First slide

Types of semiconductors

Question

A semiconductor has an electron concentration of $0.45 \times 10^{12} m^{- 3}$ and a hole concentration of $5 \times 10^{20} m^{- 3}$ . Calculate its conductivity. Given electron mobility is $0.135 m^{2} v^{- 1} s^{- 1}$ . Hole mobility is $0.048 m^{2} v^{- 1} s^{- 1}$

Moderate

A

$3.84 s m^{- 1}$
B

$3.86 {sm}^{- 1}$
C

$3.82 {sm}^{- 1}$
D

$3.80 {sm}^{- 1}$

Solution

Conductivity of semiconductor is the sum of the conductivities due to electrons and holes
$\begin{array}{l} G i v e n, n_{e} = 0.45 \times 10^{12} m^{- 3} \\ n_{h} = 5 \times 10^{20} m^{- 3} \\ μ_{e} = 0.135 m^{2} v^{- 1} s^{- 1} \end{array}$
$\begin{array}{l} μ_{h} = 0.048 m^{2} v^{- 1} s^{- 1} \\ C o n d u c t i v i t y, σ = σ_{e} + σ_{h} \end{array}$
$\Rightarrow σ = n_{e} e μ_{e} + n_{h} e μ_{h}$
$n_{e} {<<<<n}_{h} So n_{e} is negligible$
$\begin{array}{l} \Rightarrow σ = n_{h} e μ_{h} \\ σ = 5 \times 10^{20} \times 1.60 \times 10^{- 19} \times 0.048 \\ σ = 5 \times 10^{1} \times 1.6 \times 0.048 \\ σ = 5 \times 16 \times 0.048 \\ σ = 3.84 {sm}^{- 1} \end{array}$

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