--- title: "Single-phase inverter fundamental output voltage" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:single-phase-inverter-r-fundamental" source: https://www.bctengineer.com/q/semester_7/power_electronics/single-phase-inverter-r-fundamental site: BCT Engineering --- # Single-phase inverter fundamental output voltage Power Electronics · Chapter 5: Inverter - Asked in: 2082 Baishakh, 8 marks - Video: [Single Phase Full Bridge Inverter (Basics, Circuit, Working, Waveforms & Applications) Explained](https://www.youtube.com/watch?v=-WU3BxOxvII) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/single-phase-inverter-r-fundamental - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question Explain the operation of a single-phase inverter with resistive load and derive the expression for fundamental component of the output voltage. ## Short answer A single-phase full-bridge inverter with resistive load alternates between +Vs (S1,S2 ON) and -Vs (S3,S4 ON) each half cycle, producing a square wave. Fourier analysis gives odd harmonics only: v0(t) = Σ (4Vs/(nπ)) sin nωt, so peak fundamental V1m = 4Vs/π and RMS V1,rms = 4Vs/(√2 π) = 0.9003 Vs. ## Answer A single-phase inverter converts dc into ac by alternately connecting the load to the dc source in opposite directions. With a resistive load and square-wave switching, the output is a square wave of amplitude $V_s$ (full bridge) or $V_s/2$ (half bridge), and the current has the same shape. #### Circuit (full bridge) ```text +-------+-----------+ | | | | S1 S3 Vs | | | A---[ R ]---B | | v0 | | S4 S2 | | | +-------+-----------+ ``` #### Operation - **$0 < t < T/2$:** S1 and S2 ON. Point A is at $+V_s$ relative to B: $v_0 = +V_s$, $i_0 = V_s/R$. - **$T/2 < t < T$:** S3 and S4 ON. $v_0 = -V_s$, $i_0 = -V_s/R$. - S1–S4 (and S3–S2) of a leg must never be ON together. Output frequency $f = 1/T$ is set by the gating rate. - With a pure R load the current reverses exactly with the voltage, so the feedback diodes do not conduct. - **Half bridge:** two switches and a centre-tapped source give $v_0 = \pm V_s/2$. ```text v0 +Vs |________ | | 0 |--------+--------- t | |________ -Vs S1,S2 ON S3,S4 ON 0 T/2 T ``` #### RMS value $$ V_{0,rms} = \left[\frac{2}{T}\int_0^{T/2}V_s^2\,dt\right]^{1/2} = V_s $$ #### Fundamental component (Fourier analysis) The wave is odd with half-wave symmetry, so $a_0 = 0$, $a_n = 0$ and only odd sine terms remain: $$ \begin{aligned} b_n &= \frac{2}{\pi}\int_0^{\pi}V_s\sin n\theta\,d\theta = \frac{2V_s}{n\pi}\big[-\cos n\theta\big]_0^{\pi} = \frac{2V_s}{n\pi}(1 - \cos n\pi) \\ b_n &= \frac{4V_s}{n\pi}\ (n\ \text{odd}), \qquad b_n = 0\ (n\ \text{even}) \\ v_0(t) &= \sum_{n=1,3,5,\dots}\frac{4V_s}{n\pi}\sin n\omega t \end{aligned} $$ For $n = 1$: $$ \begin{aligned} V_{1m} &= \frac{4V_s}{\pi} = 1.273\,V_s \\ V_{1,rms} &= \frac{4V_s}{\sqrt2\,\pi} = 0.9003\,V_s \end{aligned} $$ For a half bridge, replace $V_s$ by $V_s/2$: $V_{1m} = 2V_s/\pi$, $V_{1,rms} = 0.45\,V_s$. Distortion: $THD = \sqrt{V_{rms}^2 - V_1^2}/V_1 = \sqrt{1 - 0.9003^2}/0.9003 = 48.3\%$. Example: $V_s = 220$ V gives $V_{1,rms} = 198.1$ V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In a full-bridge inverter with resistive load, which switches are ON in the first half cycle to give v0 = +Vs? A. S1 and S4 B. S3 and S4 C. S1 and S2 D. All four Answer: C. S1 and S2 (diagonal pair) connect A to +Vs and B to the return; S1-S4 in the same leg must never be ON together. (Key point: Operation of full bridge) 2. For the square wave of amplitude Vs, what is b_n? A. 2Vs/(n pi) for all n B. 4Vs/(n pi) for odd n, 0 for even n C. Vs/(n pi) for odd n D. 4Vs/(n pi) for all n Answer: B. b_n = (2Vs/n pi)(1 - cos n pi), which is 4Vs/(n pi) for odd n and zero for even n. (Key point: Fourier coefficient) 3. What is the rms value of the fundamental for a full bridge? A. 0.45 Vs B. Vs C. 1.273 Vs D. 0.9003 Vs Answer: D. V1,rms = 4Vs/(sqrt2 pi) = 0.9003 Vs; 1.273 Vs is the peak. (Key point: Fundamental rms) 4. How does the fundamental change for a half-bridge inverter? A. Vs is replaced by Vs/2, giving V1m = 2Vs/pi B. It is doubled C. It is unchanged D. Only odd harmonics become zero Answer: A. The half bridge output is ±Vs/2, so every harmonic amplitude is halved. 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