Q.Suppose a pure Si crystal has $5 \times 10^{28}$ atoms $\text{m}^{-3}$. It is doped by $1\ \text{ppm}$ concentration of pentavalent As. Calculate the number of electrons and holes. Given that $n_i = 1.5 \times 10^{16}\ \text{m}^{-3}$.

Concept: Intrinsic Carrier Concentration — In an extrinsic semiconductor, the product $n \cdot p = n_i^2$ always holds at equilibrium.

Step 1: Find donor concentration.

$1\ \text{ppm}$ means 1 As atom per $10^6$ Si atoms.

Step 2: Approximate electron concentration.

Since $N_D \gg n_i$, nearly all donors ionise:

Step 3: Find hole concentration using mass action law. …

Why this approach works

In a pure (intrinsic) semiconductor, the number of electrons equals the number of holes — both are $n_i$. But when we dope with a pentavalent atom like arsenic (As), which has five valence electrons, four of them bond with neighbouring silicon atoms and the fifth becomes a free electron. This makes the crystal n-type, where electrons are the majority carriers and holes are the minority carriers.

The key principle is charge neutrality: the total positive charge must equal the total negative charge. In an n-type semiconductor at room temperature, almost all donor atoms are ionised, so the electron concentration $n$ is essentially equal to the donor concentration $N_D$. Then, using the mass-action law ($n p = n_i^2$), we can find the hole concentration $p$.

Step-by-step calculation

1. Find the donor concentration from the doping level

We are told the crystal has $5 \times 10^{28}$ Si atoms per cubic metre, and it is doped with 1 ppm (parts per million) of As. This means for every million Si atoms, there is one As atom.

So the donor concentration $N_D$ is:

Since each As atom donates one free electron, $N_D$ is also the concentration of donor electrons (assuming full ionisation, which is valid at room temperature).

2. Determine the electron concentration

In an n-type semiconductor at room temperature, the electron concentration $n$ is approximately equal to the donor concentration because the intrinsic carrier concentration $n_i$ is negligible compared to $N_D$:

Why "approximately"? Because a tiny fraction of electrons come from intrinsic generation, but $n_i = 1.5 \times 10^{16}\ \text{m}^{-3}$ is six orders of magnitude smaller than $N_D$, so the approximation is excellent. …

Method: Mass Action Law for Extrinsic Semiconductors

This problem uses the mass action law combined with the charge neutrality condition for an $n$-type semiconductor.

Step 1 — Find the donor concentration

1 ppm means 1 atom of As per $10^6$ atoms of Si. So:

Step 2 — Identify the majority carrier

Pentavalent As donates an extra electron. At room temperature, nearly all donor atoms are ionised. Since $N_D \gg n_i$, the electron concentration is essentially equal to the donor concentration:

Step 3 — Apply the mass action law

For any semiconductor in thermal equilibrium:

So:

$p = \frac{n_i^2}{n} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}}$ …

Common Mistakes & How to Avoid Them

Mistake 1: Confusing ppm with a percentage

Many students treat 1 ppm as 1% — that is, they multiply the atom density by 0.01 instead of $10^{-6}$. This gives a donor concentration that is four orders of magnitude too large, which then throws off every subsequent calculation.

How to avoid: Remember that "ppm" means parts per million, i.e., $1\ \text{ppm} = 1 \times 10^{-6}$. So the donor concentration is:

Mistake 2: Forgetting that $n_i$ is given in $\text{m}^{-3}$, not $\text{cm}^{-3}$

The problem gives $n_i = 1.5 \times 10^{16}\ \text{m}^{-3}$. Some students, used to the common textbook value $1.5 \times 10^{10}\ \text{cm}^{-3}$, automatically convert or substitute the wrong number. This leads to a completely wrong hole concentration.

How to avoid: Always check the units of every given quantity before plugging into a formula. If the problem states $\text{m}^{-3}$, keep everything in $\text{m}^{-3}$. Do not "correct" the given data.

Mistake 3: Assuming $n = N_D$ without checking the doping regime

Students often write $n = N_D$ and $p = n_i^2 / N_D$ without first verifying that $N_D \gg n_i$. If $N_D$ were comparable to $n_i$, the full quadratic equation would be needed.

How to avoid: Compare $N_D$ and $n_i$ explicitly. Here:

Since $N_D$ is about 3 million times larger than $n_i$, the approximation $n \approx N_D$ is excellent. Write this comparison in your solution — examiners look for it.

Mistake 4: Using the wrong mass-action law sign or forgetting it entirely

Some students try to find $p$ by subtracting $n_i$ from $N_D$, or by using $n + p = \text{constant}$. The only correct relation is:

How to avoid: Write the mass-action law before you calculate anything. It is the bridge between electron and hole concentrations in any doped semiconductor.

Mistake 5: Arithmetic errors in the final division …