Question

Given a dc bias of 0.5mA and a bandwidth of 10kHz, calculate for the noise current

Answer 1

Solution;

Noise current is given as follows;

$I_n=\sqrt{2Iq\Delta B}$

In Which;

I=DC current (A)

$\Delta B$ =bandwidth (Hz)

q=charge of electron(c)

Given;

$I=0.5mA=0.5×10^{-3}A$

$\Delta B=10kHz=10×10^3Hz$

$q=1.602×10^{-19}c$

By substitution;

$I_n$ =$\sqrt{2×0.5×10×1.602×10^{-19}}$

$I_n=\sqrt{16.02×10^{-19}}$

$I_n=1.266×10^{-9}A$

Answer;

Noise current,$I_n$ =1.266nA