Why silicon dominates over germanium: Exercises — Hardware deep dive

Before we start, here is the toolkit, stated in plain words so nothing is assumed:

Definition The symbols you need (all in plain words)

$E_{g}$ = bandgap, the energy an electron must gain to break free from a bond and become a moving charge. Measured in electron-volts (eV): 1 eV is the energy one electron gains falling through 1 volt.
$n$ = free-electron concentration: how many free (conducting) electrons exist per cubic centimetre.
$p$ = hole concentration: how many holes — the empty bonds an escaped electron leaves behind, which act like positive movable charges — exist per cubic centimetre. In a pure crystal every freed electron leaves exactly one hole, so $n = p$ .
$n_{i}$ = intrinsic carrier concentration: in a pure crystal, the common value $n = p = n_{i}$ , i.e. how many free electrons (and equally many holes) heat alone creates per cubic centimetre. Unit: $cm^{- 3}$ = "per cubic centimetre".
$k_{B} T$ = thermal energy scale. $T$ is temperature in kelvin (K); $k_{B}$ is Boltzmann's constant. Their product $k_{B} T$ is roughly "how much energy the random jiggling of heat can hand to one particle." At room temperature (300 K), $k_{B} T \approx 0.02585 eV$ .
$exp (x)$ or $e^{x}$ = the exponential function. Here it answers one question: "what fraction of particles have enough thermal energy to clear a barrier of height $x k_{B} T$ ?" — it plummets fast as the barrier grows.
$m_{e}^{*}, m_{h}^{*}$ = effective masses of an electron and a hole — how "heavy" a carrier behaves inside the crystal's periodic pull, which controls how many quantum "seats" exist for it. Different materials have different effective masses.
$N_{D}$ = doping concentration: how many deliberately-added donor atoms per cm³ give us controllable free electrons (see Doping n-type and p-type).

Formula The one master equation for this page

$n_{i} \propto (m_{e}^{*} m_{h}^{*})^{3/4} T^{3/2} e^{- E_{g} / (2 k_{B} T)}$ Why the exponential? Because heat lifting an electron over an energy barrier is a probability question, and the probability of clearing a barrier of height $E$ scales as $e^{- E / k_{B} T}$ (see Fermi-Dirac distribution and thermal excitation). Where does the $T^{3/2}$ come from? It is not the barrier term — it counts how many "seats" exist just above the gap for a freed electron to land in. As you heat the crystal, a wider band of energy levels becomes thermally reachable, and the number of available states grows like $T^{3/2}$ (three dimensions of motion, each contributing a $T^{1/2}$ ). What is the $(m_{e}^{*} m_{h}^{*})^{3/4}$ factor? The density of those seats also depends on how heavy the carriers behave — heavier carriers pack more states into the same energy window. Since $n_{i} = n p$ , the electron mass and the hole mass each enter as a $3/4$ power. This factor is material-dependent: silicon and germanium have different effective masses, so their prefactors are not the same. We will only ever cancel it inside a same-material ratio (where it genuinely disappears), and we will flag it explicitly whenever we compare different materials. Why the $2$ in the denominator of the exponent? One broken bond makes a pair — one free electron ( $n$ ) and one hole ( $p$ ). The pair-product $n p = n_{i}^{2}$ scales as $e^{- E_{g} / k_{B} T}$ ; since $n_{i} = n_{i}^{2}$ , the square root halves the exponent → $2 k_{B} T$ .

Constants used throughout:

$E_{g}^{S i} = 1.12 eV, E_{g}^{G e} = 0.66 eV, k_{B} T_{300} = 0.02585 eV .$

Level 1 — Recognition

L1.1 — Which gate is taller?

State which of silicon and germanium has the larger bandgap, give both values, and give the difference $Δ E_{g} = E_{g}^{S i} - E_{g}^{G e}$ .

Recall Solution

Silicon: $E_{g}^{S i} = 1.12 eV$ . Germanium: $E_{g}^{G e} = 0.66 eV$ . Silicon's gap is larger. The difference is $Δ E_{g} = 1.12 - 0.66 = 0.46 eV .$ Picture (look at the figure below): each material is drawn as a wall whose height is its bandgap. The red silicon wall (1.12 eV) is nearly double the height of the black germanium wall (0.66 eV). The small black "heat" arrows are electrons trying to climb up out of the valence band (the floor). Notice: over the short germanium wall, an arrow of the same length gets much closer to the top — that is the visual meaning of "more heat-freed carriers in Ge." What to take away: wall height = bandgap = how hard it is for heat to free an electron.

Figure — Why silicon dominates over germanium

L1.2 — Name the oxide winner

Which material grows a stable, insulating native oxide, and what is wrong with the other one's oxide?

Recall Solution

Silicon grows $SiO_{2}$ — hard, insulating, chemically stable, sticks perfectly. This is the foundation of SiO2 and the planar process and the MOSFET gate. Germanium grows $GeO_{2}$ , which is water-soluble and thermally unstable — you cannot build a reliable insulator on it.

L1.3 — Read the master equation

In $n_{i} \propto T^{3/2} e^{- E_{g} / (2 k_{B} T)}$ , does raising $E_{g}$ increase or decrease $n_{i}$ ? One sentence why.

Recall Solution

It decreases $n_{i}$ . A larger $E_{g}$ makes the exponent $- E_{g} / (2 k_{B} T)$ more negative, so $e^{(more negative)}$ is smaller → fewer thermally freed carriers.

Level 2 — Application

L2.1 — Thermal energy at room temperature

Using $k_{B} = 8.617 \times 1 0^{- 5} eV/K$ and $T = 300 K$ , compute $k_{B} T$ in eV.

Recall Solution

$k_{B} T = (8.617 \times 1 0^{- 5}) (300) = 0.02585 eV .$ Why this step matters: every exponent on this page is measured in units of $k_{B} T$ . This number is the "yardstick" for how big an energy barrier looks to heat.

L2.2 — Barrier in units of heat

How many "thermal units" tall is silicon's gap? That is, compute the exponent magnitude $E_{g}^{S i} / (2 k_{B} T)$ at 300 K.

Recall Solution

$\frac{E _{g}^{S i}}{2 k _{B} T} = \frac{1.12}{2 ( 0.02585 )} = \frac{1.12}{0.0517} \approx 21.7.$ Interpretation: silicon's half-gap is about $21.7$ thermal units high — an enormous wall for heat to climb, which is why $n_{i}^{S i}$ is tiny.

L2.3 — The carrier ratio Ge vs Si

Ignoring the (material-dependent) prefactor, compute $\frac{n _{i} ( Ge )}{n _{i} ( Si )} \approx exp (\frac{E _{g}^{S i} - E _{g}^{G e}}{2 k _{B} T})$ at 300 K.

Recall Solution

Numerator: $Δ E_{g} = 0.46 eV$ . Denominator: $2 k_{B} T = 0.0517 eV$ . $\frac{0.46}{0.0517} = 8.90, e^{8.90} \approx 7.3 \times 1 0^{3} .$ Germanium has roughly 7300× more intrinsic carriers than silicon at 300 K. More carriers → more OFF leakage → more heat. This single ratio is the killer argument. Honest caveat about the cancelled prefactor: here we are comparing two different materials, so the $(m_{e}^{*} m_{h}^{*})^{3/4} T^{3/2}$ factors do not perfectly cancel — Si and Ge have different effective masses. Including them shifts this ratio by a factor of order a few (the measured room-temperature values $n_{i}^{S i} \approx 1 0^{10}$ , $n_{i}^{G e} \approx 2 \times 1 0^{13}$ give a real ratio ~ $2 \times 1 0^{3}$ , the same order of magnitude). The exponential — an ~7300× effect — dwarfs the prefactor correction, so the conclusion "thousands of times more leakage in Ge" is safe.

Level 3 — Analysis

L3.1 — Temperature sensitivity

Recompute the Ge/Si carrier ratio at $T = 400 K$ (roughly 127 °C), where $k_{B} T = 0.03447 eV$ . Does the gap between the two materials shrink or grow with heat? Explain.

Recall Solution

$2 k_{B} T = 0.06894 eV, \frac{0.46}{0.06894} = 6.673, e^{6.673} \approx 790.$ The ratio drops from ~7300 (at 300 K) to ~790 (at 400 K). The advantage shrinks with heat because a larger $k_{B} T$ makes both barriers easier to clear, so the exponent (which is $Δ E_{g} /2 k_{B} T$ ) gets smaller. But the absolute danger grows: even though the ratio narrows, both $n_{i}$ values are climbing fast, and Ge's is climbing from an already-huge base — so Ge crosses the leakage-failure threshold first.

L3.2 — When leakage swamps the dopants

A device fails when intrinsic carriers rival the deliberately-added dopants: $n_{i} ≳ N_{D}$ . Suppose $N_{D} = 1 0^{16} cm^{- 3}$ , and at 300 K $n_{i}^{S i} = 1 0^{10} cm^{- 3}$ , $n_{i}^{G e} = 2 \times 1 0^{13} cm^{- 3}$ . What "safety margin" $N_{D} / n_{i}$ does each material start with?

Recall Solution

Silicon: $\frac{1 0 ^{16}}{1 0 ^{10}} = 1 0^{6}$ — a million-to-one margin. Germanium: $\frac{1 0 ^{16}}{2 \times 1 0 ^{13}} = 5 \times 1 0^{2} = 500$ — only 500-to-one. Analysis: silicon can let $n_{i}$ grow a million-fold before the dopants are swamped; germanium only ~500-fold. Since $n_{i}$ rises exponentially with $T$ , germanium eats through its thin margin at a far lower temperature — matching the "Ge fails ~70–100 °C, Si to ~150 °C+" fact from the parent note. See Fermi-Dirac distribution and thermal excitation for why $n_{i}$ climbs this way.

L3.3 — Why mobility loses

Germanium's electron mobility (~3900 cm²/V·s) is ~2.6× silicon's (~1500). If speed alone decided the market, Ge would win. Give the physical reason Ge still loses, referencing exactly one quantity from this page.

Recall Solution

Mobility (see Carrier mobility and drift velocity) sets how fast a carrier drifts — a genuine Ge advantage. But the market is decided by leakage, driven by $n_{i}$ , and Ge's $n_{i}$ is ~7300× larger. High leakage → heat → thermal runaway → the transistor can't hold a clean OFF state. A device that is 2.6× faster but leaks 7300× more, overheats, and can't grow an oxide is unusable at scale. Manufacturability beats raw speed.

Level 4 — Synthesis

Formula Building the model used below (read before L4.1)

We want $n_{i}$ at some temperature $T$ relative to its known value at 300 K, for the same material, so the entire prefactor cancels exactly. Start from the master equation and take the ratio of the same material at two temperatures: $\frac{n _{i} ( T )}{n _{i} ( 300 )} = \frac{( m _{e}^{*} m _{h}^{*} ) ^{3/4} T ^{3/2} e ^{- E_{g} / (2 k_{B} T)}}{( m _{e}^{*} m _{h}^{*} ) ^{3/4} 30 0 ^{3/2} e ^{- E_{g} / (2 k_{B} \cdot 300)}} .$ Why a same-material ratio is honest: the effective-mass factor is identical top and bottom (same material), so it cancels with no approximation — this is the one place we are allowed to ignore it for free. The $(T /300)^{3/2}$ that survives barely moves (from 300 K to 585 K it is only a factor ~2.7), while the exponential changes by many orders of magnitude. Decision — "hold the $T^{3/2}$ fixed too": to isolate the dominant physics we set the surviving $T^{3/2}$ ratio $\approx 1$ . This under-estimates how fast $n_{i}$ grows, so the failure temperatures we compute are conservative upper bounds — the real device fails a bit sooner. Dropping it leaves only the exponentials: $\frac{n _{i} ( T )}{n _{i} ( 300 )} \approx exp [- \frac{E _{g}}{2 k _{B} T} + \frac{E _{g}}{2 k _{B} \cdot 300}] = exp [\frac{E _{g}}{2 k _{B}} (\frac{1}{300} - \frac{1}{T})] .$ That is exactly the model form used in L4.1 and L4.2.

L4.1 — Find germanium's failure temperature

Using the model just derived, $n_{i}^{G e} (T) = n_{i}^{G e} (300) exp [\frac{E _{g}^{G e}}{2 k _{B}} (\frac{1}{300} - \frac{1}{T})],$ with $n_{i}^{G e} (300) = 2 \times 1 0^{13} cm^{- 3}$ and $N_{D} = 1 0^{16} cm^{- 3}$ , find the temperature $T$ at which $n_{i}^{G e} = N_{D}$ . Use $k_{B} = 8.617 \times 1 0^{- 5} eV/K$ .

Recall Solution

WHAT we set up: we want $n_{i}^{G e} (T) = N_{D}$ , i.e. the carrier count reaches the dopant count. $\frac{N _{D}}{n _{i}^{G e} ( 300 )} = \frac{1 0 ^{16}}{2 \times 1 0 ^{13}} = 500.$ WHY a logarithm: the unknown $T$ sits inside an exponential, so we take the natural log to bring it down. $ln (500) = \frac{E _{g}^{G e}}{2 k _{B}} (\frac{1}{300} - \frac{1}{T}) .$ Compute $ln (500) = 6.215$ and $\frac{E _{g}^{G e}}{2 k _{B}} = \frac{0.66}{2 ( 8.617 \times 1 0 ^{- 5} )} = 3830 K$ . $\frac{6.215}{3830} = \frac{1}{300} - \frac{1}{T} \Rightarrow 1.623 \times 1 0^{- 3} = 3.333 \times 1 0^{- 3} - \frac{1}{T} .$ $\frac{1}{T} = 3.333 \times 1 0^{- 3} - 1.623 \times 1 0^{- 3} = 1.710 \times 1 0^{- 3} \Rightarrow T \approx 585 K .$ That is $585 - 273 \approx 312 ° C$ in this simplified fixed-prefactor, fixed-gap model. Reality check: real Ge fails much earlier (~100 °C) because three things we froze out all accelerate the collapse — the surviving $T^{3/2}$ prefactor, the mobility degradation, and (biggest of all) the bandgap shrinking with temperature, $E_{g} (T)$ , addressed in L4.3. Our answer is a conservative upper bound; the direction (small gap → early failure) is exactly right.

L4.2 — Repeat for silicon, compare

Using the same model with $E_{g}^{S i} = 1.12 eV$ and $n_{i}^{S i} (300) = 1 0^{10} cm^{- 3}$ , find silicon's crossing temperature and compare to L4.1.

Recall Solution

$\frac{N _{D}}{n _{i}^{S i} ( 300 )} = \frac{1 0 ^{16}}{1 0 ^{10}} = 1 0^{6}, ln (1 0^{6}) = 13.816.$ $\frac{E _{g}^{S i}}{2 k _{B}} = \frac{1.12}{2 ( 8.617 \times 1 0 ^{- 5} )} = 6499 K .$ $\frac{13.816}{6499} = 2.126 \times 1 0^{- 3} = 3.333 \times 1 0^{- 3} - \frac{1}{T} .$ $\frac{1}{T} = 1.207 \times 1 0^{- 3} \Rightarrow T \approx 828 K (\approx 555 ° C in-model) .$ Synthesis: silicon's in-model crossing (828 K) is ~243 K hotter than germanium's (585 K). Even with the crude fixed-prefactor, fixed-gap model, the bigger gap plus the bigger starting margin pushes silicon's failure far higher. This is the quantitative version of "silicon rides the BOAT": both levers — the steep exponent slope $E_{g} /2 k_{B}$ and the huge $1 0^{6}$ -to-one starting margin — point the same way, so no reasonable correction rescues germanium.

L4.3 — The edge case we suppressed: $E_{g}$ shrinks with heat

In reality the bandgap is not fixed — it shrinks as the crystal warms, well modelled by the Varshni form $E_{g} (T) = E_{g} (0) - \frac{α T ^{2}}{T + β} .$ For germanium ( $E_{g} (0) \approx 0.74 eV$ , $α \approx 4.8 \times 1 0^{- 4} eV/K$ , $β \approx 235 K$ ), estimate $E_{g}$ at $T = 585 K$ and argue in one sentence which direction this pushes the true failure temperature.

Recall Solution

$E_{g} (585) = 0.74 - \frac{( 4.8 \times 1 0 ^{- 4} ) ( 58 5 ^{2} )}{585 + 235} = 0.74 - \frac{164.3}{820} = 0.74 - 0.200 = 0.540 eV .$ So the wall at 585 K is not 0.66 eV but only ~0.54 eV — noticeably shorter. A smaller $E_{g}$ at high $T$ means even more thermal carriers than our fixed-gap model predicted, so the real $n_{i}$ hits $N_{D}$ sooner → the true failure temperature is lower than 585 K. Key point: $E_{g} (T)$ shrinkage always makes leakage worse, and it bites the small-gap material (Ge) hardest — reinforcing, never rescuing, silicon's advantage.

Level 5 — Mastery

L5.1 — The full engineering trade-off, defended

An engineer proposes a germanium chip "because Ge is 2.6× faster and detects infrared." Write a complete rebuttal that (a) quantifies the leakage penalty, (b) names the manufacturing showstopper, (c) concedes where Ge genuinely wins, and (d) states the deciding principle.

Recall Solution

(a) Leakage penalty (quantified): at 300 K, $n_{i}^{G e} / n_{i}^{S i} = e^{0.46/0.0517} \approx 7.3 \times 1 0^{3}$ (order-of-magnitude confirmed by measured values, ~ $2 \times 1 0^{3}$ ). Germanium leaks thousands of times more in the OFF state, and this gap in absolute $n_{i}$ only worsens as the chip heats — worse still once the temperature-shrinking gap $E_{g} (T)$ (L4.3) is included. Its safety margin $N_{D} / n_{i}$ is ~500 vs silicon's ~ $1 0^{6}$ (L3.2), so it fails at a much lower temperature (L4.1 vs L4.2). (b) Showstopper: germanium has no usable native oxide — $GeO_{2}$ is water-soluble and unstable, so the planar process and self-aligned MOSFET gate simply cannot be built. Without SiO₂, mass-produced ICs are impossible. (c) Concede honestly: Ge genuinely wins on mobility (SiGe RF front-ends) and on infrared/fiber-optic photodetection (its small gap absorbs longer wavelengths). The first transistor (1947) was germanium for good reason. (d) Deciding principle: Manufacturability and thermal reliability dominate raw speed at scale. A slightly slower material that stays cool, holds a clean OFF state, grows its own insulator, and comes from cheap sand beats a faster material that leaks, overheats, and can't be processed. Silicon rides the BOAT: Bandgap, Oxide, Abundance, Temperature.

L5.2 — Design a fair test

You want a fair head-to-head experiment showing why silicon wins on leakage, not on speed. Describe what you'd hold fixed and what you'd measure across a temperature sweep from 300 K to 450 K.

Recall Solution

Hold fixed: same doping $N_{D}$ , same device geometry, same applied OFF-state voltage. This isolates the material's intrinsic behaviour so any difference comes from $E_{g}$ (and its effective masses) alone, not from how the two devices were built. Sweep: temperature $T$ from 300 K to 450 K, stepping in convenient increments (say every 25 K). Measure: OFF-state leakage current $I_{leak}$ , which tracks $n_{i} (T) \propto (m_{e}^{*} m_{h}^{*})^{3/4} T^{3/2} e^{- E_{g} /2 k_{B} T}$ . Plot $ln (I_{leak})$ vertically against $1/ T$ horizontally — the exponential becomes a nearly straight line whose slope is $\approx - E_{g} /2 k_{B}$ . Predicted result and how to read it:

Germanium's line sits far higher on the plot (larger $n_{i}$ at every temperature) and its leakage overtakes the $N_{D}$ level near the low end of the sweep, while silicon's stays well below $N_{D}$ across the whole range — the visual proof that Ge fails first.
The slope of each line returns that material's half-gap: multiply the measured slope by $- 2 k_{B}$ to recover $E_{g}$ . You should get ~0.66 eV for Ge and ~1.12 eV for Si. If you look closely the lines are slightly curved (not perfectly straight) — that curvature is the $E_{g} (T)$ shrink from L4.3 plus the $T^{3/2}$ prefactor, a bonus signature that the real physics is even less kind to Ge. Why this is the fair test: it probes the axis that actually decides the market — leakage vs temperature — rather than switching speed (the axis Ge is strongest on). This is the experimental face of Fermi-Dirac distribution and thermal excitation: an $ln n_{i}$ vs $1/ T$ line whose steepness is the bandgap.

Connections

Why silicon dominates over germanium (index 1.3.9)
Bandgap and intrinsic carrier concentration
Intrinsic vs extrinsic semiconductors
Doping n-type and p-type
MOSFET operation and the gate oxide
SiO2 and the planar process
Carrier mobility and drift velocity
Fermi-Dirac distribution and thermal excitation