Why is the photodiode always operated under reverse bias conditions ? What would happen if it were to be operated under forward bias conditions ?
The only reason is, that a photodiode converts incident light to electric current more "effectively" in reverse bias condition than in forward bias.
why is it so? It relates to the extension of depletion region. You should note that when absorbed photons generate electron-hole pairs, only those electron-hole pairs generated in the depletion region, or very close to that, have a chance to contribute to electric current, because there is a strong electric field there to separate the two different charge carriers. The ones outside the depletion region quickly recombine and vanish.
Now, in a reverse biased pn junction, the width of depletion region increases as you increase the applied reverse bias voltage across the diode (proportional to the square root of the voltage). So, by applying a larger voltage, more of the incident photons are converted to electric current, or the efficiency increases (as long as you make sure the increased leakage current remains at a manageable level)
On the other hand, when you forward bias a pn junction, the width of the depletion region reduces, so, only a small portion of the incident photons get converted to electric current.
all answers related to the level of current are incorrect. Because you can forward bias a photodiode and keep the current level at micro amp level. It still works as a photodiode, but at a fraction of efficiency compared to a reverse biased one.
The only reason is, that a photodiode converts incident light to electric current more "effectively" in reverse bias condition than in forward bias.
why is it so? It relates to the extension of depletion region. You should note that when absorbed photons generate electron-hole pairs, only those electron-hole pairs generated in the depletion region, or very close to that, have a chance to contribute to electric current, because there is a strong electric field there to separate the two different charge carriers. The ones outside the depletion region quickly recombine and vanish.
Now, in a reverse biased pn junction, the width of depletion region increases as you increase the applied reverse bias voltage across the diode (proportional to the square root of the voltage). So, by applying a larger voltage, more of the incident photons are converted to electric current, or the efficiency increases (as long as you make sure the increased leakage current remains at a manageable level)
On the other hand, when you forward bias a pn junction, the width of the depletion region reduces, so, only a small portion of the incident photons get converted to electric current.
all answers related to the level of current are incorrect. Because you can forward bias a photodiode and keep the current level at micro amp level. It still works as a photodiode, but at a fraction of efficiency compared to a reverse biased one.
