Question

What is the reason to operate the photodiodes in reverse bias?

• A. fractional change due to the photo-effects on the majority carrier dominated forward bias current is more easily measurable than the fractional change in the reverse bias current.
• B. reverse bias is more efficient than forward bias.
• C. fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current.
• D. reverse bias configuration result in larger amount of current (in $\mu A$)

Answer 1

The correct option is C fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current.

1.In reverse biased $p-n$ junction, the width of depletion region increases as we increases the applied reverse bias voltage across the diode.

2.So,by applying a larger voltage, more of the incident photons are converted to electric current, or the efficiency increases.

3.On the other hands when we apply forward bias in the $p-n$ junction, the width of the depletion region reduces, so, only a small portion of the incident photons gets converted to electric current.

so, this means that reverse bias is more efficient than forward bias that's the reason to operate the photo diodes in reverse bias.