Question

With an irradiance of 0.001W/cm² incident on a detector of area 0.5 cm ² and with a bandwidth of 2 Hz, the ratio of the noise voltage to the signal voltage is 10. What is the noise equivalent power of the detector?

Answer 1

Solution:

Given:

Irradiance, H=0.001 W/cm²

Area, A=0.5 cm²

Bandwidth= 2Hz

Ratio of the noise voltage to the signal voltage is 10

We know that Noise to Equivalent Power ratio is given by,

$NEP= \frac{HAV_{_{N}}}{V_{S}(\Delta f)^{\frac{1}{2}}}$

Where,

H= Irradiance

A= Area

V_S= Root mean square signal voltage

V_N = Root mean square noise voltage within the measurement bandwidth ∆f

Using the above formula we get,

$NEP= \frac{0.001*0.5*10}{(2)^{\frac{1}{2}}}$

$NEP= 0.003535534 W/Hz^{\frac{1}{2}}$

$=3.5\times 10^{-3}W/Hz^{\frac{1}{2}}$