---
title: "Six-pulse converter: supply voltage and firing range"
subject: "Utilization of Electrical Energy"
chapter: 3
key: "semester_7/utilization_of_electrical_energy:six-pulse-converter-firing-range"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/six-pulse-converter-firing-range
site: BCT Engineering
---

# Six-pulse converter: supply voltage and firing range

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

- Asked in: 2073 Shrawan
- 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/six-pulse-converter-firing-range
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-3

## Question

A separately excited dc motor is fed from a three phase six pulse fully controlled bridge converter. The motor develops its full load torque at a rated speed of 1800 rpm taking a current of 60 A from a 400 V supply. Determine the rms value of supply voltage if the motor runs at its rated conditions for α = 0. What is the range of firing angles for a speed control of 1800 rpm to 900 rpm. The armature resistance is 0.5 ohm. The supply and thyristors are ideal.

## Short answer

For a six-pulse fully controlled converter, Va = 1.3505 VL cos alpha; with alpha = 0 and Va = 400 V, supply line rms voltage is 296.2 V. At 900 rpm Va = 215 V, so cos alpha = 0.5375 and alpha = 57.5°; firing angle range is 0° to 57.5°.

## Answer

Interpretation: the motor's rated armature (dc) voltage is 400 V, i.e. at rated speed and torque with $\alpha = 0$ the converter gives 400 V dc. Ideal supply and thyristors; continuous conduction.

For a 3-phase six-pulse fully controlled bridge:
$$
V_a = \frac{3\sqrt2}{\pi}V_L\cos\alpha = 1.3505\,V_L\cos\alpha
$$

#### RMS supply voltage
With $\alpha = 0$, $V_a = 400$ V:
$$
V_L = \frac{400\,\pi}{3\sqrt2} = 296.19\ \text{V (line, rms)}
$$
(phase rms $= 296.19/\sqrt3 = 171.0$ V.)

#### Back emf
$$
\begin{aligned} E_{1800} &= V_a - I_a R_a = 400 - 60\times0.5 = 370\ \text{V} \\ E_{900} &= 370 \times \frac{900}{1800} = 185\ \text{V} \end{aligned}
$$
Full-load torque is held throughout, so $I_a = 60$ A and $I_aR_a = 30$ V.

#### Firing angle at 900 rpm
$$
\begin{aligned} V_a &= 185 + 30 = 215\ \text{V} \\ \cos\alpha &= \frac{215}{400} = 0.5375 \Rightarrow \alpha = 57.49^\circ \end{aligned}
$$
At 1800 rpm, $V_a = 400$ V and $\alpha = 0^\circ$.

**Answer: supply voltage = 296.2 V (line rms); firing angle range = 0° (1800 rpm) to 57.5° (900 rpm).**

## Check yourself

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

1. For alpha = 0 and Va = 400 V, line rms supply is:
   A. 400 V
   B. 296.2 V
   C. 171 V
   D. 565 V
   Answer: B. VL = 400 pi/(3 sqrt2) = 296.2 V. (Key point: Supply voltage)

2. At 900 rpm with full-load torque, back emf is:
   A. 370 V
   B. 200 V
   C. 185 V
   D. 215 V
   Answer: C. E is proportional to speed: 370 x 900/1800 = 185 V. (Key point: Back emf scaling)

3. Firing angle range for 1800 to 900 rpm:
   A. 0° to 57.5°
   B. 0° to 90°
   C. 57.5° to 90°
   D. 30° to 60°
   Answer: A. At 900 rpm, Va = 215 V, cos alpha = 0.5375, alpha = 57.5°. (Key point: Firing range)

## Related in Utilization of Electrical Energy

- [Single-phase fully controlled converter DC motor speed](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afull-converter-dc-motor-speed) — Utilization of Electrical Energy, chapter 3
- [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×
- [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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- [Firing angles for 200 V, 875 rpm DC motor](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afull-converter-firing-angles-875rpm) — Utilization of Electrical Energy, chapter 3

## Related in other subjects

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