First slide

Types of semiconductors

Question

A semi conductor has an electron concentration of $0.60 \times 10^{11} m^{- 3}$ and a hole concentration of $6 \times 10^{19} m^{- 3}$ . Calculate its conductivity. Given electron mobility is $0.125 m^{2} v^{- 1} s^{- 1}$ , hole mobility is $0.032 m^{2} v^{- 1} s^{- 1}$

Moderate

A

$0.3052 s m^{- 1}$
B

$0.3072 s m^{- 1}$
C

$0.3062 {sm}^{- 1}$
D

$0.3082 {sm}^{- 1}$

Solution

Conductivity of semiconductor is the sum of the conductivities due to electrons and holes.
$\begin{array}{l} σ = σ_{e} + σ_{h} \\ σ = n_{e} e μ_{e} + n_{h} e μ_{h} \\ G i v e n : \\ n_{e} = 0.60 \times 10^{11} m^{- 3} \\ n_{h} = 6 \times 10^{19} m^{- 3} \\ μ_{e} = 0.125 m^{2} v^{- 1} s^{- 1} \\ μ_{h} = 0.032 m^{2} v^{- 1} s^{- 1} \\ σ = n_{e} e μ_{e} + n_{h} e μ_{h} \end{array}$
$n_{e} {<<<<n}_{h} So n_{e} is negligible$
$\begin{array}{l} σ = n_{h} e μ_{h} \\ σ = 6 \times 10^{19} \times 1.60 \times 10^{- 19} \times 0.032 \\ σ = 6 \times 1.6 \times 0.032 \\ σ = 0.3072 {sm}^{- 1} \end{array}$

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