Exercises — Thermal effects on conductivity

Silicon is a semiconductor. Heating kicks electrons across the band gap, so the carrier density $n$ grows like $e^{-E_g/(2k_{B}T)}$. More carriers ⇒ conductivity rises ⇒ resistance falls. Therefore $\alpha <0$.

The one question: does heat mostly create carriers or disturb them? For a semiconductor it creates them, so resistance drops.

• $n$ = number of charge carriers per unit volume (unit ${\text{m}}^{-3}$).
• $q$ = charge on one carrier (unit coulomb, $\text{C}$).
• $\mu$ = mobility, the drift speed a carrier reaches per unit electric field.

Mobility unit: from $\mu =v_d/E$ we get $\frac{\text{m/s}}{\text{V/m}}=\frac{{\text{m}}^2}{\text{V}\cdot \text{s}}$.

Why the linear model? Copper is a metal, so we use $R=R_0[1+\alpha \Delta T]$. $\Delta T=120-20=100^{\circ }$C. $R=25[1+(3.9\times 10^{-3})(100)]=25(1.39)=34.75\Omega .$ Resistance rose (metal, $\alpha >0$). ✔

Solve $R=R_0[1+\alpha \Delta T]$ for $\Delta T$: $\Delta T=\frac{1}{\alpha }\left(\frac{R}{R_0}-1\right)=\frac{1}{3.9\times 10^{-3}}\left(\frac{46}{25}-1\right).$ $\frac{46}{25}-1=0.84$, so $\Delta T=0.84/0.0039=215.4^{\circ }$C. Then $T=20+215.4=235.4^{\circ }$C.

This is the idea behind a resistance thermometer: measure $R$, read off $T$.

Use the ratio $\frac{n(T_2)}{n(T_1)}=\exp \left[-\frac{E_g}{2k_B}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\right]$.

• $\frac{E_g}{2k_B}=\frac{0.67}{2(8.617\times 10^{-5})}=3887.7\text{K}$.
• $\frac{1}{330}-\frac{1}{300}=3.0303\times 10^{-3}-3.3333\times 10^{-3}=-3.030\times 10^{-4}{\text{K}}^{-1}$.
• Exponent $=-3887.7\times (-3.030\times 10^{-4})=1.178$.
• Ratio $=e^{1.178}\approx 3.25$.

A mere $30$K rise triples the carriers — that's the exponential at work.

From $\rho =\frac{m}{n{q}^2\tau }$ with $n,q,m$ fixed, $\rho \propto 1/\tau$, so $\frac{{\rho }_2}{{\rho }_1}=\frac{{\tau }_1}{{\tau }_2}$. ${\rho }_2/{\rho }_1=1.15$, hence $\frac{{\tau }_2}{{\tau }_1}=\frac{1}{1.15}=0.8696$. That is a drop of about $13.0\%$ in $\tau$.

Interpretation: in a metal the resistivity change is a mobility (via $\tau$) story — carrier count barely moves.

With $\sigma \propto e^{-E_g/(2k_{B}T)}$: $\ln \frac{{\sigma }_2}{{\sigma }_1}=-\frac{E_g}{2k_B}\left(\frac{1}{T_2}-\frac{1}{T_1}\right).$

• $\ln 4=1.3863$.
• $\frac{1}{360}-\frac{1}{300}=2.7778\times 10^{-3}-3.3333\times 10^{-3}=-5.556\times 10^{-4}{\text{K}}^{-1}$.
• So $\frac{E_g}{2k_B}=\frac{-1.3863}{-5.556\times 10^{-4}}=2495.4\text{K}$.
• $E_g=2k_B\times 2495.4=2(8.617\times 10^{-5})(2495.4)=0.4301\text{eV}$.

Method note: we plotted $\ln \sigma$ against $1/T$; its slope is $-E_g/(2k_B)$. See the figure.

Carrier factor: $\frac{E_g}{2k_B}=\frac{1.1}{2(8.617\times 10^{-5})}=6382.7\text{K}$. $\frac{1}{310}-\frac{1}{300}=-1.0753\times 10^{-4}$. Exponent $=6382.7\times 1.0753\times 10^{-4}=0.6863$. $n$ factor $=e^{0.6863}=1.986$ (nearly doubles).

Mobility factor: ${\left(\frac{310}{300}\right)}^{-3/2}=(1.03333{)}^{-1.5}=0.9524$ (drops $\approx 4.8\%$).

Net conductivity factor: $1.986\times 0.9524=1.892$. Conductivity still rises $\approx 89\%$ — the exponential carrier growth crushes the mild mobility loss. ✔

Set up B's constant $C$. At $300$K: $20=C{e}^{E_g/(2k_B\cdot 300)}$. $\frac{E_g}{2k_B}=\frac{0.5}{2(8.617\times 10^{-5})}=2901.2\text{K}$, so exponent at $300$K $=2901.2/300=9.6707$. $C=20e^{-9.6707}=20\times 6.28\times 10^{-5}=1.256\times 10^{-3}\Omega$.

Equation to solve: $8[1+4\times 10^{-3}(T-300)]=1.256\times 10^{-3}{e}^{2901.2/T}$. This is transcendental — solve numerically. Try values:

• $T=380$: LHS $=8[1+0.32]=10.56$; RHS $=1.256\times 10^{-3}{e}^{7.635}=2.62$. RHS < LHS.
• $T=350$: LHS $=8[1+0.2]=9.6$; RHS $=1.256\times 10^{-3}{e}^{8.289}=4.98$.
• $T=330$: LHS $=8[1+0.12]=8.96$; RHS $=1.256\times 10^{-3}{e}^{8.792}=8.24$.
• $T=328$: LHS $=8.896$; RHS $=1.256\times 10^{-3}{e}^{8.845}=8.69$.
• $T=326$: LHS $=8.832$; RHS $=1.256\times 10^{-3}{e}^{8.899}=9.17$. Now RHS > LHS.

Crossing between $326$ and $328$K; refining gives $T\approx 327\text{K}$.

Meaning: below this temperature the semiconductor is more resistive; above it the metal overtakes — the two temperature laws cross. See the figure.

Hot resistance: $\Delta T=200$, $R=40[1+4.5\times 10^{-3}(200)]=40(1.9)=76\Omega$. Hot power: $P_{\text{hot}}=230^2/76=52900/76=695.9\text{W}$. Cold power: $P_{\text{cold}}=230^2/40=52900/40=1322.5\text{W}$.

The heater draws a big inrush when cold ($1322$W) and settles to $696$W hot — because rising resistance throttles the current. This self-limiting is why filament lamps flash bright then dim.

Write both temperatures: $n(T_2)=A{e}^{-E_g/(2k_B{T}_2)},n(T_1)=A{e}^{-E_g/(2k_B{T}_1)}.$ Divide — the $A$ cancels because it multiplies both: $\frac{n(T_2)}{n(T_1)}=\frac{e^{-E_g/(2k_B{T}_2)}}{e^{-E_g/(2k_B{T}_1)}}=e^{-E_g/(2k_B{T}_2)+E_g/(2k_B{T}_1)}.$ Factor $-\frac{E_g}{2k_B}$ out of the exponent: $=\exp \left[-\frac{E_g}{2k_B}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)\right].■$ Why $A$ vanishes: $A$ carries all the messy prefactors (density-of-states stuff) that don't change between the two temperatures. Taking a ratio isolates only the part that does change — the exponential — which is exactly why the ratio trick is so powerful in the earlier exercises.

${\mu }_e+{\mu }_h=0.135+0.048=0.183{\text{m}}^2/\text{Vs}$. $\sigma =(1.0\times 10^{16})(1.602\times 10^{-19})(0.183)=2.931\times 10^{-4}\text{S/m}.$ $\rho =1/\sigma =1/(2.931\times 10^{-4})=3411\Omega \cdot \text{m}.$ Insight: both carrier types push current the same direction (opposite charges, opposite drift), so their mobilities add — a subtlety the single-carrier formula hides.

(a) $T\to \infty$: the exponent $-E_g/(2k_{B}T)\to 0$, so $e^0=1$ and $\sigma \to {\sigma }_0$. The conductivity saturates — every possible carrier is already excited; heating can't create more. (b) $T\to 0^{+}$: the exponent $\to -\infty$, so $e^{-\infty }\to 0$ and $\sigma \to 0$. A perfect intrinsic semiconductor becomes an insulator at absolute zero — no thermal energy, no carriers across the gap. Reality check: long before saturation, the mild mobility drop ($\mu \propto T^{-3/2}$) and even melting take over, so ${\sigma }_0$ is never actually reached. The model is a room-temperature guide, not an all-temperature truth. (Contrast this with Superconductivity, where $\rho \to 0$ for a completely different reason.)