NEET UG PHYSICS MCQ: SEMI CONDUCTOR DEVICES

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Understanding +4 | -1

The value of

\left( {\frac{1}{\beta } - \frac{1}{\alpha }} \right)

is equal to :- (

\alpha

and

\beta

are current gain in CB & CE configuration).

View answer and explanation

Correct option: D

-1

Option D (-1) is correct because the relationship between alpha and beta in transistor configurations leads to this result.

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

{V_L}

\sqrt{n}.

n

p - n\;

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

P

{10^{16}}

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

N

kT =

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

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