Question

An n⁺ - n Silicon device is fabricated with uniform and non-degenerate donor doping concentrations of N_D1 = 1 × 10¹⁸ cm^-3 and N_D2 = 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 n_i = 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

Answer 1

Correct Answer - Option 4 : 0.173 V

Concept:

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:

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

Calculation:

\(n + \) \({10^{18}}/c{m^3}\)

\(n\) \({10^{15}}/c{m^3}\)

 

n⁺ = 10¹⁸/cm³

n = 10¹⁵/cm³

n_i = 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:

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

Putting on the respective values, we get:

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