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SCR half-wave rectifier numerical (200 V peak)

Short answer

Firing angle α = arcsin(100/200) = 30°, conduction angle = 150°. Average load current with no holding current is (200/(2π×100))(1+cos30°) = 0.594 A. With holding current 200 mA, off angle 174.26°, conduction 144.26°, average current ≈ 0.592 A.

A half wave rectifier circuit employing an SCR is adjusted to have a gate current of 1 mA. The forward breakdown voltage of SCR is 100V for Ig = 1mA. If sinusoidal voltage of 200V peak is applied, find: i) Firing angle ii) Conduction angle iii) Average current, assume load resistance = 100Ω iv) In this circuit if holding current IH = 200 mA, find average current in this case.

Given: Vm = 200 V, forward breakover voltage VBO = 100 V at IG = 1 mA, RL = 100 Ω.

The SCR fires when the instantaneous supply reaches VBO: Vm sin α = VBO.

i) Firing angle

sin α = VBO/Vm = 100/200 = 0.5
α = 30°

ii) Conduction angle

With zero holding current, the SCR conducts from α to 180°:

θc = 180° − α = 180° − 30° = 150°

iii) Average current (RL = 100 Ω)

Iav = (1/2π)∫α^π (Vm/RL) sin θ dθ
    = (Vm / 2πRL)(1 + cos α)
    = (200 / (2π × 100))(1 + cos 30°)
    = 0.3183 × 1.8660
Iav = 0.594 A

iv) Average current with IH = 200 mA

Now the SCR turns OFF when the load current falls to IH before the end of the half cycle:

i = (Vm/RL) sin θ = 2 sin θ  A
2 sin θ = 0.2  →  sin θ = 0.1
θ_off = 180° − 5.74° = 174.26°

(At turn-on, i = 2 sin 30° = 1 A > IH, so the SCR latches normally.)

Iav = (1/2π)∫α^θoff 2 sin θ dθ
    = (2/2π)[cos 30° − cos 174.26°]
    = (1/π)(0.8660 + 0.9950)
Iav = 0.592 A

New conduction angle = 174.26° − 30° = 144.26°.

Answer: α = 30°, conduction angle = 150°, Iav = 0.594 A (IH = 0) and Iav ≈ 0.592 A (IH = 200 mA, conduction 144.26°)

✓ Answer checked by Claude Opus 5.5 (high effort)

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