# Fourier Series Expansion of the phase current of a three phase full wave bridge rectifier

I am really struggling with this homework problem where I am asked to calculate the harmonic components as a ratio of the first fundamental harmonic of the phase current of a three phase rectifier with an RC parallel load.

Obviously with the capacitor in parallel with the load I figured there will be some phase shift on the phase voltages when their respective diodes are forward biased but I thought I could still potentially find the Fourier coefficients with the knowledge that the line to line voltages on a 3 phase delta wye connected rectifier like the one shown are as follows:

$$va(t)= \sin\omega t$$ $$vb(t) = \sin\omega t - \frac{2\pi}{3}$$ $$vc(t) = \sin\omega t - \frac{4\pi}{3}$$

My next line of reasoning was to follow this section of the course textbook explaining that the phase current through the primary phase a is active when the line to line voltage of phase A and B is highest for a given period followed by another period wherein the phase to phase voltage between phases A and C is highest. This is shown in the bottom plot below:

So I've tried to come up with the fourier expansion but it is proving difficult. I calculated that the magnitude of the signal should be 28.64 (voltage divided by impedance) using values in the first image.

Sorry guys I know I could have written this all in mathjax but I was low on time and inexperienced with this site and need to get my point across. So assuming that I am correct with the expression in the photo about the fourier series expansion, how do you evaluate this since it is 2 sine functions multiplied? Also I think the 28.64 is wrong because on the simulation the phase current has a magnitude of a little over 10A. I have no idea how because I got this by dividing the phase voltage by the impedance.

If anyone has ever worked a problem like this and can provide a hint, a link to a useful resource or anything I'd really appreciate it.

Simon.

• If you have a parallel RC load, then the current will no longer be the clean version you show in your picture, but the so-called "rabbit's ears". Still, in general, due to the dynamics, parallel loads are quite difficult to calculate analytically, even for single-phase systems, so people usually resort to either simulators, or oscilloscopes. – a concerned citizen Nov 1 '18 at 12:05

$$\i_a = \frac{2\cdot \sqrt{3}}{\pi}I_d \left( sin(\omega t - \psi) - \frac{1}{5}sin5(\omega t - \psi) \frac{1}{7}sin7(\omega t - \psi) \right) + \frac{1}{11}sin11(\omega t - \psi) + \frac{1}{13}sin13(\omega t - \psi) - \frac{1}{17}sin17(\omega t - \psi) - \frac{1}{19}sin19(\omega t - \psi) + ...\$$
This produces an RMS value of: $$\\sqrt{\frac{2}{3}}Id \approx 0.816\cdot I_d\$$