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Saturation current of cathode at 1400°C

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.

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.

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

J=B0T2exp⁡(−ΦkT),I=J×AJ = B_0 T^2 \exp\left(-\frac{\Phi}{kT}\right), \qquad I = J \times A

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

Given: l=5 cm=0.05 ml = 5\ \text{cm} = 0.05\ \text{m}, d=3 mm=3×10−3 md = 3\ \text{mm} = 3 \times 10^{-3}\ \text{m}, T=1400+273.15=1673.15 KT = 1400 + 273.15 = 1673.15\ \text{K}, B0=3 A cm−2K−2=3×104 A m−2K−2B_0 = 3\ \text{A cm}^{-2}\text{K}^{-2} = 3 \times 10^{4}\ \text{A m}^{-2}\text{K}^{-2}, Φ=2.6 eV\Phi = 2.6\ \text{eV}.

Emitting area

A=πdl=π×3×10−3×0.05=4.712×10−4 m2A = \pi d l = \pi \times 3 \times 10^{-3} \times 0.05 = 4.712 \times 10^{-4}\ \text{m}^2

Exponential term

ΦkT=2.6×1.602×10−191.381×10−23×1673.15=18.03,e−18.03=1.483×10−8\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=3×104×(1673.15)2×1.483×10−8=1.246×103 A/m2J = 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=JA=1.246×103×4.712×10−4=0.587 AI = J A = 1.246 \times 10^{3} \times 4.712 \times 10^{-4} = 0.587\ \text{A}

Answer: Saturation current I≈0.59 AI \approx 0.59\ \text{A} (J≈1.25×103 A/m2J \approx 1.25 \times 10^{3}\ \text{A/m}^2).

✓ Answer checked by Claude Opus 5.5 (high effort)

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