Question

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

Answer 1

We are given with the resistivity of pure silicon and we need to find the resistivity when the pure silicon is doped with phosphorus. When phosphorus is added, we get an n-type semiconductor as phosphorus has $5$ valence electrons. Resistivity and conductivity are inverse of each other. Use the formula to calculate the resistivity.

Complete step by step answer:
We need to calculate the resistivity after ${10^{19}}$ atoms of phosphorus are added. We know that resistivity is the inverse of conductivity.
The resistivity for pure silicon is given as
$\rho = \dfrac{1}{\sigma } = \dfrac{1}{{e\left( {{n_e}{\mu _e} + {n_h}{\mu _h}} \right)}}$
Where $\rho$ is the resistivity which is given as $\rho = 3000\Omega m$
$\sigma$ is the conductivity
$e$ is the charge on an electron which is given as $e = 1.6 \times {10^{ - 19}}\,C$
${n_e}$ is the number density of electrons
${\mu _e}$ is the mobility of electrons, value ${\mu _e} = 0.12{m^2}{v^{ - 1}}{s^{ - 1}}$
${n_h}$ is the number density of holes
${\mu _h}$ is the mobility of holes, value ${\mu _h} = 0.045{m^2}{v^{ - 1}}{s^{ - 1}}$
In order to calculate the resistivity, we need to find the values of ${n_e}$ and ${n_h}$ . But we know that for pure silicon we can have ${n_e} = {n_h} = {n_i}$ .
Substituting the given values in the above equation, we can find the value of ${n_i}$ and then using this value, we can find the resistivity of the specimen.
$\rho = \dfrac{1}{\sigma } = \dfrac{1}{{e\left( {{n_e}{\mu _e} + {n_h}{\mu _h}} \right)}}$
$\Rightarrow \rho = \dfrac{1}{{e{n_i}\left( {{\mu _e} + {\mu _h}} \right)}}$ as ${n_e} = {n_h} = {n_i}$
$\Rightarrow {n_i} = \dfrac{1}{{e\rho \left( {{\mu _e} + {\mu _h}} \right)}}$
Substituting the given values, we have;
$\Rightarrow {n_i} = \dfrac{1}{{1.6 \times {{10}^{ - 19}} \times 3000 \times \left( {0.12 + 0.045} \right)}}$
$\Rightarrow {n_i} = 1.26 \times {10^{16}}{m^{ - 3}}$
As discussed above, when ${10^{19}}$ atoms are added, the semiconductor becomes n-type, we have
$\Rightarrow {n_e} - {n_h} \approx {n_e} = {10^{19}}$
Therefore,
$\rho = \dfrac{1}{{e{n_e}{\mu _e}}}$
$\Rightarrow \rho = \dfrac{1}{{1.6 \times {{10}^{ - 19}} \times {{10}^{19}} \times 0.12}}$
$\Rightarrow \rho = 5.21\,\omega m$
The resistivity of a specimen of the material when ${10^{19}}$ atoms of phosphorus is $5.21\,\omega m$

Note:
Note that we have compared the number density of the electron and the holes after calculating the number density.
Remember that the specimen acts as n-type as phosphorus has 5 valence electrons.
Be careful with the decimal numbers and the units while calculating the values.