Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here's some nice LED's packaged in strips which can be cut apart and used for projects. Voltage is up to 12 volts, and many people are using 3S 11.1 lipo cells to power them.

Some people comment that they are too bright, or that different colors (e.g. red and green) do not match in intensity.

Can I solder a resistor in series to reduce the current flow to match the brighness? Will that improve the battery life? If I get a dial resistor, can I dial in the desired brightness and then measure the amperage flowing through the circuit to determine the needed resistance?


share|improve this question
up vote 10 down vote accepted

LED brightness varies linearly with the (average) current. If you double the resistor, that will roughly halve the brightness assuming the supply voltage is much greater than the LED voltage, if they are closer, the effect will be more significant.

A more efficient method (but complex, and likely impossible to do in an existing circuit) would be to switch the LEDs on and off, adjusting the duty cycle to adjust the average current.

Matching different LEDs in brightness will require good knowledge of the specifications and a good source that will provide the type and grade specified, not just "a bright green LED". The candela rating is weighted by the eye sensitivity, so making them match should make them appear equal at 0 degrees. Slightly different viewing angles can make the LEDs look very different as your viewer goes off-axis.


Matching a green 30 mA, 500 mcd, 3.5 Vf LED and a red, 15 mA, 3000 mcd, 1.8 Vf LED running on a 9 V source with a straight resistor ballast.

The desired current through the red LED needs to be reduced to match the green, so it should be by a factor of 500 mcd/3000 mcd = 1/6, 15 mA * 1/6 = 2.5 mA. The red ballast resistor should be (9 V (supply) - 1.8 V (red LED Vf))/2.5 mA = 2880 Ω, and the green resistor should be (9 V (supply) - 3.5 V (green LED Vf))/30 mA = 183 Ω

This assumes Vf will not change drastically with If (should be reasonably constant in a 3-30 mA range, but consult graphs in datasheet), the viewing angles are identical, and as mentioned before, that the current is proportional to the intensity (almost always is down to sub-milliamp values, but check datasheet)

share|improve this answer
Thanks, it's very helpful to see this worked out! – Mark Harrison Jan 26 '11 at 22:59

You could use a constant-current supply to set the current yourself. There are well-known ones based off of LM317. Here's another one I sketched up in answer to another question:

This also has the advantage of making the brightness independent of supply variation.

share|improve this answer
Better use an LM358, it can go down to the lower rail. Add a resistor to limit the base current. – starblue Jan 26 '11 at 20:30
LM324 also can go down to the lower rail. – markrages Jan 26 '11 at 20:33
Thanks, that's very useful! – Mark Harrison Jan 26 '11 at 22:58
Can you link to your answer to the other question that you drew this diag for? – BG100 Jan 27 '11 at 7:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.