I have a single phase 5KW generator connected to a diesel engine capable of producing approx 3500 Watts of power

I need to convert the 240V AC to DC in order to send it to my solar charge controller that can accept anything up to 600V DC. It then converts it to 48V DC to charge my battery bank. The generator is 300 feet away from the battery bank and charge controller

I understand that I need to rectify it to get DC, but I also need to smooth the ripples out as much as possible for the charge controller to be happy first.

The charge controller and battery bank are connected to a grid tied inverter. I will be exporting the power to the grid. My generator has no ability to sync to grid.

My question is what components do I need (capacitor values) in order to reduce the ripples? I'd also like to filter out as much AC noise as possible

I appreciate any insight you may have

Thanks for your time


  • \$\begingroup\$ why 380V? ~340V would be much easier... \$\endgroup\$ – PlasmaHH Nov 27 '16 at 22:56
  • \$\begingroup\$ The voltage does not really matter. I need it high enough to minimize losses on the 300' run and lower than the 600v that the charge controller needs - 340V would be just fine! \$\endgroup\$ – John Nov 27 '16 at 23:38
  • \$\begingroup\$ Are you sure, that you need very flat voltage? You could simply use a diode bridge to rectify to approx 310VDC and then see what happen. Why filtering AC as much as possible? \$\endgroup\$ – Marko Buršič Nov 27 '16 at 23:46
  • \$\begingroup\$ Marko, My charge controller cost me $1200! I'd rather be safe than sorry.. and since it is made for solar panels, I assume its expecting clean voltage. This is my charge controller - link - Conext MPPT 80 \$\endgroup\$ – John Nov 27 '16 at 23:54
  • \$\begingroup\$ PV array is not a single phase generator, second: there is no economical aspect to run diesel engine to produce mains electricity, and if so that would be a fraud, since your distributor is paying the price for clean energy, not for combustion energy. \$\endgroup\$ – Marko Buršič Nov 28 '16 at 10:22

The formula for a capacitor is I = C dU/dt . If you want the ripple to be less than say 10V all the symbols are known allowing you to calculate the capacitor. Because I don't have a calculator here, 3500W/about 300V is about 10 A. DT is 10ms, so you get: C = I dt/dU = 10A.10ms/10V = 10mF = 10000uF. Quite a lot for such a large voltage.....


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