Thermionic Emission & the Diode
Hot cathodes and one-way valves.
Lesson at a glanceShow the key points
- i = AST²e^(−φ/kT) (41.3); φ = work function.
- Diode: space-charge-limited i_p = kV_p^(3/2), then saturation; one-way conduction.
- Dynamic plate resistance r_p = ΔV_p/Δi_p (41.4).
- Half-wave and full-wave rectification.
Heat a metal enough and electrons boil off its surface. Collect them on a second electrode and you have a one-way valve for current: the diode.
Thermionic emission
At high temperature electrons escape from a metal’s surface: thermionic emission, and the escaped electrons are thermions. An escaping electron leaves the metal positively charged, which pulls it back, so only electrons with enough spare kinetic energy get out.
At ordinary temperatures almost no free electron has that much energy; heating raises their energies until some can escape. The emission current follows
An emitter has work function 4.0 eV. By what factor does its thermionic current rise between 1800 K and 2200 K?
Raising a surface from 700 K to 900 K multiplies its emission rate by 50. Find the work function.
For tungsten, A = 60 × 10⁴ A m⁻² K⁻² and φ = 4.55 eV. A cathode of area 3.0 × 10⁻⁵ m² runs at 2200 K. Find its emission current.
The diode valve
A cathode (filament) that can be heated, and an anode (plate) to collect electrons, sealed in an evacuated bulb. The cathode may be a tungsten or thoriated-tungsten wire heated directly by a current, or a nickel tube coated with barium oxide, heated by a coil inside it. The anode is a nickel cylinder around the cathode. A battery makes the anode positive by the plate voltage Vp; the resulting plate current ip flows round the circuit.
Characteristics
- The electrons between cathode and anode form a negative space charge that repels newly emitted electrons back. Raising Vp overcomes it, and ip rises: far from saturation, ip = kVp3/2 (the Langmuir–Child law).
- At large Vp every emitted electron is collected and the current saturates. A hotter cathode gives a higher saturation current.
- With the anode negative, electrons are pushed back and no current flows: the diode conducts one way only.
Raising a diode’s plate voltage from 60 V to 64 V increases its current from 30 mA to 38 mA. Find its dynamic plate resistance there.
Rectification
Turning AC into one-way current is rectification.
- Half-wave: an AC source, a diode and a resistor in series. Current flows only in the half-cycles when the anode is positive.
- Full-wave: a transformer’s secondary has a centre tap C; its ends A and B go to the anodes of two diodes, whose cathodes join one end of the load, the other end returning to C. In each half-cycle one diode conducts, and in both the current crosses the load the same way.
- 1Thermionic emission: i = AST²e^(−φ/kT) (41.3); φ is the work function.
- 2Diode: space-charge-limited i_p = kV_p^(3/2), then saturation; one-way conduction.
- 3Dynamic plate resistance r_p = ΔV_p/Δi_p (41.4).
- 4Half-wave and full-wave rectification.
Check yourself
Keep exploring
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