Calculate the ripple factor of a three-phase full-wave rectifier circuit, both analytically and using MATLAB.

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A three-phase half-wave rectifier and its output voltage are shown below

[latex]\begin{aligned}ν_A(t)=V_M \sin \omega t \ ν_B(t)=V_M \sin (\omega t -2π/3) \ ν_C(t)=V_M \sin (\omega t + 2π/3) \end{aligned}[/latex]

By symmetry, the rms voltage over the interval from 0 to T/12 will be the same as the rms voltage over the whole interval. Over that interval, the output voltage is:

[latex]\begin{aligned}& ν(t)=ν_C(t)-ν_B(t)=V_M \sin \left\lgroup\omega t+\frac{2 \pi}{3} ight group-V_M \sin \left\lgroup\omega t-\frac{2 \pi}{3} ight group \& ν(t)=V_M\left\lgroup\sin \omega t \cos \frac{2 \pi}{3}+\cos \omega t \sin \frac{2 \pi}{3} ight group-V_M\left\lgroup\sin \omega t \cos \frac{2 \pi}{3}-\cos \omega t \sin \frac{2 \pi}{3} ight group \& ν(t)=V_M\left\lgroup2 \cos \omega t \sin \frac{2 \pi}{3} ight group=\sqrt{3} \cos \omega t\end{aligned}[/latex]

Note that the period of the waveform is [latex]T=2 \pi / \omega, ext { so } T / 12 ext { is } \pi / 6 \omega[/latex] . The average voltage over the interval from 0 to T/12 is

[latex]\begin{aligned}& V_{D C}=\frac{1}{T} \int ν(t) d t=\frac{6 \omega}{\pi} \int_0^{\pi / 6 \omega} \sqrt{3} V_M \cos \omega t \ d t=\left.\frac{6 \sqrt{3}}{\pi} V_M \sin \omega t ight|_0 ^{\pi / 6 \omega} \& V_{D C}=\frac{3 \sqrt{3}}{\pi} V_M=1.6540 \ V_M\end{aligned}[/latex]

The rms voltage is

[latex]V_{ rms }=\sqrt{\frac{1}{T} \int ν(t)^2 d t}=\sqrt{\frac{6 \omega^{\pi / 6 \omega}}{\pi} \int_0^\pi 3 V_M^2 \cos ^2 \omega t d t}[/latex]

[latex]\begin{aligned}& V_{ rms }=\sqrt{\left.\frac{18 \omega}{\pi} V_M{ }^2\left\lgroup\frac{1}{2} t+\frac{1}{4 \omega} \sin 2 \omega t ight group ight|_0 ^{\pi / 6 \omega}} \& V_{ rms }=V_M \sqrt{\frac{18 \omega}{\pi}\left\lgroup\frac{\pi}{12 \omega}+\frac{1}{4 \omega} \sin \frac{\pi}{3} ight group}=V_M \sqrt{\frac{3}{2}+\frac{9 \sqrt{3}}{4 \pi}}=1.6554 \ V_M\end{aligned}[/latex]

The resulting ripple factor is

[latex]r=\sqrt{\left\lgroup\frac{V_{ rms }}{V_{ DC }} ight group^2-1} imes 100 \%=\sqrt{\left\lgroup\frac{1.6554 \ V_M}{1.6540 \ V_M} ight group^2-1} imes 100 \%=4.2 \%[/latex]

The ripple can be calculated with MATLAB using the ripple function developed in the text. We must right a new function fullwave3 to simulate the output of a three-phase half-wave rectifier. This output is just the largest voltage of [latex]ν_A(t), ν_B(t), ext { and } ν_C(t)[/latex] at any particular time. The function is shown below:

Question S1.2: Calculate the ripple factor of a three-phase full-wave recti...

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