---
title: "Half-bridge inverter RL load Fourier, THD"
subject: "Power Electronics"
chapter: 5
key: "semester_7/power_electronics:half-bridge-inverter-rl-third-order"
source: https://www.bctengineer.com/q/semester_7/power_electronics/half-bridge-inverter-rl-third-order
site: BCT Engineering
---

# Half-bridge inverter RL load Fourier, THD

Power Electronics · Chapter 5: Inverter

- Asked in: 2080 Baishakh, 8 marks
- Video: [Single phase half bridge inverter with RL-load](https://www.youtube.com/watch?v=EdBy13ZAv_k) — Anet Jose
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/half-bridge-inverter-rl-third-order
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5

## Question

A single-phase inverter with resistive load R0 = 5 ohms and L0 = 20mH and the input voltage is 200V dc with a centre point of dc source grounded. The inverter is operated to give an output ac voltage of 50 Hz. Find the time domain equation of the load current using Fourier series up to 3rd order. Also, find the Harmonic factor and THD of output voltage.

## Short answer

For a 200 V dc half-bridge with R=5 Ω, L=20 mH at 50 Hz, i0(t)=15.86 sin(314.16t−51.49°)+2.18 sin(942.48t−75.14°) A. V0,rms=100 V, V1,rms=90.03 V, HF3=33.33%, THD=48.34%.

## Answer

With the centre point of the dc source grounded, the single-phase (half-bridge) inverter gives a square wave of $\pm V_s/2$ across the load. Its Fourier series contains only odd harmonics, each of which drives a current through the RL impedance at that frequency.

**Given:** $V_s = 200$ V (so $\pm100$ V), $R = 5\ \Omega$, $L = 20$ mH, $f = 50$ Hz, $\omega = 314.16$ rad/s.

#### Fourier series of output voltage
$$
v_0 = \sum_{n=1,3,5,\dots}\frac{4(V_s/2)}{n\pi}\sin n\omega t = \sum_{n\ \text{odd}}\frac{2V_s}{n\pi}\sin n\omega t = \sum_{n\ \text{odd}}\frac{127.32}{n}\sin n\omega t
$$

#### Harmonic currents
$$
Z_n = \sqrt{R^2 + (n\omega L)^2}, \qquad \theta_n = \tan^{-1}\frac{n\omega L}{R}, \qquad I_{nm} = \frac{V_{nm}}{Z_n}
$$

| $n$ | $V_{nm}$ (V) | $n\omega L$ (Ω) | $Z_n$ (Ω) | $\theta_n$ | $I_{nm}$ (A) |
|---|---|---|---|---|---|
| 1 | 127.32 | 6.283 | 8.030 | 51.49° | 15.86 |
| 3 | 42.44 | 18.850 | 19.501 | 75.14° | 2.18 |

Load current up to the 3rd order:

$$
i_0(t) = 15.86\sin(314.16t - 51.49^\circ) + 2.18\sin(942.48t - 75.14^\circ)\ \text{A}
$$

#### Harmonic factor and THD of output voltage
$$
\begin{aligned}
V_{0,rms} &= \frac{V_s}{2} = 100\ \text{V} \\
V_{1,rms} &= \frac{127.32}{\sqrt2} = 90.03\ \text{V} \\
HF_n &= \frac{V_n}{V_1} = \frac{1}{n} \;\Rightarrow\; HF_3 = \frac{1}{3} = 33.33\%,\ HF_5 = 20\%
\end{aligned}
$$

$$
\begin{aligned}
THD &= \frac{\sqrt{V_{0,rms}^2 - V_{1,rms}^2}}{V_{1,rms}} = \frac{\sqrt{100^2 - 90.03^2}}{90.03} \\
&= \frac{43.52}{90.03} = 0.4834 = 48.34\%
\end{aligned}
$$

**Answer:** $i_0 = 15.86\sin(\omega t - 51.49^\circ) + 2.18\sin(3\omega t - 75.14^\circ)$ A; $HF_3 = 33.33\%$ (lowest-order harmonic); $THD = 48.34\%$.

## Check yourself

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

1. With the centre point of a 200 V source grounded, what is the load voltage?
   A. ±200 V
   B. ±100 V
   C. ±400 V
   D. ±50 V
   Answer: B. Each half of the source is Vs/2 = 100 V, so v0 = ±100 V. (Key point: Output amplitude)

2. For the RL load, how is the n-th harmonic current found?
   A. V_nm / sqrt(R^2 + (n omega L)^2)
   B. V_nm / R
   C. V_nm / (R + n omega L)
   D. V_nm / (n omega L)
   Answer: A. Z_n = sqrt(R^2 + (n omega L)^2), and I_nm = V_nm/Z_n. (Key point: Harmonic impedance)

3. What is the fundamental peak current (V1m = 127.32 V, Z1 = 8.03 Ω)?
   A. 25.5 A
   B. 10.6 A
   C. 18.3 A
   D. 15.86 A
   Answer: D. 127.32/8.03 = 15.86 A, lagging by tan^-1(6.283/5) = 51.49°. (Key point: Fundamental current)

4. What is the harmonic factor of the 3rd harmonic of a square wave?
   A. 33.33%
   B. 20%
   C. 48.34%
   D. 11.1%
   Answer: A. HF_n = V_n/V_1 = 1/n, so HF_3 = 1/3. (Key point: Harmonic factor)

5. How is THD found from the rms voltages?
   A. V1 / V0,rms
   B. sqrt(V0,rms^2 - V1,rms^2) / V1,rms
   C. V0,rms / V1,rms
   D. (V0,rms - V1,rms) / V0,rms
   Answer: B. The harmonic rms is sqrt(V0^2 - V1^2); dividing by V1 gives 43.52/90.03 = 48.34%. (Key point: THD)

## Related in Power Electronics

- [Full-bridge inverter RLC load, THD](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-bridge-inverter-rlc-112uf) — Power Electronics, chapter 5
- [PWM half-bridge inverter switching instants](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Apwm-inverter-switching-instants) — Power Electronics, chapter 5
- [Full-bridge square-wave inverter, 400 V, RL load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-bridge-inverter-400v-rl) — Power Electronics, chapter 5
- [Single-phase inverter fundamental output voltage](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asingle-phase-inverter-r-fundamental) — Power Electronics, chapter 5
- [Single-phase full-wave inverter with 5 Ω load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-wave-inverter-5ohm) — Power Electronics, chapter 5
- [Half-bridge square-wave inverter at 120 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asquare-wave-inverter-120hz) — Power Electronics, chapter 5

## Related in other subjects

- [Chopper definition; single-phase half bridge inverter](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Achopper-half-bridge-inverter) — Advanced Electronics, chapter 7
- [Full bridge inverter; dc-dc output power control](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Afull-bridge-inverter-dc-dc-control) — Advanced Electronics, chapter 7
- [Choppers; single-phase full bridge inverter](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Achoppers-full-bridge-inverter) — Advanced Electronics, chapter 7, asked 2×
- [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×