Solution;
Noise current is given as follows;
In=2IqΔB
In Which;
I=DC current (A)
ΔB =bandwidth (Hz)
q=charge of electron(c)
Given;
I=0.5mA=0.5×10−3A
ΔB=10kHz=10×103Hz
q=1.602×10−19c
By substitution;
In =2×0.5×10×1.602×10−19
In=16.02×10−19
In=1.266×10−9A
Answer;
Noise current,In =1.266nA
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