Visual walkthrough — Concept of carrier mobility

Before any symbol appears, let us agree on the cast of characters, in plain words and with a picture.

Look at the figure: on the left, no field — the carrier zig-zags with huge random speed but goes nowhere on average (its arrows cancel). On the right, we tilt the floor with a field $E$ (the yellow slope). Same violent zig-zag, but now the crowd creeps downhill. That tiny downhill creep is the only thing mobility measures.

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Step 1 — Put a force on one carrier

WHAT. We isolate one carrier of charge $q$ and switch on the field $E$.

WHY. Everything in electricity starts from "what force does the field apply?" A field is defined as force-per-charge, so multiplying by the charge gives the force. This is the only way to turn "a push" into a number Newton can use.

$\underset{\text{force on carrier (N)}}{\underbrace{F}}=\underset{\text{signed charge (C)}}{\underbrace{q}}\cdot \underset{\text{field (V/m)}}{\underbrace{E}}$

PICTURE. The red arrow below is the force. Because $q$ is negative for an electron, the product $qE$ points against the field; for a hole ($q$ positive) it points along the field. We carry this sign honestly and settle it fully in Step 7.

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Step 2 — Turn force into acceleration (why $m^{\ast }$, not $m_0$)

WHAT. Divide the force by the carrier's mass to get how fast it speeds up.

WHY. Newton's second law says $F=ma$, i.e. acceleration is force divided by mass. But inside a crystal the carrier is not free — the lattice's own forces are already baked in, so it responds to $E$ as if it had a effective mass $m^{\ast }$ (see Effective mass). Using $m^{\ast }$ is what makes the whole calculation match real crystals.

$\underset{\text{acceleration}}{\underbrace{a}}=\frac{F}{m^{\ast }}=\frac{\stackrel{\text{signed charge}}{\overbrace{q}}\stackrel{\text{field}}{\overbrace{E}}}{\underset{\text{effective mass}}{\underbrace{m^{\ast }}}}$

PICTURE. The steeper the slope (bigger $E$) or the lighter the carrier (smaller $m^{\ast }$), the steeper the velocity ramp. The figure shows two ramps: a light carrier climbs velocity fast (steep), a heavy one climbs slowly.

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Step 3 — The carrier does NOT accelerate forever

WHAT. We admit that the carrier can only speed up for a short while before it slams into something.

WHY. If nothing stopped it, $a$ acting forever would give infinite velocity — clearly wrong. Real carriers hit vibrating atoms and impurities (see Scattering mechanisms in semiconductors). The mean free time $\tau$ is the average time between two such collisions. It is the natural clock of the problem: it says "how long the push gets to act."

PICTURE. The velocity climbs along the ramp of Step 2, then a collision (red flash) knocks the drift back to ~zero, and the ramp restarts. The saw-tooth below shows this over and over.

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Step 4 — Velocity gained in one free flight → drift velocity

WHAT. Average "velocity = acceleration × time" over the random free-flight times.

WHY. For constant acceleration starting from zero net drift after a collision, velocity grows linearly: $v(t)=at$. A carrier that has been flying free for a time $t$ carries drift $at$; to get the population drift we must average $at$ over how long carriers have actually been flying.

Here the scattering model earns its keep:

• Naïve uniform picture. If you imagine every free flight lasts exactly $\tau$ and then resets, the velocity is a saw-tooth from $0$ to $a\tau$, whose time-average is $\frac{1}{2}a\tau$. This is the "½" you might expect.
• Correct exponential (Poisson) picture. Collisions are memoryless: at any instant a carrier is equally likely to have been flying for a short or a long time, weighted by the exponential distribution whose mean is $\tau$. Averaging $at$ against that distribution gives $a\cdot ⟨t⟩=a\tau$ — no factor of ½. The long free flights (which reach high velocity) are common enough to exactly cancel the halving.

So the physically correct drift velocity is

$\underset{\text{average net drift}}{\underbrace{v_d}}=\underset{qE/m^{\ast }}{\underbrace{a}}\cdot \underset{\text{mean free time}}{\underbrace{\tau }}=\frac{qE\tau }{m^{\ast }}$

PICTURE. The green level line in the figure is $v_d$. The saw-tooth is the naïve uniform model (average $\frac{1}{2}a\tau$); the correct exponential average sits higher, at $a\tau$, because long flights are weighted in. We use the exponential result $v_d=qE\tau /m^{\ast }$.

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Step 5 — Divide out the field: mobility is born (and is positive)

WHAT. Divide the drift velocity by the field $E$, and take the size only.

WHY. We want a number that belongs to the material, not to how hard we happened to push, and not carrying a $\pm$ sign that depends on whether we chose an electron or a hole. In $v_d=qE\tau /m^{\ast }$ the field $E$ appears once, up top. Dividing by $E$ cancels it. Then we take the magnitude of the charge, because mobility is defined as a positive "ease of sliding":

$\mu =\frac{|v_d|}{E}=\frac{|q|\tau }{m^{\ast }}=\boxed{\frac{|q|\tau }{m^{\ast }}}$

$\underset{\text{size of charge}}{\underbrace{|q|}}\underset{\text{time to build speed}}{\underbrace{\tau }}\underset{\text{inertia resisting}}{\underbrace{m^{\ast }}}$

The sign that we honestly carried since Step 1 does not disappear — it re-enters, correctly, only when we build the current in Step 6. Mobility itself is always positive: ${\mu }_n,{\mu }_p>0$.

PICTURE. Plot the magnitude of $v_d$ against $E$: it is a straight line through the origin, and the slope of that line is $\mu$. Every material is one line; a steeper line = higher mobility = an easier ride.

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Step 6 — From one carrier to a measurable current

WHAT. Count how much charge crosses an area each second when the whole crowd drifts at $v_d$.

WHY. A single carrier's crawl is invisible; a lab measures current density $J$ (current per area). If there are $n$ carriers per cubic metre, each of signed charge $q$, all creeping at $v_d=\mu E$ (in the direction fixed by the sign of $q$), the charge sweeping through unit area per second is $nq{v}_d$.

$\underset{\text{current density}}{\underbrace{J}}=\underset{{\text{carriers per m}}^3}{\underbrace{n}}\underset{\text{signed charge}}{\underbrace{q}}\underset{=\mu E}{\underbrace{v_d}}=n|q|\mu E$

The two sign flips (negative $q$ and reversed $v_d$ for an electron) multiply to a positive current along $E$, so we may write $J=n|q|\mu E$. Comparing with the definition $J=\sigma E$ reads off the conductivity (see Conductivity and Resistivity):

$\sigma =n|q|\mu ,\rho =\frac{1}{\sigma }.$

PICTURE. A slab of area $A$: in one second every carrier moves forward a distance $v_d$, so all carriers inside a box of length $v_d$ pour through the face. Count them → that's the current.

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Step 7 — The degenerate and edge cases (so nothing surprises you)

WHAT / WHY / PICTURE, all cases the formula must survive:

• $E=0$ (no field). Then $v_d=\mu \cdot 0=0$: no net drift, pure random motion, zero current. The straight line of Step 5 passes through the origin — the picture already guarantees this.
• Electron vs hole (sign of $q$). For a hole $q=+e$, drift is along $E$; for an electron $q=-e$, drift is opposite to $E$. But mobility is defined with $|q|$, so both ${\mu }_n,{\mu }_p>0$. In the current $J=n|q|\mu E$ the electron's negative $q$ and its reversed $v_d$ multiply to a current pointing the same way as the hole current — they add, they never cancel.
• $\tau \to 0$ (violent scattering). $\mu \to 0$: a carrier that is hit instantly can never build up drift. This is the "infinitely bumpy playground."
• $\tau \to \infty$ (a perfect, collision-free crystal). $\mu \to \infty$: nothing stops the acceleration, drift grows without bound. Real crystals always have phonons, so $\tau$ is finite — this limit tells you why mobility is never infinite.
• $m^{\ast }\to \infty$ (infinitely heavy carrier). $\mu \to 0$: too sluggish to respond. Heavier holes → lower ${\mu }_p$, which is why usually ${\mu }_n>{\mu }_p$.

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The one-picture summary

One figure, the whole story left-to-right: field pushes ($F=qE$, signed $q$) → carrier accelerates ($a=qE/m^{\ast }$) → collision resets it every $\tau$ → average drift $v_d=qE\tau /m^{\ast }$ → take size, divide by $E$ → $\boxed{\mu =|q|\tau /m^{\ast }}$ → scale up by $n$ → $\sigma =n|q|\mu$. The figure itself labels each box, arrow, and the sign→magnitude hand-off between the drift stage and the mobility stage.

Recall Feynman: the whole walkthrough in plain words

Picture one electron on a floor that we tilt (the field). It starts sliding — faster and faster (that's $a=qE/m^{\ast }$; a lighter electron, meaning small $m^{\ast }$, speeds up quicker). But it keeps crashing into the wobbling atoms; each crash stops its downhill progress and sends it off randomly. Because those crashes happen at random with a mean gap $\tau$, and long gaps let it reach high speed, the average downhill speed works out to $v_d=qE\tau /m^{\ast }$ — small, gentle, nothing like its frantic random speed. Now ask: for a given tilt, how easily does it slide? Divide the size of that drift by the tilt and the tilt cancels, leaving $\mu =|q|\tau /m^{\ast }$ — pure "how slippery is this material," always a positive number. Finally, if there are $n$ such electrons in each cube of material, all creeping together, the electric current per area is just $n$ times the charge times the crawl: $\sigma =n|q|\mu$. That's it — from one tilted electron all the way to the conductivity of a chip.

Recall Self-check

Why does $E$ vanish from $\mu$? ::: We divide $v_d=qE\tau /m^{\ast }$ by $E$; the field cancels, leaving only material constants $|q|,\tau ,m^{\ast }$. What sets the free-flight time $\tau$? ::: Scattering — phonons and ionized impurities. More scattering → smaller $\tau$ → smaller $\mu$. Why do electron and hole currents add rather than cancel? ::: Electron has negative $q$ and reversed drift; the two sign flips multiply to give current in the same direction as the hole current. Why is there no factor of ½ in $v_d=a\tau$? ::: Free-flight times are exponentially distributed (memoryless collisions); averaging $at$ over that distribution gives $a\tau$, not $\frac{1}{2}a\tau$. What happens to $\mu$ as $\tau \to 0$? ::: $\mu \to 0$ — a carrier hit instantly never builds up any drift.

Connections

• Parent: Concept of carrier mobility — this page is the picture-by-picture derivation of its boxed result.
• Effective mass — the $m^{\ast }$ that enters at Step 2.
• Scattering mechanisms in semiconductors — sets the $\tau$ of Step 3.
• Drift and Diffusion currents — $v_d=\mu E$ is the drift term.
• Conductivity and Resistivity — $\sigma =n|q|\mu$ from Step 6.
• Doping and carrier concentration — supplies the $n$ of Step 6.
• Einstein relation — links this $\mu$ to diffusion, $D=\mu kT/q$.