I was recently spec-ing some RGB LEDs for a project, when I noticed that the millicandela ratings on the three colors are rarely close to the same number. (i.e. 710mcd Red, 1250mcd Green, 240mcd Blue).

Does this cancel out somehow, or does this mean that the LED will always look yellowish?

Also, why do manufacturers make such unbalanced LEDs? Wouldn't it make more sense to pair up 3 LEDs of approximately the same brightness?

Example: CLY6D-FKC-CK1N1D1BB7D3D3 made by Cree

  • 5
    \$\begingroup\$ Sounds about right. To get white (6500K) using NTSC (colour TV) phosphors, the relative intensities are G=0.58, R=0.31,B=0.11 - most of the energy is in the green, least in the blue. At equal intensity, blue would appear brightest. The actual numbers will differ here (LEDs not phosphors) but the relative intensities are actually more similar than I expected.. \$\endgroup\$ Nov 28, 2015 at 19:35
  • 2
    \$\begingroup\$ @BrianDrummond Isn't the brightness in mcd already human eye luminosity-function weighted, so 100mcd should look similar brightness regardless of color? \$\endgroup\$ Nov 28, 2015 at 19:38
  • 1
    \$\begingroup\$ @Spehro ... the update got too long for a comment and turned into an answer \$\endgroup\$ Nov 28, 2015 at 20:15
  • \$\begingroup\$ Yes, interesting. Thanks. I'm going to think about this.. it's been a while.. the spectrum of solar radiation (almost black body) is not 'white' - it peaks at green-yellow yet we perceive it as white. \$\endgroup\$ Nov 28, 2015 at 20:27
  • 1
    \$\begingroup\$ Solar spectrum : en.wikipedia.org/wiki/Sun#/media/… : varies by about 20% across the visible range. Flat enough not to matter here. \$\endgroup\$ Nov 28, 2015 at 21:52

4 Answers 4


Sounds about right. To get white (6500K) using NTSC (colour TV) phosphors, the relative intensities are G=0.59, R=0.3, B=0.11 - most of the energy is in the green, least in the blue. (slightly differently rounded numbers in Wikipedia ) At equal intensity, blue would appear brightest. The actual numbers will differ here (LEDs not phosphors) but the relative intensities are actually more similar than I expected.

Spehro's interesting comment goes some way to explaining why. The Candela is a definition of luminous intensity that is weighted such that 100mcd of red, green, or blue light are perceived as equally bright.

Now as I understand the colour space conversion process - it doesn't follow from that, that mixing equal perceived intensities of R, G, B will result in what we see as white!

Indeed how can it? Our eyes are most sensitive to green. So the actual intensity of the green light is reduced in the definition of the Candela to give the same perceived intensity as red, blue (Nitpick : I believe the other intensities are increased instead). Then, to mix the three and make white, we need to increase the perceived intensity of green light to restore the correct intensity in the mixed light. (That is why the measured intensity must be greatest at the wavelength where our eyes are most sensitive. That wouldn't make sense otherwise!)

In other words, 100mcd each of red, green and blue contains much less actual energy on the green channel, whereas true white light would contain approximately equal energy in each channel - hence the definition of "white noise" in electronics.

EDIT : An interesting article places the quantum efficiencies of red and blue LEDs in the 70-80% region, well above that of (previous to 2008) green LEDs (it's a sales pitch, after all!). This makes it likely that, whatever the reason for the low intensity of the blue LEDs, it isn't that they are difficult to make.

So the relative intensities of the three LEDs in the question is the manufacturer's attempt to undo this weighting and match the LEDs so that the light generated is approximately white at rated current.

Illustration (image source) enter image description here
To my eyes at least, in the illustration above, G is by far the brightest primary, with R second and B darkest, yet when mixed, they produce a pretty good white.

  • 4
    \$\begingroup\$ After all these years, it is never the same color... \$\endgroup\$
    – PlasmaHH
    Nov 28, 2015 at 20:18
  • 2
    \$\begingroup\$ I thought NTSC (vs. PAL vs. Secam) was only an encoding scheme for the TV signal in transmission, but are these schemes actually aimed at different phosphors? \$\endgroup\$ Nov 29, 2015 at 16:07
  • \$\begingroup\$ Well, looks like I can buy a couple of them, and test them to be sure. What you say sounds reasonable, but if it turns out to be otherwise, I'll be back. \$\endgroup\$
    – Tustique
    Nov 29, 2015 at 16:28
  • 2
    \$\begingroup\$ I believe PAL and SECAM simply used the NTSC phosphors, and thus shared the RGB-YUV conversion matrix, though other parts of the encoding differed. My notes from Wood Norton (prepared in 1968 by J.R.Kirkus) called them theNTSC ptosphors though the course was PAL-oriented. \$\endgroup\$ Nov 29, 2015 at 16:49
  • \$\begingroup\$ I had to read paragraphs 2--4 several times before it clicked. "But you just said ...", no actually you didn't! I was reading one thing and interpreting another. So, please confirm this. What you are saying is: White light is actually equal actual intensity R, G and B (and of course all the others), but these diodes are weighted to show equal perceived intensity at maximum intensity. \$\endgroup\$
    – clacke
    Nov 29, 2015 at 20:44

I do not claim the other answers are wrong, but they miss two important points. One of them that I consider the most relevant.

RGB-LEDs are not meant to produce white light. They are meant to reach a certain gamut Wikipedia on gamut, i.e. the colour space that can be displayed by the LED. And they do. If the three channels are driven with an 8-bit resolution probably only less than 1% of all possible settings will yield a light mixture on the Planckian locus. Wikipedia on planckian locus, where white light can be found. So one can guess, white light is not the primary objective for a RGB LED.

The gamut is an outcome of the use case analysis a manufacturer is performing. In most cases, the use case demands high output for signal colours like red, green and yellow but only limited power when producing white light.

Even if the use case is covering the omnipresent RGB LED-strips it is neither necessary nor possible to hit the Planckian locus when driving all LED at 100%. The human eye tolerates many MacAdam-ellipses away from the Planckian locus when it has no good light source to compare and even more when the owner of the eye got the LEDs at a bargain price.

As I wrote in my comment, the die size of the three colours is usually equal which leads to a nearly equal electrical and thermal power rating for all three chips. This and the limited bandwidth of the current available epitaxial process finally prevents manufacturers to "please everybody". Hence it is extremely unlikely to get a RGB-device which hits the Planckian locus when driven at 100%. On top of that, even if there was a RGB-chip with that property, it would fail to produce the same result at an ambient temperature just 20° higher.

There's one more fact to consider if white light is desired at 100% current for all LEDs. Colour LED each produce a narrow spectrum around their so called dominant wavelength \$λ_{dom}\$. For them to imitate a white spectrum together they either have to have adjacent spectral humps or produce more light, if their dominant wavelength is far from the adjacient LEDs. For RGB, the green one is in fact in a long gap between R and B. So the output power must be increased to generate the same tristimulus as daylight. This means the green LED bears the main load in providing the flux for a light that appears white. The eye thanks to its metameric properties is rather forgiving concerning the actual "form" of the spectrum.

The outrageously abysmal colour rendering of RGB-generated white is another story.

  • 2
    \$\begingroup\$ +1. I didn't want to get into details of colorimetry and it's certainly true that the result isn't even close to white once you look critically; however I think you'd agree that it's approximately balanced while an "equal measured intensity" solution is not. A couple more points : while "ability to produce white" and appropriate gamut are not the same thing, they are somewhat related. Also good point about the narrow bands of each LED. I've been expecting higher-fi LEDs with maybe 6 or 7 dice, e.g. red/amber/yellow/green/cyan/blue/violet but it hasn't happened yet. \$\endgroup\$ Nov 29, 2015 at 11:30
  • 1
    \$\begingroup\$ BTW I think this answer could be improved with some references to the more obscure terms, or a good introductory text. \$\endgroup\$ Nov 29, 2015 at 11:30
  • \$\begingroup\$ @BrianDrummond Higher-Fi LEDs with more dies make no sense, because there is no benefit for putting them into a common housing. The explanation will be too long for this comment, so we may make a good Q&A from it? \$\endgroup\$
    – Ariser
    Nov 29, 2015 at 11:38

LEDs of different colors are made with quite different materials and processes and designs. There's no guarantee that they'll turn out to be the same brightness. It makes more sense to put more efficient LEDs in there when they are available rather than degrade the more efficient ones in order to match the least efficient color. Sure they will have to run at different currents (or duty cycles) to get a white balance, but that's not a big deal.

  • \$\begingroup\$ If process difficulties made (say) the blue LED less efficient, why wouldn't they just put a bigger blue die in the package to compensate? \$\endgroup\$ Nov 28, 2015 at 20:44
  • 2
    \$\begingroup\$ @BrianDrummond then the OP would complain the three LEDs don't have the same current rating (not even nearly so): blue would require 5 times as much current compared to green. \$\endgroup\$ Nov 28, 2015 at 21:32
  • 1
    \$\begingroup\$ That's already the case for the linked example LED : R=15ma, G=B=10ma. \$\endgroup\$ Nov 28, 2015 at 21:36
  • \$\begingroup\$ So, it would make sense to either put in a blue one, or smaller red and green ones. Assuming that the answer below were incorrect, and this balance does not result in white, then you have to cut the current to the LEDs to the point where you are effectively wasting a majority of the LED's capabilities. \$\endgroup\$
    – Tustique
    Nov 28, 2015 at 21:59
  • \$\begingroup\$ I've seen ones that have two dies of the less efficient types. \$\endgroup\$ Nov 29, 2015 at 1:23

If you pay close attention to the specs, you will notice that the mcd ratings are given with approximately equal power (30mw) applied to each LED. assuming that our eye will see "white" when the three colors have the same luminosity, one way to achieve this would be to reduce the brightness of the red and green LEDs and increase the brightness of the blue LED. Assuming that the brightness is proportional to the current, I would reduce the green LED current to 5ma, the red LED to 8.8ma, and the blue would be increased to 26ma. This would make each LED provide approximately 625 mcd. Of course, this assumes the blue LED can handle 26 ma, if not, the currents would have to be proportionally reduced based on the max current the blue LED can handle.

The answer your main question, is simply the manufacturing and price constraints. For your second question... no, it does not have to look yellowish, it just depends on the accuracy with which you balance the currents to the LEDs (and background brightness). For the third question, the answer is similar to first case, optimizing the manufacturing process dictates equal die size, deposition process, etc.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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