---
title: "Three-phase full converter at 50% of maximum output"
subject: "Power Electronics"
chapter: 3
key: "semester_7/power_electronics:three-phase-full-converter-50-percent"
source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-full-converter-50-percent
site: BCT Engineering
---

# Three-phase full converter at 50% of maximum output

Power Electronics · Chapter 3: Three phase AC to DC conversion

- Asked in: 2071 Shrawan, 8 marks
- Video: [Three-phase full-wave rectifier with R load| How to calculate average output voltage and current?](https://www.youtube.com/watch?v=3Oauas2LVVY) — power electronics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-full-converter-50-percent
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-3

## Question

A three phase full wave converter is operated from three phase star connected, 230 V, 50 Hz supply with resistance R = 10 ohm. An average output voltage of 50% of the maximum possible output voltage is required. Determine (i) the firing angle (ii) average and rms values of load current and (iii) rectification efficiency.

## Short answer

Using Vdm = 1.35 times 230 V = 310.61 V, half average voltage requires alpha = 60 degrees. With R = 10 ohm, Idc = 15.53 A and Irms = 17.61 A. Rectification efficiency is Pdc/Pac = 77.74 percent.

## Answer

A **three-phase full converter** (six-thyristor bridge) gives $V_{dc} = \frac{3\sqrt3V_m}{\pi}\cos\alpha$, where $V_m$ is the peak phase voltage. With a resistive load this holds for $\alpha \le 60^\circ$ (continuous conduction).

```text
        P (+) o-----+------+------+-----+
                    |      |      |     |
                   T1     T3     T5     |
                    |      |      |     R
 a o----------------+      |      |     |
 b o-----------------------+      |     |
 c o------------------------------+     |
                    |      |      |     |
                   T4     T6     T2     |
                    |      |      |     |
        N (-) o-----+------+------+-----+
 (a joins T1-T4, b joins T3-T6, c joins T5-T2)
```

*Assumption:* 230 V is the line voltage of the star-connected supply (as in the standard textbook problem). Then

$$
\begin{aligned}
V_{ph} &= \frac{230}{\sqrt3} = 132.79\ \text{V}, \quad V_m = \sqrt2 \times 132.79 = 187.79\ \text{V} \\
V_{dm} &= \frac{3\sqrt3V_m}{\pi} = \frac{3\sqrt3 \times 187.79}{\pi} = 310.61\ \text{V}\ (= 1.35 \times 230)
\end{aligned}
$$

#### (i) Firing angle

$$
\begin{aligned}
V_{dc} &= 0.5V_{dm} = 155.30\ \text{V} \\
\cos\alpha &= \frac{V_{dc}}{V_{dm}} = 0.5 \quad\Rightarrow\quad \alpha = 60^\circ
\end{aligned}
$$

At $\alpha = 60^\circ$ the R-load current is just continuous, so the continuous-conduction formulas are valid.

#### (ii) Average and RMS load current

$$
I_{dc} = \frac{V_{dc}}{R} = \frac{155.30}{10} = 15.53\ \text{A}
$$

$$
\begin{aligned}
V_{rms} &= \sqrt3V_m\left[\frac{1}{2} + \frac{3\sqrt3}{4\pi}\cos2\alpha\right]^{1/2} = 325.27\left[0.5 + 0.4135 \times (-0.5)\right]^{1/2} \\
&= 325.27 \times 0.5415 = 176.14\ \text{V} \\
I_{rms} &= \frac{176.14}{10} = 17.61\ \text{A}
\end{aligned}
$$

#### (iii) Rectification efficiency

$$
\begin{aligned}
P_{dc} &= V_{dc}I_{dc} = 155.30 \times 15.53 = 2412.0\ \text{W} \\
P_{ac} &= V_{rms}I_{rms} = 176.14 \times 17.61 = 3102.6\ \text{W} \\
\eta &= \frac{P_{dc}}{P_{ac}} = \frac{2412.0}{3102.6} = 77.74\ \%
\end{aligned}
$$

**Answer: $\alpha = 60^\circ$; $I_{dc} = 15.53$ A, $I_{rms} = 17.61$ A; $\eta = 77.74$ %.**

(If 230 V were taken as the phase voltage, $\alpha$ is still $60^\circ$ and $\eta$ is still 77.74%, but $I_{dc} = 26.90$ A and $I_{rms} = 30.51$ A.)

## Check yourself

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

1. For a three-phase full converter on a 230 V (line) star supply, the maximum possible average output voltage is:
   A. 268.99 V
   B. 310.61 V
   C. 155.30 V
   D. 230 V
   Answer: B. Vdm = 3 sqrt3 Vm / pi = 1.35 x 230 = 310.61 V. (Key point: Maximum output voltage)

2. For Vdc = 50% of the maximum, Vdc = Vdm cos(alpha) gives the firing angle:
   A. 30°
   B. 45°
   C. 60°
   D. 90°
   Answer: C. cos(alpha) = 0.5, so alpha = 60°, the limit where R-load current is just continuous. (Key point: Firing angle)

3. With Vdc = 155.30 V and R = 10 ohm, the average load current is:
   A. 31.06 A
   B. 26.90 A
   C. 17.61 A
   D. 15.53 A
   Answer: D. Idc = Vdc / R = 155.30 / 10 = 15.53 A. (Key point: Average load current)

4. Using P_dc = Vdc Idc and P_ac = Vrms Irms, the rectification efficiency is:
   A. 77.74 %
   B. 50 %
   C. 90 %
   D. 96.5 %
   Answer: A. P_dc = 2412 W and P_ac = 3102.6 W, so efficiency = 77.74 %. (Key point: Rectification efficiency)

## Related in Power Electronics

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