Question 10.SG.Q.28: The source of a silicon (ni = 10^{10} per cm^3) N-channel MO......
The source of a silicon (n_i = 10^{10} per cm^3) N-channel MOS transistor has an area of 1 sq mm and a depth of 1 μm. If the dopant density in the source is 10^{19}/cm^3 , the number of holes in the source region with the above volume is approximately
(a) 10^7 (b) 100
(c) 10 (d) 0
Step-by-Step
Given that A=1 \times 10^{-12} m ^2 \text { and } d= 10^{-6} m
Therefore, V=A d=10^{-18} m ^3=10^{-12} / cm ^3
Also, given that N_{ D }=n=10^{19} / cm ^3 \text { and } n_{ i }=10^{10} / cm ^3
Therefore,
p=\frac{n_{ i }^2}{N_{ D }}=\frac{10^{20}}{10^{19}}=10 / cm ^3
Therefore, holes in volume V is H=p V=10^{-11}
As the number of holes cannot be a decimal number, therefore, number of holes H = 0.
Ans. (d)
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