Exercises — Concept of carrier mobility

WHAT we are asked: mobility is drift velocity per unit field — literally the ratio. WHY this formula: $\mu =v_d/E$ is the definition, no derivation needed. $\mu =\frac{v_d}{E}=\frac{200}{2000}=0.1\frac{\text{m/s}}{\text{V/m}}=0.1\frac{{\text{m}}^2}{\text{V⋅s}}$ The units simplify because $\frac{\text{m/s}}{\text{V/m}}=\frac{\text{m}}{\text{s}}\cdot \frac{\text{m}}{\text{V}}=\frac{{\text{m}}^2}{\text{V⋅s}}$. Answer: $\mu =0.1{\text{m}}^2/\text{V⋅s}$.

WHAT: separate the "how many" factor from the "how easy" factor. $n=10^{22}{\text{m}}^{-3}$ is a count per volume — that is how many carriers. $\mu =0.135{\text{m}}^2/\text{V⋅s}$ is the mobility — that is how easily each one moves ($\mu =q\tau /m^{\ast }$). WHY it matters: they are independent knobs. Doping raises $n$ but usually lowers $\mu$. Both feed into $\sigma =nq\mu$, but they answer different physical questions.

WHAT/WHY: direct use of $v_d=\mu E$ — mobility times field gives drift speed. $v_d={\mu }_{p}E=0.048\times 5000=240\text{m/s}$ Answer: $v_d=240\text{m/s}$ — still microscopic next to the $\sim 10^5\text{m/s}$ thermal jitter.

WHY $\mu =q\tau /m^{\ast }$: more charge and more free-flight time both speed up drift; more mass resists it. $m^{\ast }=0.19\times 9.11\times 10^{-31}=1.7309\times 10^{-31}\text{kg}$ $\mu =\frac{q\tau }{m^{\ast }}=\frac{(1.6\times 10^{-19})(2\times 10^{-13})}{1.7309\times 10^{-31}}\approx 0.1849{\text{m}}^2/\text{V⋅s}$ Answer: $\mu \approx 0.185{\text{m}}^2/\text{V⋅s}$ (≈ 1849 cm²/V·s).

WHY $\sigma =nq\mu$: it converts the microscopic trio (count, charge, ease-of-motion) into a lab-measurable conductivity. $\sigma =nq{\mu }_n=(5\times 10^{21})(1.6\times 10^{-19})(0.12)=96\text{S/m}$ $\rho =\frac{1}{\sigma }=\frac{1}{96}\approx 1.042\times 10^{-2}\Omega \text{⋅m}$ Answer: $\sigma =96\text{S/m}$, $\rho \approx 1.04\times 10^{-2}\Omega \text{⋅m}$.

WHAT the proportionality means: $\frac{{\mu }_2}{{\mu }_1}={\left(\frac{T_2}{T_1}\right)}^{-3/2}$. WHY it falls: hotter lattice → more phonons → more collisions → shorter $\tau$ → smaller $\mu =q\tau /m^{\ast }$. Extra heat energy does not help directed drift. ${\mu }_2=0.14\times {\left(\frac{400}{300}\right)}^{-3/2}=0.14\times {\left(\frac{4}{3}\right)}^{-1.5}$ ${\left(\frac{4}{3}\right)}^{1.5}\approx 1.5396\Rightarrow {\mu }_2\approx \frac{0.14}{1.5396}\approx 0.0909{\text{m}}^2/\text{V⋅s}$ Answer: $\mu \approx 0.091{\text{m}}^2/\text{V⋅s}$ — a ~35% drop. See the falling curve below.

WHY reciprocals add: scattering rates ($1/\tau$) add because each mechanism is an extra independent way to get knocked off course. Since $\mu \propto \tau$, the reciprocal mobilities add too. $\frac{1}{\mu }=\frac{1}{0.20}+\frac{1}{0.05}=5+20=25\Rightarrow \mu =\frac{1}{25}=0.04{\text{m}}^2/\text{V⋅s}$ Answer: $\mu =0.04{\text{m}}^2/\text{V⋅s}$. Notice it is below the smaller of the two — the worst scatterer dominates, and having a second one can only make it worse.

WHY the ratio is just masses: with $q$ and $\tau$ equal, $\mu \propto 1/m^{\ast }$, so $\frac{{\mu }_n}{{\mu }_p}=\frac{m_p^{\ast }}{m_n^{\ast }}$. ${\mu }_n=\frac{(1.6\times 10^{-19})(10^{-13})}{0.26\times 9.11\times 10^{-31}}\approx 0.06753{\text{m}}^2/\text{V⋅s}$ ${\mu }_p=\frac{(1.6\times 10^{-19})(10^{-13})}{0.49\times 9.11\times 10^{-31}}\approx 0.03583{\text{m}}^2/\text{V⋅s}$ $\frac{{\mu }_n}{{\mu }_p}=\frac{0.49}{0.26}\approx 1.885$ Answer: electrons are ≈1.88× more mobile — lighter effective mass, same free time.

WHY both terms: electrons and holes drift in opposite directions, but they carry current in the same direction (opposite charge × opposite velocity), so their contributions add: $\sigma =q(n{\mu }_n+p{\mu }_p)$. $n{\mu }_n+p{\mu }_p=(1.5\times 10^{16})(0.135)+(1.5\times 10^{16})(0.048)$ $=1.5\times 10^{16}\times 0.183=2.745\times 10^{15}$ $\sigma =q\times 2.745\times 10^{15}=(1.6\times 10^{-19})(2.745\times 10^{15})=4.392\times 10^{-4}\text{S/m}$ $\rho =\frac{1}{\sigma }\approx 2277\Omega \text{⋅m}$ Answer: $\sigma \approx 4.39\times 10^{-4}\text{S/m}$, $\rho \approx 2.28\times 10^3\Omega \text{⋅m}$ — a poor conductor, as expected for so few carriers.

WHY this relation exists: the same random collisions that limit drift ($\mu$) also drive spreading ($D$). Einstein's relation ties the two together through the thermal energy $kT$. WHAT the group $kT/q$ is: the thermal voltage, $\frac{kT}{q}=\frac{(1.38\times 10^{-23})(300)}{1.6\times 10^{-19}}\approx 0.025875\text{V}$ (≈ 26 mV at room temperature). $D=\mu \cdot \frac{kT}{q}=0.135\times 0.025875\approx 3.493\times 10^{-3}{\text{m}}^2/\text{s}$ Answer: $D\approx 3.49\times 10^{-3}{\text{m}}^2/\text{s}$.

Step A — invert $\rho =1/(nq\mu )$ to solve for $n$. $\sigma =\frac{1}{\rho }=\frac{1}{0.01}=100\text{S/m}$ $n=\frac{\sigma }{q{\mu }_n}=\frac{100}{(1.6\times 10^{-19})(0.135)}\approx 4.630\times 10^{21}{\text{m}}^{-3}$ WHY: conductivity is the reciprocal of resistivity; then $\sigma =nq\mu$ rearranges to give the count.

Step B — the doping penalty. At that $n$, extra ionized impurities scatter carriers, so ${\mu }_n$ falls to $0.10$. Keeping the same $n$: ${\sigma }_{true}=nq{\mu }_{new}=(4.630\times 10^{21})(1.6\times 10^{-19})(0.10)\approx 74.07\text{S/m}$ ${\rho }_{true}=\frac{1}{74.07}\approx 1.35\times 10^{-2}\Omega \text{⋅m}$ Answer: you aim for $n\approx 4.63\times 10^{21}{\text{m}}^{-3}$, but because mobility droops with doping, the real resistivity is higher than target: ${\rho }_{true}\approx 0.0135\Omega \text{⋅m}$, not $0.01$. Real designs iterate to correct for this.

(a) Mobility — microscopic first: $m^{\ast }=0.25\times 9.11\times 10^{-31}=2.2775\times 10^{-31}\text{kg}$ $\mu =\frac{q\tau }{m^{\ast }}=\frac{(1.6\times 10^{-19})(1.5\times 10^{-13})}{2.2775\times 10^{-31}}\approx 0.10537{\text{m}}^2/\text{V⋅s}$ (b) Field — voltage spread over the length: $E=V/L=0.5/0.01=50\text{V/m}$. (c) Drift velocity — $v_d=\mu E=0.10537\times 50\approx 5.268\text{m/s}$. (d) Current density — $J=nq{v}_d=(2\times 10^{22})(1.6\times 10^{-19})(5.268)\approx 1.686\times 10^4{\text{A/m}}^2$. (e) Current — $I=JA=(1.686\times 10^4)(10^{-6})\approx 1.686\times 10^{-2}\text{A}\approx 16.9\text{mA}$. Answer chain: $\mu \approx 0.105$, $E=50\text{V/m}$, $v_d\approx 5.27\text{m/s}$, $J\approx 1.69\times 10^4{\text{A/m}}^2$, $I\approx 16.9\text{mA}$. Every arrow used exactly one boxed formula — this is the full pipeline from crystal microphysics to a number on an ammeter.

Given $\tau$ and $m^{\ast }$, how do you get $\mu$? ::: $\mu =q\tau /m^{\ast }$. Given $\mu$ and $E$, how do you get $v_d$? ::: $v_d=\mu E$. Given a voltage $V$ over length $L$, what is $E$? ::: $E=V/L$ (a field is volts per metre). How do two scattering mechanisms combine? ::: Add reciprocals: $1/\mu =1/{\mu }_1+1/{\mu }_2$. How do electron and hole contributions to $\sigma$ combine? ::: They add: $\sigma =q(n{\mu }_n+p{\mu }_p)$.