---
title: "Saturation current of thoriated tungsten cathode"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:vacuum-tube-1500c-saturation"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/vacuum-tube-1500c-saturation
site: BCT Engineering
---

# Saturation current of thoriated tungsten cathode

Electrical Engineering Material · Chapter 2: Free Electron Theory of Conduction in Metal

- Asked in: 2079 Bhadra, 4 marks; 2073 Shrawan, 4 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/vacuum-tube-1500c-saturation
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

A transmitter type vacuum tube operated at 1500°C has a cylindrical Thorium coated Tungsten cathode which is 5 cm long with diameter of 1.5 mm. Determine the saturation current of vacuum tube if the cathode has emission coefficient constant of 3 × 10⁴ Am⁻²K⁻² and work function of 2.6 eV.

## Short answer

Using the Richardson-Dushman equation J = B0 T² exp(−Φ/kT) with T=1773 K, Φ=2.6 eV, B0=3×10⁴ A/m²K², and emitting area A=π d l=2.356×10⁻⁴ m², the saturation current is I = J A ≈ 0.91 A (current density ≈ 3.87×10³ A/m²).

## Answer

The saturation current of a thermionic cathode is given by the Richardson-Dushman equation:

$$
J = B_0 T^2 \exp\left(-\frac{\Phi}{kT}\right), \qquad I = J \times A
$$

where $A$ is the emitting (curved) surface area of the cylinder, $A = \pi d l$.

**Given:** $T = 1500 + 273.15 = 1773.15\ \text{K}$, $l = 5\ \text{cm} = 0.05\ \text{m}$, $d = 1.5\ \text{mm} = 1.5 \times 10^{-3}\ \text{m}$, $B_0 = 3 \times 10^{4}\ \text{A m}^{-2}\text{K}^{-2}$, $\Phi = 2.6\ \text{eV}$.

#### Emitting area

$$
A = \pi d l = \pi \times 1.5 \times 10^{-3} \times 0.05 = 2.356 \times 10^{-4}\ \text{m}^2
$$

#### Exponent

$$
\frac{\Phi}{kT} = \frac{2.6 \times 1.602 \times 10^{-19}}{1.381 \times 10^{-23} \times 1773.15} = 17.01, \qquad e^{-17.01} = 4.10 \times 10^{-8}
$$

#### Current density

$$
J = 3 \times 10^{4} \times (1773.15)^2 \times 4.10 \times 10^{-8} = 3.87 \times 10^{3}\ \text{A/m}^2
$$

#### Saturation current

$$
I = J A = 3.87 \times 10^{3} \times 2.356 \times 10^{-4} = 0.911\ \text{A}
$$

**Answer:** Saturation current $I \approx 0.91\ \text{A}$ (current density $\approx 3.87 \times 10^{3}\ \text{A/m}^2$).

## Check yourself

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

1. Which equation gives the saturation current density?
   A. J = B0 T^2 exp(-Phi/kT)
   B. J = B0 T exp(Phi/kT)
   C. J = B0 T^2 exp(Phi/kT)
   D. J = B0/T^2 exp(-Phi/kT)
   Answer: A. Richardson-Dushman equation. (Key point: Equation)

2. What T should be used for 1500 C?
   A. 1773 K
   B. 1500 K
   C. 1227 K
   D. 273 K
   Answer: A. T = 1500 + 273 = 1773 K. (Key point: Temperature)

3. What area is used for the cylindrical cathode?
   A. A = pi d l
   B. A = pi d^2 / 4
   C. A = 2 pi d l
   D. A = d l
   Answer: A. Curved surface pi d l = 2.356e-4 m^2. (Key point: Emitting area)

4. What is the saturation current?
   A. 0.91 A
   B. 91 A
   C. 3.87 A
   D. 0.0091 A
   Answer: A. J = 3.87e3 A/m^2, I = J x A = 0.911 A. (Key point: Final current)

## Close variants set in other exams

- [Saturation current of a 4 m cathode](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Avacuum-tube-160c-saturation) — Electrical Engineering Material, chapter 2, asked 3×

## Related in Electrical Engineering Material

- [Saturation current of cathode at 1400°C](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Avacuum-tube-1400c-saturation) — Electrical Engineering Material, chapter 2

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