---
title: "Short note: Gunn diode"
subject: "RF and Microwave Engineering"
chapter: 4
key: "semester_7/rf_and_microwave_engineering:gunn-diode-note"
source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/gunn-diode-note
site: BCT Engineering
---

# Short note: Gunn diode

RF and Microwave Engineering · Chapter 4: RF and Microwave Components and Devices

- Asked in: 2079 Bhadra, 3 marks
- Video: [Gunn Diode (Symbol, Basics, Structure, Working, Characteristics & Applications) Explained](https://www.youtube.com/watch?v=xgfq-qAS-4M) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/gunn-diode-note
- Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-4

## Question

Write a short note on Gunn diode.

## Short answer

A Gunn diode is a two-terminal negative-resistance device made of n-type GaAs without a junction. It uses the transferred electron effect: above a threshold field (~3.2 kV/cm), electrons move to a higher-energy low-mobility valley, forming a high-field domain that drifts to the anode, yielding oscillations. Applications include local oscillators and low-power microwave sources up to 100 GHz.

## Answer

A **Gunn diode** is a two-terminal microwave source made of a single piece of n-type GaAs (or InP). It has no p-n junction. It works on the **transferred electron effect** (Ridley–Watkins–Hilsum theory), which gives the bulk material a **negative differential resistance**.

**Principle**
- The GaAs conduction band has two valleys. In the lower valley electrons are light (m* ≈ 0.068 m₀) and fast (μ ≈ 8000 cm²/V·s). The upper valley is about 0.36 eV higher, and there electrons are heavy (m* ≈ 1.2 m₀) and slow (μ ≈ 180 cm²/V·s).
- At low fields, nearly all electrons are in the lower valley, so current rises with field.
- Above a threshold field of about **3.2–3.4 kV/cm**, electrons gain energy and move to the upper valley. Average drift velocity then falls as the field rises, so dI/dV < 0.
- In this negative-resistance region a high-field **domain** forms near the cathode, drifts to the anode at about 10⁷ cm/s and disappears. Then a new domain forms. Each domain gives one current pulse, so the frequency is about f = v_d / L.

```text
 I |     peak
   |    /\
   |   /  \___  NDR region
   |  /
   | /
   +-------------- E
       E_th ~3.2 kV/cm
```

**Modes of operation:** transit-time (Gunn) mode, delayed-domain mode, quenched-domain mode and LSA (limited space-charge accumulation) mode.

**Applications:** local oscillators in receivers, police and automotive radar, microwave links, and sweep generators. Gunn diodes give mW up to about 1 W in the 1–100 GHz range, with low noise and a low bias voltage.

## Check yourself

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

1. A Gunn diode works on:
   A. Avalanche breakdown at a p-n junction
   B. The transferred electron effect giving negative differential resistance
   C. Tunnelling
   D. Stimulated emission
   Answer: B. Electrons transfer from a light, fast valley to a heavy, slow upper valley, so velocity falls as field rises. (Key point: Operating principle)

2. What is the construction of a Gunn diode?
   A. A p-n junction in silicon
   B. A single piece of n-type GaAs or InP with no junction
   C. A p-i-n structure
   D. A metal-semiconductor with p-type GaAs
   Answer: B. It is a bulk n-type device with no p-n junction. (Key point: Construction)

3. In transit-time mode the oscillation frequency is approximately:
   A. f = L/v_d
   B. f = v_d × L
   C. f = v_d/L
   D. f = 1/(v_d L)
   Answer: C. One current pulse per domain transit, so f ≈ drift velocity divided by active length. (Key point: Domain frequency)