---
title: "Built-in potential from 1 Ω cm resistivity"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:pn-builtin-resistivity-1-ohm-cm"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/pn-builtin-resistivity-1-ohm-cm
site: BCT Engineering
---

# Built-in potential from 1 Ω cm resistivity

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2069 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/pn-builtin-resistivity-1-ohm-cm
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Find the built-in potential for a p-n Si junction at room temperature if the bulk resistivity of Si is 1 Ω cm. Electron mobility in Si at room temperature is 1400 cm²V⁻¹s⁻¹; μn/μp = 3.1; ni = 1.05 × 10¹⁰ cm⁻³.

## Short answer

Hole mobility μp = 1400/3.1 = 451.6 cm²/V·s. Nd = 1/(ρ e μn) = 4.46×10^15 cm⁻³, Na = 1/(ρ e μp) = 1.38×10^16 cm⁻³. Built-in potential Vo = (kT/e) ln(Na Nd/ni²) = 0.02586 ln(5.59×10^11) ≈ 0.70 V.

## Answer

**Assumption:** both the p-side and the n-side have bulk resistivity $\rho = 1\ \Omega$ cm, room temperature $T = 300$ K, so $kT/e = 0.02586$ V. In each extrinsic region the majority carriers alone decide the conductivity.

#### Step 1: Hole mobility
$$
\mu_h = \frac{\mu_e}{3.1} = \frac{1400}{3.1} = 451.6\ \text{cm}^2\text{V}^{-1}\text{s}^{-1}
$$

#### Step 2: Donor concentration on the n-side
$\sigma = 1/\rho = N_d e \mu_e$, so
$$
\begin{aligned}
N_d &= \frac{1}{\rho\, e\, \mu_e} = \frac{1}{1 \times 1.602\times10^{-19} \times 1400} \\
&= 4.46\times10^{15}\ \text{cm}^{-3}
\end{aligned}
$$

#### Step 3: Acceptor concentration on the p-side
$$
\begin{aligned}
N_a &= \frac{1}{\rho\, e\, \mu_h} = \frac{1}{1 \times 1.602\times10^{-19} \times 451.6} \\
&= 1.38\times10^{16}\ \text{cm}^{-3}
\end{aligned}
$$

#### Step 4: Built-in potential
$$
\begin{aligned}
V_o &= \frac{kT}{e}\ln\left(\frac{N_a N_d}{n_i^2}\right) \\
&= 0.02586 \ln\left(\frac{1.38\times10^{16} \times 4.46\times10^{15}}{(1.05\times10^{10})^2}\right) \\
&= 0.02586 \ln(5.59\times10^{11}) = 0.02586 \times 27.05 \\
&= 0.70\ \text{V}
\end{aligned}
$$

**Answer:** $N_d = 4.46\times10^{15}$ cm⁻³, $N_a = 1.38\times10^{16}$ cm⁻³, built-in potential $V_o \approx 0.70$ V.

The p-side needs about 3.1 times more dopant than the n-side for the same resistivity because holes are 3.1 times less mobile than electrons.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is hole mobility found?
   A. mu_h = 3.1 x mu_e
   B. mu_h = mu_e - 3.1
   C. mu_h = 1/mu_e
   D. mu_h = mu_e/3.1, about 451.6
   Answer: D. Since mu_n/mu_p = 3.1, mu_h = 1400/3.1 = 451.6. (Key point: Hole mobility)

2. How is Nd obtained?
   A. Nd = 1/(rho e mu_e)
   B. Nd = rho e mu_e
   C. Nd = ni^2/rho
   D. Nd = rho/(e mu_e)
   Answer: A. sigma = Nd e mu_e = 1/rho, giving 4.46e15 cm^-3. (Key point: Doping from resistivity)

3. Why does the p side need more dopant than the n side for the same resistivity?
   A. Boron is heavier
   B. Holes are less mobile than electrons
   C. Holes have more charge
   D. ni differs
   Answer: B. Lower mu_h needs a larger p to give the same sigma. (Key point: Why Na > Nd)

4. Which expression and result for V0?
   A. (kT/e) ln(Na/Nd), about 0.03 V
   B. (kT/e)(Na Nd/ni^2), about 10^10 V
   C. (kT/e) ln(Na Nd/ni^2), about 0.70 V
   D. (kT/e) ln(ni^2/(Na Nd)), about -0.70 V
   Answer: C. 0.02586 x ln(5.59e11) = 0.02586 x 27.05 = 0.70 V. (Key point: Built-in potential)

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