Single-phase inverter fundamental output voltage

A single-phase inverter converts dc into ac by alternately connecting the load to the dc source in opposite directions. With a resistive load and square-wave switching, the output is a square wave of amplitude $V_s$ (full bridge) or $V_s/2$ (half bridge), and the current has the same shape.

Circuit (full bridge)

     +-------+-----------+
     |       |           |
     |      S1          S3
    Vs       |           |
     |       A---[ R ]---B
     |       |    v0     |
     |      S4          S2
     |       |           |
     +-------+-----------+

Operation

• $0 < t < T/2$: S1 and S2 ON. Point A is at $+V_s$ relative to B: $v_0 = +V_s$, $i_0 = V_s/R$.
• $T/2 < t < T$: S3 and S4 ON. $v_0 = -V_s$, $i_0 = -V_s/R$.
• S1–S4 (and S3–S2) of a leg must never be ON together. Output frequency $f = 1/T$ is set by the gating rate.
• With a pure R load the current reverses exactly with the voltage, so the feedback diodes do not conduct.
• Half bridge: two switches and a centre-tapped source give $v_0 = \pm V_s/2$.

 v0  +Vs |________
         |        |
       0 |--------+--------- t
         |        |________
     -Vs  S1,S2 ON  S3,S4 ON
         0       T/2       T

RMS value

$$V_{0,rms} = \left[\frac{2}{T}\int_0^{T/2}V_s^2\,dt\right]^{1/2} = V_s$$

Fundamental component (Fourier analysis)

The wave is odd with half-wave symmetry, so $a_0 = 0$, $a_n = 0$ and only odd sine terms remain:

$$\begin{aligned} b_n &= \frac{2}{\pi}\int_0^{\pi}V_s\sin n\theta\,d\theta = \frac{2V_s}{n\pi}\big[-\cos n\theta\big]_0^{\pi} = \frac{2V_s}{n\pi}(1 - \cos n\pi) \\ b_n &= \frac{4V_s}{n\pi}\ (n\ \text{odd}), \qquad b_n = 0\ (n\ \text{even}) \\ v_0(t) &= \sum_{n=1,3,5,\dots}\frac{4V_s}{n\pi}\sin n\omega t \end{aligned}$$

For $n = 1$:

$$\begin{aligned} V_{1m} &= \frac{4V_s}{\pi} = 1.273\,V_s \\ V_{1,rms} &= \frac{4V_s}{\sqrt2\,\pi} = 0.9003\,V_s \end{aligned}$$

For a half bridge, replace $V_s$ by $V_s/2$: $V_{1m} = 2V_s/\pi$, $V_{1,rms} = 0.45\,V_s$.

Distortion: $THD = \sqrt{V_{rms}^2 - V_1^2}/V_1 = \sqrt{1 - 0.9003^2}/0.9003 = 48.3\%$. Example: $V_s = 220$ V gives $V_{1,rms} = 198.1$ V.