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I am reviewing a homework problem where we are given three LEDs; in the blue ($450$ nm), in the green ($550$ nm) and in the red ($650$ nm). We are told that the current efficiency $$\eta=\frac{Luminance}{Current \hspace{0.1cm}Density}=\frac{L}{J}$$ is the same for all three LEDs.

We now assume that they are connected in series and we need to determine the relation between the luminance of the green LED compared to the others. (as perceived by the average human eye)

I am very confused as to how it is possible for the efficiencies to be the same? Once they are connected in series the current flowing through the LEDs will be the same and their luminance depends on the respective wavelength. (i.e. the denominator is the same for all three LEDs whereas the numerator is different).

I am not sure what I am missing. Maybe my formula for current efficiency is incorrect?

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  • $\begingroup$ Start by providing us with the exact definition of "luminance" your class is using. Also, what exactly is "current density" here? If you use the standard definition, and the LEDs have different cross-sections, then their efficiencies can be the same but luminances differ. $\endgroup$
    – Carl Witthoft
    Commented Feb 4, 2021 at 13:45
  • $\begingroup$ @CarlWitthoft Thank you for your response! In my notes, it is mentioned that a "555nm monochromatic, 1W source, emits 683 Lumens" and that Luminance is measured in "Lumen/steradian/m^2" or "cd/m^2" i.e. luminous intensity per unit area (same definitions as in Wikipedia and most other sources I've looked at). I believe the definition for current density is the standard one as well i.e. J = I/ΔS where ΔS is a surface element. $\endgroup$
    – Jim Β
    Commented Feb 4, 2021 at 14:21
  • $\begingroup$ I should note that I am not given any information about the cross-sections. All the information that is given to me in, as part of the problem, has been included in the initial question. Maybe they just want me to mention that the luminance of the green LED will be greater and that's it? The amount of points for this problem are not a lot. $\endgroup$
    – Jim Β
    Commented Feb 4, 2021 at 14:24
  • $\begingroup$ Maybe it's simpler than that: if all the $\eta$ are the same, and by definition the current is the same, then the luminances of all three are the same. Now, you could work "backwards" to determine the relative photon fluxes, which will be different. In essence, the "photons per amp of electricity" will differ significantly . $\endgroup$
    – Carl Witthoft
    Commented Feb 5, 2021 at 14:31

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