---
title: "Three-phase bridge inverter 180° and 120° numerical"
subject: "Power Electronics"
chapter: 5
key: "semester_7/power_electronics:three-phase-inverter-180-120-numerical"
source: https://www.bctengineer.com/q/semester_7/power_electronics/three-phase-inverter-180-120-numerical
site: BCT Engineering
---

# Three-phase bridge inverter 180° and 120° numerical

Power Electronics · Chapter 5: Inverter

- Asked in: 2082 Baishakh, 8 marks
- Video: [Solving Three-Phase Inverter Numericals | Power Electronics GATE Numerical 3](https://www.youtube.com/watch?v=xqCtZjNdptA) — Ekeeda GATE & ESE
- 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-inverter-180-120-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5

## Question

A three-phase bridge inverter delivers power to a resistive load from a 450 V dc source. For a star connected load of 10 Ω per phase, determine for both (a) 180° mode (b) 120° mode, (i) RMS value of load current (ii) RMS value of thyristor current (iii) Load power.

## Short answer

180° mode: Vph,rms=212.13 V, IL,rms=21.21 A, IT,rms=15.0 A, P=13.5 kW. 120° mode: Vph,rms=183.71 V, IL,rms=18.37 A, IT,rms=12.99 A, P=10.125 kW. Values obtained using phase voltage RMS formulas (√2/3 Vdc and Vdc/√6) and power P=3 I² R.

## Answer

For a star-connected resistive load, the 180° mode gives a six-step phase voltage ($V/3$, $2V/3$) and the 120° mode gives a quasi-square phase voltage ($\pm V/2$ for 120°, zero for 60°).

**Given:** $V_{dc} = 450$ V, $R = 10\ \Omega$ per phase, star connection.

#### (a) 180° conduction mode
Phase voltage is $V_{dc}/3 = 150$ V and $2V_{dc}/3 = 300$ V in 60° steps.

**(i) RMS load (phase) current**
$$
\begin{aligned}
V_{ph,rms} &= \left[\frac{1}{\pi}\left(\Big(\tfrac{V}{3}\Big)^2\tfrac{\pi}{3}\cdot2 + \Big(\tfrac{2V}{3}\Big)^2\tfrac{\pi}{3}\right)\right]^{1/2} = \frac{\sqrt2}{3}V_{dc} \\
&= 0.4714\times450 = 212.13\ \text{V} \\
I_{L,rms} &= \frac{212.13}{10} = 21.21\ \text{A}
\end{aligned}
$$

**(ii) RMS thyristor current** — each thyristor carries the phase current for 180° of every 360°:

$$
I_{T,rms} = \frac{I_{L,rms}}{\sqrt2} = \frac{21.21}{\sqrt2} = 15.0\ \text{A}
$$

**(iii) Load power**
$$
P = 3I_{L,rms}^2R = 3\times21.21^2\times10 = 13500\ \text{W} = 13.5\ \text{kW}
$$

#### (b) 120° conduction mode
Only two thyristors conduct at a time; two phases are in series across $V_{dc}$, the third is open. So during conduction:

$$
i = \frac{V_{dc}}{2R} = \frac{450}{20} = 22.5\ \text{A}, \qquad v_{ph} = \pm\frac{V_{dc}}{2} = \pm225\ \text{V}
$$

Each phase conducts for 120° out of every 180°.

**(i) RMS load current**
$$
\begin{aligned}
V_{ph,rms} &= \frac{V_{dc}}{2}\sqrt{\frac{2}{3}} = \frac{V_{dc}}{\sqrt6} = 183.71\ \text{V} \\
I_{L,rms} &= 22.5\sqrt{\tfrac{2}{3}} = 18.37\ \text{A}
\end{aligned}
$$

**(ii) RMS thyristor current** — each thyristor carries 22.5 A for 120° of 360°:

$$
I_{T,rms} = 22.5\sqrt{\tfrac{120}{360}} = \frac{22.5}{\sqrt3} = 12.99\ \text{A}
$$

**(iii) Load power**
$$
P = 3\times18.37^2\times10 = 10125\ \text{W} = 10.125\ \text{kW}
$$

| Quantity | 180° mode | 120° mode |
|---|---|---|
| $V_{ph,rms}$ | 212.13 V | 183.71 V |
| $I_{L,rms}$ | 21.21 A | 18.37 A |
| $I_{T,rms}$ | 15.0 A | 12.99 A |
| Power | 13.5 kW | 10.125 kW |

The 180° mode delivers more power (higher switch utilisation), while the 120° mode has an idle 60° in each leg that reduces the risk of shoot-through.

## Check yourself

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

1. For Vdc = 450 V in 180° mode, what is the phase rms voltage?
   A. 183.71 V
   B. 212.13 V
   C. 225 V
   D. 318.2 V
   Answer: B. V_ph,rms = (sqrt2/3)Vdc = 0.4714 x 450 = 212.13 V. (Key point: 180° phase rms voltage)

2. In 180° mode each thyristor conducts for 180°. How is its rms current related to the load rms current?
   A. I_T = I_L
   B. I_T = I_L / sqrt3
   C. I_T = I_L / sqrt2
   D. I_T = I_L / 2
   Answer: C. Conducting half of the cycle gives I_T,rms = I_L,rms/sqrt2 = 21.21/1.414 = 15.0 A. (Key point: Thyristor current 180°)

3. In 120° mode, what current flows while two phases conduct in series across 450 V with 10 Ω per phase?
   A. 45 A
   B. 22.5 A
   C. 15 A
   D. 30 A
   Answer: B. Two phases in series give 20 Ω, so i = 450/20 = 22.5 A. (Key point: 120° conduction current)

4. What is the rms thyristor current in 120° mode?
   A. 22.5 A
   B. 15.9 A
   C. 18.37 A
   D. 12.99 A
   Answer: D. Each thyristor carries 22.5 A for 120° of 360°: 22.5 x sqrt(1/3) = 12.99 A. (Key point: Thyristor rms 120°)

5. What is the load power in 180° mode?
   A. 13.5 kW
   B. 10.125 kW
   C. 4.5 kW
   D. 20.25 kW
   Answer: A. P = 3 x 21.21^2 x 10 = 13500 W; the 120° mode gives 10.125 kW. (Key point: Power)

## Related in Power Electronics

- [Three-phase bridge inverter, 120° conduction](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-inverter-120-conduction) — Power Electronics, chapter 5
- [Three-phase six-step (180° conduction) inverter](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-inverter-180-six-step) — Power Electronics, chapter 5, asked 4×
- [Switch states of three-phase VSI, 180° mode](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Avsi-180-switch-states) — Power Electronics, chapter 5
- [180° three-phase inverter and fundamental phase voltage](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ainverter-180-fundamental-per-phase) — Power Electronics, chapter 5
- [Three-phase fully controlled bridge with R load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3A3ph-full-bridge-controlled-r-load) — Power Electronics, chapter 3
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