# For the following energy band diagram determine the approximate resistivity for x > L portion of semiconductor. Eg = 1.12 eV, T = 300 K, μn = 600 cm2/V-sec, μp = 400 cm2/V-sec, ni = 1010/cm3

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1. 11.00 ohm-cm
2. 15.75 ohm-cm
3. 23 ohm-cm
4. 31.25 ohm-cm

Option 4 : 31.25 ohm-cm
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CT 1: Ratio and Proportion
2846
10 Questions 16 Marks 30 Mins

## Detailed Solution

Concept:

The resistivity of a material is the conductivity and the conductivity of a semiconductor is given by:

σ = qnμn + qpμp

n = majority carrier electron concentration

p = majority carrier hole concentration

μn = electron mobility

Also, the majority carrier hole concentration for the given intrinsic Fermi level is given by:

$$p = {n_i}e\frac{{\left( {{E_i} - {E_E}} \right)}}{{KT}}$$

Calculation:

Given semiconductor can be considered as a p-type semiconductor since the Fermi level is close to the valence band.

∴ The majority carrier hole concentration will dominate in the conductivity of the semiconductor, i.e.

σ = qpμp

Given, ni = 1010 /cm3, Eg = 1.12 eV

At n > L:

$${E_i} - {E_f} = \frac{{Eg}}{4} = \frac{{1.12}}{4}eV$$

Ei – Ef = 0.28 eV

Also at room temperature, the thermal voltage is taken as 25.9 mV

The majority carrier hole concentration for x > L will be:

$$p = {n_i}e\frac{{{E_i} - {E_f}}}{{KT}}$$

$$p = \left( {{{10}^{10}}} \right)\left( {{e^{\left( {\frac{{0.28}}{{0.026}}} \right)}}} \right)$$

P = 4.955 × 1014 / cm3

The conductivity will be:

σ = q × p × μp

= (1.6 × 10-19)(4.955 × 1014) × (400)

σ = 0.0317

∴ The resistivity will be:

$$p = \frac{1}{\sigma } = \frac{1}{{0.0317}} \approx 31.25$$