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R = (Vs − n·Vf) / If · IEC 60063

LED Resistor Calculator — Series Resistor Sizing

Series resistor for an LED — the value you can actually buy, the current it really gives, and the wattage to order.

The circuit

Resistor series

Typical forward voltages

Red1.8 – 2.2 V
Yellow / orange2.0 – 2.2 V
Green2.0 – 3.2 V
Blue / white / UV2.8 – 3.6 V

Forward voltage follows the semiconductor bandgap, so it tracks colour closely — but the spread within a colour is wide enough that these are a sanity check on a datasheet, never a substitute for one.

Resistor — nearest E24 value
360 Ω
Calculated 350 Ω, rounded up to the next E24 value. Up rather than nearest, so the current stays at or below the 20 mA you asked for.
Recommended power rating
0.5W
The resistor dissipates 136 mW; this is the smallest standard rating with the usual 2× headroom. A quarter-watt part is fine here only if it is genuinely rated 272 mW or more.
Voltage across the LEDs2.00V
Voltage across the resistor7.00V
Calculated resistance350.0Ω
Actual current at that value19.44mA
Off the target current-2.8%
Resistor dissipation136.1mW
How the value was reached

(9.0 V − 2.0 V) ÷ 20 mA = 350.0 Ω
next E24 value up = 360 Ω
I = 7.00 V ÷ 360 Ω = 19.44 mA
→ P = I²R = 136.1 mW, fit a 0.5 W part

The resistor drops 78 % of the supply, so most of the power in this circuit becomes heat rather than light. Putting more LEDs in series — or using a lower supply — is markedly more efficient if you have the choice.

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One division, and then the part you can actually buy

The resistor takes whatever the LED does not, at the current the LED wants:

R = (Vs − n × Vf) / If

An LED is a diode, not a resistor. Its forward voltage barely moves as current changes, so nothing in the LED limits the current — past the knee, a tenth of a volt more across it can double the current through it. Connect one straight across a supply and it draws whatever the supply will give until something fails. The resistor is what turns a voltage source into something the diode can live on.

Round up, never to nearest

The division almost never lands on a value anybody makes. 350 Ω is not a stocked part; 330 and 360 are. Rounding up is the rule here, and it is not a style preference: rounding down raises the current above the figure the datasheet permits, and the datasheet figure is the entire reason there is a resistor in the circuit. Rounding up costs a fraction of a milliamp of brightness that no eye can detect.

Everything downstream is then recomputed from the value you will actually fit rather than the one the division produced — the real current, the real dissipation, the wattage to order. A calculator that reports 350 Ω and 20.0 mA is describing a circuit that will not exist.

Two times the power, not one

A resistor at its nameplate wattage is within specification and also running hot enough to discolour a board over a few years. The rating assumes free air at 25 °C, which the inside of an enclosure is not. The trade builds to twice the calculated figure, and that is what the recommendation here uses — advice, not a code rule.

Series, not parallel

Several LEDs in series share one resistor quite happily: the same current flows through all of them, and that current is what sets brightness. Add up the forward voltages, subtract once, divide once.

LEDs in parallel on a single resistor is the classic mistake. Forward voltages never match exactly, even between two parts from the same reel. The one with the lowest Vf takes the largest share of the current, runs hottest, and its forward voltage falls further as it heats — which draws yet more current. It fails, and the survivors then split its current between them. Give every parallel string its own resistor.

Where this stops being the right circuit

A series resistor burns the difference between the supply and the LED as heat. On a 12 V rail feeding one 2 V LED that is 83 % of the power wasted, which is fine for an indicator and absurd for lighting. Once you are past a few hundred milliwatts, or the supply voltage is well above the LED, use a constant-current driver — it regulates the current directly instead of dropping voltage across a heater, and it holds that current as the supply sags or the LED warms.

The resistor also does nothing about temperature. Forward voltage falls roughly 2 mV per °C, so a fixed resistor delivers more current to a hot LED than a cold one. For an indicator this does not matter. For anything driven hard, it is why constant-current drive exists.

Reading the value you land on

The answer comes out as a preferred value from the E12 or E24 series, which is what the colour code can express — ask for 350 Ω and no set of bands will code it. To check the part in your hand matches, decode the bands there; to work out the voltage across the resistor or the power in it from first principles, the Ohm's law solver takes any two of V, I, R and P.

Forward voltage is a range, not a number

The calculation takes a single value for the forward voltage, and the datasheet gives a typical figure with a minimum and a maximum around it. The spread is wider than most people expect — a white LED specified as 3.2 V typical is commonly guaranteed only between 2.8 and 3.6 V.

On a supply well above the LED voltage that hardly matters. Driving one white LED from 12 V, an 0.8 V spread moves the current by less than 10 per cent. Driving the same LED from 3.7 V it is decisive: the resistor is dropping perhaps 0.5 V, so an LED at the high end of its range gets almost no current and one at the low end gets far too much. The closer the supply is to the forward voltage, the more the circuit is controlled by part tolerance rather than by your resistor.

Forward voltage also falls as the junction warms, by roughly 2 to 4 millivolts per degree. In a series resistor circuit that is self-correcting in the wrong direction — a warmer LED drops less voltage, so the resistor sees more, so the current rises, so it gets warmer still. With a healthy voltage across the resistor the loop settles harmlessly. With very little across it, thermal runaway is a real failure mode, and it is why high-power LEDs are driven by constant current sources rather than resistors.

A battery is not a fixed supply

Sizing from the nominal voltage of a battery gives a circuit that is correct for one brief moment in the discharge. A single lithium cell labelled 3.7 V leaves the charger at 4.2 V and is empty around 3.0 V. A fresh alkaline AA is 1.6 V rather than 1.5 and finishes near 0.9.

Across a three-cell alkaline pack that is a supply moving from 4.8 V down to 2.7 V. A resistor chosen at the nominal 4.5 V will overdrive the LED when the cells are new and leave it visibly dim long before they are actually flat, and the useful life of the pack ends well above the point where the energy runs out.

Size the resistor at the maximum supply voltage so the LED is never overdriven, and check what the current becomes at the minimum to see whether the brightness is still acceptable. If the answer is no across the range you need, the circuit has outgrown a resistor — a small constant-current driver holds brightness flat over the whole discharge and extends usable battery life at the same time.

Questions people ask

What series resistor does an LED need on a 9 V supply?
For a red LED at 2 V forward and 20 mA: (9 − 2) / 0.02 = 350 Ω, which nobody stocks — so round up to the next E24 value, 360 Ω. That gives 19.44 mA rather than 20, which no eye can detect, and dissipates 136 mW, so fit a 0.5 W part. Note that the resistor is dropping 78 per cent of the supply here: most of the power in this circuit becomes heat rather than light.
Should I round the resistor to the nearest value or up?
Always up, and it is not a style preference. Rounding down raises the current above the figure the datasheet permits, and that figure is the entire reason there is a resistor in the circuit. Rounding up costs a fraction of a milliamp of brightness nobody can see. Everything downstream should then be recomputed from the value you will actually fit — a calculator that reports 350 Ω and 20.0 mA is describing a circuit that will not exist.
Why can't I connect an LED straight across the supply?
Because an LED is a diode, not a resistor, and nothing in it limits current. Its forward voltage barely moves as current changes, so past the knee a tenth of a volt more across it can double the current through it. Connected straight to a supply it draws whatever the supply will give until something fails. The resistor is what turns a voltage source into something the diode can live on.
Can several LEDs share one resistor?
In series, yes — the same current flows through all of them, so add up the forward voltages, subtract once and divide once. In parallel, no, and it is the classic mistake. Forward voltages never match exactly even between two parts off the same reel, so the one with the lowest Vf takes the largest share of the current, runs hottest, and its forward voltage falls further as it heats, drawing yet more. It fails, and the survivors then split its current between them. Give every parallel string its own resistor.
When should I use a constant-current driver instead of a resistor?
Once you are past a few hundred milliwatts, or the supply is well above the LED voltage. A series resistor burns the difference as heat — on a 12 V rail feeding one 2 V LED that is 83 per cent of the power wasted, which is fine for an indicator and absurd for lighting. The resistor also does nothing about temperature: forward voltage falls roughly 2 mV per °C, so a fixed resistor delivers more current to a hot LED than a cold one. A driver regulates the current directly and holds it as the supply sags or the LED warms.