--- title: "PWM half-bridge inverter switching instants" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:pwm-inverter-switching-instants" source: https://www.bctengineer.com/q/semester_7/power_electronics/pwm-inverter-switching-instants site: BCT Engineering --- # PWM half-bridge inverter switching instants Power Electronics · Chapter 5: Inverter - Asked in: 2069 Chaitra, 8 marks - Video: [Power Electroics - PWM Inverters - Part 1](https://www.youtube.com/watch?v=yTP0xwP4nuI) — Power Electronics with Dr. K - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/pwm-inverter-switching-instants - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question The single phase inverter shown in Figure above (half-bridge inverter with S1, S2, D1, D2 and centre-tapped dc supply ±Vs/2) is operated as pulse width modulated (PWM) inverter with frequency ratio = 1 and modulation index = 0.5. The intersection between the triangular carrier wave and square wave ac modulating signal is used to determine the switching instant of switches S1 and S2. Assuming purely resistive load i) Draw the waveforms of triangular carrier wave, square wave ac modulating signal and output voltage. ii) Determine the switching instant of switches S1 and S2 iii) Determine the RMS value of output voltage ## Short answer With frequency ratio 1 and modulation index 0.5, S1 turns ON at 45° and OFF at 135°; S2 turns ON at 225° and OFF at 315°. The output voltage RMS is 0.3536 Vs. ## Answer With a frequency ratio of 1, there is one carrier triangle in each half cycle of the modulating wave, so each half cycle has a single pulse (single-pulse modulation). Reading used here: in each half cycle the triangular carrier falls from its peak $A_c$ to 0 at the middle (90°, 270°) and rises back to $A_c$, and it is compared with the magnitude of the square reference of amplitude $A_r = 0.5A_c$ ($M = A_r/A_c = 0.5$). A switch is ON while the reference exceeds the carrier. #### i) Waveforms ```text Ac \ /\ / \ / \ / 0.5Ac ---\----/----\----/--- ref |vr| \ / \ / 0 \/ \/ 0 90 180 270 360 deg vo +Vs/2 ____ 0 ___| |____ ___ -Vs/2 |____| 45 135 225 315 ``` - Positive half (reference positive): S1 is ON while the carrier is below the reference, giving $v_o = +V_s/2$. - Negative half: S2 is ON for the same interval, giving $v_o = -V_s/2$. - With a resistive load, when both switches are OFF the current is zero and $v_o = 0$. #### ii) Switching instants The carrier on $0$–$180°$ is $v_c = A_c\left(1 - \dfrac{\theta}{90°}\right)$ for $0 \le \theta \le 90°$, and rises symmetrically after $90°$. Setting $v_c = 0.5A_c$: $$ \theta_1 = 0.5 \times 90° = 45°, \qquad \theta_2 = 180° - 45° = 135° $$ Pulse width $2d = M \times 180° = 0.5 \times 180° = 90°$. | Switch | Turn ON | Turn OFF | |---|---|---| | S1 | $45°$ | $135°$ | | S2 | $225°$ | $315°$ | For example, at 120 Hz ($T = 8.333$ ms) these are: S1 ON at 1.042 ms, OFF at 3.125 ms; S2 ON at 5.208 ms, OFF at 7.292 ms. #### iii) RMS output voltage $$ \begin{aligned} V_{o,rms} &= \sqrt{\frac{1}{\pi}\int_{45°}^{135°}\left(\frac{V_s}{2}\right)^2 d\theta} = \frac{V_s}{2}\sqrt{\frac{2d}{\pi}} \\ &= \frac{V_s}{2}\sqrt{\frac{\pi/2}{\pi}} = \frac{V_s}{2}\times 0.7071 = 0.3536\,V_s \end{aligned} $$ **Answer: switching instants S1: 45° to 135°, S2: 225° to 315°; $V_{o,rms} = 0.3536\,V_s$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With frequency ratio 1, how many pulses appear per half cycle? A. One (single-pulse modulation) B. Two C. Three D. None Answer: A. One carrier triangle per half cycle of the reference gives a single pulse. (Key point: Pulses per half cycle) 2. With M = 0.5, at what angles is S1 turned ON and OFF? A. 0 and 90 degrees B. 45 and 135 degrees C. 30 and 150 degrees D. 90 and 180 degrees Answer: B. The carrier reaches 0.5Ac at 0.5 x 90 = 45 degrees and again at 180 - 45 = 135 degrees. (Key point: Switching instants) 3. At what angles does S2 conduct? A. 45 to 135 degrees B. 180 to 270 degrees C. 225 to 315 degrees D. 270 to 360 degrees Answer: C. S2 conducts for the same 90 degree width in the negative half cycle, centred at 270 degrees. (Key point: S2 instants) 4. What is Vo,rms for this half bridge? A. 0.7071 Vs B. 0.5 Vs C. 0.25 Vs D. 0.3536 Vs Answer: D. Vo = (Vs/2) sqrt(90/180) = 0.3536 Vs. 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