Foundations — Concept of carrier mobility

This page assumes you know nothing about the notation used in the parent topic. We will earn every letter before it is allowed to appear in a formula.

0 — What a "charge carrier" even is

Picture a huge dark room full of ping-pong balls flying in every direction. Each ball is a carrier. Right now they go nowhere as a group — for every ball heading left, another heads right.

Look at the figure: the many pale arrows point every which way. Their average (the short yellow arrow in the centre) is basically zero. This "average of all the little arrows" idea is the single most important picture on this page — hold onto it.

Why does the topic need this? Because current is net motion of carriers. If we can't talk about carriers and their average motion, we can't talk about current at all.

1 — The symbol $\vec{E}$: the electric field (the "tilt")

The picture: imagine the flat floor of that ping-pong room slowly tipped downhill. Balls still bounce randomly, but there is now a gentle bias — everything tends to slide the downhill way. $\vec{E}$ is that tilt, and the steepness of the tilt is the magnitude $E$.

Why the topic needs it: mobility's whole job is to convert "how hard I tilt" ($E$) into "how fast the crowd slides." Without $E$ there is nothing to convert.

2 — The symbol $q$: the carrier's charge

The picture: how "grabbable" a single ball is by the tilt. A ball with more charge feels a stronger downhill pull.

3 — Force, and the symbol $\vec{F}=q\vec{E}$

Why this combination and not another? Because $\vec{E}$ was defined as force-per-charge. Multiply the "push per charge" ($\vec{E}$) by "how much charge you have" ($q$) and you recover the actual push ($\vec{F}$). It's just undoing the division we did in Section 1.

Picture: the steeper the tilt (bigger $E$) or the more grabbable the ball (bigger $|q|$), the bigger the downhill shove $F$. (For an electron the arrow $\vec{F}$ points opposite $\vec{E}$ because $q<0$; the magnitude is still $|q|E$.)

4 — Effective mass $m^{\ast }$ and Newton's $a=F/m^{\ast }$

Why do we need a starred mass instead of the ordinary mass $m_0=9.11\times 10^{-31}\text{kg}$? Because the carrier is not in empty space — it is surrounded by a regular grid of atoms that push and pull on it. Bundling all those grid effects into one number $m^{\ast }$ lets us keep using plain Newton's law.

The tool we just reached for is Newton's second law ($\vec{F}=m^{\ast }\vec{a}$). Why this tool? Because it is exactly the machine that turns a force into a change of motion — and we have a force ($q\vec{E}$) and want to know the resulting motion. Rearranged, $\vec{a}=\vec{F}/m^{\ast }$: same push on a heavier ball gives a smaller acceleration. That is why $m^{\ast }$ sits in the denominator.

5 — Two different speeds: thermal vs drift

Here is the subtlety the whole topic hinges on. A carrier has two velocities happening at once.

Look at the figure: the wild zig-zag (blue) is thermal motion; the smooth pink arrow underneath is the slow net progress in the field's direction — that slow progress is ${\vec{v}}_d$. The zig-zag is huge; the drift is gentle. Never confuse them.

Why the topic needs both: current comes only from the drift part. The thermal part is enormous but cancels to zero, contributing no net current.

6 — Mean free time $\tau$: the "time between crashes"

Picture: a ball rolling downhill on the tilted floor, but every so often it smacks into a bump and ricochets off in a random direction, forgetting the little downhill speed it had built up. $\tau$ is the average time between two such smacks.

In the figure, each collision (yellow dot) resets the carrier's forward speed to roughly zero; between dots the field speeds it up along a straight ramp. The average height of those saw-teeth is the drift velocity.

7 — Putting it together: $v_d=a\tau$

Now every symbol is earned, so we can assemble the drift velocity. Since $\vec{a}$ and ${\vec{v}}_d$ point along the same line as $\vec{E}$, we can drop the arrows and work with magnitudes.

• WHAT we do: take the constant-acceleration rule $v=at$ (start from rest, speed after time $t$ is $at$) and set the time equal to the average free time $\tau$.
• WHY: after each collision the carrier restarts near zero net forward speed, then accelerates for a typical time $\tau$. So its typical gained speed is $a\tau$.
• WHAT IT LOOKS LIKE: the average height of the saw-teeth in the previous figure.

Every piece here we defined above: $|q|$ (§2), $E$ (§1), $m^{\ast }$ (§4), $\tau$ (§6).

8 — The last symbol: mobility $\mu$

Divide the drift velocity from §7 by $E$ and watch $E$ cancel: $\mu =\frac{v_d}{E}=\frac{1}{E}\cdot \frac{|q|\tau }{m^{\ast }}E=\frac{|q|\tau }{m^{\ast }}$

Why divide by $E$? Because $v_d$ depends on how hard you happen to push; dividing by $E$ strips your choice away and leaves a number belonging to the material alone ($|q|$, $\tau$, $m^{\ast }$). That is what makes $\mu$ a useful material property you can look up in a table. Using the magnitude $|q|$ guarantees $\mu >0$ for both electrons and holes.

Everything else in the parent — $J=n|q|\mu E$ (current), $\sigma =n|q|\mu$ (see Conductivity and Resistivity), the role of $n$ from Doping and carrier concentration, and the Einstein relation $D=\mu kT/|q|$ — is built by attaching more carriers or more uses onto this same $\mu$.

Prerequisite map

Equipment checklist