# where do i find max short circuit current of photodiode?

I'm trying to convert the current signal from a photodiode to voltage through a transimpedence amplifier.

The photodiode is put in photovoltaic mode in order to get the greatest accuracy and lowest possible dark current, which means that the cathode is connected to the invetering input of the amplifier.

Furthermore, in order to get a greater amplification and avoid "minus" voltage on the output, I intend to raise the voltage on the non-inverting input.

I'm now realizing that my photodiode only states a "typical" short circuit current at 100lx, but no maximum current.

http://www.farnell.com/datasheets/8765.pdf?_ga=2.139995315.1420390384.1543406751-57595970.1536135582

I intend to do this in a 5v circuit, and ideally ≈5v from V_Out should resemble the maximum current from the photodiode, and obviously vise-versa ≈0v the lowest.

Since I don't have a minimum and maximum output current of the photodiode in photovoltaic mode, I have a hard time understanding how i should determine the values for hence the offset voltage and the resistance for the feedback resistor?

• If solar input is > 100 klux , I wonder if junction max safe thermal rise affects the max Isc from internal self heating (I^2R) + radiant heat absoprtion.. – Tony Stewart Sunnyskyguy EE75 Dec 1 '18 at 20:00
• Are we still talking about your pulse oximeter project? Where does solar input come into it? – The Photon Dec 1 '18 at 20:22
• yes we are. I guess, that i have not talked about the solar input? – Jeppe Christensen Dec 1 '18 at 20:23
• You should try to make this question as self-contained as possible. I don't remember every detail from your earlier questions, and new readers likely won't go back and read your earlier questions to understand what you're asking. – The Photon Dec 1 '18 at 20:25
• Also for a pulse oximeter application, you should really be thinking about radiant flux (W) rather than illuminance (lux). IR interference will affect you just as much as visible light, but won't be measured by the illuminance. – The Photon Dec 1 '18 at 20:29

The photocurrent arises when a photon excites an electron-hole pair in the depletion region of the diode. So the photocurrent is generally given by

$$I = \frac{\eta e \lambda}{hc}P$$

where $$\e\$$ is the fundamental charge, $$\hc/\lambda\$$ is the photon energy, $$\P\$$ is the incident optical power, and $$\\eta\$$ is the fraction of incident photons that actually produce carrier pairs, called the quantum efficiency.

The quantum efficiency can't be higher than 1 (unless you want to talk about avalanche detectors) so from this you can find a maximum photocurrent for a given incident power.

For 850 nm, this comes out to about 0.68 A/W, or about 20% higher than the typical responsivity given in your datasheet.

• Yes and a typical Sharp 5mm IR PD with 0.5mA/mW would require an LED with 0.5mW/mA to achieve unity hFE gain in an optocoupler, which never happens for efficiency reasons. A Solar Panel, Isc, is usually done at max solar intensity of 1.? kw/m^2 or ?? kLux – Tony Stewart Sunnyskyguy EE75 Dec 1 '18 at 19:52
• hmm, that sounds rather complex. What is the formula called, just so i can check out some practical examples? – Jeppe Christensen Dec 1 '18 at 19:56
• If using solar input take W/m2 * effective area and rated A/W using 1kW/m2 then results indicate efficiency So choose S (A/W) based on your desired spectral average – Tony Stewart Sunnyskyguy EE75 Dec 1 '18 at 20:02
• I suspect that 1000 lux would be the maximum light that the photodiode is going to be exposed to... Couldn't i just then multiply the typical short circuit current at 100lx from the datasheet with 10 like so: 6.3µ*10 and then conclude that my maximum current would be 63µA ?? – Jeppe Christensen Dec 1 '18 at 20:13
• Yes that is in the linear range but in full sunlight 100klux 6.3mA might be pushing it. What tolerances specs, do you need? ( wide range? high speed? linear log output, CIE eye corrected? – Tony Stewart Sunnyskyguy EE75 Dec 1 '18 at 20:17