Rankers Physics
Topic: Semiconductor Physics
Subtopic: Properties of Semiconductors

The resistivity of a pure semiconductor is 0.5 Ωm. If the electron and hole mobility be 0.39 m²/V-s and 0.19 m²/V-s respectively then calculate the intrinsic carrier concentration.
\[2.16\times 10^{19}/m^{3}\]
\[4.32\times 10^{19}/m^{3}\]
\[10^{20}/m^{3}\]
\[10^{30}/m^{3}\]

Solution:

The resistivity () of a pure (intrinsic) semiconductor is given by the formula:

 

ρ=1q(ni)(μn+μp)\rho = \frac{1}{q (n_i) (\mu_n + \mu_p)}

Where:


  • ρ=0.5 Ωm\rho = 0.5 \, \Omega\text{m}     

    (resistivity),


  • q=1.6×10−19 Cq = 1.6 \times 10^{-19} \, \text{C}     

    (charge of an electron),


  • nin_i     

    = intrinsic carrier concentration (to be calculated),


  • μn=0.39 m2/V-s\mu_n = 0.39 \, \text{m}^2/\text{V-s}     

    (electron mobility),


  • μp=0.19 m2/V-s\mu_p = 0.19 \, \text{m}^2/\text{V-s}     

    (hole mobility).

Rearranging the formula to solve for

nin_i

:

 

ni=1qρ(μn+μp)n_i = \frac{1}{q \rho (\mu_n + \mu_p)}

Substitute the values:

 

ni=1(1.6×10−19)(0.5)(0.39+0.19)n_i = \frac{1}{(1.6 \times 10^{-19})(0.5)(0.39 + 0.19)}

ni=1(1.6×10−19)(0.5)(0.58)n_i = \frac{1}{(1.6 \times 10^{-19})(0.5)(0.58)}

 

ni=14.64×10−20=2.16×1019 m−3

 

Thus, the intrinsic carrier concentration

nin_i

is

2.16×1019 m−32.16 \times 10^{19} \, \text{m}^{-3}

 

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