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Three-phase SPWM inverter switching instants

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.

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?

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

      +----+--------+--------+
      |    |        |        |
 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: vra=Arsin⁡ωtv_{ra} = A_r\sin\omega t, vrb=Arsin⁡(ωt−120∘)v_{rb} = A_r\sin(\omega t - 120^\circ), vrc=Arsin⁡(ωt−240∘)v_{rc} = A_r\sin(\omega t - 240^\circ).
  • Carrier: triangle of peak AcA_c, frequency fc=mffrf_c = m_f f_r (mfm_f odd multiple of 3).
  • Leg a: vra>vc⇒v_{ra} > v_c \Rightarrow S1 ON, vaO=+Vdc/2v_{aO} = +V_{dc}/2; vra<vc⇒v_{ra} < v_c \Rightarrow S4 ON, vaO=−Vdc/2v_{aO} = -V_{dc}/2.
  • Line voltage vab=vaO−vbOv_{ab} = v_{aO} - v_{bO} is a train of pulses of ±Vdc\pm V_{dc} and 0 whose widths vary sinusoidally.
  • Fundamental (linear range ma=Ar/Ac≤1m_a = A_r/A_c \le 1): V^aO1=maVdc/2\hat V_{aO1} = m_a V_{dc}/2, VL1,rms=0.612 maVdcV_{L1,rms} = 0.612\,m_a V_{dc}.
 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 tkt_k:

vc(t)=−Ac+4AcTc(t−tk),tk≤t≤tk+Tc2v_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 tst_s is the root of

Arsin⁡ωts=−Ac+4AcTc(ts−tk)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 tst_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 tkt_k and held. Then the crossing is found from a straight-line equation:

duty of S1: dk=12(1+masin⁡ωtk)ton,S1=dkTc,ton,S4=(1−dk)Tc\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, ma=0.8m_a = 0.8 at ωtk=90∘\omega t_k = 90^\circ gives d=0.9d = 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 tdt_d (both OFF) is inserted at every transition to avoid a shoot-through short circuit. During tdt_d the load current flows through a diode.

The same procedure applies to legs b and c using vrbv_{rb}, vrcv_{rc}, which gives three balanced pole voltages.

✓ Answer checked by Claude Opus 5.5 (high effort)

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