You already met the valence and conduction bands and the master formula. This page is a drill hall : we list every kind of question this topic can throw at you, then work one example per kind so you never meet a scenario you haven't seen.
Intuition What we carry in from the parent noteThree tools, and nothing else. We will re-earn each symbol as we use it.
The band gap E g = E c − E v E c E v
The thermal factor e − E g / ( 2 k B T ) k B T k B T
The optical shortcut E = h c / λ h c = 1240 eV⋅nm λ
Every band-theory numeric question is one of these cells. The rest of the page fills each one.
Cell
What varies
Degenerate / limit to watch
Example
A. Gap → carriers change E g T
E g → 0 Ex 1
B. Temperature sweep fix E g T
T → 0 T → ∞ Ex 2
C. Ratio (two materials) prefactors cancel
equal gaps ⇒ ratio 1
Ex 3
D. Photon absorb / transmit compare E photon E g
E photon = E g Ex 4
E. Cutoff wavelength invert h c / λ = E g
longest λ
Ex 5
F. Full absolute n i use N c , N v
huge prefactor × tiny exponential
Ex 6
G. Word problem (real device) translate words → which cell
"leaky" Ge vs Si
Ex 7
H. Exam twist / trap conceptual, not arithmetic
full band carries no current
Ex 8
Definition Constants we will reuse (define once, use freely)
k B T = 0.02585 eV T = 300 K the number to memorise.h c = 1240 eV⋅nm Reference gaps: Si = 1.12 = 0.66 = 1.42 ≈ 5.5
Worked example Ex 1 (Cell A) — how much does shrinking the gap by 0.1 eV help?A hypothetical semiconductor has its gap engineered from E g = 1.12 1.02 T = 300
Forecast: Guess first — does a mere 0.1
Write the ratio of the two exponentials.
R = e − E g / ( 2 k B T ) e − E g ′ / ( 2 k B T ) = e ( E g − E g ′ ) / ( 2 k B T ) Why this step? Dividing two exponentials subtracts the exponents; the common prefactor cancels, isolating the effect of the gap alone.Plug numbers. E g − E g ′ = 1.12 − 1.02 = 0.10 2 k B T = 0.0517 R = e 0.10/0.0517 = e 1.934 ≈ 6.9 Why this step? 2 k B T 0.1
Verify: ln ( 6.9 ) = 1.93 ≈ 1.934 eV / eV 0.1 7× boost — exponentials are ruthless.
Worked example Ex 2 (Cell B) — heating silicon from 300 K to 400 KFor Si (E g = 1.12 T 300 400
Forecast: Higher T
Find k B T k B T T k B T ( 400 ) = 0.02585 × 300 400 = 0.03447 Why this step? k B T Ratio of exponentials.
R = e − E g / ( 2 k B T 1 ) e − E g / ( 2 k B T 2 ) = exp [ 2 E g ( k B T 1 1 − k B T 2 1 ) ] Why this step? Here T denominator , so we cannot just subtract gaps — we subtract reciprocals. Being careful about which variable moves is the whole game of this cell.Plug in. 0.02585 1 = 38.68 0.03447 1 = 29.01 R = exp [ 2 1.12 ( 38.68 − 29.01 ) ] = e 0.56 × 9.67 = e 5.42 ≈ 225
Verify: ln 225 = 5.42 225 ×
Limit check (T → 0 as T → 0 1/ ( k B T ) → ∞ → − ∞ → 0
Worked example Ex 3 (Cell C) — Ge vs Si intrinsic carriers at 300 KCompare n i ( Ge ) / n i ( Si )
Forecast: Ge's gap is smaller — more carriers. Order of magnitude?
Ratio, common prefactor cancels.
n i ( Si ) n i ( Ge ) ≈ e ( E g , Si − E g , Ge ) / ( 2 k B T ) Why this step? Same cell-A trick; because the smaller gap sits in Ge, its exponent is less negative , so Ge wins.Plug in. E g , Si − E g , Ge = 1.12 − 0.66 = 0.46 = e 0.46/0.0517 = e 8.90 ≈ 7.3 × 1 0 3
Verify: ln ( 7300 ) = 8.90 Degenerate check: if two materials had equal gaps, the exponent would be 0 1
Worked example Ex 4 (Cell D) — red light on GaAs, and the exact-cutoff edge(a) Is λ = 700 E g = 1.42 exact cutoff?
Forecast: Absorption needs E photon ≥ E g
Photon energy. E = λ h c = 700 1240 = 1.771 Why this step? h c = 1240 Compare. 1.771 eV > 1.42 eV absorbed (the electron gets lifted across the gap).
Why this step? The absorption rule is a single inequality; everything hinges on which side you land.Exact-cutoff case (b). If E photon = E g just barely promotes an electron to the very bottom of the CB — this is the absorption edge . Just below it, transparency; just above, absorption.
Verify: 1240/700 = 1.771 1.771 > 1.42
See the two regimes below — one photon clears the gap, one falls short.
Worked example Ex 5 (Cell E) — the longest wavelength silicon can still absorbFind the cutoff wavelength λ c E g = 1.12
Forecast: Visible light is 400–700 nm. Is Si transparent to visible light, or to infrared?
Set photon energy equal to the gap. At cutoff the photon has just enough energy:
λ c h c = E g ⇒ λ c = E g h c Why this step? This is the boundary of cell D — we invert the same relation instead of comparing.Plug in. λ c = 1.12 1240 = 1107 nm Why this step? h c E g λ
Verify: 1240/1107 = 1.120 1107 infrared , longer than visible — so Si absorbs all visible light (opaque, why photodiodes work) but is transparent to far infrared . Matches the parent's "cutoff ≈ 1100
Limit note: a smaller gap pushes λ c longer (Ge: 1240/0.66 = 1879
Worked example Ex 6 (Cell F) — actual intrinsic carrier density of SiUsing the full formula n i = N c N v e − E g / ( 2 k B T ) N c N v = 2.5 × 1 0 19 cm − 3 Density of states ), find n i
Forecast: The exponential is astronomically tiny (∼ 1 0 − 10 ∼ 1 0 19
Exponential factor. 2 k B T − E g = 0.0517 − 1.12 = − 21.66 e − 21.66 = 3.9 × 1 0 − 10 Why this step? This is the "how rare" number; it alone would say "no carriers," but it is only half the story.Multiply by the prefactor.
n i = 2.5 × 1 0 19 × 3.9 × 1 0 − 10 ≈ 9.8 × 1 0 9 cm − 3 Why this step? A huge density of available states times a tiny occupation probability gives a modest but usable carrier count — the reconciliation the parent flagged.
Verify: Order of magnitude ∼ 1 0 10 cm − 3 ≈ 1.0 × 1 0 10 cm − 3
Worked example Ex 7 (Cell G) — "Why did they abandon germanium transistors?"An engineer says: "Ge transistors leak too much current at high temperature." Quantify: compare the exponential factors of Ge and Si at T = 350
Forecast: We saw Ge already has ~7300× more carriers at 300 K. Does heat make it worse?
k B T 0.02585 × 300 350 = 0.03016 2 k B T = 0.06032 Why this step? This is a cell-B temperature update feeding a cell-C ratio — real problems chain cells.Ratio Ge/Si (exponential only).
n i ( Si ) n i ( Ge ) = e ( 1.12 − 0.66 ) /0.06032 = e 0.46/0.06032 = e 7.626 ≈ 2.05 × 1 0 3 Why this step? Interesting subtlety: the ratio actually shrinks as T 2 k B T But absolute leakage rises for both. Ge's own carriers climb (cell B), and Ge started far higher — so in absolute terms Ge is still "leakier" and its junctions fail thermally first.
Verify: ln ( 2050 ) = 7.626 n i
Worked example Ex 8 (Cell H) — the "full band carries no current" trapClaim on an exam: "A crystal whose valence band holds 1 0 23
Forecast: Feels true (more electrons → more current). Resist.
Identify the cell. This is conceptual: it tests the filled-band cancellation idea, not a number.
Why this step? Recognising a trap cell saves you from computing something irrelevant.Apply the cancellation rule. In a completely filled band, for every electron drifting right under a field there is an identical electron drifting left. Net current = 0 Why this step? Current needs a net imbalance of velocities; a full band is perfectly symmetric, so it self-cancels regardless of how many electrons it holds.State the correct condition. Conduction requires empty adjacent states — a partly-filled band (metal) or thermally/optically promoted electrons in the CB (semiconductor). Count of electrons is irrelevant; availability of empty states is everything.
Verify (consistency): This is exactly why insulators (full VB, empty CB, big gap) don't conduct despite being packed with electrons — self-consistent with the whole page. Answer: FALSE. See Holes as charge carriers for how the empty seats in the VB do the conducting instead.
Recall Which cell is each question?"Longest wavelength a detector responds to?" ::: Cell E — cutoff, λ c = h c / E g n i E photon E g ≥
