---
title: "Saturation current of a 4 m cathode"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:vacuum-tube-160c-saturation"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/vacuum-tube-160c-saturation
site: BCT Engineering
---

# Saturation current of a 4 m cathode

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

- Asked in: 2068 Chaitra, 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-160c-saturation
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

A transmitter type vacuum tube has a cylindrical cathode, which is 4m long and 2mm diameter. Estimate the saturation current if the tube is operated at 160°C. The emission constant A₀ = 3 × 10⁴ Am⁻²K⁻², work function φ = 2.6eV.

## Short answer

Cathode area A = π d l ≈ 0.0251 m². Current density J = A₀ T² exp(−φ/kT) ≈ 3.24×10⁻²¹ A/m², giving saturation current I = J A ≈ 8.1×10⁻²³ A, practically zero.

## Answer

The saturation current is given by the Richardson-Dushman equation:

$$
J = A_0 T^2 \exp\left(-\frac{\phi}{kT}\right), \qquad I = J \times (\pi d l)
$$

**Given (as stated):** $l = 4\ \text{m}$, $d = 2\ \text{mm}$, $T = 160 + 273.15 = 433.15\ \text{K}$, $A_0 = 3 \times 10^{4}\ \text{A m}^{-2}\text{K}^{-2}$, $\phi = 2.6\ \text{eV}$.

#### Emitting area

$$
A = \pi d l = \pi \times 2 \times 10^{-3} \times 4 = 2.51 \times 10^{-2}\ \text{m}^2
$$

#### Exponential term

$$
\frac{\phi}{kT} = \frac{2.6 \times 1.602 \times 10^{-19}}{1.381 \times 10^{-23} \times 433.15} = 69.63, \qquad e^{-69.63} = 5.75 \times 10^{-31}
$$

#### Current density and current

$$
\begin{aligned}
J &= 3 \times 10^{4} \times (433.15)^2 \times 5.75 \times 10^{-31} = 3.24 \times 10^{-21}\ \text{A/m}^2 \\
I &= J A = 3.24 \times 10^{-21} \times 2.51 \times 10^{-2} = 8.1 \times 10^{-23}\ \text{A}
\end{aligned}
$$

So at 160°C there is practically **no** thermionic emission: $kT$ (0.037 eV) is far too small compared with the 2.6 eV work function.

**Note:** a transmitter tube cathode normally runs near 1600°C and is a few cm long. With $T = 1873.15$ K and $l = 4$ cm: $\phi/kT = 16.10$, $J = 1.07 \times 10^{4}\ \text{A/m}^2$, $A = 2.51 \times 10^{-4}\ \text{m}^2$, giving $I \approx 2.69\ \text{A}$.

**Answer (data as given):** $I \approx 8.1 \times 10^{-23}\ \text{A}$, i.e. negligible. (If 1600°C and 4 cm are meant, $I \approx 2.69$ A.)

## Check yourself

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

1. At 160°C with φ = 2.6 eV the emission is
   A. About 1 A
   B. Negligible (about 1e-22 A)
   C. About 0.1 mA
   D. Large, the tube saturates
   Answer: B. φ/kT is about 70, giving e^-70 and a negligible current. (Key point: Result)

2. The reason is that
   A. kT (0.037 eV) is far smaller than φ = 2.6 eV
   B. The area is too small
   C. Richardson-Dushman does not apply to cylinders
   D. The work function is lowered at low T
   Answer: A. Very few electrons have enough energy to cross the barrier. (Key point: Why so small)

3. T must be used as
   A. 160 K
   B. 160 - 273 K
   C. 433 °C
   D. 160 + 273.15 = 433.15 K
   Answer: D. Absolute temperature is required. (Key point: Temperature)

4. If 1600°C and 4 cm are meant, the current is about
   A. 8.1e-23 A
   B. 2.69 mA
   C. 2.69 A
   D. 269 A
   Answer: C. At 1873 K, J = 1.07e4 A/m^2 and A = 2.51e-4 m^2. (Key point: Corrected case)

## Close variants set in other exams

- [Saturation current of thoriated tungsten cathode](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Avacuum-tube-1500c-saturation) — Electrical Engineering Material, chapter 2, asked 3×

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