Compound semiconductors (GaN, GaAs, SiC) overview: Worked examples

Before we start, let us re-earn every symbol so a reader who skipped the parent note can still follow.

Definition Every symbol on this page, in plain words

$E_{g}$ = bandgap energy, measured in electron-volts (eV). It is the energy a single electron must gain to jump from the "stuck / bonded" state to the "free / conducting" state. Think of it as the height of a cliff the electron must climb. Bigger $E_{g}$ = taller cliff. See Energy bands and bandgap.
$λ$ = wavelength of a photon (a particle of light), in nanometres (nm), one nm = one billionth of a metre. Small $λ$ = bluer/more energetic light; big $λ$ = redder/infrared.
$μ_{e}$ = electron mobility, in $cm^{2} / (V\cdots)$ : how fast electrons drift for a given push. Big $μ_{e}$ = fast transistor. See Doping and carrier concentration.
$E_{cr i t}$ = breakdown field in MV/cm (millions of volts per centimetre): the field strength at which the material stops insulating and arcs.
$a$ = lattice constant: the repeat spacing of atoms in the crystal, in ångström (Å) = 0.1 nm. See Silicon crystal structure.
$f$ = lattice mismatch, a pure number (often written as a $%$ ): how much wider or narrower the deposited layer's atomic spacing is compared with the substrate it sits on, $f = (a_{layer} - a_{sub}) / a_{sub}$ . Positive $f$ = layer wants to be wider; negative $f$ = layer wants to be narrower. See Epitaxy and crystal growth.

The scenario matrix

Every case this topic can throw at you, and where we cover it:

#	Case class	Concrete instance	Covered by
A	Small $E_{g}$ → long (infrared) $λ$	GaAs, $E_{g} = 1.42$ eV	Example 1
B	Large $E_{g}$ → short (visible/UV) $λ$	GaN, $E_{g} = 3.4$ eV	Example 2
C	Inverse direction (given $λ$ , find $E_{g}$ )	want green 530 nm	Example 3
D	Degenerate / limiting input ( $E_{g} \to 0$ and $E_{g} \to \infty$ )	metals & insulators	Example 4
E	The "direct vs indirect" trap (formula gives a number that is physically wrong)	SiC 3.26 eV	Example 5
F	Breakdown scaling, ratio form	GaN vs Si field ratio	Example 6
G	Sign of $f$ : layer bigger vs smaller than substrate	both signs	Example 7
H	Real-world word problem (mobility included)	pick a material for an EV inverter	Example 8
I	Exam-style twist (multi-step, combines two formulas)	ternary $Al_{x} Ga_{1 - x} As$ colour tuning	Example 9

Worked examples

Figure below plots the whole see-saw so you can see where each example lands before we compute it.

Figure — Compound semiconductors (GaN, GaAs, SiC) overview

The navy curve is $λ = 1240/ E_{g}$ . The shaded band is the visible range (380–700 nm). Notice the dots: GaAs (Example 1) sits below the red edge in infrared; GaN/SiC (Examples 2 & 5) sit at the far-left violet/UV corner. As you slide right (bigger gap) the curve dives — that steep drop is the whole story of why colour depends so sharply on gap. Keep this figure open: each example below points back to a spot on this curve.

Cell A — small gap, infrared light

Worked example Example 1 · GaAs emission wavelength

Statement: GaAs has $E_{g} = 1.42$ eV. What wavelength does it emit, and is it visible?

Forecast: Guess before computing — is this bluer or redder than red light (~700 nm)?

Write the see-saw. $λ = 1240/ E_{g}$ . Why this step? $E_{g} \cdot λ = 1240$ is the only relation linking gap to colour; we solve for the unknown $λ$ .
Plug in. $λ = 1240/1.42 = 873 nm$ . Why this step? Direct substitution of the given gap.
Classify the colour. Human vision stops at ~700 nm. 873 nm > 700 nm → infrared, invisible. Why this step? The question asks "visible?", so we compare against the visible ceiling.

Locate it on the figure: the magenta GaAs dot sits at the far right-low of the see-saw — below the orange 700 nm dashed line, confirming infrared by eye.

Verify: Reverse it — $1240/873 = 1.42$ eV ✓. Units: eV·nm / eV = nm ✓. And GaAs is famously the 850 nm IR emitter in TV remotes and fibre lasers — matches LEDs and laser diodes.

Cell B — large gap, visible/UV light

Worked example Example 2 · GaN emission wavelength (the blue-LED win)

Statement: GaN has $E_{g} = 3.4$ eV. What wavelength? Which colour?

Forecast: Bigger gap than GaAs, so predict a smaller $λ$ — blue or violet?

$λ = 1240/3.4 = 365 nm$ . Why this step? Same see-saw; larger denominator ⇒ shorter wavelength.
Classify: 365 nm sits just below violet (~380 nm) → near-ultraviolet / deep blue edge. Why this step? We locate it on the spectrum to name the colour.
Physical note: real blue LEDs run ~450 nm because a bit of indium is alloyed in to narrow the gap slightly — see Example 9 for why alloying moves colour. Why this step? Connects the ideal number to the real device.

Locate it on the figure: the violet GaN dot is at the far left corner of the curve — high on the vertical axis is where the wavelength is smallest; here the curve has plunged, landing GaN near the top of the violet band.

Verify: $1240/365 = 3.40$ eV ✓. Bigger $E_{g}$ gave smaller $λ$ than Example 1 (365 < 873) — the see-saw behaved. This is precisely why blue light needs a wide direct gap: silicon's 1.12 eV can never reach here.

Cell C — inverse direction

Worked example Example 3 · What gap do I need for a green LED?

Statement: I want to emit green light at $λ = 530$ nm. What bandgap must my material have?

Forecast: Green sits between GaAs-IR and GaN-UV — guess a gap between 1.4 and 3.4 eV.

Rearrange the see-saw. From $λ = 1240/ E_{g}$ we get $E_{g} = 1240/ λ$ . Why this step? Now $λ$ is the known and $E_{g}$ the unknown, so we flip the equation.
Plug in. $E_{g} = 1240/530 = 2.34 eV$ . Why this step? Substitute the target wavelength.
Match to a material. 2.34 eV is between GaP (2.26) and the GaN alloys — so real green LEDs use $InGaN$ or $GaP$ . Why this step? Turns the abstract number into a shopping decision.

Locate it on the figure: here we run the see-saw backwards — enter the curve at $λ = 530$ nm (middle of the shaded band) and read down to the horizontal axis; you land between the GaAs and GaN dots, exactly as the bracket predicted.

Verify: $1240/2.34 = 530$ nm ✓, and $1.42 < 2.34 < 3.4$ — the forecast bracket held.

Cell D — degenerate & limiting inputs

Worked example Example 4 · The two extremes:

$E_{g} \to 0$ and $E_{g} \to \infty$ Statement: What does the formula say as the gap shrinks to zero, and as it grows huge? What real materials do those limits describe?

Forecast: Since $λ = 1240/ E_{g}$ , guess what happens to a fraction when the denominator goes to 0, and when it goes to $\infty$ .

Limit $E_{g} \to 0$ : $λ = 1240/ E_{g} \to \infty$ . Why this step? Dividing a fixed number by something approaching zero blows up without bound. Physically a zero-gap material is a metal — electrons are always free, there is no "cliff", so there is no clean emission wavelength. The formula's divergence is the maths screaming "this isn't a light-emitter."
Limit $E_{g} \to \infty$ : $λ = 1240/ E_{g} \to 0$ . Why this step? Denominator huge ⇒ fraction vanishes. A very wide gap (like diamond, 5.5 eV → 225 nm) emits deep UV; push further and you leave the useful optical range. The formula stays finite and well-behaved — no breakdown here.
Degenerate check — negative $E_{g}$ ? A negative gap is unphysical (you cannot have a negative energy cliff). If your algebra ever produces $E_{g} < 0$ , you made a sign error, not a discovery. Why this step? Ruling out the nonsense input protects you in exams.

Locate it on the figure: trace the navy curve to the left edge — it shoots up toward infinity (the $E_{g} \to 0$ metal limit). Trace it to the right edge — it flattens toward zero (the wide-gap limit). The two arms of the curve are the two limits.

Verify: $E_{g} = 5.5 \Rightarrow λ = 1240/5.5 = 225$ nm ✓ (diamond, deep-UV). The two limits behave as reciprocal-function limits must: one arm $\to \infty$ , the other $\to 0$ .

Cell E — the direct/indirect trap

Worked example Example 5 · SiC: the formula lies if you forget the gap

type Statement: SiC (4H) has $E_{g} = 3.26$ eV. Compute $λ$ . Would SiC make a good blue LED?

Forecast: The number will look great (near blue). Will the device work? Recall Energy bands and bandgap.

$λ = 1240/3.26 = 380 nm$ . Why this step? Plain application — and it lands right at violet, temptingly close to GaN.
Apply the physics filter. SiC is indirect-gap: the electron and hole sit at different momenta, so recombination also needs a lattice vibration (phonon) to conserve momentum. That triple-coincidence is rare → emission efficiency is tiny. Why this step? The wavelength formula only tells you what colour if it emits, never whether it emits. The gap type is the missing gate.
Conclusion: SiC would emit ~380 nm light in principle but so inefficiently it is useless as an LED. Its real job is power/heat (see Example 8). Why this step? Matches the parent mistake box: "ask direct-or-indirect before wide-or-narrow."

Locate it on the figure: the orange SiC dot sits almost on top of the violet GaN dot — the curve cannot tell them apart, because the see-saw only knows the gap size, not its type. That visual near-overlap is the whole trap.

Verify: $1240/380 = 3.26$ eV ✓ arithmetically. Physically the answer to "good LED?" is no — the number is right, the device is wrong. That gap between arithmetic and reality is the whole lesson.

Cell F — breakdown scaling in ratio form

Worked example Example 6 · How much thinner can a GaN device be than silicon?

Statement: Using $E_{cr i t} \propto E_{g}^{2}$ , estimate the breakdown-field ratio of GaN ( $E_{g} = 3.4$ ) to Si ( $E_{g} = 1.12$ ). Then say how it affects device thickness — see Power electronics & MOSFETs.

Forecast: GaN's gap is about 3× Si's. Squaring 3 gives ~9. Guess the field ratio ≈ 9–11×.

Take the gap ratio. $E_{g}^{GaN} / E_{g}^{Si} = 3.4/1.12 = 3.04$ . Why this step? Ratio form cancels the unknown proportionality constant, so we never need it.
Square it (the $n \approx 2$ law). $E_{cr i t}$ ratio $\approx 3.0 4^{2} = 9.2$ . Why this step? Impact-ionization argument from the parent: to knock an electron across a bigger cliff you need a proportionally bigger field, and the energy scales roughly with the square.
Turn field into thickness. Voltage blocked $\approx E_{cr i t} \times$ thickness. Same voltage at ~9× the field means ~1/9 the drift-region thickness → far lower on-resistance and heat. Why this step? Answers the "so what" — this is the entire commercial reason for GaN/SiC power devices.

Verify: $3.0 4^{2} = 9.24$ ✓. The observed lab ratio is ~11× (real $n$ is a bit above 2), so our estimate is the right order of magnitude — an honest scaling law, not an exact identity.

Cell G — sign of the lattice mismatch $f$

Worked example Example 7 · Both signs of lattice mismatch

Statement: Compute $f = (a_{layer} - a_{sub}) / a_{sub}$ for two cases: (a) GaN layer ( $a = 3.19$ Å) on sapphire substrate ( $a = 2.75$ Å along the matching direction); (b) a hypothetical layer ( $a = 2.60$ Å) on the same sapphire ( $a = 2.75$ Å). What does each sign mean? See Epitaxy and crystal growth.

Forecast: In (a) the layer atoms are spaced wider than the substrate; in (b) narrower. Guess the sign of $f$ in each before computing.

Case (a): $f = (3.19 - 2.75) /2.75 = 0.44/2.75 = + 0.16 = + 16%$ . Why this step? Layer bigger than substrate ⇒ numerator positive ⇒ $f > 0$ . Positive $f$ = the film is squeezed (compressive strain) to fit the smaller template.
Case (b): $f = (2.60 - 2.75) /2.75 = - 0.15/2.75 = - 0.055 = - 5.5%$ . Why this step? Layer smaller ⇒ numerator negative ⇒ $f < 0$ . Negative $f$ = the film is stretched (tensile strain) to match the wider template.
Read the magnitude. It is $∣ f ∣$ , not the sign, that predicts cracking: $∣ + 16% ∣$ is enormous → GaN-on-sapphire genuinely needs buffer layers; $∣ - 5.5%∣$ is milder but still risky. Why this step? The sign tells you which way the strain pulls; the magnitude tells you whether it breaks.

Verify: (a) $0.44/2.75 = 0.16$ ✓; (b) $- 0.15/2.75 = - 0.0545$ ✓. Opposite signs from layer-bigger vs layer-smaller — exactly the two cases the definition must cover.

Cell H — real-world word problem

Worked example Example 8 · Choosing a material for a 900 V EV inverter

Statement: An electric-car inverter must block 900 V, run at 150 °C, and dump lots of heat. Your candidates are Si ( $E_{g} = 1.12$ eV, $E_{cr i t} = 0.3$ MV/cm, $μ_{e} = 1400$ ), GaAs ( $1.42$ eV, $0.4$ MV/cm, $μ_{e} = 8500$ ), and SiC ( $3.26$ eV, $3.0$ MV/cm, $μ_{e} = 900$ ). Which do you pick, and why not the others?

Forecast: High voltage + high heat. Which of the three superpowers (Runs / Lights / Fights, defined in the mnemonic above) matches?

Estimate drift thickness for 900 V using thickness $\approx V / E_{cr i t}$ . First convert each field to $V/cm$ (1 MV/cm $= 1 0^{6}$ V/cm).
- Si: $t = \frac{900 V}{0.3 \times 1 0 ^{6} V/cm} = 3.0 \times 1 0^{- 3} cm$ . Convert to microns: $3.0 \times 1 0^{- 3} cm \times 1 0^{4} \frac{μ m}{cm} = 30 μ m$ .
- GaAs: $t = \frac{900}{0.4 \times 1 0 ^{6}} = 2.25 \times 1 0^{- 3} cm = 22.5 μ m$ .
- SiC: $t = \frac{900}{3.0 \times 1 0 ^{6}} = 3.0 \times 1 0^{- 4} cm = 3.0 μ m$ . Why this step? Thicker drift region = higher resistance = more heat. SiC needs 10× less thickness than Si and GaAs — the direct payoff of its wide gap and high $E_{cr i t}$ .
Bring in mobility $μ_{e}$ (the "Runs" number). GaAs has the biggest $μ_{e} = 8500$ — but mobility only helps switching speed, not voltage blocking or heat. For a slow, hot, high-voltage inverter, $μ_{e}$ is nearly irrelevant, and SiC's modest $μ_{e} = 900$ is perfectly fine because the win comes from the 10× thinner drift region. Why this step? This is exactly the parent's "highest mobility ≠ best everywhere" mistake — we must check whether the biggest number even matters for this job.
Eliminate GaAs. $E_{cr i t} = 0.4$ MV/cm is barely above Si → 22.5 μm drift region, almost as thick as silicon, and its small gap plus poor thermal behaviour make it unfit for high power. GaAs is the "Runs" sprinter (RF), not a "Fights" material. Why this step? Its one advantage (mobility) does not serve this job, and every relevant number is weak.
Eliminate Si on heat. At 150 °C Si's small gap lets thermal carriers leak (intrinsic conduction), degrading its ability to stay "off". SiC's wide gap keeps carriers frozen out when hot. Why this step? Wide gap = fewer thermally freed carriers = survives heat.
Pick SiC. Wide gap + high $E_{cr i t}$ (3 μm drift, 10× thinner) + excellent thermal conductivity ⇒ the "Fights" material. Its low mobility is a non-issue at inverter switching speeds. Why this step? SiC uniquely satisfies all three constraints — voltage, heat, and low on-resistance.

Verify: Thicknesses: Si $30 μ$ m, GaAs $22.5 μ$ m, SiC $3.0 μ$ m ✓. SiC/Si thickness ratio $= 3/30 = 1/10$ , matching the field ratio $0.3/3.0 = 1/10$ ✓. Decision (SiC) consistent with the "SiC → Fights" mnemonic.

Cell I — exam-style twist (two formulas chained)

Worked example Example 9 · Tuning colour with a ternary alloy

Statement: $Al_{x} Ga_{1 - x} As$ has a bandgap that rises roughly linearly with the aluminium fraction $x$ : $E_{g} (x) = 1.42 + 1.25 x$ eV (valid for $x$ up to ~0.45, where it stays direct). What $x$ gives red light at $λ = 650$ nm? What colour does $x = 0$ give?

Forecast: Red needs a bigger gap than plain GaAs's IR. So we expect $x > 0$ . How big?

Find the required gap (formula 1). $E_{g} = 1240/650 = 1.91 eV$ . Why this step? The see-saw converts the target colour into a target gap first, before we touch the alloy law.
Solve the alloy law (formula 2). Set $1.91 = 1.42 + 1.25 x$ , so $1.25 x = 1.91 - 1.42 = 0.49$ , giving $x = 0.49/1.25 = 0.39$ . Why this step? We chain the gap requirement into the composition-vs-gap line to get the recipe fraction $x$ — this is the "two formulas combined" twist the exam is testing.
Check the validity window. $x = 0.39 < 0.45$ → the alloy is still direct-gap, so it will actually emit efficiently. ✓ Why this step? An exam twist hides an out-of-range trap; always confirm the formula's stated domain before trusting the answer.
Baseline $x = 0$ : $E_{g} (0) = 1.42$ eV → $λ = 1240/1.42 = 873$ nm (infrared) — exactly Example 1, as it must be, since $x = 0$ is pure GaAs. Why this step? Cross-checks the alloy law against a case we already trust: a good sanity anchor.

Locate it on the figure: as $x$ climbs from 0 to 0.39, the operating point slides left and up along the see-saw — starting at the magenta GaAs dot (873 nm, IR) and climbing into the red edge of the visible band (650 nm). Alloying literally walks you along the curve.

Verify: $1240/650 = 1.908$ eV ✓; $(1.908 - 1.42) /1.25 = 0.390$ ✓; $E_{g} (0) = 1.42 \Rightarrow 1240/1.42 = 873$ nm ✓ (agrees with Example 1). Two formulas chained, domain respected, baseline consistent.

Active recall

Given

E_{g} = 2.0

eV, what wavelength (nm)?

1240/2.0 = 620

nm (orange-red).

Green at 530 nm needs what bandgap?

E_{g} = 1240/530 = 2.34

eV.

E_{g} \to 0

, what does

λ

do, and what material is that?

λ \to \infty

; it's a metal (no gap, no clean emission).

Sign of

f

when the layer's lattice is larger than the substrate's?

Positive (compressive strain in the film).

SiC has a 3.26 eV gap — why is it still a poor LED?

It is indirect-gap, so emission needs a phonon and is very inefficient.

GaN/Si breakdown-field ratio from

E_{cr i t} \propto E_{g}^{2}

(3.4/1.12)^{2} \approx 9

For a 900 V inverter, why does GaAs's huge mobility not save it?

Mobility only helps switching speed; the inverter needs high

E_{cr i t}

and heat tolerance, where GaAs is weak.

What does each material's superpower stand for?

GaAs Runs (mobility), GaN Lights (direct gap), SiC Fights (power/heat).