How can I find the right resistor to make 50% brightness in an led? I know how to find the right resistor to get the brightest but I'm wanting to find the perfect middle.

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    \$\begingroup\$ Somewhat rhetorical question: what does "50% brightness" mean? Do you mean half as many photons? That won't look half as bright. Do you mean looks half as bright? Then we'd need to know the ambient light level to know what "dark" is, in order to pick a perceived brightness halfway between that and full brightness. \$\endgroup\$ – Phil Frost Jul 11 '14 at 18:00

Chances are that for an perceived "half as bright" LED you'll need to increase the resistor by around (5:1), which will yield around 20% of the lumens.

Compare the popular North American auto tail light bulb the 1157, which has two filaments- brake and running light. The filaments are about 28W and about 8W, but the lumens are about 10:1 (402 vs. 38 lumens). Yet the brake lights are not perceived as ten times brighter- maybe 3 times.

This is because your eye has approximately a logarithmic response over some range, so in fact a difference of 20% or 25% in lumens is barely perceived.

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    \$\begingroup\$ That is such a succinct and great answer and yet reveals so much information. You explain physics of electricity transformed to light and then the biology of how light is perceived. Thanks for a great answer. \$\endgroup\$ – raddevus Jul 11 '14 at 12:37

You will never be able to get 50%...the properties of resistors and other semiconductors might change with temperature and every resistor has a tolerance of how much not exact is it's resistance.

Good LEDs have in their datasheet a drop like graph that shows the spread of light, so the brightness also depends on the angle that you look at it.

In any case, the datasheet should show you the relationship between lumens and current and you can get some estimative about it:

Get the maximum current the LED can handle and with it you get the maximum lumen, than you can calculate easily the resistor, assuming half this maximum current, using V = R*I.

Remember the relationship will not be that much linear, and only a good LED will give you all the parameters to calculate half the brightness, but in practice, assuming half the brightness for half the maximum current is usually a good approximation.

Also get in mind that the human eye doesn't respond to all colors in the same way, some colors will appear brighter than others for the same lumens, that's why there are several color correction curves and calibrations for light related equipments.

  • \$\begingroup\$ If the LED's output is specified in lumens, that already accounts for the spectral response of the human eye. At least for a laboratory standard eye. \$\endgroup\$ – The Photon Jul 11 '14 at 5:16
  • \$\begingroup\$ @ThePhoton it is called Photopic vision, which lumens are related to. Humans actually have more than one vision mode, and swap between them at night and day. \$\endgroup\$ – KyranF Jul 11 '14 at 10:50
  • \$\begingroup\$ I am not a specialist in "light" science, I have studied about colors and human perception some years a go, so I might not be 100% correct in here, I do know humans have a logarithmic perception of light and different for Red Green and Blue since the eye Cone cells respond differently to them (I am not going to point out Rod cells since they are more related to "night vision" than "color vision")...now how the lumen unit relate to that I am not 100% sure... \$\endgroup\$ – mFeinstein Jul 12 '14 at 23:07

If you want 50% brightness, you will probably find it easier to use PWM, with a 50% duty cycle.

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    \$\begingroup\$ I don't know....picking a different resistor is pretty damn easy. Also a 50% duty cycle results in half the irradiance from the LED, but that doesn't equate to half the perceived brightness due to the logarithmic perception of light. \$\endgroup\$ – Phil Frost Jul 11 '14 at 18:03
  • \$\begingroup\$ Good point about the Weber–Fechner law. getting a different resistor is easy, but getting one, or a combination of, that is exactly correct would be hard. \$\endgroup\$ – BanksySan Jul 12 '14 at 14:13
  • \$\begingroup\$ For that matter, achieving exactly 50% duty cycle is hard, too. There's temperature variation, clock jitter... (and there's the bigger problem that 50% duty cycle still doesn't get you 50% brightness) \$\endgroup\$ – Phil Frost Jul 12 '14 at 14:37
  • \$\begingroup\$ i didn't want to point out PWM because it appeared to me that the Question was really related to basic electronics, using just regular resistors....if would prefer PWM so this way the current can be tuned to 50% light perception to later find out the Resistor...but I am not sure the OP has this kind of electronics knowledge yet... \$\endgroup\$ – mFeinstein Jul 12 '14 at 23:09
  • \$\begingroup\$ @PhilFrost Would those things be negligible compared to picking a resistor? I might be wrong there. \$\endgroup\$ – BanksySan Jul 13 '14 at 12:40

Most LED's will have a datasheet showing you two important graphs: the forward current vs luminous intensity. Plus another graph showing the forward voltage veruss forward current.

An example is this datasheet: http://www.farnell.com/datasheets/1626761.pdf at the bottom of page 2.

You can see that 100% intensity is at 30mA, and 50% at 15mA.

Then looking at the other graph, 15mA responds to 1.8V as opposed to 30mA which is around 1.9V.

Knowing the forward voltage voltage and current, you can now easily calculate the resistor required. In this example at 5V supply, 100% intensity is: (5-1.9)/0.030 = 103 ohms, and for 50% intensity: (5-1.8)/0.015 = 213.33ohm.

Generally speaking the relativionship between luminous intensity and current is always linear. While the effect of forward voltage on current is little. So a general rule of thumb is you just half the current to half the luminous intensity. Using that rule of the thumb you would have calculated half brightness requiring: 5-1.9/0.015 = 206.66 ohm.

Of course brightness is a subjective property. 50% luminus intensity may not appear 'half as bright' to an eye.


How about hooking it up to a variable resistor (pot) and playing with it until you get the desired brightness? Then you could measure the pot's resistance and get/build a 1% or 5% resistor (even 10% might do) to replicate that resistance.

If it's going to be a fixed resistance, play with it under various ambient lighting conditions to make sure a given brightness is adequate under all conditions.

  • \$\begingroup\$ This is exactly how I'd do it. You could do a bunch of calculations (if you even have the appropriate datasheet -- I suspect most indicator LEDs are in a jar without even the part number known) but this is way simpler and accounts for all the non-linearity in our perception and the LED. \$\endgroup\$ – Phil Frost Jul 11 '14 at 18:06

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