Exercises — Drift current and electric field

WHAT tool? Drift velocity is the first tool: $v_d=\mu E$. No collisions or densities appear, so nothing else is needed. $v_d={\mu }_{n}E=8500\times 200=1.7\times 10^6\text{cm/s}$ Sanity check: this is below the thermal speed ($\sim 10^7\text{cm/s}$), so the drift is still a small bias — consistent with the drunk-walking-downhill picture.

WHAT tool? Conductivity: $\sigma =e(n{\mu }_n+p{\mu }_p)$. Holes are negligible, so drop the $p{\mu }_p$ term (this is the 80/20 simplification from the parent note). $\sigma =en{\mu }_n=(1.6\times 10^{-19})(5\times 10^{15})(1350)=1.08(\Omega \text{⋅cm}{)}^{-1}$

(a) Conductivity — tool 3: $\sigma =en{\mu }_n=(1.6\times 10^{-19})(10^{16})(1350)=2.16(\Omega \text{⋅cm}{)}^{-1}$ (b) Current density — microscopic Ohm's law $J=\sigma E$: $J=2.16\times 500=1080{\text{A/cm}}^2$ (c) Current — multiply by area only at the very end (density is size-free, current is not): $I=JA=1080\times 2\times 10^{-4}=0.216\text{A}$

WHAT tool? Invert the microscopic mobility $\mu =\frac{q\tau }{m^{\ast }}$ to solve for $\tau =\frac{\mu m^{\ast }}{q}$. WHY convert units? $m^{\ast }$ is in kilograms, so we must work in SI here. Convert $\mu$: $1350{\text{cm}}^2/(\text{V⋅s})=1350\times 10^{-4}{\text{m}}^2/(\text{V⋅s})=0.135{\text{m}}^2/(\text{V⋅s})$. $m^{\ast }=0.26\times 9.11\times 10^{-31}=2.369\times 10^{-31}\text{kg}$ $\tau =\frac{\mu m^{\ast }}{q}=\frac{(0.135)(2.369\times 10^{-31})}{1.6\times 10^{-19}}=1.999\times 10^{-13}\text{s}\approx 0.2\text{ps}$ That is a fifth of a picosecond between collisions — enormously frequent, which is exactly why drift stays a small steady bias rather than runaway acceleration.

Both carriers are present, so we keep the full tool 3: $\sigma =e{n}_i({\mu }_n+{\mu }_p)=(1.6\times 10^{-19})(1.5\times 10^{10})(1350+480)$ $\sigma =(1.6\times 10^{-19})(1.5\times 10^{10})(1830)=4.392\times 10^{-6}(\Omega \text{⋅cm}{)}^{-1}$ Why they add: electrons ($-e$) drift against $E$; holes ($+e$) drift along $E$. Two sign flips — one in charge, one in direction — cancel, so both conventional currents point the same way (along $E$) and their conductivities sum. (This is the steel-manned mistake from the parent note.) Analysis takeaway: compare to the doped sample in L2.1 ($\sigma =2.16$). Intrinsic silicon is about $500,000\times$ less conductive — doping is what makes semiconductors useful.

WHY does $\sigma$ fall when it's hotter? Carrier count $n$ is fixed (donors already all ionised), so only mobility changes. More heat $\Rightarrow$ more lattice vibrations $\Rightarrow$ more collisions $\Rightarrow$ smaller $\tau$ $\Rightarrow$ smaller $\mu$. ${\mu }_n(T_2)=1350{\left(\frac{400}{300}\right)}^{-3/2}=1350\times (1.3333{)}^{-1.5}=1350\times 0.6495=876.9{\text{cm}}^2/(\text{V⋅s})$ $\sigma (T_2)=en{\mu }_n(T_2)=(1.6\times 10^{-19})(10^{16})(876.9)=1.403(\Omega \text{⋅cm}{)}^{-1}$ Down from $2.16$ at 300 K — the sample got more resistive when heated. This is the counter-intuitive behaviour of a doped semiconductor: mobility loses the tug-of-war because carrier count isn't allowed to grow. See Scattering Mechanisms and Mean Free Time.

(a) Set ${\mu }_{n}E=v_{sat}$ and solve for $E$: $E=\frac{v_{sat}}{{\mu }_n}=\frac{1.0\times 10^7}{1350}=7407\text{V/cm}\approx 7.4\times 10^3\text{V/cm}$ (b) The linear law assumes $\mu$ is constant — but $\mu =q\tau /m^{\ast }$ hides an assumption that $\tau$ doesn't depend on how fast carriers move. At high field, carriers gain enough energy between collisions to excite extra scattering (optical phonons), so $\tau$ drops as the field rises. The result: $v_d$ stops climbing and levels off at $v_{sat}$. Beyond this point $v_d=\mu E$ over-predicts, and you must use Velocity Saturation and High-Field Effects. Look at the red curve in the figure: the dashed magenta line is the naive $v_d=\mu E$; the solid violet curve is reality — they agree only in the shaded low-field region.

Bridge drift physics to circuit language. Resistivity $\rho =1/\sigma$, and $R=\rho L/A$: $\rho =\frac{1}{2.16}=0.4630\Omega \text{⋅cm}$ $R=\frac{\rho L}{A}=\frac{0.4630\times 0.5}{2\times 10^{-4}}=1157.4\Omega$ Voltage — Ohm's law at the macroscopic level, $V=IR$: $V=IR=0.216\times 1157.4=250.0\text{V}$ Consistency check: $V/L=250/0.5=500\text{V/cm}$ — exactly the field in L2.1. The microscopic law $J=\sigma E$ and the macroscopic law $V=IR$ are the same statement in two costumes.

Step 1 — net doping. Donors give electrons, acceptors take them. Net electron surplus: $n\approx N_D-N_A=8\times 10^{15}-3\times 10^{15}=5\times 10^{15}{\text{cm}}^{-3}$ Step 2 — minority holes from the mass-action law $np=n_i^2$: $p=\frac{n_i^2}{n}=\frac{(1.5\times 10^{10}{)}^2}{5\times 10^{15}}=\frac{2.25\times 10^{20}}{5\times 10^{15}}=4.5\times 10^4{\text{cm}}^{-3}$ Step 3 — is the hole term negligible? Compare $n{\mu }_n=5\times 10^{15}\times 1350=6.75\times 10^{18}$ against $p{\mu }_p=4.5\times 10^4\times 480=2.16\times 10^7$ — eleven orders of magnitude smaller. Drop it. Step 4 — conductivity: $\sigma \approx en{\mu }_n=(1.6\times 10^{-19})(5\times 10^{15})(1350)=1.08(\Omega \text{⋅cm}{)}^{-1}$ Mastery insight: compensation cancels dopants against each other — the electrical behaviour is set by the difference $N_D-N_A$, not the totals. Even though there are $11\times 10^{15}$ impurities, the sample behaves like one with only $5\times 10^{15}$ net donors.

Step 1 — recover $J$ from the measured current: $J=\frac{I}{A}=\frac{5\times 10^{-3}}{10^{-3}}=5{\text{A/cm}}^2$ Step 2 — invert $J=en{\mu }_{n}E$ to solve for $n$: $n=\frac{J}{e{\mu }_{n}E}=\frac{5}{(1.6\times 10^{-19})(1350)(100)}=2.315\times 10^{14}{\text{cm}}^{-3}$ Step 3 — drift velocity (tool 1, independent of $n$): $v_d={\mu }_{n}E=1350\times 100=1.35\times 10^5\text{cm/s}$ Cross-check: $J=en{v}_d=(1.6\times 10^{-19})(2.315\times 10^{14})(1.35\times 10^5)=5.0{\text{A/cm}}^2$ ✓ — the density and drift velocity multiply back to the measured current density.

Which tool converts field to drift velocity? ::: $v_d=\mu E$ (tool 1). When must you switch to SI (metres/kilograms)? ::: The moment $m^{\ast }$ in kg enters, e.g. computing $\tau =\mu m^{\ast }/q$. For fixed $n$, does heating raise or lower $\sigma$? ::: Lower — phonon scattering cuts $\mu \propto T^{-3/2}$. What sets the electrical behaviour of a compensated sample? ::: The net doping $N_D-N_A$. Above roughly what field does $v_d=\mu E$ fail for Si? ::: A few $\text{kV/cm}$, where velocity saturates at $v_{sat}\approx 10^7\text{cm/s}$.