Question

The donor and accepter impurities in an abrupt junction silicon diode are 1 × 10¹⁶ cm^-3 and 5 × 10¹⁸ cm^-3, respectively. Assume that the intrinsic carrier concentration is silicon n_i = 1.5 × 10¹⁰ cm^-3 at \(300\;K,\;\frac{{kT}}{q} = 26\;mV\) and the permittivity of silicon ϵ_si = 1.04 × 10^-12 F/cm. The built-in potential and the depletion width of the diode under thermal equilibrium conditions, respectively, are
1. 0.7 V and 1 × 10^-4 cm
2. 0.86 V and 1 × 10^-4 cm
3. 0.7 V and 3.3 × 10^-5 cm
4. 0.86 V and 3.3 × 10^-5 cm

Answer 1

Correct Answer - Option 4 : 0.86 V and 3.3 × 10^-5 cm

Concept:

The built-in potential barrier for a PN junction with given doping concentration Na, Nd, is given by:-

\( {{V}_{bi}}=\frac{kT}{q}\ln \left( \frac{{{N}_{a}}{{N}_{d}}}{n_{i}^{2}} \right)\)

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:

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

Calculation:

With N_a = 5 × 10¹⁸ cm^-3

N_d = 1 × 10¹⁶ cm^-3

\(\frac{{KT}}{q} = 26\;mV\)

\({V_{bi}} = 26\;m\ln \left( {\frac{{{{10}^{16}} × 5 × {{10}^{18}}}}{{{{\left( {1.5 × {{10}^{10}}} \right)}^2}}}} \right)\)

= 26 m×ln (2.22 × 10¹⁴)

= 26 m × 33.03

V_bi ≃ 0.86 V

Now, the depletion width will be:

\(W = \sqrt {\frac{{2 × 1.04 × {{10}^{ - 12}} × 0.86}}{{1.6 × {{10}^{ - 19}}}} × \frac{{\left( {{{10}^{16}} + 5 × {{10}^{18}}} \right)}}{{{{10}^{16}} × 5 × {{10}^{18}}}}} \)

\( = \sqrt {\frac{{2 × 1.04 × {{10}^{ - 12}} × 0.86}}{{1.6 × {{10}^{ - 19}}}} × \frac{1}{{{{10}^{16}}}}} \)

Solving for the above, we get:

W = 3.3 × 10^-5 cm