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The formula I see everywhere is:

$$ I_{sat} = \frac{V_{bias}}{R_L} $$

But this seems incomplete for the following reason:

The larger the resistance is the smaller the Saturation Current, it stands to reason then that if it was zero, we would have the greatest current the device can physically produce, but if you see the short circuit current of any photodiode like this one (second table in the data sheet) it will be very low, micro Amperes in this case, instead of micro Amperes as one would expect.

How then can I calculate the highest saturation current of a photodiode and how is it different from the saturation current?

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  • \$\begingroup\$ Where specifically do you see that formula? \$\endgroup\$ – Andy aka Feb 1 at 12:58
  • \$\begingroup\$ For instance at chapter 2.3.5 of "Optoelectronic Sensors" by Didier Decoster and Joseph Harari imgur.com/a/7QB8O95 Also, in this Thorlabs guide thorlabs.com/images/TabImages/Photodetector_Lab.pdf \$\endgroup\$ – Fernando Franco Félix Feb 1 at 13:40
  • \$\begingroup\$ Your formula for Isat clearly doesn't apply when the photodiode is used in photovoltaic mode. \$\endgroup\$ – The Photon Feb 1 at 16:22
  • \$\begingroup\$ On what page of that thorlabs document did you see the formula in your question? \$\endgroup\$ – Andy aka Feb 1 at 17:17
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Saturation current is the current at which increasing optical power does not produce increasing current. Combined with the sensitivity curve, it establishes the maximum optical power which the PD can measure.

Short circuit current, on the other hand, is specified at a particular optical power - in the case of your link, 100 lux.

So, short circuit current provides an indication of how much current you'll get (for proper detection circuitry) at some specified optical power. Saturation current determines the maximum optical power you can measure.

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  • \$\begingroup\$ Just to make sure I'm not massively misunderstanding something, this is a completely different meaning of saturation current than that used in the Shockley diode equation, right? \$\endgroup\$ – Hearth Feb 1 at 19:13
  • \$\begingroup\$ @Hearth - Yes. See, for instance, thorlabs.com/images/TabImages/Photodetector_Lab.pdf \$\endgroup\$ – WhatRoughBeast Feb 1 at 19:17
  • \$\begingroup\$ Okay. Good to know I'm not going even further insane, then. \$\endgroup\$ – Hearth Feb 1 at 19:18
  • \$\begingroup\$ @Hearth - Saturation current in the Shockley sense simply does not apply to how PDs are used. A standard diode simply does not produce current in the absence of an applied voltage - but a PD does. So the Shockley diode does not apply (operationally - it does apply for a PD in the dark, but that's not how you use them). \$\endgroup\$ – WhatRoughBeast Feb 1 at 19:21
  • \$\begingroup\$ I don't know that much about photodiodes, so I wouldn't have been too surprised if that saturation current showed up somewhere in the equation for photocurrent. That said, now I know that it does not! \$\endgroup\$ – Hearth Feb 1 at 19:23

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