, not just to run bright.
mA
A whole number from 1 to 100. In series they share one resistor and one current; in parallel each gets its own resistor.
How are they wired?
Resistor needed
145
The exact resistance, in ohms, that gives your target current: the voltage left after the LEDs divided by the current.
Stock resistor to use
150 Ω
Current you actually get
19.3333 mA
Power in each resistor
0.0561 W
Resistor power rating to buy
1/8 W — each resistor dissipates 0.05607 W, and this is the smallest stocked rating that covers it with the 2x headroom this page assumes. That assumes a room-temperature ambient: a resistor's rating is quoted at a stated temperature (Vishay rate theirs at 70 °C) and falls away above it, so use the resistor wattage page if this sits in a hot enclosure.
Power in the LEDs
0.0406 W
Total drawn from the supply
0.0967 W
Fraction reaching the LEDs
42.00
Design check
Use 150 Ω: the next E24 value at or above the calculated 145 Ω. Rounding up is deliberate — it lowers the current to 19.33 mA, 3.3 % under your 20 mA target, which is the safe direction for an LED. Generous headroom: the resistor drops 58 % of the supply, so more than half the power is being turned into heat in the resistor rather than light in the LED. Electrically safe and very repeatable, but wasteful — on battery power, put more LEDs in series or use a lower supply.

Background.

A red indicator LED rated at 2.1 V and 20 mA, run from a 5 V supply, needs (5 − 2.1) ÷ 0.02 = 145 Ω in series with it. Nobody sells 145 Ω, so you fit the next standard value up, 150 Ω, and the LED runs at 19.33 mA instead of 20: a shade under its rating, which is where you want it. What follows the arithmetic decides whether the circuit keeps working once it leaves the bench.

The resistor is there because an LED cannot limit its own current. A resistor's current rises in proportion to the voltage across it; a diode's rises exponentially. In an ideal diode at room temperature each extra 60 mV or so multiplies the current about tenfold, and real LEDs, though less steep, behave the same way in kind. Connected straight to a supply, an LED has nothing to stop the current climbing until something fails. With a resistor in series, the resistor takes up the difference between the supply and the LED's forward voltage, and Ohm's law fixes the current.

For one LED or a string of up to 100, you get the exact resistance, the E24 resistor to buy, the current it really delivers, the heat in each part, a power rating with margin, the share of the supply that reaches the LEDs, and a headroom verdict on whether the design will behave the same with the next batch of parts. Two wirings are offered: one series string behind a single resistor, or parallel LEDs each with a resistor of their own. Several parallel LEDs sharing one resistor is left out on purpose, because the LED with the lowest forward voltage takes most of the current.

The model is a DC indicator circuit on a steady supply at room temperature. Power LEDs drawing hundreds of milliamps belong on a constant-current driver, and currents above 5 A are refused for that reason.

What is led resistor calculator?

A current-limiting resistor is a resistor wired in series with an LED so that the resistor, not the diode, sets the current. The loop obeys Kirchhoff's voltage law: the supply voltage equals the LED's forward voltage plus the voltage across the resistor. Because the forward voltage changes only a little over the useful current range, the resistor's share is nearly fixed, and the current is that share divided by the resistance.

Forward voltage is the voltage an LED drops while conducting at a given current. It is set mostly by the semiconductor's band gap, the same property that sets the colour of the light, so red LEDs sit lower than green, blue and white ones. It also rises with current, falls as the die warms and varies between parts of the same type. Kingbright, for example, quote a ±0.1 V tolerance on the forward voltages in their datasheets and specify different values for the red, green and blue dice inside a single multi-colour lamp. That is why forward voltage is an input to read from your own part's datasheet, at the current you plan to use, rather than a number assumed for a colour.

How to use this calculator.

  1. Enter the Supply voltage the circuit really sees, at its highest. A car's 12 V system runs at around 14 V with the engine on, and a 9 V battery starts above 9 V and sags as it discharges.
  2. Enter the LED forward voltage (Vf) from the datasheet at your chosen current. If the datasheet gives typical and maximum figures, start with typical and re-run with the extremes.
  3. Enter Forward current you want in milliamps. 20 mA is the usual rating point for a 5 mm indicator, but the absolute maximum on your part is the ceiling, and modern high-efficiency indicators are often bright enough at a few milliamps.
  4. Set How many LEDs? and How are they wired?: in series, one string behind one resistor; in parallel, every LED with its own resistor.
  5. Fit the Stock resistor to use rather than the exact value, and check Current you actually get.
  6. Read the Design check and the power-rating line. Fix tight headroom before anything else, and remember that a parallel array needs one resistor of that rating per LED.

The formula.

series: R = (V_s − n·V_f)/I_f parallel: R = (V_s − V_f)/I_f, one per LED

Around a series loop the voltage drops add up to the supply. With n LEDs in one string the LEDs take n × V_f between them and the resistor takes the rest, so V_R = V_s − n·V_f and R = (V_s − n·V_f) ÷ I_f. In the parallel wiring each LED sits in its own loop with its own resistor, so n drops out: R = (V_s − V_f) ÷ I_f for each one. The LED drop must stay below the supply. As n·V_f approaches V_s the required resistance falls towards zero, and once they meet the string cannot light at all, so that case is reported as an error instead of producing a negative resistance.

The exact value is then rounded up to the next value in the IEC 60063 E24 series, the 24 preferred values per decade sold as ±5% parts: 10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82 and 91, times a power of ten. Rounding up, never to the nearest, keeps the current at or below your target. The real current is recomputed with the stock part as I = V_R ÷ R_stock.

Power follows from the same loop. Each resistor dissipates V_R × I. The LEDs together take n × V_f × I. The supply delivers V_s times the total current, which is I in series and n × I in parallel, and the resistor and LED figures add back to it exactly. The share reaching the LEDs reduces to n·V_f ÷ V_s in series and V_f ÷ V_s in parallel: the current cancels, so efficiency depends only on how much of the supply voltage the LEDs occupy. The suggested rating is the smallest common one — 1/16, 1/10, 1/8, 1/4 or 1/2 W, then 1, 2, 3, 5 or 10 W — that is at least twice the dissipation, assuming a room-temperature ambient.

The headroom verdict sorts the resistor's share of the supply, V_R ÷ V_s, into four bands: under 10% very tight, 10–20% tight, 20–50% comfortable, and 50% or more generous but wasteful. The bands reflect sensitivity. A shift ΔV_f in each LED changes the current by the fraction n·ΔV_f ÷ V_R, so the smaller the resistor's voltage, the more the LEDs rather than the resistor decide the current. The calculation holds V_f at the value you enter; in reality it creeps up with current and down with temperature, which is why margin matters.

A worked example.

Example

Three red LEDs, each specified at 2.1 V and 20 mA, are wired in one string on a 12 V supply. Together they drop 3 × 2.1 = 6.3 V, leaving 12 − 6.3 = 5.7 V for the resistor, and 5.7 ÷ 0.020 = 285 Ω gives exactly 20 mA. 285 Ω is not an E24 value. The next one up is 300 Ω, which gives 5.7 ÷ 300 = 0.019 A, or 19 mA — 5% under target and indistinguishable to the eye. The nearest value below, 270 Ω, would have pushed 21.1 mA through LEDs rated at 20. The resistor dissipates 5.7 × 0.019 = 0.1083 W. Doubling that for margin gives 0.2166 W, so a 1/4 W part is the smallest common rating that covers it. The LEDs take 3 × 2.1 × 0.019 = 0.1197 W, and the supply delivers 12 × 0.019 = 0.228 W, the two added together. The share reaching the LEDs is 0.1197 ÷ 0.228 = 52.5%, the same as 6.3 ÷ 12. The resistor drops 5.7 ÷ 12 = 47.5% of the supply, inside the comfortable band. If every LED in the string came in 0.1 V above nominal, the string would drop 6.6 V, the resistor would see 5.4 V and the current would ease to 18 mA: a modest change, because the resistor rather than the LEDs is setting the current.

forward Current Ma20
led Count3
forward Voltage2.1
supply Voltage12
wiringseries

Frequently asked questions.

What resistor does a single LED need on a 5 V supply?
It depends on the LED's forward voltage. A red LED at 2.1 V and 20 mA needs 145 Ω, so fit 150 Ω, which gives 19.33 mA and dissipates about 0.056 W; a 1/8 W resistor covers that twice over. A white LED at 3.2 V leaves only 1.8 V for the resistor, so it needs 90 Ω; fit 91 Ω for 19.78 mA. In both cases an ordinary 1/4 W through-hole resistor is more than enough.
Can several LEDs share one resistor?
Only in series. A string carries one current through every LED, so one resistor sets it for all of them. In parallel, no: LEDs of the same type still differ slightly in forward voltage, and on an exponential curve the one with the lowest voltage takes the largest share of the current. It then runs hotter, its forward voltage falls further and it takes more still. The usual outcome is uneven brightness, and sometimes one failed LED. Give each parallel LED its own resistor.
Should I wire my LEDs in series or in parallel?
In series whenever the supply allows it. Three red LEDs on 12 V in series need one 300 Ω resistor and draw 0.228 W in total. Wired in parallel, each needs its own 510 Ω resistor dissipating 0.1922 W, enough to call for a 1/2 W part, and the array draws 0.6988 W, with only 17.5% reaching the LEDs. Series strings also match in brightness, since every LED carries the same current. Parallel strings make sense once one string can no longer fit under the supply voltage.
How many LEDs can go in one string?
As many as leave the resistor at least about a fifth of the supply. On 12 V with 2.1 V red LEDs, four drop 8.4 V and leave 3.6 V, 30% of the supply, for a 180 Ω resistor; five drop 10.5 V and leave 1.5 V, 12.5%, which is rated tight. With 3.2 V white LEDs, three drop 9.6 V and leave 2.4 V, right on the 20% line, and a fourth needs 12.8 V, more than the supply, so the string will not light.
Why is a resistor a poor way to run a white LED from 3.3 V?
Because almost nothing is left for the resistor. A 3.2 V white LED on a 3.3 V rail leaves 0.1 V, which calls for 5.1 Ω and gives 19.61 mA on paper, with the resistor dropping 3% of the supply. If the actual part comes in at 3.0 V, that same resistor sees 0.3 V and would allow about 58.8 mA, ignoring the rise in forward voltage at higher current. The current is being set by the LED's tolerance, not the resistor. Use a 5 V rail or a constant-current driver.
What voltage should I design for in a car?
The highest one the circuit will see. A nominal 12 V system runs at around 14.4 V while the alternator is charging. Three red LEDs sized at 12 V get a 300 Ω resistor, but at 14.4 V that resistor passes (14.4 − 6.3) ÷ 300 = 27 mA, well over a 20 mA rating. Sized at 14.4 V the exact value is 405 Ω and the stock part 430 Ω, which gives 18.84 mA when charging and about 13.26 mA with the engine off, slightly dimmer but always within rating.
What power rating does the resistor need, and why twice the dissipation?
A resistor's rating is the most it can shed at a stated ambient temperature, and running it near that limit makes it hot. Vishay, for example, rate their MRS and CRCW families at 70 °C, with a 155 °C limit on the resistive film, and the permitted dissipation falls as the surroundings warm. Buying at least twice the calculated dissipation keeps the part cool on an open board. In a closed or hot enclosure, check the derating with the resistor wattage calculator, and in a parallel array remember that every LED's resistor carries that load.
Why is my LED dimmer than I expected?
Usually not because of the resistor. A few percent less current than the target makes no visible difference, since perceived brightness changes far more slowly than current. More likely causes are viewing angle, as a narrow-beam LED looks dim off-axis and glaring on-axis; the spread in luminous intensity between parts, which Kingbright quote at ±15%; and a supply lower than the one you entered. Check the actual current figure first, then the part.
Is a resistor good enough for a high-power LED?
Not at hundreds of milliamps and up. The resistor's wasted power becomes a large slice of the total, and the temperature behaviour works against you: as the die warms its forward voltage falls, the resistor's share of the voltage rises, the current goes up and the die warms further. A resistor slows that loop without breaking it. Power LEDs are run from constant-current drivers, which hold the current steady whatever the forward voltage and temperature do, and forward currents above 5 A are not accepted here.
How do I size the resistor for a PWM-dimmed LED or an AC supply?
For PWM, size the resistance for the current while the LED is switched on, the peak, because that is what flows during each pulse; dimming comes from the duty cycle, not from a larger resistor. The average heat in the resistor falls with the duty cycle, so the power rating can be smaller than the peak suggests. On AC an LED needs rectification, and usually protection against reverse voltage, because LEDs tolerate only a small reverse voltage. This calculation covers the DC case only.

References& sources.

  1. [1]Kingbright, "WP154A4SUREQBFZGW T-1 3/4 (5mm) Full Color LED Lamp" datasheet. Read directly from the PDF on 2026-07-29. Confirms the design decisions on this page: electrical and optical characteristics and absolute maximum ratings are both specified at TA = 25 °C; forward voltage carries a stated tolerance of ±0.1 V; a single package contains hyper red, green and blue dice with separately specified characteristics — which is why forward voltage is an input here rather than a constant; and the datasheet warns that "excess driving current and / or operating temperature higher than recommended conditions may result in severe light degradation or premature failure", the reason the stock resistor is always rounded up. Luminous intensity is quoted with a ±15 % spread.
  2. [2]OpenStax (Rice University), "University Physics Volume 2", §10.2 "Resistors in Series and Parallel". Equation 10.2 and the common-current constraint behind it are what make a series LED string tractable: every element carries the identical current and the voltage drops add, which is the Kirchhoff's-voltage-law statement this page's formula is derived from. Retrieved and read 2026-07-29.
  3. [3]IEC 60063:2015, "Preferred number series for resistors and capacitors", third edition, 27 March 2015. Defines the E24 series of twenty-four values per decade, the ±5 % set this page rounds up into when choosing a stock resistor. Retrieved 2026-07-29 via the reference summary of the series values and their standard tolerances.
  4. [4]Vishay Draloric / Beyschlag / BCcomponents, "Resistor Color Card", document VMN-MS6212-1501 (2015). Read directly from the PDF on 2026-07-29. Prints the E24 preferred-value table under the heading "According to IEC 60063"; the twenty-four mantissas used by this calculator's round-up were transcribed from it.
  5. [5]Vishay BCcomponents, "MRS16, MRS25 Professional Thin Film Leaded Resistors", document 28724, revision 07-Mar-16. Read directly from the PDF on 2026-07-29. Source of the caveat attached to the power-rating recommendation: rated dissipation is quoted as P70 — 0.4 W and 0.6 W at 70 °C ambient — with a peak permissible film temperature of 155 °C, and "the rated dissipation applies only if the permitted film temperature is not exceeded".
  6. [6]Vishay, "D/CRCW e3 Standard Thick Film Chip Resistors", document 20035, revision 14-Apr-2026. Read directly from the PDF on 2026-07-29. Confirms the same P70 rating convention and 155 °C film limit on surface-mount parts, and notes that the temperature rise depends on "the thermal resistance of the assembled resistor together with the printed circuit board" — which is why the rating on this page is qualified as a room-temperature figure.

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Method
series: R = (V_s − n·V_f)/I_f parallel: R = (V_s − V_f)/I_f, one per LED
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