--- title: "Single-phase square-wave inverter rms and fundamental" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:square-wave-inverter-rms-fundamental" source: https://www.bctengineer.com/q/semester_7/power_electronics/square-wave-inverter-rms-fundamental site: BCT Engineering --- # Single-phase square-wave inverter rms and fundamental Power Electronics · Chapter 5: Inverter - Asked in: 2072 Chaitra, 8 marks - Video: [Single Phase Full Bridge Inverter | Power Electronics | Lecture 77](https://www.youtube.com/watch?v=Nl3saO5AG4A) — Electronics Maddy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/square-wave-inverter-rms-fundamental - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question Explain the operation of single phase square wave inverter. Derive the expression for rms value of output voltage and fundamental component of output voltage. ## Short answer A full-bridge inverter switches the load between +Vs and -Vs each half-cycle, producing a square wave. The rms output voltage is Vs. The fundamental component has peak 4Vs/π and rms value 0.9003 Vs. ## Answer A single-phase square-wave inverter switches the load alternately to $+V_s$ and $-V_s$ for equal half periods, giving a square-wave ac output whose frequency is set by the switching rate and whose amplitude is fixed by $V_s$. #### Circuit and operation (full bridge) ```text +-------+-----------+ | | | | S1 D1 S3 D3 Vs | | | A---[Load]--B | | v0 | | S4 D4 S2 D2 | | | +-------+-----------+ ``` - $0 < t < T/2$: S1, S2 ON, $v_0 = +V_s$. - $T/2 < t < T$: S3, S4 ON, $v_0 = -V_s$. - With an inductive load, current continues after each reversal through the feedback diodes (D3, D4 or D1, D2) until it changes sign. - **Half bridge** version (two switches, centre-tapped source): $v_0 = \pm V_s/2$. ```text v0 +Vs |________ | | 0 |--------+--------- t | |________ -Vs S1,S2 S3,S4 0 T/2 T ``` #### RMS value of output voltage $$ V_{0,rms} = \left[\frac{2}{T}\int_0^{T/2}V_s^2\,dt\right]^{1/2} = V_s $$ (Half bridge: $V_s/2$.) #### Fundamental component The wave is odd with half-wave symmetry, so $a_0 = a_n = 0$: $$ \begin{aligned} b_n &= \frac{2}{\pi}\int_0^{\pi}V_s\sin n\theta\,d\theta = \frac{2V_s}{n\pi}(1 - \cos n\pi) \\ &= \frac{4V_s}{n\pi}\ \ (n\ \text{odd}), \qquad 0\ \ (n\ \text{even}) \\ v_0(t) &= \frac{4V_s}{\pi}\Big[\sin\omega t + \frac{1}{3}\sin3\omega t + \frac{1}{5}\sin5\omega t + \dots\Big] \end{aligned} $$ $$ \begin{aligned} V_{1m} &= \frac{4V_s}{\pi} = 1.273\,V_s \\ V_{1,rms} &= \frac{4V_s}{\sqrt2\,\pi} = 0.9003\,V_s \end{aligned} $$ Half bridge: $V_{1,rms} = \dfrac{2V_s}{\sqrt2\,\pi} = 0.45V_s$. #### Performance - $n$th harmonic: $V_n = V_1/n$ (HF$_3$ = 33.3%). - $THD = \sqrt{V_{0,rms}^2 - V_{1,rms}^2}/V_{1,rms} = \sqrt{1 - 0.9003^2}/0.9003 = 48.3\%$. - Output voltage cannot be varied without changing $V_s$; PWM is used for that. Example: $V_s = 230$ V gives $V_{0,rms} = 230$ V and $V_{1,rms} = 207.1$ V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the full-bridge square-wave inverter, which switches give +Vs and which give -Vs? A. S1, S4 give +Vs; S2, S3 give -Vs B. S1, S2 give +Vs; S3, S4 give -Vs C. S1, S3 give +Vs; S2, S4 give -Vs D. S1 alone gives +Vs; S2 alone gives -Vs Answer: B. S1, S2 ON for 0 to T/2 gives +Vs; S3, S4 ON for T/2 to T gives -Vs. (Key point: Switching operation) 2. What is the rms value of the full-bridge square-wave output? A. Vs/sqrt(2) B. 0.9003 Vs C. Vs/2 D. Vs Answer: D. The magnitude is Vs at all times, so V0,rms = Vs (half bridge: Vs/2). (Key point: RMS value) 3. Why is the output given by sum of 4Vs/(n pi) sin(n w t) for odd n only? A. Odd symmetry and half-wave symmetry give a0 = an = 0 and b_n = 2Vs(1 - cos n pi)/(n pi), which vanishes for even n B. The wave is even, so only cosines appear C. The wave has a dc component that removes even terms D. Feedback diodes remove even harmonics Answer: A. 1 - cos(n pi) = 2 for odd n and 0 for even n. (Key point: Fourier coefficient) 4. What is the rms fundamental output voltage of the full-bridge inverter? A. 1.273 Vs B. 0.9003 Vs C. 0.45 Vs D. 0.707 Vs Answer: B. V1,rms = 4Vs/(sqrt(2) pi) = 0.9003 Vs. (Key point: Fundamental rms) ## Related in Power Electronics - [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 - [Square-wave inverter ±100 V: fundamental load current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ainverter-100v-rl-fundamental) — Power Electronics, chapter 5, asked 2× - [Square-wave inverter ±48 V: fundamental and third harmonic](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Ainverter-48v-rl-fundamental-third) — Power Electronics, chapter 5 - [Single-phase PWM inverter rms and Fourier series](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Asingle-phase-pwm-inverter-fourier) — 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 - [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 ## Related in other subjects - [Voltage follower square-wave output with slew rate](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Avoltage-follower-square-wave) — Advanced Electronics, chapter 2 - [Variable inverter output; SCR full-wave average](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ainverter-variable-output-scr-fwr) — Advanced Electronics, chapter 7