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Single-pulse PWM inverter control and rms

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.

Explain the operation of single PWM techniques for inverter control and therefore determine the rms value of output voltage.

Single-pulse width modulation controls the inverter output voltage by giving one pulse per half cycle whose width δ\delta is varied; the pulse is placed symmetrically about π/2\pi/2 (and about 3π/23\pi/2 for the negative half).

Operation

 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 ArA_r is compared with a triangular carrier of amplitude AcA_c and frequency 2f2f (ff = 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 δ=Mπ\delta = M\pi, where the modulation index M=Ar/AcM = A_r/A_c. Varying ArA_r from 0 to AcA_c varies δ\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 π−δ2\frac{\pi-\delta}{2} to π+δ2\frac{\pi+\delta}{2} with amplitude VsV_s:

V0,rms=[22π∫(π−δ)/2(π+δ)/2Vs2 d(ωt)]1/2=[Vs2π⋅δ]1/2=Vsδπ=VsM\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

v0=∑n=1,3,5,…4Vsnπsin⁡nδ2sin⁡nωt,V1,rms=0.9Vssin⁡δ2v_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}
  • δ=180∘\delta = 180^\circ: V0,rms=VsV_{0,rms} = V_s (square wave).
  • δ=120∘\delta = 120^\circ: V0,rms=0.816VsV_{0,rms} = 0.816V_s and the 3rd harmonic is eliminated (sin⁡180∘=0\sin 180^\circ = 0).
  • δ=90∘\delta = 90^\circ: V0,rms=0.707VsV_{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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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