The previous two problems dealt with pure semiconductors, also called intrinsic semiconductors. Useful semiconductor devices are instead made from doped semiconductors, which contain substantial numbers of impurity atoms. One example of a doped semiconductor was treated in Problem 7.5. Let us now consider that system again. (Note that in Problem 7.5 we measured all energies relative to the bottom of the conduction band, Ee. We also neglected the distinction between $g_0$and $g_{0c}$; this simplification happens to be ok for conduction electrons in silicon.)

(a) Calculate and plot the chemical potential as afunction of temperature, for silicon doped with ${10}^{17}$phosphorus atoms per $c{m}^3$(as in Problem 7.5). Continue to assume that the conduction electrons can be treated as an ordinary ideal gas.

(b) Discuss whether it is legitimate to assume for this system that the conduction electrons can be treated asan ordinary ideal gas, as opposed to a Fermi gas. Give some numerical examples.

(c)Estimate the temperature at which the number of valence electrons excitedto the conduction band would become comparable to the number ofconduction electrons from donor impurities. Which source of conductionelectrons is more important at room temperature?

We have been given silicon doped with phosphorus atoms.

Chemical potential is given by$\mu =-kT\ln \left(\frac{V{Z}_{int}}{N_c{v}_Q}\right)$

Due to two spin states $\mu =-kT\ln \left(\frac{2V}{N_c{v}_Q}\right)$

Ratio of Conduction electrons to number of doner atoms is $\frac{N_c}{N_d}=x=1-2e^{I/kT}\left(\frac{v_Q{N}_d}{2V}\right)$

where $\mu =-kT\ln \left(\frac{2V}{x{N}_d{v}_Q}\right)$and $x=1-2e^{I/kT}\left(\frac{v_Q{N}_d}{2V}\right)$

Let $t=\frac{kT}{I}$

Therefore Quantum volume $v_Q={\left(\frac{h^2}{2\pi mI}\right)}^{3/2}\frac{1}{t^{3/2}}$

Putting the vlaues of universal constants $m,handI.$$v_Q={\left(\frac{{\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)}^2}{2\pi \left(9.11\times {10}^{-31}\mathrm{kg}\right)\left(0.044\times 1.6\times {10}^{-19}\mathrm{J}\right)}\right)}^{3/2}\frac{1}{t^{3/2}}$$=\left(3.6\times {10}^{-26}\right)t^{-3/2}$

Also it is known $\frac{N_d}{V}=1\times {10}^{23}{\mathrm{m}}^3$

Evaluting$\frac{\mu }{I}=-t\ln \left(\frac{2}{x\left(1\times {10}^{23}{\mathrm{m}}^3\right)\left(3.6\times {10}^{-26}\right)t^{-3/2}}\right)$

Therefore$\frac{\mu }{I}=-t\ln \left(\frac{2t^{3/2}}{0.0036x}\right)andx=1-\frac{(0.0036)e^{1/t}}{t^{3/2}}$.

Plotting a graph between $\raisebox{1ex}{$\mu $}\!\left/ \!\raisebox{-1ex}{$I$}\right.$

We have to find whether the conduction elements can be treated as an ordinary ideal gas.

Considering the denominator of Fermi Dirac that is much greater than 1 we treat it as gas.

Therefore energies of the conduction band $ϵ-\mu \gg kT$

The difference between edge energy of conduction band is 3, therefore localid="1650464758608" $ϵ-\mu \approx 3I$.

Solving $\frac{I}{kT}=\frac{0.044\mathrm{eV}}{0.026\mathrm{eV}}=1.695$

$\frac{ϵ-\mu }{kT}\approx \frac{3I}{kT}=3\times 1.695\approx 5$

Therefore the value of exponential $e^5\approx 150$which is much greater than 1.

Therefore approximating the Fermi Dirac distribution by Boltzmann distribution should be accurate.

We have to find out the temperature for which number of valence electrons excited to the conduction band would become comparable to the number of conduction electrons from donor impurities.

Saturation point of number of conduction electrons${10}^{17}\mathrm{P}\text{atoms per}{\mathrm{cm}}^3$

i.e$\frac{N_d}{V}=1\times {10}^{23}{\mathrm{m}}^3$

For temperature $\frac{N_C}{V}=\frac{2}{v_Q}{e}^{-\Delta ϵ/2kT}$

Let this expression be equal to ${10}^{23}$; Solving$-\frac{\Delta ϵ}{2kT}=\ln \left(\frac{v_Q{N}_C}{2V}\right)$

$\frac{1}{T}=-\frac{2k}{\Delta ϵ}\ln \left(\frac{v_Q{N}_C}{2V}\right)$

Substituting $\frac{1}{T}=-\frac{2\left(8.62\times {10}^{-5}\mathrm{eV}\right)}{1.11\mathrm{eV}}\ln \left(\frac{v_Q{N}_C}{2V}\right)$

Quantum volume is given by $v_Q={\left(\frac{{\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)}^2}{2\pi \left(9.11\times {10}^{-31}\mathrm{kg}\right)\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)}\right)}^{3/2}\frac{1}{T^{3/2}}$

$=\left(3.6\times {10}^{-20}\right)T^{-3/2}$

Therefore

$\frac{1}{T}=-\left(1.553\times {10}^{-4}\right)\ln \left((0.0018)T^{-3/2}\right)$

Intersection in graph occurs at $T=520K$