I have been doing some experimenting with a transimpedance amp (LF353) and a PIN
>photodiode. So far I have been using only a simple 1M feedback resistor with
>zero bias on the diode. Then I noticed in the datasheet for the PIN photodiode
>(NEC PH302) it shows a graph of photodiode current as a function of reverse
>bias. According to that graph, the photodiode current should be about double
>with 2 volts of reverse bias on the diode as compared with zero bias. When I
>tried it, the gain of the system remained the same (I have a squarewave light
>signal generated by a LED). It did improve the transient response quite a bit,
>especially when I rasied the bias to 7 volts (due to the reduced diode
>capacitance, I guess) but I saw no evidence of increased photodiode current. Is
>this bogus? I have not seen this stated anywhere else besides the NEC PH302
>datasheet, but Graeme in his classic book on Photodiode Amplifiers does say that
>PIN diodes are usually used with bias. Why is that? Why PINs and not all
>photodiodes?
>
>
>Robert Scott
>Ypsilanti, Michigan
I think current can be lower at zero bias because some photon-smacked hole-electron pairs recombine and are lost. With a reverse-bias field, they are swept up and collected quickly.
For the first few volts of reverse bias, junction capacitance drops rapidly, so the *circuit* gets faster. That effect flattens out quickly, but higher bias sweeps carriers out faster and continues to improve *diode* speed.
At very high voltages (50? 100?) you can get avalanche multiplication that greatly increases gain (and noise.) Some diodes are designed to do this well.
Something like that.
John