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Full changelogEngine 1.18.3

LED Series Resistor

Dividers & Interfacing

Series resistor for a target LED current, and how far the current moves as Vf varies.

Inputs

V
Red 1.8–2.2 V, green 2.8–3.4 V, blue or white 2.9–3.6 V
V
Lowest bin, hottest temperature
V
Highest bin, coldest temperature
V
A
10 mA

Results

1 passed
Nearest E24
300Ω
Pass

Checked: headroom 2.9 V, 58 % of VS: 35 % or more is good, 20 to 35 % marginal, under 20 % needs a constant-current sink.
Show working
  1. Headroom across the resistorV_R = V_S − n × V_f = 5 − 1 × 2.1 = 2.9 V
  2. Ideal resistanceR = V_R / I_f = 2.9 V / 10 mA = 290 Ω
  3. Current with the nearest E24I_f = V_R / R = 2.9 V / 300 Ω = 9.667 mA
  4. Resistor dissipationP_R = I_f² × R = 9.67 mA² × 300 Ω = 28.03 mW
Actual current If

-3.3 % from the 10 mA target
9.667mA
Current spread over Vf range

± 9 %
8.67mA … 10.3 mA
Resistor dissipation PR

0402/0603 fine
28.03mW
All results (2)
Ideal resistance
290Ω
Efficiency (LED / total)

20.3 mW in the LEDs, 28.03 mW in the resistor
42%
Headroom is 58 % of the supply. The Vf range moves the current by 17 %.

Never share one resistor between LEDs in parallel: the lowest-Vf LED takes most of the current, heats up and takes more. Give each string its own resistor.

If over the Vf range

Accuracy

Verified against

1 independent case. See the working.

This is a design aid. The engineer remains responsible for the design and for checking the standard itself.

Parameter sweep

Vary one input over a range and see the answer and verdict at each step, as a table and a curve.Pro

Worst-case corners

Put a tolerance on each input and get the worst-case band around the answer.Pro

Circuit

The principle

An LED's current rises exponentially with its forward voltage, so it cannot be driven straight from a voltage rail. A series resistor sets the current. The more voltage the resistor drops, the less the current depends on Vf.

Series resistor
R=VS−nVfIfR = \frac{V_S - n V_f}{I_f}
Resulting current
If=VS−nVfRI_f = \frac{V_S - n V_f}{R}
Sensitivity to V_f
ΔIfIf=−n ΔVfVS−nVf=−n ΔVfVR\frac{\Delta I_f}{I_f} = \frac{-n\,\Delta V_f}{V_S - n V_f} = \frac{-n\,\Delta V_f}{V_R}
A Vf change is divided by the headroom VR: small headroom, large current swing.
Dissipation
PR=If2RPLED=nVfIfP_R = I_f^{2} R \qquad P_{LED} = n V_f I_f
  • VSV_Ssupply voltage
  • VfV_fLED forward voltage at If
  • nnnumber of LEDs in series
  • VRV_Rvoltage across the resistor (the headroom)
  • IfI_fforward current
More detail

Why Vf varies

  • Bins. One reel can span 300–400 mV.
  • Temperature. Vf falls about 2 mV/°C per LED, so the current rises as the LED warms.
  • Current. The datasheet Vf is quoted at one If; at 1 mA it can be 200–300 mV lower than at 20 mA.
Design rule. Keep at least 35 % of the supply across the resistor. From 20 to 35 % it suits an indicator; below 20 %, and for power LEDs (>100 mA), use a constant-current driver.
Related: Junction Temperature & Safe Limits, for the thermal path of a power LED.

Engine version ⁨1.18.3⁩