First slide

Types of semiconductors

Question

The band gap in germanium is $Δ E = 0.68 eV$ . Assuming that the number of hole – electron pairs is proportional to $e^{- \frac{Δ E}{2 k T}}$ , find the percentage increase in the number of charge carriers in pure germanium as the temperature is increased from 300 K to 320 K

Difficult

A

120%
B

127%
C

138%
D

130%

Solution

The number of charge carriers in an intrinsic semiconductor is double the number of hole – electron pairs (No of holes = No of electrons). If $N_{1}$ be the number of charge carriers at temperature    $T_{1}$ and $N_{2}$ at    $T_{2}$ , we have
$\begin{array}{l} N_{1} = N_{0} e^{- ∆ E / 2 k T_{1}} \end{array}$
$N_{2} = N_{0} e^{- ∆ E / 2 k T_{2}}$
The percentage increase as the temperature is raised from $T_{1}$ to $T_{2}$ is
$f = \frac{N_{2} - N_{1}}{N_{1}} \times 100 = [\frac{N_{2}}{N_{1}} - 1] 100$
$= 100 [e^{\frac{Δ E}{2 k} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]} - 1]$
$N o w; \frac{Δ E}{2 k} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$
$\begin{array}{l} = \frac{0.68 \times 1.6 \times 10^{- 19}}{2 \times 1.38 \times 10^{- 23}} [\frac{1}{300} - \frac{1}{320}] \\ = 0.82 \end{array}$
$T h u s, f = 100 \times [e^{0.82} - 1] ≃ 127 (∵ e^{0.82} = 2.27)$
$∴$ The number of charge carriers increases by about 127 %.

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