Is the photocurrent doe to absorption in the depletion region of a photodiode considered as one or two flow particles?

[Ziv Glasser]

As from what I learned, when a photon is absorbed in the depletion region it excites an electron, causing a pair of free charge carriers, i.e. an electron and a hole which flow in opposite directions, Therefor contributing two charges to the total photocurrent. However in sum publications I find the equation that describes the photocurrent in depletion region (assuming uniform illumination) as: I=q*A*W*g, were I is the depletion-photocurrent, q the electron charge, A the diodes area, W depletion width and g the photo-generation rate. My question is isn't this equation missing a factor of 2 since every absorbed photon creates a pair of free charges?

 

[Mark Oude Alink]

No: a current flows in a closed loop. Assume the electron drifts to contact A, and the hole to contact B. The hole movement is identical to an electron moving from contact B towards the point of the generated electron-hole pair. The external circuit moves the generated electron from contact A to contact B. The current loop is closed, and it consists of only 1 elementary charge.

 

[Ziv Glasser]

Thank you Mark, although I feel a little foolish but I'm still not convinced. How come in some articles such as: G. Lucovsky et.al. "Transit-time Considerations in p-i-n Diodes."(1963) -attached-, the photocurrent is calculated by summing both contributions of electrons and holes separately?

 

[Mark Oude Alink]

Electron-hole pairs are generated all over the diode, in particular in the intrinsic region of the p-i-n diode. For all of the individual eh-pairs you can use my reasoning: a single eh-pair generates one 'quantum of current' throughout the circuit. If you look at any particular point within your diode, you'll therefore have to sum the individual electron and hole currents (or current densities) to arrive at the total current (or current density).