---
title: "SCR half-wave rectifier numerical (220 V, 150 Ω)"
subject: "Advanced Electronics"
chapter: 7
key: "semester_4/advanced_electronics:scr-hwr-125ma-110v"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/scr-hwr-125ma-110v
site: BCT Engineering
---

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

Advanced Electronics · Chapter 7: Introduction to Power Electronics

- Asked in: 2068 Chaitra, 2+1+2+2 marks
- Video: [Thyristors: Silicon Controlled Rectifier SCR - Application and Solution to Sample Problems](https://www.youtube.com/watch?v=J_m9E-0VpH4) — ECEmadeEasy - by Engr. Lian
- 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-125ma-110v
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-7

## Question

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

## 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.

## Answer

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:

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

#### b) Conduction angle

With zero holding current, it conducts until 180°:

```text
θc = 180° − 30° = 150°
```

#### c) Average output voltage

```text
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

```text
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.

## Check yourself

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

1. In the SCR half-wave rectifier, at what instant does the SCR fire?
   A. When the supply voltage reaches the forward breakover voltage VBO
   B. When the supply voltage reaches zero
   C. When the supply voltage reaches its peak value
   D. When the gate current falls to zero
   Answer: A. The SCR turns on when the instantaneous supply Vm sin α equals VBO (for the set gate current), so sin α = VBO/Vm. (Key point: Firing angle condition)

2. With Vm = 220 V (peak) and VBO = 110 V, what is the firing angle?
   A. 60°
   B. 45°
   C. 30°
   D. 15°
   Answer: C. sin α = 110/220 = 0.5, so α = 30°. (Key point: Firing angle value)

3. With zero holding current, how long does the SCR conduct after firing at 30°?
   A. From 30° to 360°, so 330°
   B. From 30° to 180°, so 150°
   C. From 0° to 150°, so 150° before firing
   D. Only until 90°, so 60°
   Answer: B. With IH = 0 the SCR stays on until the supply returns to zero at 180°, so conduction angle = 180° − 30° = 150°. (Key point: Conduction angle)

4. Which expression gives the average output voltage of this half-wave controlled rectifier?
   A. (Vm/π)(1 + cos α)
   B. (Vm/2π)(1 − cos α)
   C. (Vm/2π)(1 + cos α)
   D. Vm/2 sin α
   Answer: C. Vav = (1/2π)∫ from α to π of Vm sin θ dθ = (Vm/2π)(1 + cos α) = 65.34 V for α = 30°. (Key point: Average output voltage formula)

## Related in Advanced Electronics

- [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
- [SCR half-wave rectifier fired at 30°](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ascr-hwr-30-degree) — Advanced Electronics, chapter 7
- [SCR analogy; half-wave rectifier for 1 A average](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ascr-analogy-hwr-120v) — Advanced Electronics, chapter 7

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