Intrinsic carrier density versus temperature

As temperature rises, the number of electrons thermally excited from the valence band to the conduction band grows exponentially. Enter a material and temperature to calculate the intrinsic carrier density ni and check up to what temperature a doped semiconductor keeps the properties set by its dopants.

Inputs

Results

Intrinsic carrier density ni
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Band gap at this temperature Eg(T)
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Thermal energy kT
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Ratio to ni at 300 K
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Majority-carrier (electron) density n
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Minority-carrier (hole) density p
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Temperature where ni = ND (onset of intrinsic behavior)
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The vertical axis is logarithmic. The solid line is the selected material; the dashed lines are the other two (from top: Ge, Si, GaAs). The red dot is the current value.

Uses representative ni values at 300 K of Si 1.0×10¹⁰, Ge 2.0×10¹³ and GaAs 2.1×10⁶ cm⁻³ (the same as the PN junction calculator), with the effective densities of states Nc and Nv proportional to T^1.5. The temperature dependence of the band gap follows the Varshni equation Eg(T) = Eg(0) − αT²/(T + β) with representative coefficients (Si: 1.17 eV, 4.73×10⁻⁴ eV/K, 636 K; Ge: 0.7437, 4.774×10⁻⁴, 235; GaAs: 1.519, 5.405×10⁻⁴, 204). Majority and minority carriers are calculated for n-type material with all donors ionized. Incomplete donor ionization at low temperature (freeze-out) is not included.

Intrinsic carrier density formula

In a semiconductor with no impurities, electrons thermally excited into the conduction band and the holes left in the valence band form in equal numbers, and their density is the intrinsic carrier density ni = √(Nc·Nv)·exp(−Eg/2kT). Nc and Nv are the effective densities of states of the conduction and valence bands and are proportional to T^1.5. Because the band gap sits in the exponent, ni rises very steeply with temperature, and at the same temperature it is orders of magnitude smaller in materials with a larger band gap. This tool takes the representative value at 300 K as the reference and calculates ni(T) = ni(300 K) × (T/300)^1.5 × exp[−(Eg(T)/kT − Eg(300 K)/(k·300 K))/2].

Assumptions

The effective masses are treated as independent of temperature, so the only temperature dependence of Nc and Nv is T^1.5. The temperature variation of the band gap is included through the Varshni equation. For n-type material with all donors ionized, the majority and minority carriers follow from charge neutrality n = p + ND and the mass-action law n·p = ni²: n = ND/2 + √((ND/2)² + ni²) and p = ni²/n. The results do not apply at low temperatures where donor ionization is incomplete (roughly below 100 K for Si) or at concentrations high enough to cause degeneracy.

Worked example

With the default Si at 300 K, ni = 1.0×10¹⁰ cm⁻³. At 100 °C (373 K) it rises to about 1.3×10¹² cm⁻³, about 130 times higher, and at 400 K to about 5.3×10¹² cm⁻³, about 530 times higher. For n-type with ND = 10¹⁵ cm⁻³ at 300 K, the electron density is almost exactly 10¹⁵ cm⁻³ and the hole density is ni²/ND = 10⁵ cm⁻³. As the temperature is raised, ni catches up with ND at about 547 K (about 274 °C); above this, thermally generated carriers outnumber those from the dopants and the material increasingly loses its n-type character. Under the same conditions Ge reaches this point at about 391 K (about 118 °C) and GaAs at about 771 K (about 498 °C), which shows why materials with larger band gaps can operate at higher temperatures.

Values used and their sources

The ni values at 300 K and the Varshni coefficients are representative material values from semiconductor device textbooks and materials handbooks. Reported values of ni for Si range from 9.65×10⁹ to 1.5×10¹⁰ cm⁻³, and about 1.0×10¹⁰ cm⁻³ is commonly used today. Ge ranges from about 2.0 to 2.4×10¹³ cm⁻³ and GaAs from about 1.8 to 2.1×10⁶ cm⁻³. The Boltzmann constant is k = 8.617×10⁻⁵ eV/K. Use the results to get the order of magnitude and the trend with temperature.

Practice semiconductor physics questions

What happens to the intrinsic carrier density when the temperature of an intrinsic semiconductor is raised?

  1. AIt increases
  2. BIt decreases
  3. CIt does not change
Show answer and explanation

Answer: A. It increases

As temperature rises, valence electrons more easily cross the band gap with thermal energy, creating more electron-hole pairs. The intrinsic carrier density follows ni ∝ T^(3/2)·exp(−Eg/2kT), and in silicon near room temperature it roughly doubles for every 10 °C rise. This steep increase is why PN-junction reverse leakage grows at high temperature, and why wide-bandgap SiC and GaN handle high-temperature operation well.

Which value is closest to the intrinsic carrier density of silicon at room temperature?

  1. AAbout 1×10⁴ cm⁻³
  2. BAbout 1×10¹⁰ cm⁻³
  3. CAbout 1×10²² cm⁻³
Show answer and explanation

Answer: B. About 1×10¹⁰ cm⁻³

Silicon's intrinsic carrier density is about 1×10¹⁰ cm⁻³ at room temperature (1.0–1.5×10¹⁰ depending on the source). Compared with the atomic density of about 5×10²² cm⁻³, only about one atom in 10¹² contributes a free carrier. Germanium is about 2×10¹³ and GaAs about 2×10⁶ cm⁻³, so ni drops by orders of magnitude as the band gap widens. This small ni is why tiny amounts of doping can control conductivity so precisely.

In a semiconductor in thermal equilibrium, what is the product of electron density n and hole density p (n×p)? (ni: intrinsic carrier density)

  1. AIt is always 0
  2. BIt equals ni squared
  3. CIt equals the square of the impurity concentration
Show answer and explanation

Answer: B. It equals ni squared

In thermal equilibrium, n × p = ni² holds even with doping (the law of mass action): adding electrons gives holes more partners to recombine with, so holes decrease correspondingly. For example, adding 1×10¹⁶ cm⁻³ donors to silicon (ni ≈ 1×10¹⁰ cm⁻³) gives a hole density of 1×10⁴ cm⁻³. It does not hold out of equilibrium, such as under illumination or current injection, where np > ni².

What is the approximate hole density in n-type silicon with a donor concentration of 1×10¹⁶ cm⁻³ (all donors ionized, intrinsic carrier density ni = 1×10¹⁰ cm⁻³)?

  1. A1×10⁴ cm⁻³
  2. B1×10⁶ cm⁻³
  3. C1×10¹⁰ cm⁻³
Show answer and explanation

Answer: A. 1×10⁴ cm⁻³

With all donors ionized, the electron density is n ≈ 1×10¹⁶ cm⁻³. From the law of mass action n × p = ni², p = (1×10¹⁰)² ÷ 1×10¹⁶ = 1×10⁴ cm⁻³. These minority carriers are a trillionth of the majority carriers, yet they carry the reverse current of a PN junction and the operation of a bipolar transistor. Since ni rises steeply with temperature, the minority carrier density also grows sharply when hot.

FAQ

By how much does ni of Si increase for a 10 K rise in temperature?

Near room temperature, by about 2.2 times (300 K → 310 K). Quantities proportional to ni², such as the reverse saturation current, increase by about 5 times, which is why PN junction leakage current is so sensitive to temperature.

What is ni used for?

It is used for the minority-carrier density p = ni²/ND, the built-in potential of a PN junction (kT/q)ln(NA·ND/ni²), the reverse saturation current of a diode (proportional to ni²), and estimates of leakage current at high temperature. Through the temperature dependence of ni, all of these determine how a device behaves with temperature.

Why do SiC and GaN power devices tolerate high temperatures?

Their band gaps are large, about 3 eV or more, so their ni is dozens of orders of magnitude smaller than that of silicon. Even at several hundred °C, thermally generated carriers do not catch up with those from the dopants, so the material does not easily become intrinsic, leakage current stays small, and the device keeps working at high temperature.

Related terms

Intrinsic carrier density Band gap Effective mass Fermi level Donor Wide-bandgap semiconductor

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