A $p - n\;$ junction has acceptor impurity concentration of ${10^{17}}{\rm{c}}{{\rm{m}}^{ - 3}}$ in the $P$ side and donor impurity concentration of ${10^{16}}$ ${\rm{c}}{{\rm{m}}^{ - 3}}$ in the $N$ side. What is the contact potential at the junction? ($kT = $ thermal energy, intrinsic carrier concentration ${n_i} = 1.4 \times {10^{10}}$ ${\rm{c}}{{\rm{m}}^{ - 3}}$ )

$\left( {\frac{{kT}}{e}} \right)$ ln ($2.5 \times {10^{23}}$)

$\left( {\frac{{kT}}{e}} \right)$ ln (${10^{23}}$)

$\left( {\frac{{kT}}{e}} \right)$ ln (${10^9}$)

JEE PHYSICS MCQ: SEMI CONDUCTOR DEVICES

1 Hard

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Application +4 | -1

p - n\;

junction has acceptor impurity concentration of

{10^{17}}\mathrm{c}{\mathrm{m}^{ - 3}}

in the

P

side and donor impurity concentration of

{10^{16}}

\mathrm{c}{\mathrm{m}^{ - 3}}

in the

N

side. What is the contact potential at the junction? (

kT =

thermal energy, intrinsic carrier concentration

{n_i} = 1.4 \times {10^{10}}

\mathrm{c}{\mathrm{m}^{ - 3}}

)

View answer and explanation

Correct option: A

\left( {\frac{{kT}}{e}} \right)

ln (

4 \times {10^{12}}

)

Option A (

\left( {\frac{{kT}}{e}} \right)

ln (

4 \times {10^{12}}

)) is correct because it is calculated using the formula for contact potential, considering the acceptor and donor concentrations and intrinsic carrier concentration.

JEE PHYSICS MCQ: SEMI CONDUCTOR DEVICES

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