Correct Answer - Option 1 :

\(\sqrt {\frac{{2\epsilon_s{V_{bi}}}}{{qN_B}}} \)
__Depletion width__ for an unbiased abrupt pn junction diode is given by:

\(W = \sqrt {\frac{{2ϵ}}{q}\left( {\frac{1}{{{N_D}}} + \frac{1}{{{N_A}}}} \right){V_{bi}}} \)

This can be written as:

\(W = \sqrt {\frac{{2ϵ}}{q} (\frac{N_A+N_D}{N_AN_D}){V_{bi}}} \)

ND = Donor concentration on n side

NA = Acceptor concentration on p side

Vbi = Contact potential or the Built-in Potential

ϵ = Permittivity of the semiconductor

A one-sided pn junction is a junction

The one-sided p-n junction is defined as the junction in which one region is highly doped and another region is doped lower to form the p and n regions.

In this junction, the concentration of one side impurity is considered, as it dominates the other, i.e. for a one-sided junction with N_{A} >> N_{D}, the depletion width can be written as:

\(W=\sqrt {\frac{{2\epsilon_s(N_A){V_{bi}}}}{{qN_B.(N_A.N_D)}}} \)

\(W=\sqrt {\frac{{2\epsilon_s{V_{bi}}}}{{qN_D}}} \)

Similarly, if N_{D} >> N_{A}, the depletion layer width becomes:

\(W=\sqrt {\frac{{2\epsilon_s{V_{bi}}}}{{qN_A}}} \)

Generalizing this for any one-sided junction we can write:

\(\sqrt {\frac{{2\epsilon_s{V_{bi}}}}{{qN_B}}} \)

Where N_{B} = N_{A} or N_{B} depending upon the doping profile.