The conductivity of a semiconductor increases with increase in temperature because (a) number density of free current carriers increases. (b) relaxation time increases. (c) both number density of carriers and relaxation time increase. (d) number density of current carriers increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density.

> [!TLDR] > In an N-type semiconductor, doping creates extra electrons, but at higher temperatures, more covalent bonds break, dramatically increasing the number of free charge carriers. The relaxation time (average time between collisions) actually decreases with temperature due to increased lattice vibrations. The net effect is a rise in conductivity because the increase in carrier density far outweighs the decrease in relaxation time. The correct option is **(D)**. ## Why This Question Tests a Key Insight Many students memorise that "conductivity increases with temperature for semiconductors" but miss the *why*. The trap is thinking that both factors — number of carriers and relaxation time — move in the same direction. They don't. Let's unpack the physics. In an N-type semiconductor at room temperature, most conduction comes from donor electrons (from pentavalent impurities like phosphorus). But as temperature rises, something more dramatic happens: thermal energy becomes large enough to break covalent bonds in the silicon or germanium lattice itself. Each broken bond creates an electron-hole pair. This is called **intrinsic carrier generation**, and it floods the material with both electrons and holes. So the number density of free carriers ($n$) rises sharply with temperature. That alone would increase conductivity. But what about relaxation time ($\tau$)? That's the average time an electron travels between collisions. As temperature increases, lattice atoms vibrate more vigorously (phonon scattering increases). Electrons bump into these vibrations more often, so $\tau$ *decreases*. The conductivity formula ties both together: > [!FORMULA] > $$ \sigma = n e \mu = n e \left( \frac{e \tau}{m^*} \right) = \frac{n e^2 \tau}{m^*} $$ Here $\mu$ is mobility, $e$ is electron charge, $m^*$ is effective mass. So $\sigma \propto n \tau$. If $n$ increases and $\tau$ decreases, which one wins? ## Step-by-Step Reasoning 1. **At low to moderate temperatures**, the

The conductivity of a semiconductor increases with increase in temperature because (a) number density of free

NCERT Exemplar · Q1

Q. The conductivity of a semiconductor increases with increase in temperature because

(a) number density of free current carriers increases.

(b) relaxation time increases.

(d) number density of current carriers increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density.

Uttar Pradesh UpmspMCQ· 1mImportance★★★★★

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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.
Device limits: In a p-n junction, leakage current depends on $n_i^2$ .

5. Quick check for understanding

Question: If you heat a pure silicon crystal from $300\text{ K}$ to $400\text{ K}$ , what happens to $n_i$ ?

Answer: It increases dramatically — the exponential term $e^{-E_g/(2kT)}$ becomes much larger because $T$ is in the denominator of the exponent. For silicon, $n_i$ rises from $\approx 1.5\times10^{10}$ to roughly $\approx 5\times10^{12}\ \text{cm}^{-3}$ (about 2-3 orders of magnitude, not 4).

Bottom line: Intrinsic carrier concentration is the natural electron-hole population in a pure semiconductor — a fundamental property that governs all semiconductor behaviour.

"Intrinsic carrier concentration formula semiconductor" and "semiconductor electronics class 12 physics ncert" are commonly searched phrases, both anchored in the Semiconductor Electronics chapter of the NCERT/CBSE Class 12 Physics curriculum. This baseline electron-hole concentration also underlies several JEE Main and NEET p-n junction questions.

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}

Important

The factor of $1/2$ in the exponent comes directly from the square root — it's not an arbitrary fudge. Physically, it reflects that creating an electron-hole pair requires energy $E_g$ , but the probability of that event involves both an electron being excited and a hole being left behind, each contributing half the Boltzmann factor.

Why this formula makes physical sense

Bandgap $E_g$ : A larger gap means fewer electrons can be thermally excited across it — $n_i$ drops exponentially.
Temperature $T$ : Higher temperature gives more thermal energy, so $n_i$ rises sharply (the exponential dominates).
Effective masses (through $N_c$ and $N_v$ ): Materials with heavier carriers have more states near the band edges, so $n_i$ is larger.

Watch out

A common mistake is to think $n_i$ depends on doping. It does not — $n_i$ is a material property at a given temperature. Doping changes $n$ and $p$ individually, but their product $np$ always equals $n_i^2$ at equilibrium.

The temperature dependence in practice

For silicon at 300 K, $n_i \approx 1.5 \times 10^{10} \text{ cm}^{-3}$ . For germanium, it's about $2.4 \times 10^{13} \text{ cm}^{-3}$ — the smaller bandgap (0.67 eV vs 1.12 eV) makes a huge difference. For gallium arsenide (1.43 eV), $n_i$ is only about $2 \times 10^6 \text{ cm}^{-3}$ .

The formula $n_i = \sqrt{N_c N_v} \, e^{-E_g/2kT}$ is the foundation for understanding pn junctions, transistors, and essentially all semiconductor device physics. It's not just a memorised equation — it's the direct consequence of thermal equilibrium and the requirement that charge neutrality holds in a pure crystal.

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Q.The conductivity of a semiconductor increases with increase in temperature because