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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 in silicon is 1.5 × 10¹⁰ cm⁻³ at 300 K, kT/q = 26 mV and the permittivity of silicon εsi = 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

Correct answer: 0.86 V and 3.3 × 10⁻⁵ cm

Solution

The built-in potential is calculated using the concentration of donor and acceptor impurities, leading to a value of 0.86 V, which reflects the energy barrier formed at the junction. The depletion width is determined by the charge distribution and the permittivity of silicon, resulting in a narrower width of 3.3 × 10⁻⁵ cm due to the high acceptor concentration.

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