--- title: "Single-pulse PWM inverter control and rms" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:single-pwm-rms" source: https://www.bctengineer.com/q/semester_7/power_electronics/single-pwm-rms site: BCT Engineering --- # Single-pulse PWM inverter control and rms Power Electronics · Chapter 5: Inverter - Asked in: 2075 Asoj, 8 marks - Video: [54. Single Pulse-Width Modulation Technique](https://www.youtube.com/watch?v=d1LlPGYr8Cs) — Electronic Waves - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/single-pwm-rms - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question Explain the operation of single PWM techniques for inverter control and therefore determine the rms value of output voltage. ## Short answer Single-pulse PWM gives one pulse per half cycle, width delta, by comparing a dc reference with a triangular carrier at twice output frequency. RMS output voltage is Vs times square root of delta over pi, equal to Vs times square root of M, where M is Ar over Ac. Varying M controls output; delta of 180 degrees gives a square wave. ## Answer Single-pulse width modulation controls the inverter output voltage by giving one pulse per half cycle whose width $\delta$ is varied; the pulse is placed symmetrically about $\pi/2$ (and about $3\pi/2$ for the negative half). #### Operation ```text vc /\ /\ /\ carrier, 2f Ar -/--\------/--\------/--\--- reference / \ / \ / \ g1 |====| gate T1,T2 g4 |====| gate T3,T4 v0 +Vs ____ 0 --' '---. .---- wt -Vs '--' |<-d->| 0 pi/2 pi 3pi/2 2pi ``` 1. A dc (rectangular) reference signal of amplitude $A_r$ is compared with a triangular carrier of amplitude $A_c$ and frequency $2f$ ($f$ = output frequency). 2. While the reference exceeds the carrier, gate pulses are given: to T1, T2 in the positive half cycle and to T3, T4 in the negative half cycle of a full bridge. 3. The pulse width is $\delta = M\pi$, where the modulation index $M = A_r/A_c$. Varying $A_r$ from 0 to $A_c$ varies $\delta$ from 0 to 180°, and so varies the output voltage. 4. The output frequency is changed by changing the carrier (and so the gate) frequency. #### RMS value of output voltage The pulse extends from $\frac{\pi-\delta}{2}$ to $\frac{\pi+\delta}{2}$ with amplitude $V_s$: $$ \begin{aligned} V_{0,rms} &= \left[\frac{2}{2\pi}\int_{(\pi-\delta)/2}^{(\pi+\delta)/2}V_s^2\,d(\omega t)\right]^{1/2} \\ &= \left[\frac{V_s^2}{\pi}\cdot\delta\right]^{1/2} = V_s\sqrt{\frac{\delta}{\pi}} = V_s\sqrt{M} \end{aligned} $$ #### Harmonics $$ v_0 = \sum_{n=1,3,5,\dots}\frac{4V_s}{n\pi}\sin\frac{n\delta}{2}\sin n\omega t, \qquad V_{1,rms} = 0.9V_s\sin\frac{\delta}{2} $$ - $\delta = 180^\circ$: $V_{0,rms} = V_s$ (square wave). - $\delta = 120^\circ$: $V_{0,rms} = 0.816V_s$ and the 3rd harmonic is eliminated ($\sin 180^\circ = 0$). - $\delta = 90^\circ$: $V_{0,rms} = 0.707V_s$. **Limitation:** at low output voltage (narrow pulse) the harmonic content is very high; the lowest harmonic (3rd) is close to the fundamental and hard to filter. Multiple-pulse and sinusoidal PWM overcome this. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are gate pulses generated in single-pulse PWM? A. A sine reference is compared with a carrier at 10 times f B. A dc reference A_r is compared with a triangular carrier at 2f C. The carrier is at f and the reference a sine D. A comparator is not needed Answer: B. A rectangular (dc) reference is compared with a triangle at twice the output frequency; the pulse is ON when reference exceeds carrier. (Key point: Gate generation) 2. What is V0,rms for a pulse of width delta and amplitude Vs? A. Vs delta / pi B. Vs sqrt(delta/pi) C. Vs sqrt(delta) D. Vs sin(delta/2) Answer: B. Integrating Vs^2 over the pulse gives V0,rms = Vs sqrt(delta/pi) = Vs sqrt(M). (Key point: RMS expression) 3. What is V0,rms for delta = 120°? A. 0.707 Vs B. Vs C. 0.5 Vs D. 0.816 Vs Answer: D. sqrt(120/180) = 0.816; the 3rd harmonic is also eliminated. (Key point: Value at 120°) 4. What is the main limitation of single-pulse PWM? A. Output cannot be varied B. Harmonic content is very high at low output voltage and the 3rd harmonic is hard to filter C. Needs a sinusoidal source D. Output frequency is fixed Answer: B. At narrow pulse widths the low-order harmonics are close to the fundamental, motivating multiple-pulse and sinusoidal PWM. 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