Question

The human eye can easily detect a flux of 1000 [photons/second] at a wavelength of 500 [\(\mu\)m].

Calculate the current this would produce in a photodiode, assuming each photon produces an electron / hole pair (i.e. a quantum efficiency of 1).

Answer 1

\[\begin{align*}
I &= \Phi\cdot\eta \\
&= 1000\,[\gamma/\text s]\times1\,[\text e/\gamma]
\times1.602\times10^{-19}\,[\text C/\text e] \\
&= 1.602\times10^{-16}\,[\text A]
\end{align*}\]

Answer 2

is this right?

Answer 3

it makes sense that a photodiode should not receive a strong signal when the incident flux is at the lower limit of a human eyes capabilities

Answer 4

Yes, I would expect the current to be very small at that level of light intensity.

Shouldn't there be a factor of 2 in there - each photon produces an electron hole pair ?

Answer 5

That makes sense, @ProfBrainstorm
so \[\begin{align*}
I &= 2\eta\,\Phi \\
&= 2\times1\,[{\text e/\gamma}]
\times1000\,[{\gamma/\text s}]
\times1.602\times10^{-19}\,[\text C/{\text e}]\\
&= 3.204\times10^{-16}\,[\text A]
\end{align*}\]