---
title: "Built-in potential and depletion in forward bias"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:forward-bias-depletion"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/forward-bias-depletion
site: BCT Engineering
---

# Built-in potential and depletion in forward bias

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2068 Shrawan, 8 marks
- Video: [Diode under Forward bias & Breakdown | Electronic Devices | Electronics & Communication Engineering](https://www.youtube.com/watch?v=vl0rhD4VF2o) — IMS GATE ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/forward-bias-depletion
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

Derive the expression of built in potential and width of depletion layer in forward biased P-N junction.

## Short answer

Built-in potential Vo = (kT/e) ln(Na Nd/ni²). Equilibrium depletion width W = √[2ε(Na+Nd)Vo/(e Na Nd)]. Under forward bias V, barrier drops to Vo−V, so W_fwd = √[2ε(Na+Nd)(Vo−V)/(e Na Nd)]; the depletion layer narrows.

## Answer

When p and n regions are joined, electrons diffuse to the p-side and holes to the n-side, leaving behind fixed ionized donors ($+$) and acceptors ($-$). This **depletion (space charge) region** creates an electric field and a potential barrier, the **built-in potential** $V_o$.

```text
        p-side   |  W  |   n-side
   o o o o o o |- - | + +| e e e e e
   o o o o o o |- - | + +| e e e e e
              -Wp   0   Wn
         E-field points n -> p
```

#### Built-in potential
In equilibrium the Fermi level is flat across the junction. On the p-side, far from the junction, $p_{p0} = N_a$; on the n-side, $n_{n0} = N_d$ and the minority hole density is $p_{n0} = n_i^2/N_d$. The Boltzmann relation between hole densities on the two sides of the barrier is
$$
\frac{p_{n0}}{p_{p0}} = \exp\left(-\frac{eV_o}{kT}\right)
$$
(the same result follows from setting hole drift current = hole diffusion current). Therefore
$$
V_o = \frac{kT}{e}\ln\frac{p_{p0}}{p_{n0}} = \frac{kT}{e}\ln\left(\frac{N_a N_d}{n_i^2}\right)
$$

#### Depletion width at equilibrium
Assume an abrupt junction and full depletion. Charge neutrality: $N_a W_p = N_d W_n$.
Poisson's equation $\dfrac{d^2V}{dx^2} = -\dfrac{\rho(x)}{\varepsilon}$, with $\rho = -eN_a$ for $-W_p<x<0$ and $\rho = eN_d$ for $0<x<W_n$. Integrating once gives a triangular field with maximum at $x = 0$:
$$
E_o = -\frac{eN_d W_n}{\varepsilon} = -\frac{eN_a W_p}{\varepsilon}
$$
Integrating again, the potential drop equals the area of the field triangle:
$$
\begin{aligned}
V_o &= \frac{1}{2}|E_o| W = \frac{e N_a N_d W^2}{2\varepsilon (N_a + N_d)} \\
W &= W_p + W_n = \sqrt{\frac{2\varepsilon (N_a + N_d) V_o}{e N_a N_d}}
\end{aligned}
$$
where $\varepsilon = \varepsilon_0\varepsilon_r$.

#### Forward bias
With forward voltage $V$ (p positive), the applied voltage opposes the built-in field. Almost all of $V$ drops across the high-resistance depletion region, so the barrier falls from $V_o$ to $V_o - V$:
$$
W_{fwd} = \sqrt{\frac{2\varepsilon (N_a + N_d)(V_o - V)}{e N_a N_d}}
$$
- The depletion layer becomes **narrower** as $V$ increases ($W \propto \sqrt{V_o - V}$).
- The lower barrier lets majority carriers cross; minority carrier density at the edges rises by $e^{eV/kT}$, giving the diode current $I = I_o(e^{eV/kT} - 1)$.
- For reverse bias, $V$ is replaced by $-V_r$, and $W$ widens.

For a one-sided junction ($N_a \gg N_d$), $W \approx \sqrt{2\varepsilon (V_o - V)/(eN_d)}$, so the depletion region lies almost entirely in the lightly doped side.

## Check yourself

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

1. What is V0 for an abrupt pn junction?
   A. (kT/e)(Na + Nd)
   B. (kT/e) ln(ni^2/(Na Nd))
   C. kT ln(Na/Nd)
   D. (kT/e) ln(Na Nd/ni^2)
   Answer: D. Boltzmann balance gives V0 = (kT/e) ln(p_p0/p_n0). (Key point: Built-in potential)

2. Which equation expresses charge neutrality of the depletion region?
   A. Na Wp = Nd Wn
   B. Na Wn = Nd Wp
   C. Na + Nd = Wp + Wn
   D. Wp = Wn always
   Answer: A. Total negative and positive charge per area are equal. (Key point: Charge neutrality)

3. What is the shape of the electric field in the depletion region?
   A. Uniform
   B. Triangular, maximum at the junction
   C. Zero at junction
   D. Rectangular
   Answer: B. Constant charge density gives linearly varying field peaking at x = 0. (Key point: Field shape)

4. How does the depletion width change under forward bias V?
   A. Same with (V0 + V), wider
   B. Unchanged
   C. W = sqrt(2 eps (Na+Nd)(V0 - V)/(e Na Nd)), narrower
   D. W proportional to V
   Answer: C. V0 is replaced by V0 - V, so W shrinks. (Key point: Forward bias width)

## Related in Electrical Engineering Material

- [PN junction built-in potential and depletion layer](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-junction-builtin-depletion) — Electrical Engineering Material, chapter 6, asked 2×
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