---
title: "SCR half-wave rectifier fired at 30°"
subject: "Advanced Electronics"
chapter: 7
key: "semester_4/advanced_electronics:scr-hwr-30-degree"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/scr-hwr-30-degree
site: BCT Engineering
---

# SCR half-wave rectifier fired at 30°

Advanced Electronics · Chapter 7: Introduction to Power Electronics

- Asked in: 2072 Kartik, 8 marks
- Video: [Single Phase Half Wave Controlled Rectifier with R Load | Power Electronics | Lecture 38](https://www.youtube.com/watch?v=-_iWld95KBY) — Electronics Maddy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/scr-hwr-30-degree
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-7

## Question

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.

## 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.

## Answer

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

```text
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

```text
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)

```text
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

```text
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**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula and value give the average load voltage?
   A. Vdc = (Vm/π)(1 + cos α) = 29.7 V
   B. Vdc = (Vm/2π)(1 + cos α) = 14.85 V
   C. Vdc = Vm/π = 15.9 V
   D. Vdc = (Vm/2π)(1 − cos α) = 1.06 V
   Answer: B. For a half-wave controlled rectifier, integrate from α to π and divide by 2π. (Key point: Average voltage)

2. What is the RMS load voltage?
   A. 35.36 V
   B. 14.85 V
   C. 24.64 V
   D. 25 V
   Answer: C. Vrms = (Vm/2)·sqrt[(π−α)/π + sin2α/(2π)] = 25 × 0.9855. (Key point: RMS voltage)

3. How is the rectification efficiency obtained?
   A. η = Vdc/Vrms = 60 %
   B. η = Vrms/Vm
   C. η = 1 − α/π
   D. η = Pdc/Pac = (Vdc/Vrms)^2 = 36.33 %
   Answer: D. Both powers are across the same R, so the ratio of squares is used. (Key point: Efficiency)

4. How is the ripple factor found?
   A. RF = sqrt(FF^2 − 1) with FF = Vrms/Vdc = 1.659, so RF = 1.324
   B. RF = Vdc/Vrms
   C. RF = FF − 1
   D. RF = sqrt(1 − FF^2)
   Answer: A. Ripple factor relates to the form factor through sqrt(FF^2 − 1). (Key point: Ripple factor)

## Related in Advanced Electronics

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- [SCR half-wave rectifier numerical (200 V peak)](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ascr-hwr-1ma-200v) — Advanced Electronics, chapter 7
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## Related in other subjects

- [Full controlled rectifier, 25 A, α = 30°, derivation](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-controlled-rectifier-25a-30deg-derive) — Power Electronics, chapter 2
- [Half-controlled bridge with R load, α = 45°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ahalf-controlled-bridge-45deg) — Power Electronics, chapter 2
- [Full-wave thyristor rectifier, 220 V, 50 Ω, 45°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-wave-thyristor-220v-50ohm-45) — Power Electronics, chapter 2
- [RMS and average of half-wave rectified sine](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Arms-average-half-wave) — Fundamental of Electrical and Electronics Engineering, chapter 2