---
title: "Saturation current of cathode at 1400°C"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:vacuum-tube-1400c-saturation"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/vacuum-tube-1400c-saturation
site: BCT Engineering
---

# Saturation current of cathode at 1400°C

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

- Asked in: 2078 Kartik, 4 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/vacuum-tube-1400c-saturation
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

A vacuum tube with a cylindrical Thorium coated Tungsten cathode is 5 cm long and 3 mm in diameter. Estimate the saturation current if the tube is operated at 1400°C. Given, the Emission constant is 3 Acm⁻²K⁻² and work function for Thorium coated Tungsten is 2.6 eV.

## Short answer

Using J = B0 T^2 exp(-Phi/kT) with T=1673 K, kT≈0.144 eV, Phi=2.6 eV, B0=3e4 A/m^2 K^2 gives J = 1.25e3 A/m^2. Emitting area A = pi × 0.003 m × 0.05 m = 4.71e-4 m^2. Saturation current I = J A = 0.59 A.

## Answer

The saturation (thermionic) current is given by the Richardson-Dushman equation

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

where $A = \pi d l$ is the curved emitting surface of the cylindrical cathode.

**Given:** $l = 5\ \text{cm} = 0.05\ \text{m}$, $d = 3\ \text{mm} = 3 \times 10^{-3}\ \text{m}$, $T = 1400 + 273.15 = 1673.15\ \text{K}$, $B_0 = 3\ \text{A cm}^{-2}\text{K}^{-2} = 3 \times 10^{4}\ \text{A m}^{-2}\text{K}^{-2}$, $\Phi = 2.6\ \text{eV}$.

#### Emitting area

$$
A = \pi d l = \pi \times 3 \times 10^{-3} \times 0.05 = 4.712 \times 10^{-4}\ \text{m}^2
$$

#### Exponential term

$$
\frac{\Phi}{kT} = \frac{2.6 \times 1.602 \times 10^{-19}}{1.381 \times 10^{-23} \times 1673.15} = 18.03, \qquad e^{-18.03} = 1.483 \times 10^{-8}
$$

#### Current density

$$
J = 3 \times 10^{4} \times (1673.15)^2 \times 1.483 \times 10^{-8} = 1.246 \times 10^{3}\ \text{A/m}^2
$$

#### Saturation current

$$
I = J A = 1.246 \times 10^{3} \times 4.712 \times 10^{-4} = 0.587\ \text{A}
$$

**Answer:** Saturation current $I \approx 0.59\ \text{A}$ ($J \approx 1.25 \times 10^{3}\ \text{A/m}^2$).

## Check yourself

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

1. The saturation current of the cathode is
   A. I = J / A
   B. I = J x A, with J from Richardson-Dushman
   C. I = J x T
   D. I = B0 exp(-Φ/kT) only
   Answer: B. Current is current density times emitting area. (Key point: Saturation current)

2. The emitting area of the cylindrical cathode is
   A. π d l (curved surface)
   B. π d^2 / 4
   C. π d^2 l / 4
   D. 2 π d l
   Answer: A. The side surface of a cylinder is circumference times length. (Key point: Emitting area)

3. The temperature to use in J is
   A. 1400 K from °C directly
   B. 1400 - 273 = 1127 K
   C. 1400 x 273 K
   D. 1400 + 273 = 1673 K
   Answer: D. Richardson-Dushman needs absolute temperature. (Key point: Temperature)

4. B0 = 3 A cm^-2 K^-2 equals
   A. 3e-4 A m^-2 K^-2
   B. 3e2 A m^-2 K^-2
   C. 3e4 A m^-2 K^-2
   D. 3 A m^-2 K^-2
   Answer: C. 1 cm^-2 = 1e4 m^-2. The result is I about 0.59 A. (Key point: Unit of constant)

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