1 electron = 1 photon?

Started Apr 12, 2013 | Discussions thread
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Jack Hogan
Veteran MemberPosts: 4,508
Re: 1 electron = 1 photon?
In reply to John Siward, Apr 12, 2013

John Siward wrote:

Optical power in Watts = photon energy * no. of photons / sec.

Photocurrent in Amps = electron charge * no. of electrons generated / sec. = electron charge * (no. of photons / sec) * efficiency.

Response (A / W) as plotted = photocurrent (A) / optical power (W)

= electron charge * efficiency / photon energy

Photon energy = h * frequency = h * c / wavelength (where h is Planck's constant and c is speed of light)

So you can use that to work out the efficiency from the curve you showed.

Silicon is what they call an "indirect bandgap" material, which means that an electron jumping from the valence band to the conduction band also exchanges momentum with the lattice in the form of a phonon.  This represents a loss of energy, so you can't get to 100% efficiency.  This is in contrast to "direct bandgap" materials such as GaAs, InP, etc.

Your basic point is right: a red photon is much more likely to penetrate deeper into the Si than a blue photon.  This is the principle of operation of Sigma's "Foveon" sensor which stacks three photodiodes vertically to allow the sensor to measure colour without using a colour filter array.


Thank you for walking me through it, John, quite clear.  So could we rearrange the Responsivity diagram's x-axis to represent logarithmic penetration depth, the curve then representing QE at the relative depth?  If so, I would have expected more discontinuities in it, per this interesting introduction.

PS Re: Feynman

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