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I saw a diagram of the photosensitivity (Current per Power) of a photodiode. Photodiode sensitivity

So there is this diagonal stating the 100% quantum efficiency. I wondered why the sensitivity for bluer light lower than that of redder light.

Is it because a blue photon has a higher energy? Thus fewer blue photons are needed to achieve the same power as with red photons? The photocurrent is proportional to the number of converted photons. So a 1 watt beam in red has more photons, which generate more photo-electrons, thus more current per watt.

Is this reasoning correct and explains the curve in the diagram?

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1 Answer 1

I don't know the physics of this device, but let's think it through.

Did you notice the units of the vertical scale? Its per Watt, not per photon, but the interaction occurs on a per-photon basis.

For a given incidence power, the number of photons is inversely proportional to the wavelength, which leads to a conjecture: what if each photon over threshold releases the same number of electrons. Then we expect $$ \text{current} \propto \frac{\text{# of photons}}{\text{second}} = \frac{\text{power}}{\text{energy per photon}} \quad.$$

Rearranging that and noting that photon energy goes by the inverse wavelength we get $$ \frac{\text{current}}{\text{power}} \propto \text{wavelength} \quad, $$ which is in reasonably good agreement with the central slope of the graphs.

The other features are probably related to multiple emission (at the left end of the graph) and threshold (on the right).

So, to address the question in the title the sensitivity is not a strong function of wavelength if understood on a per-photon basis.


BTW: blue is low wavelength (high frequency), and red is high wavelength (low frequency).

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so you agree to my reasoning, under the conjecture that each photon over a trheshold releases the same numer of electrons. Good! –  elcojon Jan 24 '13 at 19:09

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