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

# 3-phase half controlled bridge: no-load speed, firing angle

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

- Asked in: 2072 Chaitra, 8 marks; 2069 Chaitra (asked 2 times)
- Video: [Three Phase Semi Converter](https://www.youtube.com/watch?v=LNKT9ZoZLZc) — Power Electronics & Drives
- 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/half-bridge-motor-no-load-speed
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3

## Question

A three phase half controlled thyristor bridge with 400V, 3-phase, 50Hz supply is feeding a separately excited dc motor. Armature resistance is 0.2Ω, 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 5A and firing angle is 45°. Also determine firing angle to obtain a speed of 1500 rpm at rated current.

## Short answer

For a three-phase semiconverter, Va = 270.09(1+cos alpha). At alpha=45 and Ia=5A, Eb=460.08V, giving no-load speed 1840 rpm. For 1500 rpm at 100A, Va=395V and alpha = 62.5 degrees.

## Answer

For a three-phase half-controlled (semi-converter) bridge with line voltage $V_L$ and continuous current:

$$
V_a = \frac{3\sqrt2\,V_L}{2\pi}(1 + \cos\alpha) = \frac{3V_{ml}}{2\pi}(1+\cos\alpha), \qquad V_{ml} = \sqrt2\,V_L
$$

$$
\frac{3\sqrt2\times400}{2\pi} = 270.09\ \text{V}
$$

**Data:** $R_a = 0.2\ \Omega$, $K = 0.25$ V/rpm ($E_b = 0.25N$).

#### No-load speed ($I_a = 5$ A, $\alpha = 45^\circ$)

$$
\begin{aligned}
V_a &= 270.09\,(1 + \cos45^\circ) = 270.09\times1.7071 = 461.08\ \text{V} \\
E_b &= V_a - I_aR_a = 461.08 - 5\times0.2 = 460.08\ \text{V} \\
N_0 &= \frac{460.08}{0.25} = 1840.3\ \text{rpm}
\end{aligned}
$$

#### Firing angle for 1500 rpm at rated current (100 A)

$$
\begin{aligned}
E_b &= 0.25\times1500 = 375\ \text{V} \\
V_a &= 375 + 100\times0.2 = 395\ \text{V} \\
1 + \cos\alpha &= \frac{395}{270.09} = 1.4624 \\
\cos\alpha &= 0.4624 \;\Rightarrow\; \alpha = 62.45^\circ
\end{aligned}
$$

**Answer:** no-load speed $\approx 1840$ rpm; firing angle for 1500 rpm at 100 A $\approx 62.5^\circ$.

## Check yourself

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

1. Average output of three-phase semiconverter:
   A. (3 sqrt2 VL/2pi)(1 + cos alpha)
   B. (3 sqrt2 VL/pi) cos alpha
   C. VL cos alpha
   D. (VL/pi)(1 - cos alpha)
   Answer: A. With VL = 400, coefficient = 270.09 V. (Key point: Formula)

2. No-load speed at 45° is about:
   A. 1840 rpm
   B. 1500 rpm
   C. 1000 rpm
   D. 2000 rpm
   Answer: A. Va = 461.08 V, Eb = 460.08 V, N = 460.08/0.25 = 1840 rpm. (Key point: No-load speed)

3. For 1500 rpm at 100 A, Va is:
   A. 375 V
   B. 395 V
   C. 415 V
   D. 355 V
   Answer: B. Eb = 375 V plus Ia Ra = 20 V. (Key point: Required Va)

4. Firing angle for 1500 rpm is about:
   A. 45°
   B. 62.5°
   C. 30°
   D. 75°
   Answer: B. cos alpha = 395/270.09 - 1 = 0.4624, alpha = 62.45°. (Key point: Firing angle)

## Close variants set in other exams

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

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