---
title: "3-phase thyristor bridge: no-load speed and firing angle"
subject: "Utilization of Electrical Energy"
chapter: 3
key: "semester_7/utilization_of_electrical_energy:full-bridge-motor-no-load-speed"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/full-bridge-motor-no-load-speed
site: BCT Engineering
---

# 3-phase thyristor bridge: no-load speed and firing angle

Utilization of Electrical Energy · Chapter 3: Control of Electric Drive

- Asked in: 2080 Baisakh, 8 marks
- Video: [3 Phase Full Wave Controlled Rectifier with R load | Power Electronics | Lecture 57](https://www.youtube.com/watch?v=4NhE-EMCDcM) — Electronics Maddy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/full-bridge-motor-no-load-speed
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3

## Question

A 3 phase fully controlled Thyristor Bridge with 400 V, 3 phase, 50 Hz supply is feeding a separately excited DC motor. Armature resistance is 0.2 Ω, armature rated current is 100 A and back emf constant is 0.25 V/rpm. Determine no-load speed if no-load armature current is 5 A and firing angle is 45°. Also determine firing angle to obtain a speed of 1500 rpm at rated current.

## Short answer

No-load speed at α = 45° and 5 A is 1523.9 rpm. Firing angle to obtain 1500 rpm at rated armature current (100 A) is 43.0°.

## Answer

Data: $V_L = 400$ V, $R_a = 0.2\ \Omega$, $I_{rated} = 100$ A, $K = 0.25$ V/rpm.

For a three-phase fully controlled bridge (continuous conduction):
$$
V_a = \frac{3\sqrt2 V_L}{\pi}\cos\alpha = \frac{3\sqrt2\times400}{\pi}\cos\alpha = 540.19\cos\alpha
$$

#### No-load speed at $\alpha = 45^\circ$, $I_a = 5$ A
$$
\begin{aligned} V_a &= 540.19 \cos45^\circ = 381.97\ \text{V} \\ E &= V_a - I_a R_a = 381.97 - 5\times0.2 = 380.97\ \text{V} \\ N_0 &= \frac{E}{K} = \frac{380.97}{0.25} = 1523.9\ \text{rpm} \end{aligned}
$$

#### Firing angle for 1500 rpm at rated current
$$
\begin{aligned} E &= 0.25 \times 1500 = 375\ \text{V} \\ V_a &= 375 + 100\times0.2 = 395\ \text{V} \\ \cos\alpha &= \frac{395}{540.19} = 0.7312 \Rightarrow \alpha = 43.01^\circ \end{aligned}
$$

**Answer: No-load speed = 1523.9 rpm; firing angle for 1500 rpm at 100 A = 43.0°.**

## Check yourself

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

1. Va of a 3-phase fully controlled bridge is:
   A. (3sqrt2 VL/pi) cos alpha = 540.19 cos alpha
   B. (sqrt2 VL/pi) cos alpha
   C. VL cos alpha
   D. (2 Vm/pi) cos alpha
   Answer: A. For VL = 400 V, 1.35 x 400 = 540.19 V. (Key point: Converter output)

2. No-load speed at 45° and Ia = 5 A:
   A. 1527.9 rpm
   B. 1500 rpm
   C. 1523.9 rpm
   D. 1600 rpm
   Answer: C. E = 381.97 - 1 = 380.97 V; N = 380.97/0.25 = 1523.9 rpm. (Key point: No-load speed)

3. For 1500 rpm at 100 A, required Va is:
   A. 375 V
   B. 395 V
   C. 355 V
   D. 415 V
   Answer: B. Va = E + Ia Ra = 375 + 20 = 395 V; cos alpha = 0.7312, alpha = 43.0°. (Key point: Firing angle)

## Close variants set in other exams

- [3-phase half controlled bridge: no-load speed, firing angle](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ahalf-bridge-motor-no-load-speed) — Utilization of Electrical Energy, chapter 3, asked 3×

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