Communication System Question 19
Question: In semiconductor the concentrations of electrons and holes are $ 8\times 10^{18}/m^{3} $ and $ 5\times 10^{18}/{{m}^{{}}} $ respectively. If the( mobilities of electrons and hole are $ 2.3m^{2} $ /volt-sec and $ 0.01m^{2} $ /volt-sec respectively, then semiconductor is
Options:
A) $ N $ -type and its resistivity is 0.34 ohm-metre
B) $ P $ -type and its resistivity is 0.034 ohm-metre
C) $ N $ -type and its resistivity is 0.034 ohm-metre
D) $ P $ -type and its resistivity is 3.40 ohm-metre
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ n _{e}=8\times 10^{18}/m^{3},n _{h}=5\times 10^{18}/m^{3} $
$ {\mu _{e}}=2.3\frac{m^{2}}{volt\text{-}\sec },{\mu _{h}}=0.01\frac{m^{2}}{volt\text{-}\sec } $
$ \because n _{e}>n _{h} $ so semiconductor is N-type Also conductivity $ \sigma =\frac{1}{\text{Resistivity},(\rho )}=e(n _{e}{\mu _{e}}+n _{h}{\mu _{h}}) $
$ \Rightarrow \frac{1}{\rho }=1.6\times {{10}^{-19}}[8\times 10^{18}\times 2.3+5\times 10^{18}\times 0.01] $
$ \Rightarrow \rho =0.34,\Omega \text{-}m. $
BETA
