Question

A 650-nm continuous-wave laser beam strikes a photodetector. The photon arrival rate is 4x10^16 photons per second, resulting in 6x10^16 electrons per second being collected at the photodetector's positive contact (i.e.,anode).

1. What is the average received optical power in mW?

2. What is the average photocurrent in the external circuit?

3. What is the quantum efficiency of the photodetector?

4. What is the responsivity?

5. Does this photodetector exhibit photocurrent gain?

Answer 1

3.quantum efficiency(QE)=electrons collected (per second)/photons incident (per second)

So QE =(6×(10*16))/(4×(10*16))

QE=1.5 OR 150%

4.RESPONSIVITY=QE×(elementary charge /energy (hv))

So if we calculate the responsivity using the above formula

h=planks constant, elementary charge=1.6×(10*-19)

v=frequency which can be calculated by c=v(frequency)×wavelength. And remaining all variables are available.after calculation we get responsivity as 790mA/W.

5.here QE is more than 100% this is called as cross relaxation.this can only happen when the photon energy of laser is lower than half that of light.yes it can.

1&2 questions are correlated.photo current= totalcharge×frequency.

Total charge =(1.6×(10*-19)) ×(6×(10*16))

Frequency we already have .so from that u get current.from that power is obtained from. The below formula

P=photo current/responsivity.

Please do let me know if u have any doubts