--- title: "Three-phase SPWM inverter switching instants" subject: "Power Electronics" chapter: 5 key: "semester_7/power_electronics:spwm-switching-instants" source: https://www.bctengineer.com/q/semester_7/power_electronics/spwm-switching-instants site: BCT Engineering --- # Three-phase SPWM inverter switching instants Power Electronics · Chapter 5: Inverter - Asked in: 2076 Chaitra, 8 marks; 2073 Chaitra, 8 marks; 2070 Asar, 8 marks (asked 3 times) - Video: [SPWM (Sinusoidal PWM) Three Phase Inverters](https://www.youtube.com/watch?v=NXAAhArHang) — Ozan Keysan - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/spwm-switching-instants - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-5 ## Question Explain the operation of a three phase Sinusoidal PWM inverter with neat circuit diagram and associated waveforms. How switching instants for inverter switch pair of a phase are determined? ## Short answer In three-phase SPWM, each leg switches when its sinusoidal reference crosses a common triangular carrier. Switching instants are found either by solving the transcendental crossing equation (natural sampling) or by sampling the reference once per carrier period to compute duty cycle d=0.5(1+ma sin ωt_k) (regular sampling). Dead time is added to avoid shoot-through. ## Answer A three-phase sinusoidal PWM inverter is a six-switch bridge in which each leg is switched by comparing its own sinusoidal reference (120° apart from the others) with a common triangular carrier. The switching instants are the crossing points of the reference and the carrier. #### Circuit ```text +----+--------+--------+ | | | | Vdc/2 S1 D1 S3 D3 S5 D5 --- | | | | O +-a +--b +--c | --- | | | | Vdc/2 S4 D4 S6 D6 S2 D2 | | | | +----+--------+--------+ a, b, c -> 3-phase load ``` #### Operation - References: $v_{ra} = A_r\sin\omega t$, $v_{rb} = A_r\sin(\omega t - 120^\circ)$, $v_{rc} = A_r\sin(\omega t - 240^\circ)$. - Carrier: triangle of peak $A_c$, frequency $f_c = m_f f_r$ ($m_f$ odd multiple of 3). - Leg a: $v_{ra} > v_c \Rightarrow$ S1 ON, $v_{aO} = +V_{dc}/2$; $v_{ra} < v_c \Rightarrow$ S4 ON, $v_{aO} = -V_{dc}/2$. - Line voltage $v_{ab} = v_{aO} - v_{bO}$ is a train of pulses of $\pm V_{dc}$ and 0 whose widths vary sinusoidally. - Fundamental (linear range $m_a = A_r/A_c \le 1$): $\hat V_{aO1} = m_a V_{dc}/2$, $V_{L1,rms} = 0.612\,m_a V_{dc}$. ```text vc /\ /\ /\ /\ /\ /\ / \/ \/ \/ \/ \/ \ vra .--''''--. ------'----------'---------- vaO _ ___ _____ ___ _ + _| || | | || |_ - (S1 ON when vra>vc) ``` #### How the switching instants of one leg are found **1. Natural sampling (analogue comparison).** In one carrier half-period the carrier is a straight line. For a rising slope starting at $t_k$: $$ v_c(t) = -A_c + \frac{4A_c}{T_c}(t - t_k), \qquad t_k \le t \le t_k + \frac{T_c}{2} $$ The switching instant $t_s$ is the root of $$ A_r\sin\omega t_s = -A_c + \frac{4A_c}{T_c}(t_s - t_k) $$ which is transcendental and is solved numerically (or by an analogue comparator in hardware). At $t_s$ the leg changes state: S1 OFF → S4 ON on the rising slope (as the carrier rises above the reference), and S4 OFF → S1 ON on the falling slope. **2. Regular (uniform) sampling (used in microcontrollers/DSP).** The reference is sampled once per carrier period at $t_k$ and held. Then the crossing is found from a straight-line equation: $$ \begin{aligned} \text{duty of S1: } d_k &= \frac{1}{2}\left(1 + m_a\sin\omega t_k\right) \\ t_{on,S1} &= d_k T_c, \qquad t_{on,S4} = (1-d_k)T_c \end{aligned} $$ The ON pulse of S1 is centred in the carrier period (symmetric sampling). For example, $m_a = 0.8$ at $\omega t_k = 90^\circ$ gives $d = 0.9$: S1 is ON for 90% of that carrier period. **3. Complementary gating with dead time.** The pair in a leg (S1–S4) receives complementary signals; a short dead time $t_d$ (both OFF) is inserted at every transition to avoid a shoot-through short circuit. During $t_d$ the load current flows through a diode. The same procedure applies to legs b and c using $v_{rb}$, $v_{rc}$, which gives three balanced pole voltages. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the switching signals of a three-phase sinusoidal PWM inverter generated? A. Each leg's own sinusoidal reference is compared with a common triangular carrier B. A common sine is compared with three carriers C. A rectangular reference is compared with a carrier at twice the output frequency D. Switches are fired every 60° in sequence Answer: A. Three references 120° apart are each compared with one common triangular carrier; crossings give the switching instants. (Key point: Principle of SPWM) 2. In leg a, when v_ra > v_c, what happens? A. S4 is ON and v_aO = -Vdc/2 B. Both S1 and S4 are ON C. S1 is ON and v_aO = +Vdc/2 D. Both are OFF and the output is zero Answer: C. When the reference exceeds the carrier S1 is ON, giving +Vdc/2; otherwise S4 is ON. (Key point: Leg switching rule) 3. With regular sampling, m_a = 0.8 at wt_k = 90°. What is the duty of S1 in that carrier period? A. 0.5 B. 0.8 C. 0.1 D. 0.9 Answer: D. d = 0.5(1 + m_a sin wt_k) = 0.5(1 + 0.8) = 0.9. (Key point: Regular sampling duty) 4. Why are the natural-sampling switching instants found numerically or by an analogue comparator? A. The carrier is not linear B. The equation A_r sin(wt) = carrier line is transcendental C. The reference is rectangular D. Dead time prevents a direct solution Answer: B. Equating a sine with a straight-line carrier segment is transcendental and has no closed-form root. 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