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Controlled Rectifiers (SCR)

Problem

A 120 V RMS source feeds a half-wave SCR rectifier with a resistive load. The firing angle is α = 60°. Find the average output voltage.

Step 1 — Peak Voltage

Vm=120×2=169.7 VV_m = 120 \times \sqrt{2} = 169.7\,\text{V}

Step 2 — Average Output Voltage

Vavg=169.72π(1+cos⁡60°)=169.76.283×1.5=40.5 VV_{avg} = \frac{169.7}{2\pi}(1 + \cos 60°) = \frac{169.7}{6.283} \times 1.5 = 40.5\,\text{V}

Step 3 — Compare with Full Conduction

Vavg(α=0°)=169.72π×2=54.0 VV_{avg}(\alpha = 0°) = \frac{169.7}{2\pi} \times 2 = 54.0\,\text{V}
α = 60°InputOutput
Figure 1. Output waveform at α = 60° — conduction region shaded
At α = 60°, the output is 75% of the maximum (40.5 / 54.0). The firing angle gives smooth, lossless control of the average output.