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Derive an Expression for responsively of PIN photodiode.
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It is defined as:

$R = \dfrac {I_p}{P_0 } (AW^{-1}) ---- Eq. (1)$

Where $I_p$ is the output photocurrent in amperes and $P_0$ is the incident optical power in watts.

The responsivity is a useful parameter as it gives the transfer characteristic of the detector (i.e. photocurrent per unit incident optical power).

Considering the energy of a photon E = hf. Thus the incident photon rate $r_p$ may be written in terms of incident optical power and the photonenergy as:

$r_p= \dfrac {P_0}{hf} ---- Eq. (2)$

Since quantum efficiency is given by,

$η = \dfrac {r_e}{r_p }$

The electron rate is given by:

$ηr_p = r_e$

Substituting from Eq. (2) we obtain:

$η \dfrac {P_0}{hf} = r_e$

Therefore, the output photocurrent is:

$I_p=\dfrac{ηeP_0}{hf}$

Where e is the charge on an electron. Thus from Eq. (1) the responsivity may be written as:

$R = \dfrac {ηe}{hf} ---- Eq. (3)$

The frequency f of the incident photons is related to their wavelength λ and the velocity of light in air c, by:

$f = \dfrac cλ$

Substituting into Eq. (3) a final expression for the responsivity is given by:

$R =\dfrac {ηeλ}{hc}$