Question

If resistivity of pure silicon is $3000\Omega m$ and the electron and hole mobilities are $0.12m^2{v}^{-1}{s}^{-1}$ and $0.045m^2{v}^{-1}{s}^{-1}$ respectively. Determine the resistivity of a specimen of the material when ${10}^{19}$ atoms of phosphorous are added per $m^3$ Given e= $1.6\times {10}^{-19}C$, $p=3000\Omega$, ${\mu }_e=0.12m^2{v}^{-1}{s}^{-1}$, ${\mu }_h=0.045m^2{v}^{-1}{s}^{-1}.$

Answer 1

Step 1. Given data:

Resistivity of Silicon $\left(\rho \right)$ = $3000\Omega m$

Electron mobility $\left({\mu }_e\right)$ = $0.12m^2{v}^{-1}{s}^{-1}$

Hole mobility $\left({\mu }_h\right)$ = $0.045m^2{v}^{-1}{s}^{-1}$

No. of phosphorous atoms = ${10}^{19}{m}^{-3}$

Charge on electrons $\left(e\right)$ = $1.6\times {10}^{-19}C$

Step 2. Calculations:

The resistivity of silicon (Si) metal is given by,

$\rho =\frac{1}{\sigma }=\frac{1}{e\left(n_e{\mu }_e+n_h{\mu }_h\right)}$

Where $n_e$=number of electrons.

$n_h$=number of holes

For pure Silicon, as given in the question,

$n_e=n_h=n_i$

So the formula for resistivity becomes,

$\rho =\frac{1}{e{n}_i\left({\mu }_e+{\mu }_h\right)}$

Or $n_i=\frac{1}{e\rho \left({\mu }_i+{\mu }_h\right)}$, Putting all the given values, we get

When ${10}^{19}$atoms of phosphorous are added per $m^3$, the semiconductor becomes $n$-type semiconductor.

Or it becomes pentavalent as the atoms of phosphorus completely occupy the holes as ${10}^{19}>1.26\times {10}^{16}$

Hence, in that case, we will find the resistivity of $n_e={10}^{19}$

Resistivity $\left(\rho \right)$ = $\frac{1}{n_e{\mu }_{e}e}$=$\frac{1}{1.6\times {10}^{-19}\times {10}^{19}\times 0.12}$=$5.21\Omega m$

Hence, the resistivity of a specimen is $5.21\Omega m$.