I am trying to estimate photocurrent from a photodiode in a general sense. When I read literature on the subject, I see something along the lines of: enter image description here

Text and equation from Here

I assume they are referring to spectral flux from what they describe as "energy spectrum of incident light" due to similar units.

Photodiodes typically report a responsivity graph and LEDs report relative spectral emissions. How would I calculate spectral flux (if that is the correct unit)?

The spectral emission and responsivity do not include distance from the emitter to detector, but if I move my LED closer to a photodiode then I expect to see a larger photocurrent, but I have not been able to find an example showing this. Any help would be appreciated.

  • \$\begingroup\$ Some sensors (with a photodiode inside) are calibrated for the amount of lux for the light spectrum that equals the human eye. For example the TSL2561. With just a photodiode it is not possible to determine the lux, since there are leak-currents and voltage and temperature dependencies. You could make a conversion table and calibrate the photodiode yourself, or you could buy a TSL2561. \$\endgroup\$ – Jot Feb 16 '17 at 11:40

You mostly just need to count the number of photons hitting the photodiode (that have an energy larger than the band gap.. (check the data sheet.)) It's not perfect but you can assume that each photon makes one electron hole pair, so one electrons worth of current. Hold your LED closer to the PD and you get more photons -> more current.

  • \$\begingroup\$ No, it is definitely not true that all it takes is \$E>E_{gap}\$. Look e.g. at spectral sensitivity graph in Fig. 7 in this datasheet. It is far from not depending on wavelength (for \$E>E_{gap}\$). Spectral sensitivity goes even down to 0 for shorter wavelength (= higher photon energy). The reason is that higher energy photons just go through the sensitive semiconductor material without getting absorbed. Also: the formula explicitely states that \$A(..)\$ is a function of \$\lambda\$. \$\endgroup\$ – Curd Feb 16 '17 at 16:55
  • \$\begingroup\$ @Curd, That curve gives photocurrent vs incident power, not photocurrent vs photon flux. If you account for the fact that photon energy depends on wavelength, the curve for current vs photon flux will be very flat over a wide range. \$\endgroup\$ – The Photon Feb 16 '17 at 17:42
  • \$\begingroup\$ @The Photon: No, that doesn't help: just multiply the the graph by \$1/\lambda\$ but that won't make it flat: in the range 300nm to 900nm it will just cause a factor of 3 while the sensitivity varies by more than a factor of 10. And for still shorter wavelength it goes down to 0 much more than \$1/\lambda\$ goes up. \$\endgroup\$ – Curd Feb 16 '17 at 17:48
  • \$\begingroup\$ @Curd, Over the range of emission of an LED, it will be close enough for many engineering calculations. \$\endgroup\$ – The Photon Feb 16 '17 at 17:52
  • \$\begingroup\$ @The Photon: !?!? So if the answer doesn't fit the problem you change the problem? ;-) Of course a LED is more or less "monochromatic" (not compared to a laser but compared to a light bulb or daylight)...but that is not the given situation here. The given formula explicitely considers the fact that there are many different wavelength for which \$A(..)\$ varies. That's the reason why there is an integral. \$\endgroup\$ – Curd Feb 16 '17 at 17:58

if I move my LED closer to a photodiode then I expect to see a larger photocurrent,

This is accounted for when they said the photodiode response depends on the incident light.

Only light that actually reaches the photodiode from the source is considered "incident" on the photodiode.

To calculate what fraction of the source's output light actually reaches the photodiode is in general a nontrivial problem, depending on what optics (lenses, apertures, diffraction gratings...) are placed between the two devices, what reflective paths allow light to reach the photodiode, etc. You'd typically use a ray-tracing program like Zemax to solve this.

If you just want to handle the case where the source and detector are placed facing each other in free space, with no other objects present, then you can figure out what solid angle the detector subtends when seen from the source, and use the angular emission profile of the source (which will be provided for higher-priced LEDs) to determine what fraction of the source power is emitted into that solid angle.


How would I calculate spectral flux (if that is the correct unit)?

It depends on your type of source.
A light bulb could be approximated by a black body and you can use Planck's Law to calculate it.
If it is the sun or sky you have to use some measured spectral distribution or model function.

The spectral emission and responsivity do not include distance from the emitter to detector

Not in all cases there is a distance dependency: imagine your photodiode was a hollow sphere enclosing the source. In that case there is no distance dependency. All the enegy radiated by the source would be swallowed by the photodiode, no matter how close the source is to the diode.

Of course this is just a hypothetical configuration; I mentioned it to make clear that in general intensity depends on the whole setup (not only on the distance). In most cases you have distance dependency because source and detector appear under a smaller solid angle if they are more distant.
You can use inverse square law if distance compared to size of source and detector is large.

The \$F(\lambda)\$ in your formula already contains the factor depending on the particular setup (i.e. even if sources are the same \$F(\lambda)\$'s of two different setups differ by a constant (i.e. not dependent on \$\lambda\$) factor).


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