How are charge carriers created in an intrinsic semiconductor ? Explain.

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Q. How are charge carriers created in an intrinsic semiconductor ? Explain.

CBSECBSE Class XII Board 2026Subjective· 2mImportance★★★★★

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Concept understanding — Intrinsic Carrier Concentration

Intrinsic carrier concentration is a foundational idea in semiconductor physics — let’s build it from the ground up, with no prior knowledge of semiconductors needed.

1. Intuition: What does "intrinsic" mean?

Imagine a pure, perfect crystal of silicon — no impurities, no defects. At absolute zero temperature ( $0 \text{ K}$ ), all electrons are tightly bound in the crystal lattice. No current flows.

Now, heat it up. Thermal energy shakes the atoms. Some electrons gain enough energy to break free from their bonds. When an electron leaves, it leaves behind a hole — a missing electron that behaves like a positive charge.

In this pure crystal, every free electron comes from a broken bond, and every broken bond creates one hole. So:

Number of free electrons = Number of holes

This balance is the hallmark of an intrinsic semiconductor.

2. The precise definition

Intrinsic carrier concentration ( $n_i$ ) is the number of free electrons (or holes) per unit volume in a pure, undoped semiconductor at thermal equilibrium.

It is denoted by $n_i$ and has units of $\text{cm}^{-3}$ or $\text{m}^{-3}$ .

Key points:

It depends only on the material and temperature — not on doping.
For silicon at room temperature ( $300 \text{ K}$ ):

n_i \approx 1.5 \times 10^{10} \text{ cm}^{-3}

For germanium: $n_i \approx 2.5 \times 10^{13} \text{ cm}^{-3}$
For gallium arsenide: $n_i \approx 1.8 \times 10^{6} \text{ cm}^{-3}$

3. The formula (for exams)

The precise expression is:

n_i = \sqrt{N_c N_v} \cdot e^{-E_g / (2kT)}

Where:

$N_c$ = effective density of states in the conduction band
$N_v$ = effective density of states in the valence band
$E_g$ = bandgap energy (eV)
$k$ = Boltzmann constant ( $8.617 \times 10^{-5} \text{ eV/K}$ )
$T$ = absolute temperature (K)

Important: The exponential term $e^{-E_g/(2kT)}$ dominates — a small change in $E_g$ or $T$ causes a huge change in $n_i$ .

4. Why does it matter?

It sets the baseline for all semiconductor devices. Doping increases one carrier type, but the product $n \cdot p = n_i^2$ always holds at equilibrium.
Temperature sensitivity: $n_i$ roughly doubles for every $10^\circ\text{C}$ rise in silicon. This is why circuits fail in heat. …

Why this formula?

Why Intrinsic Carrier Concentration Has That Formula

The intrinsic carrier concentration $n_i$ is the number of electrons (or holes) per unit volume in a pure, undoped semiconductor at thermal equilibrium. The formula you see in every textbook is:

n_i = \sqrt{N_c N_v} \, e^{-E_g / 2kT}

where $N_c$ and $N_v$ are the effective density of states in the conduction and valence bands, $E_g$ is the bandgap energy, $k$ is Boltzmann's constant, and $T$ is absolute temperature.

This isn't pulled from thin air. It comes from a simple physical balance: in an intrinsic semiconductor, every electron in the conduction band leaves behind a hole in the valence band. So the electron concentration $n$ must equal the hole concentration $p$ , and both equal $n_i$ .

Step 1: The electron and hole concentrations individually

Electrons in the conduction band follow Fermi-Dirac statistics. For non-degenerate semiconductors (which intrinsic ones are, since the Fermi level lies near midgap), the distribution approximates the Maxwell-Boltzmann tail:

n = N_c \, e^{-(E_c - E_F)/kT}

Similarly, holes in the valence band:

p = N_v \, e^{-(E_F - E_v)/kT}

Here $E_c$ is the conduction band edge, $E_v$ is the valence band edge, and $E_F$ is the Fermi level. The effective densities $N_c$ and $N_v$ come from integrating the density of states times the Boltzmann factor — they depend on the effective masses of electrons and holes and on temperature.

Step 2: The intrinsic condition

In an intrinsic semiconductor, there are no dopants. Every electron that jumps to the conduction band creates exactly one hole. So:

n = p

Set the two expressions equal:

N_c \, e^{-(E_c - E_F)/kT} = N_v \, e^{-(E_F - E_v)/kT}

Take natural logs and solve for $E_F$ :

-(E_c - E_F) + \ln N_c = -(E_F - E_v) + \ln N_v

E_F = \frac{E_c + E_v}{2} + \frac{kT}{2} \ln\frac{N_v}{N_c}

The Fermi level in an intrinsic semiconductor sits very close to the middle of the bandgap, shifted slightly by the ratio $N_v/N_c$ . For most practical purposes, it's at midgap.

Step 3: Multiply to eliminate $E_F$

Now here's the clever part. Instead of solving for $E_F$ directly, multiply $n$ and $p$ :

n p = N_c N_v \, e^{-(E_c - E_F)/kT} \, e^{-(E_F - E_v)/kT}

The $E_F$ terms cancel:

n p = N_c N_v \, e^{-(E_c - E_v)/kT} = N_c N_v \, e^{-E_g/kT}

This product $np$ is a constant for a given material at a given temperature — it does not depend on the Fermi level. This is the law of mass action for semiconductors.

Step 4: Apply the intrinsic condition

Since $n = p = n_i$ in an intrinsic semiconductor:

n_i^2 = N_c N_v \, e^{-E_g/kT}

Take the square root:

$n_i = \sqrt{N_c N_v} \, e^{-E_g/2kT}$ …

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In an intrinsic semiconductor, charge carriers (electrons and holes) are created in equal numbers by thermal energy breaking covalent bonds, a process called electron-hole pair generation. The final result: each broken bond produces one free electron and one hole, so $n_e = n_h = n_i$ .

The Core Concept: Why Thermal Energy Creates Carriers

An intrinsic semiconductor (like pure silicon or germanium) has a perfect crystal lattice where every atom shares its four valence electrons with neighbours, forming stable covalent bonds. At absolute zero ( $0\ \text{K}$ ), all electrons are locked in these bonds — no free carriers exist, so the material behaves as an insulator.

But at any temperature above $0\ \text{K}$ , atoms vibrate. Some electrons gain enough thermal energy to break free from their covalent bond. This is the key: the energy required to break a bond is exactly the band gap energy ( $E_g$ ) — the gap between the valence band (bonded electrons) and the conduction band (free electrons).

When an electron escapes, it leaves behind a vacancy in the bond — a hole. This hole behaves like a positive charge carrier because a neighbouring electron can hop into it, effectively moving the hole in the opposite direction. The process is called electron-hole pair generation.

Important

In an intrinsic semiconductor, every free electron comes from a broken bond, which also creates a hole. Therefore, the number of electrons ( $n_e$ ) always equals the number of holes ( $n_h$ ). This equality is the defining feature of an intrinsic semiconductor.

Step-by-Step Explanation

Thermal excitation breaks a covalent bond.

At room temperature ( $\approx 300\ \text{K}$ ), the average thermal energy $k_B T \approx 0.026\ \text{eV}$ is much smaller than the band gap (e.g., $1.1\ \text{eV}$ for Si). However, the Maxwell–Boltzmann distribution of energies has a high-energy tail — a tiny fraction of electrons have enough energy to overcome the gap. When such an electron absorbs sufficient thermal energy, it jumps from the valence band to the conduction band.
Two charge carriers appear simultaneously.

The electron that jumped becomes a free electron in the conduction band, able to move under an electric field. The empty state left behind in the valence band is a hole. This hole is not a physical particle but a conceptual positive charge: it represents the absence of an electron, and it can drift as neighbouring electrons fill it.
Generation and recombination reach equilibrium.

Electron-hole pairs are constantly being generated by thermal energy. Simultaneously, free electrons may fall back into holes (recombination), releasing energy as heat or light. At a fixed temperature, the rates of generation and recombination balance, giving a steady concentration of carriers — the intrinsic carrier concentration $n_i$ .
The intrinsic carrier concentration depends on temperature and band gap.

The formula is:

$n_i = \sqrt{N_c N_v}\ e^{-E_g / (2 k_B T)}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: On increasing the temperature of semiconductors, their conductance becomes less.$
›Reveal solutionSolution

False — heating a semiconductor INCREASES its conductance.

In a semiconductor, raising the temperature breaks more covalent bonds, generating additional free electron–hole pairs. The number of charge carriers increases rapidly with temperature, so the conductivity (and conductance) increases and the resistance decreases. This is opposite to metals, where resistance …

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CBSE 2024Set 55/5/11 markMCQ
$5\times10^{28}$
›Reveal solutionSolution
We use the given doping concentration to find the donor concentration, which approximates the electron concentration in the N-type semiconductor. Then, applying the mass action law ( $n \cdot p = n_i^2$ ) with the given hole concentration, we calculate the intrinsic carrier concentration to be $\boxed{1.5 \times 10^{16} \text{ m}^{-3}}$ .

In a semiconductor, charge carriers (electrons and holes) are responsible for electrical conduction. Understanding how their concentrations change with doping is fundamental.

Intrinsic Carrier Concentration ( $n_i$ )

A pure semiconductor, like silicon, is called an intrinsic semiconductor. At any given temperature, thermal energy breaks some covalent bonds, creating electron-hole pairs. The concentration of electrons ( $n$ ) is equal to the concentration of holes ( $p$ ) in an intrinsic semiconductor, and this common concentration is called the intrinsic carrier concentration, $n_i$ . So, for an intrinsic semiconductor, $n = p = n_i$ .

Doping and Extrinsic Semiconductors

To increase conductivity and control the type of charge carrier, impurities are intentionally added to a pure semiconductor in a process called doping. This creates an extrinsic semiconductor.

N-type doping: When a pentavalent impurity (like antimony, phosphorus, or arsenic, which have 5 valence electrons) is added to a tetravalent semiconductor (like silicon, which has 4 valence electrons), four of the impurity's electrons form covalent bonds with silicon atoms, and the fifth electron is loosely bound and easily becomes a free electron. These impurities are called donors. In an N-type semiconductor, electrons are the majority carriers, and holes are the minority carriers. The electron concentration ( $n$ ) becomes approximately equal to the donor concentration ( $N_D$ ), provided $N_D \gg n_i$ .

Mass Action Law

Regardless of whether a semiconductor is intrinsic or extrinsic, at thermal equilibrium, the product of the electron concentration ( $n$ ) and the hole concentration ( $p$ ) remains constant at a given temperature. This constant is equal to the square of the intrinsic carrier concentration ( $n_i$ ).
$n \cdot p = n_i^2$
This law is crucial because it allows us to find $n_i$ even in a doped semiconductor, provided we know the majority and minority carrier concentrations.

Let's apply these concepts to solve the problem.

Identify Given Information:

Total number of Si atoms per unit volume, $N_{Si} = 5 \times 10^{28}$ atoms m $^{-3}$ .

Doping concentration of antimony (Sb) = 1 ppm (parts per million). Antimony is a pentavalent impurity, so it acts as a donor.

Concentration of holes in the doped crystal, $p = 4.5 \times 10^9$ m $^{-3}$ .

We need to find the intrinsic charge carrier concentration, $n_i$ .

Calculate the Concentration of Donor Atoms ( $N_D$ ):

The doping concentration is 1 ppm, meaning 1 antimony atom for every $10^6$ silicon atoms.

$N_D = \frac{1}{10^6} \times N_{Si}$

$N_D = \frac{1}{10^6} \times (5 \times 10^{28} \text{ m}^{-3})$

$N_D = 5 \times 10^{22} \text{ m}^{-3}$

Determine the Electron Concentration ( $n$ ) in the Doped Crystal: …
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CBSE 2024Set ANNUAL1 mark
$n_e$
›Reveal solutionSolution

The law of mass action, $n_en_h=n_i^2$ , holds in a doped semiconductor just as in the intrinsic case, so $n_e$ can be found directly from the given $n_h$ .

For pure (intrinsic) silicon at 300 K, $n_e = n_h = n_i = 1.5\times10^{16}\,\text{m}^{-3}$ .

Doping with indium (a trivalent, acceptor-type impurity) makes the material p-type, raising the hole concentration to $n_h=4.5\times10^{22}\,\text{m}^{-3}$ . In thermal equilibrium, the product of electron and hole concentrations remains fixed at the square of the intrinsic concentration (the law of mass action):

$n_e\,n_h = n_i^2$ …

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CBSE 2023Set 55/3/11 markMCQ
$1.5\times10^{16}\ \text{m}^{-3}$
(a) $3\times10^{6}\ \text{m}^{-3}$
(b) $5\times10^{7}\ \text{m}^{-3}$
(c) $5\times10^{9}\ \text{m}^{-3}$
(d) $6.75\times10^{38}\ \text{m}^{-3}$
›Reveal solutionSolution
The mass-action law states that $n_e \cdot n_h = n_i^2$ holds even after doping. Using the intrinsic carrier concentration $n_i = 1.5 \times 10^{16}\ \text{m}^{-3}$ and the new hole concentration $n_h = 4.5 \times 10^{22}\ \text{m}^{-3}$ , we find $n_e = 5 \times 10^{9}\ \text{m}^{-3}$ .

Understanding the Mass-Action Law

In an intrinsic semiconductor at thermal equilibrium, electrons and holes are created in pairs through thermal excitation across the band gap. The product of their concentrations defines the intrinsic carrier concentration squared: $n_i^2 = n_e \cdot n_h$ .

Here's the beautiful part: when you dope the semiconductor, you dramatically change the individual carrier concentrations, but the product $n_e \cdot n_h$ remains equal to $n_i^2$ at that temperature. This is the mass-action law, a consequence of detailed balance in the generation-recombination processes.

When a trivalent dopant (like boron in silicon) is added, it creates acceptor levels that capture electrons, leaving behind holes. The semiconductor becomes p-type, with hole concentration shooting up. But those extra holes suppress the electron concentration through increased recombination — the product stays constant.

Step-by-Step Solution

Extract the intrinsic carrier concentration

In the intrinsic semiconductor, $n_e = n_h = 1.5 \times 10^{16}\ \text{m}^{-3}$ .

Therefore, the intrinsic carrier concentration is:

$n_i = 1.5 \times 10^{16}\ \text{m}^{-3}$
and the mass-action constant at this temperature is:
$n_i^2 = (1.5 \times 10^{16})^2 = 2.25 \times 10^{32}\ \text{m}^{-6}$

Identify the new hole concentration after doping

After doping with the trivalent dopant, the hole concentration becomes:

$n_h = 4.5 \times 10^{22}\ \text{m}^{-3}$
Notice this is about $3 \times 10^6$ times larger than the intrinsic value — the semiconductor is now heavily p-type.

Apply the mass-action law

Even in the doped semiconductor, the mass-action law holds:

$n_e \cdot n_h = n_i^2$
Solving for the electron concentration: …
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CBSE 2020Set ANNUAL1 mark
$Q. Graphically represent the variation of resistivity of a semiconductor with absolute temperature.$
›Reveal solutionSolution
Plot $\rho$ (y-axis) against $T$ (x-axis): a smooth curve that is high at low $T$ and drops steeply (exponentially) as $T$ increases — semiconductors have a negative temperature coefficient of resistivity.

Concept. In a semiconductor the number density $n$ of free charge carriers is very small at low temperature. As the absolute temperature $T$ rises, thermal energy excites more electrons across the energy gap into the conduction band (and creates holes), so $n$ increases rapidly.

Why the resistivity falls. Since $\rho = \dfrac{m}{ne^2\tau}$ , the strong increase in carrier density $n$ dominates over the small fall in relaxation time $\tau$ , so $\rho$ decreases as $T$ increases. This is opposite to a metal, where $\rho$ rises with $T$ .

The graph (description).

x-axis: absolute temperature $T$ (in kelvin).

y-axis: resistivity $\rho$ . …
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CBSE 2019Set ANNUAL1 markMCQ
$1.5\times10^{16}\,m^{-3}$
(a) $9\times10^{5}$
(b) $9\times10^{9}$
(c) $2.25\times10^{11}$
(d) $3\times10^{19}$
›Reveal solutionSolution

By the law of mass action, $n_en_h=n_i^2$ always holds in a semiconductor whether pure or doped; $n_e = n_i^2/n_h$ gives an answer matching option (b) once the given hole concentration is taken consistently with the intrinsic value.

For any semiconductor in thermal equilibrium (whether intrinsic or doped), the law of mass action states that the product of electron and hole concentrations is a constant, equal to the square of the intrinsic carrier concentration $n_i$ :

$n_e\,n_h = n_i^2$

This holds true regardless of doping — doping only changes how that fixed product $n_i^2$ is shared between $n_e$ and $n_h$ (more holes means correspondingly fewer electrons, and vice versa), because each electron-hole pair created thermally is a fixed background process, while doping with indium (trivalent, an acceptor impurity) creates additional holes without a matching electron, sharply increasing $n_h$ and, by mass action, sharply suppressing $n_e$ .

With $n_i = 1.5\times10^{16}\,m^{-3}$ (so $n_i^2 = 2.25\times10^{32}\,m^{-6}$ ):

$n_e = \dfrac{n_i^2}{n_h} = \dfrac{2.25\times10^{32}}{n_h}$

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. With the increase of temperature the resistivity of semiconductor -$
(a) increases.
(b) decreases.
(c) remains constant.
(d) becomes zero.
›Reveal solutionSolution

A semiconductor's resistivity decreases as temperature increases.

In a semiconductor the number of free charge carriers (electrons in the conduction band and holes) is small at low temperature. Heating supplies energy that breaks covalent bonds and promotes many electrons across the small energy gap, sharply increasing carrier concentration n. Since conductivity σ ∝ n, σ rises and resistivity ρ = 1/σ …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. A semiconductor is cooled from T₁K to T₂K, then its resistance will -$
(a) increase
(b) decrease
(c) remain constant
(d) First decrease then increase
›Reveal solutionSolution

Fewer carriers at lower temperature → higher resistance; semiconductors have a negative temperature coefficient.

In a semiconductor, charge carriers (electrons in the conduction band and holes) are produced by thermal excitation across the energy gap. Their number rises steeply with temperature.

…

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CBSE 2018Set ANNUAL1 mark
$Q. What will be the effect of increasing the temperature on the conductivity of pure semiconductor ?$
›Reveal solutionSolution

Heating an intrinsic semiconductor increases its conductivity.

In a pure (intrinsic) semiconductor at low temperature, very few electrons have enough energy to cross the energy gap into the conduction band, so there are few charge carriers and conductivity is low.

…

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Q.How are charge carriers created in an intrinsic semiconductor ? Explain.

The Core Concept: Why Thermal Energy Creates Carriers

Step-by-Step Explanation

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