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Single-phase square-wave inverter rms and fundamental

Short answer

A full-bridge inverter switches the load between +Vs and -Vs each half-cycle, producing a square wave. The rms output voltage is Vs. The fundamental component has peak 4Vs/π and rms value 0.9003 Vs.

Explain the operation of single phase square wave inverter. Derive the expression for rms value of output voltage and fundamental component of output voltage.

A single-phase square-wave inverter switches the load alternately to +Vs+V_s and −Vs-V_s for equal half periods, giving a square-wave ac output whose frequency is set by the switching rate and whose amplitude is fixed by VsV_s.

Circuit and operation (full bridge)

     +-------+-----------+
     |       |           |
     |    S1 D1       S3 D3
    Vs       |           |
     |       A---[Load]--B
     |       |    v0     |
     |    S4 D4       S2 D2
     |       |           |
     +-------+-----------+
  • 0<t<T/20 < t < T/2: S1, S2 ON, v0=+Vsv_0 = +V_s.
  • T/2<t<TT/2 < t < T: S3, S4 ON, v0=−Vsv_0 = -V_s.
  • With an inductive load, current continues after each reversal through the feedback diodes (D3, D4 or D1, D2) until it changes sign.
  • Half bridge version (two switches, centre-tapped source): v0=±Vs/2v_0 = \pm V_s/2.
 v0  +Vs |________
         |        |
       0 |--------+--------- t
         |        |________
     -Vs   S1,S2     S3,S4
         0       T/2       T

RMS value of output voltage

V0,rms=[2T∫0T/2Vs2 dt]1/2=VsV_{0,rms} = \left[\frac{2}{T}\int_0^{T/2}V_s^2\,dt\right]^{1/2} = V_s

(Half bridge: Vs/2V_s/2.)

Fundamental component

The wave is odd with half-wave symmetry, so a0=an=0a_0 = a_n = 0:

bn=2π∫0πVssin⁡nθ dθ=2Vsnπ(1−cos⁡nπ)=4Vsnπ  (n odd),0  (n even)v0(t)=4Vsπ[sin⁡ωt+13sin⁡3ωt+15sin⁡5ωt+…]\begin{aligned} b_n &= \frac{2}{\pi}\int_0^{\pi}V_s\sin n\theta\,d\theta = \frac{2V_s}{n\pi}(1 - \cos n\pi) \\ &= \frac{4V_s}{n\pi}\ \ (n\ \text{odd}), \qquad 0\ \ (n\ \text{even}) \\ v_0(t) &= \frac{4V_s}{\pi}\Big[\sin\omega t + \frac{1}{3}\sin3\omega t + \frac{1}{5}\sin5\omega t + \dots\Big] \end{aligned} V1m=4Vsπ=1.273 VsV1,rms=4Vs2 π=0.9003 Vs\begin{aligned} V_{1m} &= \frac{4V_s}{\pi} = 1.273\,V_s \\ V_{1,rms} &= \frac{4V_s}{\sqrt2\,\pi} = 0.9003\,V_s \end{aligned}

Half bridge: V1,rms=2Vs2 π=0.45VsV_{1,rms} = \dfrac{2V_s}{\sqrt2\,\pi} = 0.45V_s.

Performance

  • nnth harmonic: Vn=V1/nV_n = V_1/n (HF3_3 = 33.3%).
  • THD=V0,rms2−V1,rms2/V1,rms=1−0.90032/0.9003=48.3%THD = \sqrt{V_{0,rms}^2 - V_{1,rms}^2}/V_{1,rms} = \sqrt{1 - 0.9003^2}/0.9003 = 48.3\%.
  • Output voltage cannot be varied without changing VsV_s; PWM is used for that.

Example: Vs=230V_s = 230 V gives V0,rms=230V_{0,rms} = 230 V and V1,rms=207.1V_{1,rms} = 207.1 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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