Single-phase PWM inverter rms and Fourier series

A single-phase PWM inverter controls its output voltage (and reduces low-order harmonics) by chopping each half cycle of the output into one or more pulses whose widths are varied. The most common form in exam answers is multiple-pulse (uniform) PWM, of which single-pulse PWM is the special case $p = 1$; sinusoidal PWM is the same idea with pulse widths varying sinusoidally.

Circuit

     +-------+-----------+
     |       |           |
     |    T1 D1       T3 D3
    Vs       |           |
     |       A---[Load]--B
     |       |    v0     |
     |    T4 D4       T2 D2
     |       |           |
     +-------+-----------+

Operation (multiple-pulse PWM)

1. A rectangular reference of amplitude $A_r$ (at output frequency $f$) is compared with a triangular carrier of amplitude $A_c$ and frequency $f_c$.
2. Number of pulses per half cycle: $p = \dfrac{f_c}{2f}$.
3. While $A_r >$ carrier, the switch pair is ON: T1, T2 in the positive half ($v_0 = +V_s$), T3, T4 in the negative half ($v_0 = -V_s$). Otherwise $v_0 = 0$ (one switch of the pair is kept ON and current freewheels via a diode).
4. Modulation index $M = A_r/A_c$; each pulse width $\delta = \dfrac{M\pi}{p}$. Changing $M$ changes all pulse widths and so the output voltage.

 vc   /\/\/\/\/\/\/\/\  carrier (p = 2 shown)
 Ar  ----------------- reference
 v0  +Vs  _    _
          | |  | |
      0 --' '--' '--.  .--.  .---
                    | |  | |
     -Vs            |_|  |_|
         0       pi        2pi

RMS value of output voltage

Each half cycle contains $p$ pulses of height $V_s$ and width $\delta$:

$$\begin{aligned} V_{0,rms} &= \left[\frac{2p}{2\pi}\int_{\gamma-\delta/2}^{\gamma+\delta/2}V_s^2\,d(\omega t)\right]^{1/2} = \left[\frac{p\,\delta\,V_s^2}{\pi}\right]^{1/2} \\ V_{0,rms} &= V_s\sqrt{\frac{p\,\delta}{\pi}} = V_s\sqrt{M} \end{aligned}$$

For $p = 1$ this gives the single-pulse result $V_s\sqrt{\delta/\pi}$.

Fourier series

Let the $m$th pulse in the positive half cycle be centred at $\gamma_m$ (for uniform PWM, $\gamma_m = \frac{(2m-1)\pi}{2p}$, $m = 1,\dots,p$), with the negative half a mirror image. The wave has half-wave symmetry, so only odd $n$ exist. For one pulse:

$$\begin{aligned} b_{n,m} &= \frac{2}{\pi}\int_{\gamma_m-\delta/2}^{\gamma_m+\delta/2}V_s\sin n\theta\,d\theta = \frac{2V_s}{n\pi}\Big[\cos n\big(\gamma_m - \tfrac{\delta}{2}\big) - \cos n\big(\gamma_m + \tfrac{\delta}{2}\big)\Big] \\ &= \frac{4V_s}{n\pi}\sin n\gamma_m\,\sin\frac{n\delta}{2} \end{aligned}$$

The cosine terms of pulses placed symmetrically about $\pi/2$ cancel ($a_n = 0$). Adding all pulses:

$$v_0(t) = \sum_{n=1,3,5,\dots}\left[\frac{4V_s}{n\pi}\sin\frac{n\delta}{2}\sum_{m=1}^{p}\sin n\gamma_m\right]\sin n\omega t$$

Fundamental:

$$V_{1m} = \frac{4V_s}{\pi}\sin\frac{\delta}{2}\sum_{m=1}^{p}\sin\gamma_m$$

Example

$V_s = 200$ V, $p = 2$, $\delta = 45^\circ$ (pulses centred at 45° and 135°):

$$\begin{aligned} V_{0,rms} &= 200\sqrt{\frac{2\times45^\circ}{180^\circ}} = 141.42\ \text{V} \\ V_{1m} &= \frac{4\times200}{\pi}\sin22.5^\circ\,(\sin45^\circ + \sin135^\circ) = 254.65\times0.3827\times1.4142 = 137.81\ \text{V} \end{aligned}$$

Sinusoidal PWM (note)

If a sinusoidal reference is used instead, the pulse widths $\delta_m$ vary sinusoidally; then $V_{0,rms} = V_s\sqrt{\sum_m \delta_m/\pi}$ and $b_n = \sum_m \frac{4V_s}{n\pi}\sin n\gamma_m\sin\frac{n\delta_m}{2}$. Low-order harmonics are almost eliminated; the dominant harmonics move to around $2p$ (the carrier frequency), which are easily filtered.

Advantages of PWM

• Output voltage control inside the inverter, with a fixed dc input.
• Lower low-order harmonics, smaller filter.
• Drawback: more switchings per cycle, so higher switching loss.