How would you make a RGB LED equally bright to a broad-spectrum white LED?

Question

Say you had a broad spectrum white LED (all visible wavelengths) and you wanted to make a RGB LED (just red-green-blue wavelengths) that could produce white light of equal brightness to the broad-spec one.

Would you need the wavelengths in the RGB LED to be of a higher intensity than the wavelengths in the broad-spec LED to make up for the missing wavelengths (violet, yellow, etc)?

Fewer wavelengths means a less bright light? So because you have fewer wavelengths you need to increase their intensity?

Answer 1

There are many interesting aspects to this question. I will explain the spectrum of a broad white LED and then I will explain the spectrum of a RGB white LED. But before that, it is important to understand how the eye perceives colour. It is not as simple as 3 distinct wavelengths seen by 3 cones.

Colour vision:

The retina of the eye has three different cones, red green and blue.

This image$^1$ shows the sensitivity of different cones to different wavelengths of light. Different colours are seen due to different levels of stimulation of the three different cones. The colour white is seen when all three cones are equally stimulated. This does not require all wavelengths of visible light to be present. Instead a simple combination of yellow and blue light will stimulate all three cones equally. In fact this is the technique used to create a broad spectrum white LED.

Broad spectrum white LED:

This is the common spectrum$^2$ of a white LED. It is a combination of yellow and blue light. This combination appears as white. While white LEDs can be made with a combination of many wavelengths this is the most commonly used type.

RGB white LED:

In an RGB white LED a combination of three wavelengths are used to stimulate the cones equally.

This is the spectrum$^3$ of a RGB white LED.

Now coming to your question, the luminosity of the entire spectrum can be obtained integrating the spectral luminosity over the entire range of wavelengths. For equal brightness luminosity must be equal. So for the two described LEDs, for equal luminosity the spectral luminosity of the wavelengths in the RGB LED must be greater since it encompasses fewer wavelengths.

Note:

1.In order to have equal luminosity the spectral radiant intensity of the red and blue wavelengths must be greater than the green wavelength due to the difference in sensitivity of eye to different colours.

2.As pointed out by @Ruslan

Presence of a hue in white is extremely subjective. It depends on chromatic adaptation, and that, in its turn, depends on individual variation, in particular, age (which results in yellowing of eye lens) and amount and distribution of macular pigmentation. Actually, one might even disagree with himself on this question depending on whether he looks at the object with fovea or more eccentric part of the retina.

However as it is not possible to account for all possible variations, the luminosity function which is the average of such variations has been used to answer this question.

Sources:

1. Cone sensitivity to different wavelengths

2. Broad spectrum white LED

3. RGB white LED