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SCR half-wave rectifier fired at 30°

Short answer

Average voltage Vdc=(50/(2π))(1+cos30°)=14.85 V. RMS voltage = (50/2)√((π−α)/π+sin60°/(2π))=24.64 V. Efficiency = (14.85/24.64)²=36.33%. Ripple factor = √((24.64/14.85)²−1)=1.324.

A SCR used in half-wave controlled rectifier is fired at the angle of 30°. Input voltage for the rectifier is 50 sin(2π50t) and the load resistance is 100Ω. Find average load voltage, rms load voltage, efficiency and ripple factor for the rectifier.

Given: vs = 50 sin(2π·50t), so Vm = 50 V; α = 30°; R = 100 Ω; half-wave controlled rectifier with resistive load.

Average load voltage

Vdc = (1/2π)∫α^π Vm sin ωt d(ωt)
    = (Vm/2π)(1 + cos α)
    = (50/2π)(1 + 0.8660)
Vdc = 14.85 V
Idc = Vdc/R = 0.1485 A

RMS load voltage

Vrms = [ (1/2π)∫α^π Vm² sin² ωt d(ωt) ]^(1/2)
     = (Vm/2)·√[ (π − α)/π + sin2α/(2π) ]
     = 25 × √[ 0.8333 + 0.1378 ]
     = 25 × 0.9855
Vrms = 24.64 V
Irms = 0.2464 A

Efficiency (rectification efficiency)

Pdc = Vdc²/R = 14.85²/100 = 2.205 W
Pac = Vrms²/R = 24.64²/100 = 6.070 W
η = Pdc/Pac = (Vdc/Vrms)²
  = (14.85/24.64)² = 0.3633
η = 36.33 %

Ripple factor

Form factor FF = Vrms/Vdc = 24.64/14.85 = 1.659
RF = √(FF² − 1) = √(1.659² − 1)
RF = 1.324

Answer: Vdc = 14.85 V, Vrms = 24.64 V, η = 36.33 %, ripple factor = 1.324

✓ Answer checked by Claude Opus 5.5 (high effort)

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