High power photodiodes

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What do you need a high current photodiode for? Can't you just attenuate the input light, or amplify the output?

If you live in the USA (this seller don't ship outside USA, apparently), probably this one is the better deal you can find (it's not cheap, but considering the commercial prices of these units, it's probably still a deal).

This other seller have similar units, and cheap too, but those are apparently NOT standard PIN10D units, and they may have spectral responses and electrical characteristics different from standard products ..... if you need somenting with a certified response, the first one is the better, if instead you only want one for do experiments, the second one can be good too (also if, in this case, you need to derive the spectral response yourself with experiments, and this is not an easy thing to do, usually)

Would a small photovoltaic cell work instead?

kukurio said:

Thanks for the tips, HIMNL9. How can you tell that those are high power? Are you basically assuming that high area -> high power? .....

Click to expand...

Normally, large sensitive area diodes are made for laser / power measurement, so they can take more power than normal / small photodiodes ..... also, can't say for the second one, but the first one is a standard PIN10D model, so you can see in the datasheet its characteristics HERE or HERE (sorry, i know that datasheetarchive is a site made in a very wrong way, but sometimes can be found something helpful also there ) ..... you can see the characteristics of the diodes, that one is the serie PIN10D(P) with the BNC, with 100mm^2 of sensitive area .....

Due to its structure, it can hold more power than a small size diode (defocused, ofcourse ..... it burn like any other photodiode, if you shoot it with a focused beam ). ..... anyway, if instead is the output current, that you need high, it's always better that you use an op-amp as output stage ..... photodiodes are not photocells, regardless the surface, they cannot give you too high currents ..... so, for a direct high current output, is better to use a small photovoltaic cell, instead a photodiode, also if the photodiode is more precise.

As far as I know a photovoltaic cell and a photodiode are the same in the working principle, only the construction differs. What would make the difference between the two in maximum current?
But anyway, above a few mA you can just attenuate the light source, much easier. The noise level of a good photodiode is extremely low, your electrical measurement system will be much more noisier than the shot noise limit. HP makes optical sensors capable of detecting 10 femtowatt, which is the best I've seen so far, but it will be nearly impossible to measure 10mA with around 10fA accuracy, that's an SNR of 120dB. And the shot noise limit is probably not noticable at such high powers, my guess is a 10mW laser has less than 1pW shot noise.

The optical shot noise is sqrt(N), with N the number of photons, which scales linearly with power, so the noise goes with sqrt(P). The scaling of the optical power to current is linear, so P ~ I. The current shot noise by the optical signal is then sqrt(I). This scales equal to the current shot noise, which also scales with sqrt(I). The noise vs noise will then always be equal. At 10mA, the current noise will be 57pA, much larger than the optical shot noise in the signal.

That why people can't count photons in all but the lowest powers, and at the lowest powers they use devices with gain, like PMT's or APD, to avoid the current shot noise. But above all the resistor noise of the current measurement system will be large.