Skip to main content

SCR half-wave rectifier numerical (220 V, 150 Ω)

Short answer

With 220 V peak, sin α = 110/220 gives α = 30°. Conduction angle = 180° − 30° = 150°. Average output voltage Vav = (220/(2π))(1+cos30°) = 65.34 V, average current Iav = 65.34/150 = 0.436 A.

The gate current in a SCR half wave rectifier is adjusted to 1.25mA and the forward break-down voltage of SCR corresponding to this gate current is 110V. The applied voltage is 220V, the load resistance is 150Ω and holding current is zero. Determine: a) Firing angle b) Conduction angle c) Average output voltage d) Average current

Given: IG = 1.25 mA, forward breakover voltage VBO = 110 V, applied voltage 220 V, RL = 150 Ω, IH = 0.

Assumption: the 220 V applied voltage is taken as the peak value (Vm = 220 V), as is usual in this type of textbook problem.

a) Firing angle

The SCR fires when the supply reaches VBO:

Vm sin α = VBO
sin α = 110/220 = 0.5
α = 30°

b) Conduction angle

With zero holding current, it conducts until 180°:

θc = 180° − 30° = 150°

c) Average output voltage

Vav = (1/2π)∫α^π Vm sin θ dθ = (Vm/2π)(1 + cos α)
    = (220/2π)(1 + 0.8660)
    = 35.01 × 1.8660
Vav = 65.34 V

d) Average current

Iav = Vav/RL = 65.34/150
Iav = 0.4356 A ≈ 435.6 mA

Answer: α = 30°, conduction angle = 150°, Vav = 65.34 V, Iav = 0.436 A

If 220 V is instead taken as RMS (Vm = 311.13 V): α = 20.70°, conduction angle = 159.30°, Vav = 95.84 V, Iav = 0.639 A.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Advanced Electronics chapter 7 questions →