Compare band gaps - conductor - semiconductor - insulator: Worked examples

Before we start, we earn every symbol in the master equation so you never meet one cold.

Definition The quantities in every example

$E_{g}$ = the band gap, measured in electron-volts (eV). One eV is the energy an electron gains crossing a 1-volt drop. Think of it as the height of the jump an electron must make from the full valence band to the empty conduction band.
$T$ = absolute temperature in kelvin (K). $0 K$ is the coldest possible; room temperature is $300 K$ .
$k$ = Boltzmann's constant $= 8.617 \times 1 0^{- 5} eV/K$ . It is the exchange rate turning temperature into energy: the typical thermal "kick" an electron has is about $k T$ . At $300 K$ , $k T \approx 0.02585 eV \approx 26 meV$ .
$n_{i}$ = the intrinsic carrier concentration — how many free electrons (per cubic centimetre) have jumped the gap in a pure, undoped material. This is the number that decides whether a material conducts. Bigger $n_{i}$ → more conduction.
$N_{c}, N_{v}$ = the effective densities of states in the conduction band and valence band. Loosely, how many "seats" are available for electrons at the bottom of the CB ( $N_{c}$ ) and holes at the top of the VB ( $N_{v}$ ). Their geometric mean $N_{c} N_{v}$ is the prefactor that scales $n_{i}$ — it counts available seats before the exponential decides how many are filled.

The scenario matrix

Every problem this topic can throw is one of these cells. The examples below are tagged to the cell they hit.

#	Cell (case class)	What makes it distinct	Example
A	Zero-gap / degenerate ( $E_{g} = 0$ , metal)	exponential collapses to $e^{0} = 1$	Ex 1
B	Small gap, one point (Si at 300 K)	compute an actual $n_{i}$	Ex 2
C	Ratio of two materials (prefactor cancels)	gap difference in exponent	Ex 3
D	Same material, two temperatures (heating)	inverse-temperature difference	Ex 4
E	Cooling toward $T \to 0$ (limiting behaviour)	exponent $\to - \infty$ , carriers $\to 0$	Ex 5
F	High- $T$ / small-gap limit ( $E_{g} ≪ k T$ )	approximation breaks, sign check	Ex 6
G	Real-world word problem (sensor / thermistor)	translate device behaviour to formula	Ex 7
H	Exam twist — the factor-of-2 trap	wrong exponent → wrong by $1 0^{many}$	Ex 8
I	Insulator sanity check (huge gap)	show why "no carriers"	Ex 9

We now sweep the whole table.

Example 1 — Cell A: the zero-gap metal

Worked example The degenerate case

$E_{g} = 0$ A metal has valence and conduction bands that overlap, so effectively $E_{g} = 0$ . Evaluate the thermal suppression factor $e^{- E_{g} /2 k T}$ and explain what happens to carrier count as $T$ changes.

Forecast: guess before reading — will heating a metal create more free electrons?

Step 1. Substitute $E_{g} = 0$ into the exponent. $- \frac{E _{g}}{2 k T} = - \frac{0}{2 k T} = 0$ Why this step? This is the degenerate input — the one value that makes the formula collapse. Always test it.

Step 2. Evaluate the exponential. $e^{0} = 1$ Why this step? Anything to the power zero is 1; there is no suppression at any temperature.

Step 3. Interpret. The carrier count is not governed by the exponential — it's fixed by how many electrons already sit in the overlapping band. Why this step? Because the exponential is a flat 1, temperature can't change carrier number; it can only change how often carriers scatter off lattice vibrations (see Conductivity and Mobility).

Verify: $e^{0} = 1$ exactly, independent of $T$ — so raising $T$ from 300 K to 600 K changes the factor by $1/1 = 1$ (no change in carrier number). Heating a metal therefore cannot add carriers; resistance rises instead. Matches Temperature Dependence of Resistance in Metals. Forecast answer: no, metals do not gain carriers when heated. ✓

The figure below makes the contrast physical: plot the suppression factor against temperature for a metal and a semiconductor. Look at the flat coral line — for the metal it never budges from 1 no matter how hot, while the lavender curve for the semiconductor climbs steeply. This picture is the whole of Cell A versus Cell D in one glance.

Figure — Compare band gaps - conductor - semiconductor - insulator

Example 2 — Cell B: an actual number for silicon

Worked example Compute

$n_{i}$ for Si at 300 K For silicon: $E_{g} = 1.12 eV$ , $T = 300 K$ , and the prefactor $N_{c} N_{v} = 2.5 \times 1 0^{19} cm^{- 3}$ (the count of available seats, defined at the top of this page). Find the intrinsic carrier concentration $n_{i}$ .

Forecast: silicon is a "decent" semiconductor. Guess the order of magnitude of $n_{i}$ in $cm^{- 3}$ (a metal has $\sim 1 0^{22}$ ).

Step 1. Compute $k T$ . $k T = (8.617 \times 1 0^{- 5}) (300) = 0.02585 eV$ Why this step? Every exponent needs the thermal budget in the same energy units as $E_{g}$ .

Step 2. Build the exponent $E_{g} /2 k T$ . $\frac{E _{g}}{2 k T} = \frac{1.12}{2 ( 0.02585 )} = \frac{1.12}{0.0517} = 21.66$ Why this step? This is the "how many budgets per carrier" number. 21.66 is large → strong suppression.

Step 3. Apply the exponential. $e^{- 21.66} = 3.9 \times 1 0^{- 10}$ Why this step? This fraction is what survives the gap.

Step 4. Multiply by the prefactor. $n_{i} = (2.5 \times 1 0^{19}) (3.9 \times 1 0^{- 10}) \approx 9.8 \times 1 0^{9} cm^{- 3}$ Why this step? The prefactor $N_{c} N_{v}$ counts available seats; the exponential counts what fraction is filled. Seats × fraction = carriers.

Verify: The textbook value for Si is $n_{i} \approx 1.0 \times 1 0^{10} cm^{- 3}$ — our $9.8 \times 1 0^{9}$ lands right on it. ✓ Forecast: notice a metal ( $1 0^{22}$ ) has a trillion times more carriers — that's the semiconductor/conductor split in one number.

Example 3 — Cell C: ratio Si vs diamond (prefactor cancels)

Worked example How much more conductive is Si than diamond?

Compare intrinsic carriers: Si ( $E_{g} = 1.12 eV$ ) vs diamond ( $E_{g} = 5.5 eV$ ) at $T = 300 K$ , $k T = 0.02585 eV$ . Assume the prefactors are equal.

Forecast: the gap is ~5× bigger for diamond. Will carriers differ by 5×, or something wildly larger?

Step 1. Write the ratio; equal prefactors cancel. $\frac{n _{i}^{Si}}{n _{i}^{dia}} = e^{- (E_{g}^{Si} - E_{g}^{dia}) /2 k T}$ Why this step? Taking a ratio is the clean trick when you don't trust the prefactors — the $N_{c} N_{v}$ terms divide out, leaving only the gap difference.

Step 2. Plug the gaps. $= e^{- (1.12 - 5.5) /0.0517} = e^{4.38/0.0517} = e^{84.72}$ Why this step? The Si gap is smaller, so the difference is negative, and the double-negative flips it positive → Si wins by a huge factor.

Step 3. Convert $e^{84.72}$ to base 10. $lo g_{10} (e^{84.72}) = \frac{84.72}{2.303} = 36.8 \Rightarrow \approx 1 0^{36.8}$ Why this step? $ln 10 = 2.303$ turns natural-log exponents into powers of ten we can read.

Verify: $\approx 1 0^{37}$ — matches the parent note. A ~5× gap difference becomes a $1 0^{37}$ carrier difference. That is the exponential doing its violence. ✓

Example 4 — Cell D: heat Si from 300 K to 600 K

Worked example Carrier factor on doubling temperature

By what factor does $n_{i}$ for Si ( $E_{g} = 1.12 eV$ ) grow when heated from $300 K$ to $600 K$ ? Use $k = 8.617 \times 1 0^{- 5} eV/K$ .

Forecast: temperature doubled. Guess: does $n_{i}$ merely double, or explode?

Step 1. Ratio of the two exponentials (prefactor treated as roughly constant). $\frac{n _{i} ( 600 )}{n _{i} ( 300 )} = exp [\frac{E _{g}}{2 k} (\frac{1}{300} - \frac{1}{600})]$ Why this step? Here $T$ changes and $E_{g}$ stays fixed, so $T$ moves in the denominator — the difference of reciprocals is what matters.

Step 2. Compute $E_{g} /2 k$ (units: kelvin). $\frac{E _{g}}{2 k} = \frac{1.12}{2 ( 8.617 \times 1 0 ^{- 5} )} = 6499 K$ Why this step? Grouping $E_{g} /2 k$ once keeps arithmetic clean.

Step 3. Compute the reciprocal difference. $\frac{1}{300} - \frac{1}{600} = 0.003333 - 0.001667 = 0.001667 K^{- 1}$ Why this step? Cooler → less budget; hotter → more. This positive gap is the extra budget gained.

Step 4. Multiply and exponentiate. $exp [6499 \times 0.001667] = e^{10.83} \approx 5.1 \times 1 0^{4}$ Why this step? The surviving exponent is the carrier boost.

Verify: $\approx 50, 000 \times$ — matches the parent note's "~ $5 \times 1 0^{4}$ ". Doubling $T$ multiplies carriers ~50,000×; this is opposite to a metal (Ex 1). ✓ Forecast: it explodes, not doubles.

The figure plots this ratio $n_{i} (T) / n_{i} (300)$ on a logarithmic vertical axis. Follow the mint curve from the coral dot at 300 K (where the ratio is 1 by construction) up to the coral dot at 600 K: it has climbed by nearly five orders of magnitude. A straight-looking rise on a log axis is an exponential — the picture is literally the master equation's shape.

Example 5 — Cell E: the cold limit $T \to 0$

Worked example What happens as we freeze a semiconductor toward absolute zero?

Take Si ( $E_{g} = 1.12 eV$ ) and let $T \to 0^{+}$ . Find the limiting value of the suppression factor $e^{- E_{g} /2 k T}$ .

Forecast: at absolute zero, is a pure semiconductor a conductor or a perfect insulator?

Step 1. Examine the exponent as $T \to 0^{+}$ . $\frac{E _{g}}{2 k T} ⟶ \frac{1.12}{2 k \cdot 0 ^{+}} ⟶ + \infty$ Why this step? Dividing a fixed positive number by something shrinking to zero blows it up. This is the limiting-value cell — you probe the boundary, not a middle point.

Step 2. Feed that into the exponential. $e^{- \infty} = 0$ Why this step? A hugely negative exponent drives the factor to zero.

Step 3. Numerical taste at $T = 30 K$ (ten times colder than room): $\frac{1.12}{2 ( 8.617 \times 1 0 ^{- 5} ) ( 30 )} = 216.6, e^{- 216.6} \approx 1 0^{- 94}$ Why this step? A concrete cold point shows the collapse is already brutal well before $T = 0$ .

Verify: at 30 K the factor is $\sim 1 0^{- 94}$ , effectively zero. So a pure (intrinsic) semiconductor becomes a perfect insulator when frozen — exactly why real cold electronics rely on doping, whose carriers don't need the full gap. ✓ Forecast: perfect insulator at absolute zero.

Example 6 — Cell F: the small-gap / high- $T$ limit (approximation breaks)

Worked example When does the "

$- E_{g} /2 k T$ " picture stop being safe? The parent used the Boltzmann approximation $f (E) \approx e^{- (E - E_{F}) / k T}$ , valid only when $E - E_{F} ≫ k T$ . For a narrow-gap material with $E_{g} = 0.1 eV$ at $T = 300 K$ , check whether the approximation is trustworthy.

Forecast: small gap, warm. Is "much greater than $k T$ " still true?

Step 1. The energy cost per carrier is $E_{g} /2$ . $\frac{E _{g}}{2} = \frac{0.1}{2} = 0.05 eV$ Why this step? We compare this against the budget $k T$ to test the " $≫$ " condition.

Step 2. Compare with $k T = 0.02585 eV$ . $\frac{E _{g} /2}{k T} = \frac{0.05}{0.02585} = 1.93$ Why this step? The approximation demands this ratio $≫ 1$ ; here it's only ~2. The "+1" in Fermi–Dirac is not negligible.

Step 3. Estimate the error. The exact factor is $\frac{1}{1 + e ^{1.93}}$ vs the approximation $e^{- 1.93}$ . $exact = \frac{1}{1 + 6.89} = 0.1268, approx = e^{- 1.93} = 0.1451$ Why this step? Direct comparison shows a real ~14% discrepancy — not the tiny error we get for Si (where the ratio was 21.66).

Verify: relative error $= (0.1451 - 0.1268) /0.1268 = 0.144 \approx 14%$ . For Si (ratio 21.66) the same error would be $e^{- 21.66}$ -sized, utterly negligible. So the neat exponential is a large-gap tool; for $E_{g} \sim k T$ you must keep full Fermi–Dirac. ✓ Forecast: no, the approximation fails here.

Example 7 — Cell G: real-world thermistor sensor

Worked example Designing a temperature sensor

A thermistor is a semiconductor whose resistance drops as it heats (more carriers = less resistance). Its resistance follows $R (T) = R_{0} e^{+ E_{g} /2 k T}$ , where $R_{0}$ is a fixed constant of the device (its resistance in the hypothetical infinite-temperature limit, set by geometry and material — it simply cancels whenever we take a ratio). A sensor made from a material with $E_{g} = 0.5 eV$ reads $R = 10 k Ω$ at $300 K$ . What resistance does it show at $350 K$ ?

Forecast: hotter → more carriers → lower resistance. By roughly how much — a few percent, or a factor of several?

Step 1. Ratio form (the $R_{0}$ cancels). $\frac{R ( 350 )}{R ( 300 )} = exp [\frac{E _{g}}{2 k} (\frac{1}{350} - \frac{1}{300})]$ Why this step? Resistance carries the positive exponent, so we keep the sign straight: heating makes the bracket negative, shrinking $R$ . Taking the ratio removes the unknown $R_{0}$ .

Step 2. $E_{g} /2 k$ . $\frac{0.5}{2 ( 8.617 \times 1 0 ^{- 5} )} = 2901 K$ Why this step? Same grouping trick as Ex 4.

Step 3. Reciprocal difference. $\frac{1}{350} - \frac{1}{300} = 0.002857 - 0.003333 = - 4.76 \times 1 0^{- 4} K^{- 1}$ Why this step? Negative because 350 K is hotter → its reciprocal is smaller.

Step 4. Combine. $\frac{R ( 350 )}{R ( 300 )} = e^{2901 \times (- 4.76 \times 1 0^{- 4})} = e^{- 1.381} = 0.2513$ $R (350) = 10 k Ω \times 0.2513 = 2.51 k Ω$ Why this step? Multiply the ratio by the known 300 K resistance.

Verify: resistance dropped from $10 k Ω$ to $2.51 k Ω$ — a 4× fall for just a 50 K rise. Units: $k Ω \times$ (dimensionless) $= k Ω$ . ✓ Forecast: not a few percent — a factor of four, which is exactly why thermistors are sensitive sensors.

The figure shows $R (T)$ falling as the material heats. Trace the coral curve downward and note the two lavender dots: 10 kΩ at 300 K collapsing to 2.5 kΩ at 350 K. The steepness of that fall — a 4× drop over a mere 50 K — is the visual payoff of putting $E_{g} /2 k T$ in an exponent, and it is precisely what makes a thermistor a usable thermometer.

Example 8 — Cell H: the exam twist (factor-of-2 trap)

Worked example Spot the error that costs

$1 0^{many}$ A student computes the carrier ratio of Ge ( $E_{g} = 0.67 eV$ ) to Si ( $E_{g} = 1.12 eV$ ) at 300 K but forgets the factor of 2, using $e^{- Δ E_{g} / k T}$ instead of $e^{- Δ E_{g} /2 k T}$ . Compute both and find how many orders of magnitude the mistake introduces.

Forecast: just a missing 2 in a denominator. How bad can it be — a factor of 2?

Step 1. Gap difference ( $Δ E_{g} = E_{g}^{Si} - E_{g}^{Ge}$ , so Ge has more carriers). $Δ E_{g} = 1.12 - 0.67 = 0.45 eV$ Why this step? Fix which material is on top so the sign is unambiguous.

Step 2. Correct exponent (with the 2). $\frac{Δ E _{g}}{2 k T} = \frac{0.45}{0.0517} = 8.70 \Rightarrow \frac{n _{i}^{Ge}}{n _{i}^{Si}} = e^{8.70} = 6.0 \times 1 0^{3}$ Why this step? This is the physically right answer — Ge has ~6000× more intrinsic carriers than Si, using the correct mid-gap cost $E_{g} /2$ .

Step 3. Wrong exponent (no 2). $\frac{Δ E _{g}}{k T} = \frac{0.45}{0.02585} = 17.41 \Rightarrow e^{17.41} = 3.6 \times 1 0^{7}$ Why this step? This is the buggy calculation — using the full gap $E_{g}$ as the cost per carrier doubles the exponent.

Step 4. Size of the blunder — divide wrong by correct. $\frac{3.6 \times 1 0 ^{7}}{6.0 \times 1 0 ^{3}} = 6.0 \times 1 0^{3}$ Why this step? Because $e^{2 x} / e^{x} = e^{x}$ , dropping the 2 squares the true answer — so the error factor equals the true ratio itself.

Verify: the mistake inflates the answer by ~ $6 \times 1 0^{3}$ — nearly four orders of magnitude from one missing digit. ✓ Forecast: catastrophic, not a factor of 2. Remember the mnemonic: "Half in the exponent."

Example 9 — Cell I: insulator sanity check

Worked example Why diamond carries essentially nothing

Diamond: $E_{g} = 5.5 eV$ , $T = 300 K$ , prefactor $N_{c} N_{v} \approx 2.5 \times 1 0^{19} cm^{- 3}$ . Estimate its intrinsic carrier concentration and count how many carriers sit in a $1 cm^{3}$ crystal.

Forecast: "insulator" — do you expect thousands of carriers, one, or less than one per crystal?

Step 1. Exponent. $\frac{E _{g}}{2 k T} = \frac{5.5}{0.0517} = 106.4$ Why this step? Huge gap → huge exponent → savage suppression.

Step 2. Exponential and $n_{i}$ . $e^{- 106.4} = 6.2 \times 1 0^{- 47}$ $n_{i} = (2.5 \times 1 0^{19}) (6.2 \times 1 0^{- 47}) \approx 1.5 \times 1 0^{- 27} cm^{- 3}$ Why this step? Multiply available seats ( $N_{c} N_{v}$ ) by the surviving fraction ( $e^{- E_{g} /2 k T}$ ).

Step 3. Carriers in $1 cm^{3}$ : it's just $n_{i} \times 1 cm^{3} = 1.5 \times 1 0^{- 27}$ . Why this step? Concentration times volume = count.

Verify: $1.5 \times 1 0^{- 27}$ carriers per cm³ means you would need $\sim 1 0^{27} cm^{3}$ of diamond — a cube ~ $1 0^{9} m$ on a side, bigger than the Sun — to expect one thermally excited carrier. ✓ Forecast: far less than one — that is what "insulator" really means. Same equation, same physics as Si (Ex 2), just a bigger $E_{g}$ (see Intrinsic vs Extrinsic Semiconductors).

Active Recall

What does the suppression factor

e^{- E_{g} /2 k T}

equal for a metal, and why?

e^{0} = 1

— because

E_{g} = 0

, so there's no thermal suppression; carrier count is fixed.

What is

n_{i}

, and what does the prefactor

N_{c} N_{v}

represent?

n_{i}

is the intrinsic carrier concentration (free carriers per cm³ in a pure material);

N_{c} N_{v}

counts the available "seats" (effective densities of states) before the exponential sets how many are filled.

In the ratio of two materials'

n_{i}

, why can we ignore the prefactors?

The

N_{c} N_{v}

terms roughly cancel in a ratio, leaving only the exponent with the gap difference.

When you heat a semiconductor, does

T

appear in the numerator or denominator of the exponent?

Denominator — so you take a difference of reciprocals

1/ T_{1} - 1/ T_{2}

T \to 0^{+}

, what happens to intrinsic carriers and why?

E_{g} /2 k T \to + \infty

e^{- E_{g} /2 k T} \to 0

; a pure semiconductor becomes a perfect insulator.

When does the Boltzmann approximation

e^{- (E - E_{F}) / k T}

fail?

When

E_{g} /2

is not

≫ k T

(narrow gap / high

T

); then the "+1" in Fermi–Dirac matters.

What does dropping the factor of 2 in the exponent do to a carrier ratio?

It squares the ratio — introducing an error equal to the true ratio itself (many orders of magnitude).

Connections

Parent: 2.1.3 Compare band gaps
Fermi–Dirac Distribution
Intrinsic vs Extrinsic Semiconductors
Density of States
Doping and the Fermi Level
Conductivity and Mobility
Temperature Dependence of Resistance in Metals

Worked examples — Compare band gaps - conductor - semiconductor - insulator