NEET UG PHYSICS MCQ: SEMI CONDUCTOR DEVICES

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In a p-n junction, the depletion region is 400nm wide and an electric field of

5 \times {10^5}V/m

exists in it. What should be the minimum kinetic energy of a conduction electron which can diffuse from the inside to the p-side?

View answer and explanation

Correct option: C

0.2 eV

Option C (0.2 eV) is correct because the minimum kinetic energy required is equal to the potential difference across the depletion region, which can be calculated as

E = qV = qEd

, where

E

is the electric field and

d

is the width of the depletion region.

More Questions from SEMI CONDUCTOR DEVICES

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Following is the relation between current and charge

I = A{T^2}{e^{q/{V_L}}}

, then value of

{V_L}

will be To measure light intensity we use The ratio of RMS output voltage of full wave rectifier and half wave rectifier is

\sqrt{n}.

The value of

n

is? (The input voltage is sine wave). The truth table shown in figure is for
A    0   0   1   1
B    0   1   0   1
Y    1   0   0   1  The average value of output direct current in a half wave rectifier is A

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}}

)