Question

Electric Field of 1 V/m is applied to a Boron doped Silicon semiconductor slab having doping density of 1016 atoms/cm3 at 300K temperature. Determine the approximate resistivity of the slab. (Consider intrinsic carrier concentration of Silicon at 300 K = 1.5 × 1010 / cm3 Hole Mobility = 500 cm2/Vs at 300 K; Electron Mobility = 1300 cm2/Vs at 300 K).

Answer 1

Answer:

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Answer 2

Given :

Electric field E = 1 Volt per meter.

Doping density in silicon semiconductor (of boron) N :

${10}^{16} \frac{atoms}{ {cm}^{3} }$

Intrinsic carrier concentration of silicon n :

$1.5 \times {10}^{10} {cm}^{ - 3}$

Hole mobility :

$\bf \mu _{h} \: = 500 \: \frac{ {cm}^{2} }{V}$

Electron mobility :

$\bf \mu _{n} \: = 1300 \: \frac{ {cm}^{2} }{V}$

Temperature = 300 K

To find :

Resistivity of slab.

Solution :

This is a p-type semiconductor because boron atoms of group III are doped in intrinsic semiconductor.

The dopant atoms can enter into lattice kf silicon atoms and replace one silicon atom to bind with four nearby atoms.

When we inject some impurity in intrinsic semiconductor its conductivity increases.

As conductivity of semiconductor :

$\bf \: \sigma = e(( \mu _{n} \times \: n \:) + (\mu _{h} \times \: p) )$

where

σ = conductivity of semiconductor.

p = concentration of holes.

In case of p - type semiconductor

$\frac{p}{n} = \frac{n}{N} \\ \\ so \\ p = \frac{ {n}^{2} }{N} \:$

putting thins value of p in formula of conductivity

$\bf \: \sigma = e(( \mu _{n} \times \: n \:) + (\mu _{h} \times \frac{ {n}^{2} }{N} ) ) \:$

putting all the values from above

we get the value of conductivity :

$\bf \: \sigma = (1.6 \times {10}^{ - 19)} (( 1300\times 1.5 \times {10}^{10} \: ) + (500 \times \frac{ { 1.5 \times {10}^{10} }^{2} }{ {10}^{16} } ) ) \:$

on solving we get

$\sigma \: = 3.12 \times {10}^{ - 6}$

as we know that

Resistivity :

$\rho \: = \frac{1}{ \sigma}$

where

ρ= Resistivity of material

So

Resistivity of semiconductor slab :

$\rho \: = \frac{1}{ 3.12 \times {10}^{ - 6} } = 320512.821 \: \Omega m$

So

Resistivity of slab : ρ= 320512.821 Ωm