Thermal effects on conductivity: Worked examples — Hardware deep dive

This page is the "no surprises" drill for Thermal effects on conductivity. We first lay out every kind of question the topic can ask, then work one example per cell so you never meet a case you haven't already seen.

Definition The names every example uses (read this first)

These symbols recur in every worked example, so we pin them down once, up front:

$R$ = resistance of a specific object (a whole wire), measured in ohms $Ω$ . This is what a multimeter reads.
$ρ$ = resistivity of the material itself, independent of shape, in $Ω \cdot$ m. Related by $R = ρ L / A$ (length over cross-section).
When to use which: if the problem gives a resistance in $Ω$ (Ex 1–3, 9) work in $R$ ; if it talks about the material's property or compares two materials (Ex 4, 10) work in $ρ$ . Both obey the same temperature laws because $R$ and $ρ$ differ only by the fixed geometry factor $L / A$ .
$R_{0}$ = reference resistance: the resistance measured at a chosen baseline temperature. $T_{0}$ = reference temperature: that chosen baseline (often $0^{\circ}$ C or $2 0^{\circ}$ C). The linear law measures every change relative to this pair.
$Δ T = T - T_{0}$ = change in temperature from the reference. It is positive when we heat above $T_{0}$ and negative when we cool below it — the sign matters and is never dropped.

The scenario matrix

Before symbols scare anyone: below, $T$ is temperature, $R$ is resistance, $ρ$ is resistivity (how much a material resists per unit size), $α$ is the "resistance-per-degree" number, $E_{g}$ is the band gap (the energy price to free one carrier in a semiconductor), and $k_{B}$ is Boltzmann's constant (the exchange rate between temperature and energy).

#	Case class	What's special about it	Example
A	Metal, heating up ( $Δ T > 0$ )	linear law, $R$ rises	Ex 1
B	Metal, cooling down ( $Δ T < 0$ )	linear law, sign flips, $R$ falls	Ex 2
C	Metal, find $α$ from two measurements	invert the linear law	Ex 3
D	Metal limiting case: $T \to 0$ (residual resistance)	law breaks, floor appears	Ex 4
E	Semiconductor ratio $n (T_{2}) / n (T_{1})$	exponential, factor-of-2 gap	Ex 5
F	Semiconductor, negative $Δ T$ (cooling)	exponent flips sign, $n$ collapses	Ex 6
G	Find $E_{g}$ from a measured conductivity ratio	invert the exponential (needs $ln$ )	Ex 7
H	Common-error case: "heat a metal, does $n$ jump too?"	$n$ ≈ constant, don't use the exp law	Ex 8
I	Real-world word problem (thermometer / thermistor)	pick the right law from the story	Ex 9
J	Exam twist: metal and semiconductor, crossover $T$	set $ρ$ metal $= ρ$ semi	Ex 10
K	Extrinsic (doped) semiconductor, saturation & freeze-out	$n$ set by impurities, not the gap	Ex 11

(Cell H is labelled a common-error case, not a "trap": it is the scenario where the natural but wrong instinct is to reuse the semiconductor exponential on a metal. We work it to show why that instinct fails.)

Case A — Metal heated

Case B — Metal cooled

Figure — Thermal effects on conductivity

What the figure shows (in words): the metal linear law $R (T)$ drawn as a single straight blue line through the reference point $(T_{0}, R_{0})$ marked with a yellow dot. Heating (going right, $Δ T > 0$ , Ex 1) climbs the line to the red point above $R_{0}$ ; cooling (going left, $Δ T < 0$ , Ex 2) slides down to the green point below $R_{0}$ . The two example points sit symmetrically on either side of the yellow reference dot.

Case C — Recover $α$

Case D — Metal limiting case $T \to 0$

What the figure shows (in words): two curves of resistivity $ρ$ against temperature $T$ . The dashed yellow line is the idealized vibration-only model — a straight line through the origin that (wrongly) dives to $ρ = 0$ at $T = 0$ . The solid blue curve is the real metal: it runs parallel to the yellow line at high $T$ but, as $T \to 0$ , levels off onto a flat red floor at a small positive value $ρ_{res}$ instead of reaching zero.

Worked example Ex 4 (Cell D) — Where the linear law dies

Near room temperature the metal law is $ρ (T) = ρ_{0} [1 + α (T - T_{0})]$ . Expand the bracket: $ρ (T) = constant ρ_{0} (1 - α T_{0}) + a ρ_{0} α T$ . So the slope of the $T$ -part is exactly $a = ρ_{0} α$ , and the constant is the floor. If you keep only the vibration-driven (sloped) part referenced to $T_{0} = 0$ , it is $ρ_{vib} (T) = a T$ with $a = ρ_{0} α$ , which would vanish at $T = 0$ . A real copper sample instead flattens to a floor $ρ_{res} = 2 \times 1 0^{- 11} Ω \cdot$ m. If the vibration part at $300$ K is $ρ_{vib} = 1.7 \times 1 0^{- 8} Ω \cdot$ m, what is the total $ρ$ at $300$ K, and what does the graph look like as $T \to 0$ ?

Forecast: does the curve hit zero, or stop somewhere above it?

Step 1. Split the resistivity into its two physical pieces via Matthiessen's rule: $ρ (T) = ρ_{res} + ρ_{vib} (T)$ with $ρ_{vib} (T) = a T$ and $a = ρ_{0} α$ . Why this step? Electrons scatter off two independent things: frozen impurities (temperature-independent, the constant $ρ_{res}$ ) and lattice vibrations (the $\propto T$ part). Separating them lets us see which survives as $T \to 0$ . It also explains why the room-temperature linear law has a non-zero intercept: that intercept is $ρ_{res}$ . Step 2. At $300$ K: $ρ = 2 \times 1 0^{- 11} + 1.7 \times 1 0^{- 8} = 1.702 \times 1 0^{- 8} Ω \cdot$ m. Why this step? We simply add the two contributions at the temperature asked, to get the measurable total. Step 3. As $T \to 0$ : $ρ_{vib} = a T \to 0$ but $ρ_{res}$ stays. So $ρ \to 2 \times 1 0^{- 11} Ω \cdot$ m, not zero — the flat blue floor in the figure. Why this step? Taking the limit isolates the survivor; it is the whole point of this "limiting-case" cell.

Verify: The vibration part dominates at room temp ( $1.7 \times 1 0^{- 8}$ vs $2 \times 1 0^{- 11}$ — a factor $\sim 850$ ), so the total is barely above the pure-vibration value. ✔ The limit is finite: the " $ρ \to 0$ " prediction was the failure of the pure-slope piece, exactly the case this cell warns about. (See Superconductivity for the truly-zero case, which is a different mechanism.)

Case E — Semiconductor carrier ratio

Worked example Ex 5 (Cell E) — Silicon, 300K → 350K

Silicon has $E_{g} = 1.1$ eV. By what factor does the carrier density $n$ change from $300$ K to $350$ K? Use $k_{B} = 8.617 \times 1 0^{- 5}$ eV/K.

Forecast: a mere $50$ K rise — factor of 2? 5? 20?

Step 1. Ratio $= \frac{n ( T _{2} )}{n ( T _{1} )} = exp [- \frac{E _{g}}{2 k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})]$ . Why this step? The full intrinsic law is $n \propto T^{3/2} e^{- E_{g} / (2 k_{B} T)}$ ; the $T^{3/2}$ prefactor comes from counting available quantum states (the density of states). We drop it here because the exponential dominates so overwhelmingly — over $300 \to 350$ K the $T^{3/2}$ factor contributes only $(350/300)^{3/2} = 1.26 \times$ , tiny beside the $\sim 21 \times$ from the exponential. The exponential is the right tool because carrier creation is a probability of clearing an energy barrier, and such Boltzmann probabilities are exponential in $- energy / k_{B} T$ . Step 2. Compute the constant $\frac{E _{g}}{2 k _{B}} = \frac{1.1}{2 \times 8.617 \times 1 0 ^{- 5}} = 6383$ K. Why this step? Bundling $E_{g} / (2 k_{B})$ into a single number (in kelvin) makes the exponent a clean product and exposes it as a characteristic temperature scale for this material. Step 3. Compute the temperature bracket $\frac{1}{350} - \frac{1}{300} = - 4.762 \times 1 0^{- 4}$ K $^{- 1}$ . Why this step? This is the actual driver of the change; its negative sign (because the hotter $T_{2}$ has the smaller reciprocal) is what will make the final ratio exceed 1. Step 4. Exponent $= - 6383 \times (- 4.762 \times 1 0^{- 4}) = + 3.04$ . Ratio $= e^{3.04} \approx 20.9$ (exponential part only). Why this step? Multiplying the two pieces and exponentiating gives the physical answer; keeping only the exponential is justified by Step 1.

Verify: $T_{2} > T_{1}$ makes $1/ T_{2} < 1/ T_{1}$ , so the bracket is negative, and the leading minus flips it positive → ratio $> 1$ , i.e. $n$ grows on heating. ✔ Matches the parent note's "~21×". Including the $T^{3/2}$ prefactor would bump it to $\approx 20.9 \times 1.26 \approx 26$ , so quoting "~21×" (exponential only) is a slight under-estimate but captures the physics. Compare with a metal (Ex 1) where $n$ barely moved — this is the Intrinsic vs Extrinsic Semiconductors signature.

Case F — Semiconductor cooled

Worked example Ex 6 (Cell F) — Same silicon cooled 300K → 250K

By what factor does $n$ change on cooling from $300$ K to $250$ K? (Exponential part only, as in Ex 5.)

Forecast: cooling a semiconductor should starve it of carriers — factor above or below 1?

Step 1. Same ratio formula, now $T_{2} = 250$ K, $T_{1} = 300$ K. Why this step? The formula doesn't care which direction; the sign of the bracket handles it. Step 2. Compute the temperature bracket $\frac{1}{250} - \frac{1}{300} = + 6.667 \times 1 0^{- 4}$ K $^{- 1}$ (now positive, since $250 < 300$ ). Why this step? The sign is the whole story here: cooling makes the destination reciprocal $1/ T_{2}$ the larger one, flipping the bracket positive — the mirror of Ex 5's negative bracket. Step 3. Exponent $= - 6383 \times (6.667 \times 1 0^{- 4}) = - 4.26$ . Ratio $= e^{- 4.26} \approx 0.0141$ . Why this step? The leading minus times a positive bracket gives a negative exponent, so the ratio drops below 1 — quantifying how badly carriers are starved.

Verify: Cooling → fewer thermally-freed carriers → $n$ drops to about $1.4%$ of its value. Ratio $< 1$ . ✔ This is why cold intrinsic semiconductors become near-insulators — the mirror image of Ex 5. (Doped ones behave differently at low $T$ ; see Ex 11.)

Case G — Recover the band gap

Worked example Ex 7 (Cell G) — Measure

$E_{g}$ from a conductivity ratio A semiconductor's intrinsic conductivity rises by a factor of $8.0$ when heated from $300$ K to $340$ K. Estimate $E_{g}$ (ignore the small mobility change and the $T^{3/2}$ prefactor). $k_{B} = 8.617 \times 1 0^{- 5}$ eV/K.

Forecast: eyeballing — a small gap (< 0.5 eV) or a big gap (~1 eV)?

Step 1. Start from $\frac{σ _{2}}{σ _{1}} = exp [- \frac{E _{g}}{2 k _{B}} (\frac{1}{T _{2}} - \frac{1}{T _{1}})]$ and take $ln$ of both sides. Why this step? $E_{g}$ is trapped inside an exponential; the natural log is the exact inverse of $exp$ , the only tool that frees it. Step 2. Left side: $ln (8.0) = 2.079$ . Right-side bracket: $\frac{1}{340} - \frac{1}{300} = - 3.922 \times 1 0^{- 4}$ K $^{- 1}$ . Why this step? We evaluate both known numbers now so that after the log the only unknown remaining is $E_{g}$ ; the negative bracket combined with the measured increase will force $E_{g}$ positive. Step 3. Solve for the unknown: $E_{g} = - \frac{2 k _{B} ln 8.0}{( 1/340 - 1/300 )} = - \frac{2 ( 8.617 \times 1 0 ^{- 5} ) ( 2.079 )}{- 3.922 \times 1 0 ^{- 4}}$ . Why this step? Algebraically isolating $E_{g}$ is the goal of the whole "recover the gap" cell; the two leading minus signs cancel to give a positive energy. Step 4. $E_{g} = \frac{3.583 \times 1 0 ^{- 4}}{3.922 \times 1 0 ^{- 4}} = 0.914$ eV. Why this step? Dividing the assembled numerator by the bracket delivers the physical band gap in eV.

Verify: Round-trip it — with $E_{g} = 0.914$ : exponent $= - \frac{0.914}{2 ( 8.617 \times 1 0 ^{- 5} )} (- 3.922 \times 1 0^{- 4}) = 2.079 = ln 8$ , so $σ$ ratio $= 8$ . ✔ The gap ( $\sim 0.9$ eV) sits between Ge ( $0.67$ ) and Si ( $1.1$ ) — physically reasonable. Links to Band Theory & Energy Gaps.

Case H — The common-error case

Worked example Ex 8 (Cell H) — "Does copper's

$n$ jump like silicon's?" Someone tries to use $n \propto e^{- E_{g} / (2 k_{B} T)}$ on copper heated $300$ K → $350$ K, claiming $n$ should rise ~21× like silicon. What's wrong, and what actually happens to copper's $n$ ?

Forecast: is the exponential law even applicable here?

Step 1. Check the premise: does copper have a band gap for its conduction electrons? No — in a metal the conduction band is already partly filled (Free Electron Model of Metals), so $E_{g} \approx 0$ for freeing carriers. Why this step? A formula built for "cross the gap" carriers is meaningless when there is no gap to cross. Step 2. With $E_{g} \to 0$ , $e^{- E_{g} / (2 k_{B} T)} \to e^{0} = 1$ — no temperature dependence of $n$ from that mechanism. Step 3. So copper's $n$ is essentially constant; the resistance change comes entirely from $μ$ (via $τ$ ), i.e. use the metal linear law, not the exponential.

Verify: Ex 1 already gave copper's real behaviour — a modest $\sim 20%$ resistance rise, not a 20× carrier explosion. Using the wrong law would over-predict by orders of magnitude. ✔ This is the Doping and Carrier Concentration vs metallic-sea distinction.

Case I — Real-world word problem

Worked example Ex 9 (Cell I) — Thermistor thermometer

A thermistor (a semiconductor sensor) has resistance $R_{1} = 12$ k $Ω$ at $T_{1} = 298$ K and behaves as $R \propto e^{+ B / T}$ with $B = 3500$ K. What is $R$ at body temperature $T_{2} = 310$ K, and is this a positive or negative temperature coefficient device?

Forecast: blood is warmer than room — will the reading go up or down?

Step 1. For a semiconductor resistance rises as $1/ T$ shrinks, hence the model $R = R_{1} e^{B (1/ T_{2} - 1/ T_{1})}$ . Why this step? Resistance is $1/ σ$ ; since $σ \propto e^{- B / T}$ , we get $R \propto e^{+ B / T}$ . The single constant $B$ packages $E_{g} / (2 k_{B})$ so we don't need $E_{g}$ separately. Step 2. Evaluate the exponent's bracket: $\frac{1}{310} - \frac{1}{298} = - 1.299 \times 1 0^{- 4}$ K $^{- 1}$ , so $B (1/ T_{2} - 1/ T_{1}) = 3500 \times (- 1.299 \times 1 0^{- 4}) = - 0.4546$ (dimensionless, since K $\times$ K $^{- 1}$ cancels). Why this step? We must reduce the exponent to a plain number before exponentiating; its negative value already tells us $R$ will fall. Step 3. Raise $e$ to that exponent and multiply by $R_{1}$ : $R_{2} = R_{1} e^{- 0.4546} = 12000 Ω \times 0.6347 = 7616 Ω \approx 7.6$ k $Ω$ . Why this step? Applying the exponential factor to the reference resistance $R_{1}$ gives the actual reading at body temperature; the unit $Ω$ rides along with $R_{1}$ while $0.6347$ is a pure number.

Verify: Heating a semiconductor → resistance falls ( $7.6 k < 12 k$ ), so $d R / d T < 0$ → this is a negative temperature coefficient (NTC) device. ✔ That is exactly how a fever is detected: a warmer body gives a lower resistance, a clean measurable signal.

Case J — Exam twist: metal–semiconductor crossover

What the figure shows (in words): resistivity $ρ$ versus temperature $T$ for two materials on the same axes. The blue metal curve is a straight line rising gently with $T$ (vibrations worsen it). The green semiconductor curve starts high on the left and falls steeply as $T$ increases (its carrier count explodes). The two curves cross at one point — the yellow dot — marked as the crossover $T_{cross}$ . Left of the dot the metal conducts better (lower $ρ$ ); right of it the semiconductor wins.

Worked example Ex 10 (Cell J) — At what temperature do they cross?

A metal has $ρ_{M} (T) = a T$ with $a = 5 \times 1 0^{- 11} Ω \cdot$ m/K. A semiconductor has $ρ_{S} (T) = c e^{E_{g} / (2 k_{B} T)}$ with $c = 1 \times 1 0^{- 9} Ω \cdot$ m and $E_{g} = 0.30$ eV. Find the crossover temperature $T_{cross}$ where $ρ_{M} = ρ_{S}$ near room temperature. ( $k_{B} = 8.617 \times 1 0^{- 5}$ eV/K.)

Forecast: as $T$ rises, the metal gets worse (line up) and the semiconductor gets better (curve down) — where do the two meet?

Step 1. Set $ρ_{M} = ρ_{S}$ : $a T = c e^{E_{g} / (2 k_{B} T)}$ . This equates a straight line with a decaying exponential, so it is transcendental and we solve numerically. Why this step? "Crossover" literally means the two $ρ (T)$ curves are equal; writing that equality is the definition of the unknown $T_{cross}$ . Step 2. Compute the exponent constant $E_{g} / (2 k_{B}) = 0.30/ (2 \times 8.617 \times 1 0^{- 5}) = 1741$ K, and define $f (T) = a T - c e^{1741/ T}$ ; we want the root $f (T_{cross}) = 0$ . Why this step? Bundling the material constant into one kelvin-valued number lets us evaluate $f$ quickly at trial temperatures. Step 3. Bracket the root by trying values. $T = 400$ K: $ρ_{M} = 2.0 \times 1 0^{- 8}$ ; $ρ_{S} = 1 0^{- 9} e^{4.352} = 7.77 \times 1 0^{- 8}$ → semiconductor still higher. $T = 600$ K: $ρ_{M} = 3.0 \times 1 0^{- 8}$ ; $ρ_{S} = 1 0^{- 9} e^{2.902} = 1.82 \times 1 0^{- 8}$ → metal now higher → root lies between. $T = 530$ K: $ρ_{M} = 2.65 \times 1 0^{- 8}$ ; $ρ_{S} = 1 0^{- 9} e^{3.285} = 2.68 \times 1 0^{- 8}$ → essentially equal. Why this step? Watching $f$ change sign traps the crossover; refining the guess to $530$ K makes the two resistivities match.

Verify: At $T_{cross} \approx 530$ K, $ρ_{M} \approx 2.65 \times 1 0^{- 8}$ and $ρ_{S} = 1 0^{- 9} e^{1741/530} = 1 0^{- 9} e^{3.285} = 2.68 \times 1 0^{- 8}$ — they match to $\sim 1%$ . ✔ Physical meaning: hot enough, the semiconductor's carrier explosion (Ex 5) makes it beat the metal, whose vibrations only worsen. The green curve dives under the blue line at the intersection in the figure.

Case K — Extrinsic (doped) semiconductor: saturation & freeze-out

What the figure shows (in words): carrier density $n$ (on a logarithmic vertical axis) versus temperature $T$ , for a doped semiconductor, split into three coloured zones. On the left (low $T$ ) a red rising curve labelled freeze-out climbs from near zero. In the middle a flat green plateau labelled saturation sits at the dopant level $N_{D}$ . On the right (high $T$ ) a blue curve labelled intrinsic shoots steeply upward above $N_{D}$ . A yellow dashed horizontal line marks $n = N_{D}$ .

Worked example Ex 11 (Cell K) — Doped silicon across three temperature zones

A silicon sample is doped with $N_{D} = 1 \times 1 0^{16}$ donor atoms per cm³ (each donor can release one electron). A donor's electron is bound by only $E_{d} = 0.045$ eV (far less than the $1.1$ eV gap). Describe what $n$ does across (i) very low $T$ , (ii) room temperature, (iii) very high $T$ , and estimate the carrier density in the middle zone.

Forecast: at room temperature, is $n$ set by the tiny donor energy, the big band gap, or the fixed dopant count?

Step 1. Identify the three regimes shown in the figure. In a doped semiconductor the carriers can come from two sources: donors (cheap, $E_{d}$ ) or the intrinsic gap (expensive, $E_{g}$ ). Which wins depends on $T$ . Why this step? Unlike the intrinsic Cases E–G, here $n$ is not one clean exponential — the source of carriers changes with temperature, so we must locate which zone the sample is in before choosing a formula. Step 2 — (i) Freeze-out, very low $T$ (red curve). Thermal energy $k_{B} T$ is too small even to ionize the shallow donors, so electrons stay stuck on their donor atoms: $n$ drops far below $N_{D}$ as $T \to 0$ , governed by $n \propto e^{- E_{d} / (2 k_{B} T)}$ with the tiny $E_{d}$ . Why this step? It shows the low- $T$ limiting behaviour, the doped analogue of Case F, and explains the rising red curve. Step 3 — (ii) Saturation / extrinsic plateau, room $T$ (green line). Every donor is ionized but the gap is still far too big to excite intrinsic pairs, so $n \approx N_{D} = 1 \times 1 0^{16}$ cm⁻³ — flat, temperature-independent. This is the zone electronics are designed to sit in. Why this step? Here $n$ is pinned by the dopant count, not by any exponential — a qualitatively different behaviour we must flag so readers don't misapply the gap law. Step 4 — (iii) Intrinsic takeover, very high $T$ (blue curve). Eventually $k_{B} T$ is large enough that intrinsic pair creation ( $n_{i} \propto e^{- E_{g} / (2 k_{B} T)}$ ) overtakes $N_{D}$ ; then $n$ climbs exponentially again as in Ex 5. Why this step? It closes the picture, connecting the far-right of the graph back to the intrinsic law of Case E.

Verify: Check the ordering of energies: $E_{d} = 0.045$ eV is $\sim 24 \times$ smaller than $E_{g} = 1.1$ eV, so donors ionize at much lower $T$ than the gap — that is exactly why a wide flat plateau ( $n \approx N_{D}$ ) exists between freeze-out and intrinsic. ✔ At room temperature $k_{B} T \approx 0.0259$ eV is comfortably above $E_{d}$ (donors ionized) but far below the gap $E_{g} = 1.1$ eV (no intrinsic pairs yet), confirming the sample is in the saturation zone with $n \approx 1 \times 1 0^{16}$ cm⁻³. See Doping and Carrier Concentration and Intrinsic vs Extrinsic Semiconductors.

Recall Which law for which cell?

Metal linear vs semiconductor exponential — how do you decide? ::: Ask "is $n$ fixed?" If the material is a metal (no gap to cross, Free Electron Model of Metals) use $R = R_{0} [1 + α Δ T]$ ; if it's an intrinsic semiconductor use the $e^{- E_{g} /2 k_{B} T}$ carrier law; if it's doped, check which of the three zones (freeze-out / saturation / intrinsic) you're in. Why does a negative $Δ T$ never need a new formula? ::: The linear and exponential laws are already signed in $Δ T$ and $1/ T$ ; cooling just flips the sign of the bracket (Ex 2, Ex 6). What tool inverts the exponential to recover $E_{g}$ ? ::: The natural log $ln$ , the exact inverse of $exp$ (Ex 7). What's the limiting-case gotcha for metals near $T = 0$ ? ::: Resistivity flattens to a residual floor $ρ_{res}$ (impurity scattering, Matthiessen), it does not reach zero — see Ex 4 and contrast Superconductivity. What happens to a doped semiconductor as $T \to 0$ ? ::: Freeze-out — carriers re-trap onto donors and $n$ falls well below $N_{D}$ (Ex 11), unlike the flat saturation plateau at room temperature. When do I work in $R$ and when in $ρ$ ? ::: Use $R$ (ohms) when a specific object is measured (Ex 1–3, 9); use $ρ$ (ohm-metre) when comparing materials or their intrinsic property (Ex 4, 10). Both obey the same temperature laws.

Connections

Parent: Thermal effects on conductivity
Free Electron Model of Metals — why metals have a fixed carrier sea (Cells A, H).
Band Theory & Energy Gaps — the gap $E_{g}$ behind Cells E–G, J.
Intrinsic vs Extrinsic Semiconductors — whose $n (T)$ explodes vs saturates (Cells E, K).
Mobility and Drift Velocity — the $μ = q τ / m$ that fades with heat.
Doping and Carrier Concentration — controlling $n$ independently of $T$ (Cell K).
Superconductivity — the true $ρ \to 0$ limit vs the residual floor of Cell D.