---
title: "Reverse saturation current in pn junction"
subject: "Electrical Engineering Material"
chapter: 6
key: "semester_3/electrical_engineering_material:reverse-saturation-current"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/reverse-saturation-current
site: BCT Engineering
---

# Reverse saturation current in pn junction

Electrical Engineering Material · Chapter 6: Semiconductors

- Asked in: 2070 Chaitra, 4 marks
- Video: [Reverse biasing a PN junction | Class 12 (India) | Physics | Khan Academy](https://www.youtube.com/watch?v=TtF3bNyAm7k) — Khan Academy India - English
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/reverse-saturation-current
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-6

## Question

What is reverse saturation current in pn junction semiconductor?

## Short answer

It is the small, nearly constant reverse current in a pn junction due to minority carriers thermally generated near the depletion region and swept across. Its magnitude is I0∝ni² and is strongly temperature dependent, nearly independent of reverse voltage.

## Answer

**Reverse saturation current** ($I_0$ or $I_s$) is the small, nearly constant current that flows through a reverse-biased pn junction. It is caused by **minority carriers** that are thermally generated near the junction and swept across by the junction field.

#### How it arises

- Under reverse bias the barrier rises from $eV_0$ to $e(V_0+V_r)$; majority carriers cannot cross it, so diffusion current becomes almost zero.
- Minority carriers — holes in the n region and electrons in the p region — that reach the depletion edge by diffusion are **pulled across** by the field (drift).
- Their number depends only on how fast they are thermally generated, not on the applied voltage. So once $V_r$ exceeds a few $kT/e$ (about 0.1 V), the current **saturates**.

#### Expression

From the ideal diode (Shockley) equation $I = I_0(e^{eV/kT}-1)$, for large reverse $V$: $I \to -I_0$, where

$$
I_0 = Ae n_i^2\left(\frac{D_h}{L_hN_d} + \frac{D_e}{L_eN_a}\right)
$$

$A$ = junction area, $D$ = diffusion coefficients, $L$ = diffusion lengths. There is also a thermal generation current in the depletion layer, proportional to $n_iW/\tau$, which is important in Si.

```text
     I
     |        /  forward
     |       /
 ----+------/------- V
 ____|_____ -I0  (reverse, flat)
     |
```

#### Main features

- Very small: nA for Si, µA for Ge (Ge has a smaller $E_g$, hence larger $n_i$).
- **Strongly temperature dependent** because $I_0 \propto n_i^2 \propto T^3e^{-E_g/kT}$; roughly doubles every 10°C.
- Nearly independent of reverse voltage until breakdown.
- Increases with junction area and decreases with heavier doping.
- Light shining on the junction generates more minority carriers and increases it (photodiode principle).

## Check yourself

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

1. What carries the reverse saturation current?
   A. Majority carriers crossing the barrier
   B. Thermally generated minority carriers swept across by the junction field
   C. Ions moving in the lattice
   D. Carriers tunnelling from the contacts
   Answer: B. Minority carriers reaching the depletion edge are pulled across by the field. (Key point: Origin of I_0)

2. Why does the reverse current saturate?
   A. Its size depends on the thermal generation rate, not the applied voltage
   B. The barrier disappears
   C. Resistance falls with voltage
   D. Majority carriers increase with voltage
   Answer: A. Once V_r exceeds a few kT/e, all available minority carriers are collected, so the current is constant. (Key point: Voltage independence)

3. How does I_0 vary with temperature?
   A. It is constant
   B. It decreases as n_i^2 falls
   C. It rises strongly since I_0 is proportional to n_i^2
   D. It falls linearly
   Answer: C. I_0 is proportional to n_i^2, which increases as T^3 exp(-Eg/kT); roughly doubling every 10 C. (Key point: Temperature dependence)

4. Why is I_0 larger in Ge than Si?
   A. Ge has a larger band gap
   B. Ge has heavier doping
   C. Ge has fewer minority carriers
   D. Ge has a smaller E_g, hence larger n_i
   Answer: D. I_0 scales with n_i^2, and smaller E_g gives larger n_i. (Key point: Si versus Ge)

## Related in Electrical Engineering Material

- [Forward and reverse biased pn junction](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aforward-reverse-bias) — Electrical Engineering Material, chapter 6
- [pn junction from 5 Ω cm resistivity](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-resistivity-5-ohm-cm) — Electrical Engineering Material, chapter 6, asked 2×
- [PN junction formation and built-in potential](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-formation-builtin) — Electrical Engineering Material, chapter 6
- [pn junction 10¹⁸/10¹⁶: built-in potential and widths](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-1e18-1e16-depletion) — Electrical Engineering Material, chapter 6, asked 3×
- [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×
- [Built-in potential of a pn junction](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apn-builtin-potential-derivation) — Electrical Engineering Material, chapter 6, asked 2×

## Related in other subjects

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