Understand conductors, insulators, and semiconductors: Worked examples

This page belongs under the parent topic and pushes the numbers all the way through. Nothing here contradicts the parent — we just walk every case the topic can throw at you, one worked example per cell.

Before we begin, one promise: every symbol below was defined in the parent, but we re-anchor each one in plain words the first time it appears here, so you never meet a letter you haven't been introduced to.

The scenario matrix

Think of this whole topic as a machine with a few knobs. The knobs are: which material (governed by carrier density $n$ — re-anchored just below), temperature $T$ , geometry ( $L$ , $A$ ), and doping. Each cell below is one twist of a knob into a corner case. The examples that follow hit every cell.

Three symbols show up in the matrix (rows G, H and beyond), so let's re-anchor them before you meet them in the table:

Definition Carrier density

$n$ , conductivity $σ$ , mobility $μ$ (plain words)

==Carrier density $n$ == (units $m^{- 3}$ , "how many per cubic metre") is the head-count of free charges — the electrons (and holes) actually able to move and carry current inside one cubic metre of the material. This single number is what separates the three classes: conductors have a huge $n$ , insulators almost none, semiconductors a small but tunable $n$ .
==Conductivity $σ$ == (units $S/m$ , "siemens per metre") is a material's willingness to carry current — literally the inverse of resistivity, $σ = 1/ ρ$ . Big $σ$ = easy flow.
==Mobility $μ$ == (units $m^{2} / (V \cdot s)$ ) is how fast a carrier drifts per unit of electric push — nimble carriers have high $μ$ .
Together they build the master formula $σ = n e μ$ , where $e$ is the charge on one carrier.

Cell	What is being tested	Corner it probes	Example
A	Classify by resistivity $ρ$	tiny $ρ$ (conductor edge)	Ex 1
B	Classify by resistivity $ρ$	huge $ρ$ (insulator edge)	Ex 2
C	Geometry: $R = ρ L / A$	ordinary shaped wire	Ex 3
D	Geometry limits	degenerate: both $R \to 0$ and $R \to \infty$	Ex 4
E	Temperature — semiconductor	$T$ up → $ρ$ down	Ex 5
F	Temperature — metal	$T$ up → $ρ$ up (opposite sign)	Ex 6
G	Doping ratio	$n$ jumps by orders of magnitude	Ex 7
H	Microscopic $σ = n e μ$ (conductivity from $n, e, μ$ )	build $σ$ from scratch	Ex 8
I	Real-world word problem	pick a material for a job	Ex 9
J	Exam twist / zero input	$n \to 0$ limiting behaviour	Ex 10

Two more quantities do the geometry work, so let's re-anchor them once:

Prerequisite links if you need them: Ohm's Law and Resistance, Electric Charge and Current, Energy Bands in Solids, Temperature Coefficient of Resistance, Semiconductor Diodes and Transistors.

Cell A — classify a tiny resistivity

Cell B — classify a huge resistivity

Cell C — resistance of an ordinary wire (with figure)

Figure — Understand conductors, insulators, and semiconductors

Worked example Example 3 — Aluminium wire resistance

Aluminium, $ρ = 2.8 \times 1 0^{- 8} Ω \cdot m$ , length $L = 10 m$ , cross-section area $A = 2 \times 1 0^{- 6} m^{2}$ . Find $R$ . Forecast: Bigger or smaller than 1 ohm? Guess before computing.

Step 1 — Choose the geometry formula. Use $R = ρ \frac{L}{A}$ . Why this step? We asked about this object, not the raw material, so shape ( $L$ and $A$ ) must enter. Look at the figure: the current runs along the red length $L$ and squeezes through the mint face $A$ .

Step 2 — Substitute. $R = \frac{2.8 \times 1 0 ^{- 8} \times 10}{2 \times 1 0 ^{- 6}} = \frac{2.8 \times 1 0 ^{- 7}}{2 \times 1 0 ^{- 6}} = 0.14 Ω.$ Why this step? Longer wire (bigger $L$ ) pushes $R$ up; fatter wire (bigger $A$ ) pulls $R$ down — the formula encodes both, exactly as the figure's arrows suggest.

Answer: $R = 0.14 Ω$ .

Verify (units): $\frac{( Ω \cdot m ) ( m )}{m ^{2}} = Ω$ . ✓ A tenth of an ohm is tiny — correct for a metal.

Cell D — degenerate geometry (all four limits)

Worked example Example 4 — Push the geometry to both extremes

Use copper, $ρ = 1.7 \times 1 0^{- 8} Ω \cdot m$ , throughout. Ask four things: what does $R = ρ L / A$ do as (i) $L \to 0$ , (ii) $A \to \infty$ , (iii) $L \to \infty$ , (iv) $A \to 0$ ? Forecast: Two of these send $R$ to zero and two send $R$ to infinity — sort them before reading on.

Step 1 — The two $R \to 0$ limits. In $R = ρ L / A$ , length $L$ is in the numerator and area $A$ in the denominator. So $L \to 0$ gives $R \to 0$ , and $A \to \infty$ gives $R \to 0$ . Why this step? No length means no distance for collisions to accumulate; a huge area is like millions of wires in parallel. Both make resistance vanish — this is why chip interconnects are short and power buses are fat.

Step 2 — The two $R \to \infty$ limits. Flip each: $L \to \infty$ puts an unbounded numerator over a fixed denominator, so $R \to \infty$ ; $A \to 0$ puts a fixed numerator over a vanishing denominator, so $R \to \infty$ . Why this step? An infinitely long wire is endless collisions; a needle-thin wire squeezes all the current through almost nothing. Both choke the flow completely.

Step 3 — A concrete point between the extremes. For $ρ = 1.7 \times 1 0^{- 8}$ , $L = 0.001 m$ , $A = 1 \times 1 0^{- 6} m^{2}$ : $R = \frac{1.7 \times 1 0 ^{- 8} \times 0.001}{1 \times 1 0 ^{- 6}} = 1.7 \times 1 0^{- 5} Ω.$ Why this step? A short, ordinary-area copper wire lands very close to the $R \to 0$ corner — a real number to anchor the limits.

Step 4 — Cross-check against a longer wire (self-contained). Take the same copper but with $L = 2 m$ , $A = 1 \times 1 0^{- 6} m^{2}$ : $R = \frac{1.7 \times 1 0 ^{- 8} \times 2}{1 \times 1 0 ^{- 6}} = 3.4 \times 1 0^{- 2} Ω = 0.034 Ω.$ Why this step? Comparing two wires that differ only in length isolates the $L$ dependence cleanly.

Answer: $L \to 0$ and $A \to \infty$ give $R \to 0$ ; $L \to \infty$ and $A \to 0$ give $R \to \infty$ . The concrete short wire is $1.7 \times 1 0^{- 5} Ω$ ; the 2 m wire is $0.034 Ω$ .

Verify: $0.034 Ω/1.7 \times 1 0^{- 5} Ω = 2000$ , exactly the length ratio $2 m /0.001 m = 2000$ . ✓ Longer = proportionally more resistive, confirming the $L \to \infty$ direction.

Cell E — temperature raises carriers in a semiconductor (with figure)

The figure plots relative carrier density $n$ (vertical axis, normalised so $n = 1$ at 300 K) against temperature $T$ in kelvin (horizontal axis). The lavender curve is $n \propto e^{- E_{g} / (2 k_{B} T)}$ : notice how flat it is at low $T$ (almost no electrons have climbed the gap) and how it swings sharply upward as $T$ rises — that steep bend is the whole story of thermal excitation across the band gap. The mint dot marks the reference point at 300 K ( $n = 1$ ); the coral dot marks 350 K, sitting about 21× higher. The shaded region beneath the curve is the growing population of freed electron–hole pairs. Read it as: warm the crystal a little, and the head-count of mobile carriers balloons — so resistivity falls.

Worked example Example 5 — Heat silicon from 300 K to 350 K

Silicon's band gap $E_{g} = 1.1 eV$ . Carrier density follows $n \propto e^{- E_{g} / (2 k_{B} T)}$ , where $k_{B}$ is Boltzmann's constant. In electron-volt units $k_{B} = 8.617 \times 1 0^{- 5} eV/K$ . By what factor does $n$ grow going from $T_{1} = 300 K$ to $T_{2} = 350 K$ ? Forecast: A modest $50 K$ warm-up — factor of ~2? ~10? ~100? Guess.

Step 1 — Why the exponential $e^{- E_{g} / (2 k_{B} T)}$ ? Electrons must be thermally kicked across the gap. The exponential is the tool that answers "what fraction of particles has enough thermal energy to clear an energy wall?" — nothing linear captures that steepness. The factor of 2 in the exponent is because each freed electron leaves a hole, so the gap energy is shared between the electron–hole pair. See the figure: the tiny climbing population balloons with $T$ .

Step 2 — Form the ratio to kill the unknown constant. $\frac{n _{2}}{n _{1}} = exp [- \frac{E _{g}}{2 k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})] .$ Why this step? The proportionality constant is unknown, but it cancels in a ratio — a standard trick.

Step 3 — Plug numbers. $\frac{E _{g}}{2 k _{B}} = \frac{1.1}{2 ( 8.617 \times 1 0 ^{- 5} )} = 6383 K .$ $\frac{1}{T _{2}} - \frac{1}{T _{1}} = \frac{1}{350} - \frac{1}{300} = - 4.762 \times 1 0^{- 4} K^{- 1} .$ $\frac{n _{2}}{n _{1}} = exp [- 6383 \times (- 4.762 \times 1 0^{- 4})] = exp (3.039) \approx 20.9.$

Answer: Carriers jump about 21-fold for a mere 50 K rise.

Verify: Since $σ = n e μ$ scales with $n$ , conductivity rises ~21× → resistivity falls ~21×. Correct sign for a semiconductor.

Cell F — the opposite sign: a metal gets more resistive when hot

Worked example Example 6 — Copper heated from 20 °C to 100 °C

Copper's temperature coefficient of resistance is $α = 0.0039 K^{- 1}$ , using the linear law $ρ (T) = ρ_{0} [1 + α (T - T_{0})]$ . Start at $T_{0} = 20^{\circ} C$ with $ρ_{0} = 1.7 \times 1 0^{- 8} Ω \cdot m$ . Find $ρ$ at $100^{\circ} C$ . Forecast: Does copper's $ρ$ go up or down? (Careful — this is the trap case.)

Step 1 — Choose the linear metal law, not the exponential. For a metal, $n$ is fixed (bands already overlap); heat only makes ions vibrate more, so electrons collide more, shortening the average collision time $τ$ . Since $μ = e τ / m$ , mobility drops and $ρ$ rises linearly over ordinary ranges. Why this step? Wrong law = wrong sign. Metal ↑T → $ρ$ ↑; semiconductor ↑T → $ρ$ ↓. See Temperature Coefficient of Resistance.

Step 2 — Substitute $Δ T = 80 K$ . $ρ = 1.7 \times 1 0^{- 8} [1 + 0.0039 \times 80] = 1.7 \times 1 0^{- 8} \times 1.312 = 2.23 \times 1 0^{- 8} Ω \cdot m .$

Answer: $ρ \approx 2.23 \times 1 0^{- 8} Ω \cdot m$ — higher than the cold value.

Verify: Increase factor $= 1 + 0.0039 \times 80 = 1.312$ , i.e. a $31.2%$ rise. Opposite direction to the silicon in Ex 5 — exactly the contrast the parent warns about.

Cell G — doping multiplies carriers

Worked example Example 7 — Doping silicon

Intrinsic (pure) silicon has $n_{i} = 1.5 \times 1 0^{16} m^{- 3}$ free carriers. Dope it with phosphorus to $n_{d} = 5 \times 1 0^{22} m^{- 3}$ . By what factor does conductivity rise? (Assume $e$ and mobility $μ$ roughly unchanged.) Forecast: Thousands? Millions? Guess the order of magnitude.

Step 1 — Use $σ = n e μ$ . With $e$ and $μ$ fixed, $σ$ is directly proportional to $n$ . Why this step? Doping's whole job is to change only $n$ — the knob that separates the three classes.

Step 2 — Take the ratio. $\frac{σ _{d}}{σ _{i}} = \frac{n _{d}}{n _{i}} = \frac{5 \times 1 0 ^{22}}{1.5 \times 1 0 ^{16}} \approx 3.33 \times 1 0^{6} .$

Answer: Conductivity jumps about 3.3 million-fold from a pinch of impurity.

Verify: $lo g_{10} (3.33 \times 1 0^{6}) \approx 6.52$ — a leap of over six decades, dwarfing anything temperature alone (Ex 5's ~21×) could do. That controllability is why silicon computes, not copper. See Semiconductor Diodes and Transistors.

Cell H — build conductivity from scratch (microscopic)

Worked example Example 8 — Compute

$σ$ and $ρ$ of doped silicon from $n, e, μ$ Take $n = 5 \times 1 0^{22} m^{- 3}$ , elementary charge $e = 1.6 \times 1 0^{- 19} C$ , electron mobility $μ = 0.14 m^{2} / (V \cdot s)$ . Find $σ$ , then $ρ$ . Forecast: Will $ρ$ land in the semiconductor band ( $1 0^{- 4}$ – $1 0^{3}$ )?

Step 1 — Assemble $σ = n e μ$ . Why this step? This is the master formula — each factor is a physical count: $n$ carriers, each of charge $e$ , each moving with mobility $μ$ (drift speed per unit field). $σ = (5 \times 1 0^{22}) (1.6 \times 1 0^{- 19}) (0.14) = 1120 S/m .$

Step 2 — Invert for resistivity. $ρ = 1/ σ$ . Why this step? Resistivity is just the reciprocal — "stubbornness" is the inverse of "willingness." $ρ = \frac{1}{1120} \approx 8.93 \times 1 0^{- 4} Ω \cdot m .$

Answer: $σ \approx 1120 S/m$ , $ρ \approx 8.9 \times 1 0^{- 4} Ω \cdot m$ .

Verify: $8.9 \times 1 0^{- 4}$ sits at the low edge of the semiconductor band ( $1 0^{- 4}$ – $1 0^{3}$ ) — sensible for heavily doped silicon. Units: $m^{- 3} \cdot C \cdot m^{2} V^{- 1} s^{- 1} = \frac{C}{V \cdot s \cdot m} = \frac{A}{V \cdot m} = \frac{S}{m}$ . ✓

Cell I — real-world word problem

Worked example Example 9 — Pick the material for a phone charger cable core vs. its outer sheath

A charger cable must (a) carry current through its core with almost no heat loss, and (b) be safe to touch on the outside. Choose $ρ$ -appropriate materials for each part, and justify with power loss $P = I^{2} R$ . Use copper for the core, $ρ_{Cu} = 1.7 \times 1 0^{- 8} Ω \cdot m$ , with length $L = 10 m$ and cross-section $A = 2 \times 1 0^{- 6} m^{2}$ , carrying $I = 2 A$ . Forecast: Which class goes inside, which goes outside?

Step 1 — Core: minimise $R$ to minimise heat. Power wasted as heat is $P = I^{2} R$ , and $R = ρ L / A$ . To keep $P$ low we need tiny $ρ$ → a conductor; we pick copper ( $ρ_{Cu} = 1.7 \times 1 0^{- 8}$ ). Why this step? Heat loss scales linearly with $R$ , hence with $ρ$ ; conductors minimise it.

Step 2 — Sheath: maximise $R$ so no current reaches your hand. We want $ρ$ enormous → an insulator (rubber/PVC, $ρ \sim 1 0^{13}$ ). Why this step? A large gap ( $> 5$ eV) keeps electrons stuck, blocking leakage current.

Step 3 — Quantify the copper core loss. First the core resistance, using the copper $ρ$ from the statement: $R = ρ_{Cu} \frac{L}{A} = \frac{1.7 \times 1 0 ^{- 8} \times 10}{2 \times 1 0 ^{- 6}} = 0.085 Ω.$ Then the wasted power: $P = I^{2} R = (2)^{2} (0.085) = 0.34 W .$ Why this step? We now use a single, consistent material (copper) end-to-end — no material swap mid-problem.

Answer: Copper core (low $ρ$ ), rubber sheath (high $ρ$ ); core resistance $0.085 Ω$ , core loss $\approx 0.34 W$ .

Verify: If we foolishly used a semiconductor core with $ρ = 8.9 \times 1 0^{- 4}$ (Ex 8) at the same geometry, its $R$ would be $8.9 \times 1 0^{- 4} /1.7 \times 1 0^{- 8} \approx 5.2 \times 1 0^{4}$ times larger — dumping tens of kilowatts as heat. That is exactly why we choose a conductor.

Cell J — the zero / limiting input (exam twist)

Worked example Example 10 — What is

$ρ$ of a perfect insulator as $n \to 0$ ? An idealised insulator has essentially no free carriers, $n \to 0$ . Using $ρ = 1/ (n e μ)$ , what happens to $ρ$ ? And what does $σ$ do? Forecast: Does $ρ$ go to $0$ or to $\infty$ ?

Step 1 — Track $σ = n e μ$ as $n \to 0$ . With $n \to 0$ , the product $n e μ \to 0$ , so $σ \to 0$ . Why this step? Conductivity is a count of moving charges; no carriers means no conduction.

Step 2 — Invert. $ρ = 1/ σ \to 1/0 \to + \infty$ . Why this step? Reciprocal of zero blows up — a perfect insulator has infinite resistivity in this idealisation. Sanity on the physics: real insulators aren't literally infinite; a few carriers always sneak across the gap, so $ρ$ is huge but finite ( $1 0^{10}$ – $1 0^{16}$ ).

Step 3 — Contrast the conductor limit. As $n \to \infty$ , $σ \to \infty$ and $ρ \to 0$ — the perfect conductor.

Answer: $n \to 0 \Rightarrow σ \to 0, ρ \to \infty$ (ideal insulator); $n \to \infty \Rightarrow ρ \to 0$ (ideal conductor).

Verify: Test with a small-but-nonzero $n = 1 0^{6} m^{- 3}$ , $e = 1.6 \times 1 0^{- 19}$ , $μ = 0.1$ : $σ = 1.6 \times 1 0^{- 14} S/m$ , so $ρ = 6.25 \times 1 0^{13} Ω \cdot m$ — squarely in the insulator band. The limit direction is confirmed.

Recall Quick self-test on the matrix

Metal heated: $ρ$ up or down? ::: Up (fixed $n$ , more collisions, $τ$ falls) Pure semiconductor heated: $ρ$ up or down? ::: Down ( $n \propto e^{- E_{g} /2 k_{B} T}$ grows fast) Which knob does doping turn? ::: The carrier density $n$ (raises $σ = n e μ$ ) As $n \to 0$ , resistivity $ρ$ tends to? ::: Infinity (ideal insulator) As $A \to 0$ or $L \to \infty$ , resistance $R$ tends to? ::: Infinity Does $R = ρ L / A$ care about material only? ::: No — it also depends on shape ( $L$ , $A$ )