I want to detect a laser using a normal photodiode (Datasheet), but I am not sure whether it is robust enough, as laser (lower than 5mW) emits high energy photons.

Can I shine on such a photo diode constantly with a laser (<5mW) without being afraid of damaging the sensor? What would happen, if the light would have to much energy? Would it just stop working or could it even burn?

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    \$\begingroup\$ What kind of laser are you dealing with? Photon energy: E=hc/lambda. I doubt that a 5 mW laser, even UV (lambda ~= 350 nm) will damage your photodiode. You don't have a gamma ray laser, do you? :) \$\endgroup\$ – uint128_t Dec 29 '16 at 21:05
  • \$\begingroup\$ Right:). I have a red laser. \$\endgroup\$ – black Dec 29 '16 at 21:06
  • \$\begingroup\$ The device you linked to is designed to detect visible wavelengths. It won't have any trouble with a red laser. But why use this device (which has 3 different sensors to distinguish colors) if you only need to detect a signal from a monochromatic source? \$\endgroup\$ – The Photon Dec 29 '16 at 21:36
  • \$\begingroup\$ Response will also depend on whether you focus the light into a small enough spot to all fall on the (red part of the) sensor...which could be challenging if you're on a budget. \$\endgroup\$ – The Photon Dec 29 '16 at 21:44
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    \$\begingroup\$ What parameters do you want to measure in your design spec? Power , Luminous Intensity, Turn on/off time? What tolerance? What \$\lambda\$ ? This query is too vague. I have some RGB lasers >100mW I measure with a diffuse large area sensor. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 29 '16 at 21:54

The photons from a red laser aren't any more energetic than those from a flashlight. Red light corresponds to a photon energy of 2 eV, which isn't much in the grand scheme of things.

Ultimately, the limiting factor with most photodiodes is simply thermal dissipation. In your case, you are dissipating < 5 mW on an area that is perhaps 4 mm^2, and the temperature rise will be negligible. The photon energy doesn't much matter for visible (and IR/UV) light; all that matters is the power density and thermal dissipation of the target (the photodiode).

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No it won't burn but you could be heavily saturating any amplifier following the photodiode.

In the photodiode data sheet it shows a graph of output current for an input light power density of 10 mW per square cm and I would suggest that any higher than this value will be causing saturation of any signal amplifier you might use. It's just something to be aware of.

Your laser may pack all the light (5 mW) into 1 square mm and that means 500 mW per square cm equivalent and way-above the graphs in the photodiode data sheet.

So, I'm advising to design your photodiode amplifier with some care.

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  • \$\begingroup\$ Ah, okay, deleting my comment. \$\endgroup\$ – The Photon Dec 30 '16 at 16:52
  • \$\begingroup\$ Me too but happy new year anyway! \$\endgroup\$ – Andy aka Dec 30 '16 at 16:53

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