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An n- n Silicon device is fabricated with uniform and non-degenerate donor doping concentrations of ND1 = 1 × 1018 cm-3 and ND2 = 1 × 1015 cm-3 corresponding to the n+ and n regions respectively. At the operational temperature T, assume complete impurity ionization, kT/q = 25 mV, and intrinsic carrier concentration to be ni = 1 × 1010 cm-3. What is the magnitude of the built – in potential of this device?

1. 0.748 V
2. 0.460 V
3. 0.288 V
4. 0.173 V

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Correct Answer - Option 4 : 0.173 V


The Built-in Potential of a p-n junction diode is given by:

\({V_o} = \frac{{kT}}{q}\ln \left( {\frac{{{N_A}{N_D}}}{{n_i^2}}} \right)\)

Where NA = doping density of holes

ND = doping density of electrons

Also, the law of mass-Action states that:

\(n_i^2 = {N_A}{N_D}\)


\(n + \)





n+ = 1018/cm3

n = 1015/cm3

ni = 1010/cm3

The n+ Junction here has more electrons than the n Junction [Analogous to p-n Junction]

Hole concertation in n – region using mass action law will be:

 \(p= \frac{{n_i^2}}{N} = \frac{{{{10}^{20}}}}{{{{10}^{15}}}}\) = NA

Putting on the respective values, we get:

\({V_o} = \frac{{KT}}{2}{l_n}\left( {\frac{{N_D^ + + {N_A}}}{{n_i^2}}} \right) = 0.173V\)

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