Concept of carrier mobility: Worked examples — Hardware deep dive

Before anything, let us re-anchor every symbol we lean on, in plain words:

Definition The cast of characters (say them out loud)

$E$ = the electric field — "how hard we tilt the playground." Units volts per metre, $V/m$ .
$v_{d}$ = the drift velocity — "average sliding speed of the whole crowd," metres per second, $m/s$ .
$μ$ = the mobility — "how much sliding speed we get per unit of tilt," $μ = v_{d} / E$ , units $m^{2} / (V\cdots)$ .
$τ$ = mean free time — average seconds a carrier flies before a collision.
$m^{*}$ = effective mass — the "felt" mass of a carrier inside the crystal.
$q$ = the carrier's charge magnitude, $1.6 \times 1 0^{- 19} C$ .
$n$ = electrons per cubic metre; $p$ = holes per cubic metre.
$J$ = current density — "how much electric current flows through each square metre of cross-section," units amperes per square metre, $A/m^{2}$ . It counts the charge sliding past per second per unit area.
$σ$ = conductivity — "how easily this material carries current for a given field," the constant in $J = σ E$ , units siemens per metre, $S/m$ . Big $σ$ = good conductor.

The two engines we will drive: $μ = \frac{q τ}{m ^{*}}, v_{d} = μ E, J = n q v_{d} = n q μ E, σ = n q μ .$

The scenario matrix

Every mobility question is really one of these cells. The examples below are labelled by cell so you can see the whole space is covered.

Cell	What is special about it	Which example
A. Baseline forward	Ordinary numbers, find $v_{d}$ from $μ, E$	Ex 1
B. Reverse solve	Given $v_{d}$ , back out $τ$ or $m^{*}$	Ex 2
C. Electron vs hole sign	Same $E$ , opposite drift direction, both $μ > 0$	Ex 3
D. Both carriers add	$σ = q (n μ_{n} + p μ_{p})$ , intrinsic vs doped	Ex 4
E. Zero / degenerate input	$E = 0$ , $τ \to 0$ , $n \to 0$ — what collapses?	Ex 5
F. Limiting / large input	$τ \to \infty$ (perfect crystal), field huge	Ex 6
G. Temperature twist	$μ$ goes down when heated (lattice regime)	Ex 7
H. Combine two scatterings	Matthiessen's rule, smallest $μ$ wins	Ex 8
I. Real-world word problem	Copper wire, "how long to cross a wire?"	Ex 9
J. Exam twist	Mixing up $μ$ and $n$ ; catch the trap	Ex 10

The examples

Cell B — Reverse solve

Worked example Ex 2 — Back out the mean free time

A material shows $μ = 0.045 m^{2} / (V\cdots)$ with effective mass $m^{*} = 0.40 m_{0}$ , $m_{0} = 9.11 \times 1 0^{- 31} kg$ . Find $τ$ .

Forecast: will $τ$ be nanoseconds, picoseconds, or femtoseconds? (Semiconductor scattering is fast.)

Start from $μ = \frac{q τ}{m ^{*}}$ and rearrange to $τ = \frac{μ m ^{*}}{q}$ . Why this step? We know everything except $τ$ , so isolate it — algebra, no new physics.
$m^{*} = 0.40 \times 9.11 \times 1 0^{- 31} = 3.644 \times 1 0^{- 31} kg$ . Why this step? Effective mass is quoted as a multiple of the free mass; convert to kg first.
$τ = \frac{0.045 \times 3.644 \times 1 0 ^{- 31}}{1.6 \times 1 0 ^{- 19}} \approx 1.025 \times 1 0^{- 13} s$ . Why this step? Substitute the numbers into the rearranged formula from step 1; dividing the numerator (mobility × mass) by the charge leaves seconds, which is exactly what $τ$ should be.

Verify: $\approx 0.10 ps$ — a fraction of a picosecond, exactly the scattering timescale expected in a semiconductor ✓. Plug back: $\frac{q τ}{m ^{*}} = \frac{1.6 \times 1 0 ^{- 19} \times 1.025 \times 1 0 ^{- 13}}{3.644 \times 1 0 ^{- 31}} \approx 0.045$ ✓.

Cell C — Electron vs hole sign

What to observe: one red field arrow points right (

+ x

). The white circle marked "

+

" is a hole and its black drift arrow points right; the white circle marked "

-

" is an electron and its black drift arrow points left — opposite ways. Yet the dashed arrows at the bottom (the currents

J

) both point right. The takeaway you should read off the picture: opposite drifts, but currents that add.

Worked example Ex 3 — Same field, opposite drift, both

$μ > 0$ A field $E = 1000 V/m$ points to the right ( $+ x$ ). Electrons have $μ_{n} = 0.135$ , holes $μ_{p} = 0.048 m^{2} / (V\cdots)$ . Find each drift velocity and its direction.

Forecast: which way does each go, and does the current from each add or cancel?

Magnitudes: $v_{d, n} = μ_{n} E = 0.135 \times 1000 = 135 m/s$ ; $v_{d, p} = μ_{p} E = 0.048 \times 1000 = 48 m/s$ . Why this step? Mobility is kept positive for both carriers by convention, so we use it purely for the magnitude of the drift and decide direction separately.
Directions (look at the figure): the hole (charge $+ q$ ) feels force $+ q E$ , drifts right, $+ x$ . The electron (charge $- q$ ) feels force $- q E$ , drifts left, $- x$ . Why this step? The sign of the charge — not the sign of $μ$ — sets the direction of drift.
Currents: conventional current is charge-flow direction. A $+$ charge going right = current right. A $-$ charge going left is also current to the right. So both currents point $+ x$ and add. Why this step? This is why $σ = q (n μ_{n} + p μ_{p})$ is a sum, never a difference.

Verify: In the figure the two drift arrows are antiparallel, yet the resulting $J$ arrows are parallel — consistent with $J = n q μ_{n} E + pq μ_{p} E > 0$ ✓.

Cell D — Both carriers add

Worked example Ex 4 — Intrinsic vs doped conductivity

Intrinsic silicon: $n = p = n_{i} = 1.5 \times 1 0^{16} m^{- 3}$ , $μ_{n} = 0.135$ , $μ_{p} = 0.048 m^{2} / (V\cdots)$ . Find $σ$ .

Forecast: with both carriers, does the electron term or the hole term dominate?

Engine: $σ = q (n μ_{n} + p μ_{p})$ . Why this step? Two independent carrier populations conduct in parallel; their conductivities add.
$n μ_{n} + p μ_{p} = 1.5 \times 1 0^{16} (0.135 + 0.048) = 1.5 \times 1 0^{16} \times 0.183 = 2.745 \times 1 0^{15}$ . Why this step? Since $n = p$ here, factor them out — the bracket is just the sum of the two mobilities.
$σ = 1.6 \times 1 0^{- 19} \times 2.745 \times 1 0^{15} \approx 4.39 \times 1 0^{- 4} S/m$ . Why this step? Multiply the bracketed sum from step 2 by the charge $q$ to finish the formula; charge (C) times the $n μ$ group $(m^{- 3} \cdot m^{2} / V\cdots)$ gives siemens per metre, the unit of $σ$ .

Verify: Electron term $/$ hole term $= 0.135/0.048 \approx 2.8$ , so electrons carry most current — sensible since $μ_{n} > μ_{p}$ ✓. Order of magnitude ( $1 0^{- 4} S/m$ ) is the textbook value for intrinsic Si ✓.

Cell E — Zero / degenerate input

Worked example Ex 5 — What happens at the edges?

Test three degenerate inputs for the same electron sample ( $μ_{n} = 0.135$ , $n = 1 0^{22} m^{- 3}$ ): (a) $E = 0$ , (b) $τ \to 0$ , (c) $n \to 0$ . What does each do to $v_{d}$ , $μ$ , $J$ ?

Forecast: which one kills the drift but keeps mobility alive, and which kills mobility itself?

(a) $E = 0$ : $v_{d} = μ E = 0.135 \times 0 = 0$ , so $J = n q v_{d} = 0$ . But $μ = 0.135$ is unchanged. Why this step? $μ$ is a material property; no field means no bias, but the material's willingness to slide is the same. No tilt → the crowd just mills about.
(b) $τ \to 0$ : $μ = \frac{q τ}{m ^{*}} \to 0$ , hence $v_{d} \to 0$ and $J \to 0$ for any field. Why this step? Infinite collisions ( $τ = 0$ ) reset drift instantly — the carrier can never build up net speed. This is a genuine material death, unlike (a).
(c) $n \to 0$ : $μ$ stays $0.135$ and a lone carrier still drifts at $v_{d} = μ E$ , but $J = n q v_{d} \to 0$ because there is no crowd to carry current. Why this step? $μ$ and $n$ are independent — mobility is how easily each carrier moves, $n$ is how many carriers exist. An empty band conducts nothing even with high mobility.

Verify: All three give $J = 0$ but by three different mechanisms — (a) no push, (b) no free time, (c) no carriers. The formula $J = n q μ E$ correctly zeros out if any of $E$ , $μ$ (via $τ$ ), or $n$ hits zero ✓.

Cell F — Limiting / large input

Worked example Ex 6 — The perfect crystal and the strong field

(a) Idealised defect-free, absolute-zero crystal so scattering vanishes, $τ \to \infty$ . What is $μ$ , and is $v_{d}$ physical? (b) In a real device $μ_{n} = 0.135$ but $E = 5 \times 1 0^{7} V/m$ (a strong field). Naive $v_{d}$ ?

Forecast: does $v_{d}$ really grow without limit, or does nature step in?

(a) $μ = \frac{q τ}{m ^{*}} \to \infty$ as $τ \to \infty$ , so $v_{d} = μ E \to \infty$ . Why this step? Mathematically the model has no ceiling — with no collisions the carrier accelerates forever.
Reality check: an electron cannot exceed thermal-ish drift indefinitely; real crystals always have some phonons above $0 K$ , capping $τ$ . So $τ \to \infty$ is a limit, not an achievable state. Why this step? Covering the limiting case means also stating why the limit is never reached — the model is linear only for modest fields.
(b) Naive $v_{d} = 0.135 \times 5 \times 1 0^{7} = 6.75 \times 1 0^{6} m/s$ . Why this step? Plugging in mechanically gives a speed above the thermal speed $1 0^{5} m/s$ — a red flag.

Verify: $6.75 \times 1 0^{6} m/s$ exceeds thermal speed by $\sim 70 \times$ , which is unphysical: at high fields carriers hit velocity saturation ( $\sim 1 0^{5} m/s$ ) and $μ$ is no longer constant. So the linear law $v_{d} = μ E$ is a low-field law ✓.

Cell G — Temperature twist

What to observe: the red curve is mobility versus temperature and it falls as you move right (hotter). The two black dots mark our worked points at

300 K

and

400 K

: the second dot sits lower than the first. The takeaway you should read off the picture: heating a lattice-limited sample reduces mobility, because more heat means more phonons and more collisions.

Worked example Ex 7 — Heating

lowers mobility (lattice regime) A lightly-doped sample is lattice(phonon)-limited, so $μ \propto T^{- 3/2}$ . At $T_{1} = 300 K$ , $μ_{1} = 0.135 m^{2} / (V\cdots)$ . Find $μ$ at $T_{2} = 400 K$ .

Forecast: more heat = more energy — will $μ$ go up or down? (Intuition trap!)

Ratio form: $\frac{μ _{2}}{μ _{1}} = (\frac{T _{2}}{T _{1}})^{- 3/2}$ . Why this step? Using a ratio cancels the unknown proportionality constant — we only need the two temperatures.
$\frac{T _{2}}{T _{1}} = \frac{400}{300} = 1.3333$ , so $(1.3333)^{- 3/2} = 0.6495$ . Why this step? Negative exponent means bigger $T$ → smaller $μ$ ; the $3/2$ power comes from phonon-scattering physics.
$μ_{2} = 0.135 \times 0.6495 \approx 0.0877 m^{2} / (V\cdots)$ . Why this step? Multiply the base mobility $μ_{1}$ by the dimensionless ratio from step 2 to scale it to the new temperature; the ratio being less than 1 is what makes the answer smaller than $μ_{1}$ .

Verify: $μ$ dropped from $0.135$ to $0.088$ despite heating — because more phonons = more collisions = smaller $τ$ (look at the falling red curve). Extra thermal energy is random, so it does not help directed drift ✓.

Cell H — Combine two scatterings

Worked example Ex 8 — Matthiessen's rule

A sample has lattice-limited mobility $μ_{L} = 0.135$ and impurity-limited $μ_{I} = 0.060 m^{2} / (V\cdots)$ acting together. Find the net $μ$ .

Forecast: will the net mobility be near the bigger one, the smaller one, or between?

Rule: $\frac{1}{μ} = \frac{1}{μ _{L}} + \frac{1}{μ _{I}}$ . Why this step? Scattering rates ( $1/ τ$ ) add, and $μ \propto τ$ , so inverse mobilities add — like parallel resistors.
$\frac{1}{μ} = \frac{1}{0.135} + \frac{1}{0.060} = 7.407 + 16.667 = 24.074 (V\cdots) / m^{2}$ . Why this step? Add the two rates numerically.
$μ = 1/24.074 \approx 0.04154 m^{2} / (V\cdots)$ . Why this step? Step 2 gave us $1/ μ$ (the summed rate), so we flip it (take the reciprocal) to recover the mobility $μ$ itself — the same last move you make when finishing a parallel-resistor sum.

Verify: $0.0415 < 0.060 < 0.135$ — the net is below the smaller of the two, exactly as "the worst scatterer dominates" predicts ✓.

Cell I — Real-world word problem

Worked example Ex 9 — How long to drift across a copper wire's length?

A $2.0 m$ copper wire carries a field $E = 0.10 V/m$ . Copper electron mobility $μ \approx 4.4 \times 1 0^{- 3} m^{2} / (V\cdots)$ . How long for one electron's drift to cross the wire?

Forecast: seconds? minutes? hours? (Signals travel near light-speed — but the electrons crawl.)

Drift speed: $v_{d} = μ E = 4.4 \times 1 0^{- 3} \times 0.10 = 4.4 \times 1 0^{- 4} m/s$ . Why this step? Same baseline law $v_{d} = μ E$ , now for a metal.
Time to cross: $t = \frac{L}{v _{d}} = \frac{2.0}{4.4 \times 1 0 ^{- 4}} \approx 4545 s$ . Why this step? Distance over speed — the electron physically crawls the length.
$4545 s \approx 76 minutes$ . Why this step? Convert seconds to minutes (divide by 60) so the answer is human-readable; $4545/60 \approx 76$ makes the "electrons crawl" point vivid.

Verify: Over an hour to physically cross $2 m$ ! ✓ This is the famous point that drift is slow; the electrical signal (field) propagates near light speed while the carriers themselves barely move.

Cell J — Exam twist

Worked example Ex 10 — The

$μ$ -vs- $n$ trap Sample A: $n_{A} = 1 0^{21} m^{- 3}$ , $μ_{A} = 0.20$ . Sample B: $n_{B} = 1 0^{22} m^{- 3}$ , $μ_{B} = 0.05 m^{2} / (V\cdots)$ . "B is more heavily doped so it must conduct better." True? Compute both $σ$ .

Forecast: more carriers but lower mobility — who wins?

$σ_{A} = n_{A} q μ_{A} = 1 0^{21} \times 1.6 \times 1 0^{- 19} \times 0.20 = 32 S/m$ . Why this step? Apply $σ = n q μ$ to each sample independently.
$σ_{B} = n_{B} q μ_{B} = 1 0^{22} \times 1.6 \times 1 0^{- 19} \times 0.05 = 80 S/m$ . Why this step? Same formula; note B's $10 \times$ carriers but $4 \times$ lower mobility.
Compare: $σ_{B} = 80 > σ_{A} = 32$ . So B does win here — but only because the $10 \times$ gain in $n$ beat the $4 \times$ loss in $μ$ . The claim "more doping ⇒ better" is not automatic. Why this step? $μ$ and $n$ are independent factors and conductivity is their product $σ = n q μ$ ; you must multiply both to see who wins, never judge on carrier count alone.

Conclusion: B is the better conductor in this case, yet the reasoning "more doping, therefore better" is a lucky guess, not a law. Had B's mobility instead dropped by a factor of $20$ (to $μ_{B} = 0.01$ ), we would get $σ_{B} = 1 0^{22} \times 1.6 \times 1 0^{- 19} \times 0.01 = 16 S/m < σ_{A} = 32$ , and the more heavily doped sample would conduct worse. Pedagogical point: always compute the product $n μ$ ; carrier count and mobility trade off, and only their product decides conductivity.

Verify: $σ_{B} / σ_{A} = 80/32 = 2.5$ . Cross-check via factors: $(n_{B} / n_{A}) \times (μ_{B} / μ_{A}) = 10 \times 0.25 = 2.5$ ✓ — consistent, and shows exactly which factor did the heavy lifting.

Recall Quick self-test across the matrix

Setting $E = 0$ kills current but not mobility — why? ::: $μ$ is a material property ( $q τ / m^{*}$ ); no field just means no bias, so $v_{d} = μ E = 0$ while $μ$ is unchanged. Electron and hole drift in opposite directions, yet their currents add — why? ::: A negative charge moving left is the same conventional current as a positive charge moving right, so $σ = q (n μ_{n} + p μ_{p})$ is a sum. Heating a lattice-limited sample lowers $μ$ — why? ::: More phonons → more scattering → smaller $τ$ ; and $μ \propto τ$ . With two scattering mechanisms, is net $μ$ above or below both? ::: Below the smaller one, because inverse mobilities (rates) add: $1/ μ = 1/ μ_{L} + 1/ μ_{I}$ . "More doping always conducts better" — is it true? ::: No; conductivity is $n q μ$ , and doping raises $n$ but lowers $μ$ — multiply both.

Connections

Concept of carrier mobility — parent note with the derivations.
Drift and Diffusion currents — Ex 1, 5, 9 are pure drift problems.
Conductivity and Resistivity — Ex 4, 10 use $σ = n q μ$ .
Effective mass — the $m^{*}$ reversed in Ex 2.
Scattering mechanisms in semiconductors — Ex 7, 8 set $τ$ .
Doping and carrier concentration — the $n$ -vs- $μ$ tradeoff in Ex 10.
Einstein relation — links these $μ$ values to diffusion.

Worked examples — Concept of carrier mobility