1
\$\begingroup\$

I am designing a power management unit (PMU), which takes the power from an alternator (alternator output: three phases, 30-85VAC, 35-130Hz, Max current: 20A).

I want to reduce the ripple using capacitors (Between the rectifier and a buck converter).

enter image description here

I would like to reduce the ripple to 500 mV. I found a formula for three phase rectifiers, but I am not sure if it is correct, because the value is too high.

The formula I found is:

C = I/6fVpp, where 'I' is the current, 'f' is the line frequency, and 'C' is the filter capacitance.

The most unfavorable situation is with the lowest frequency and maximum current:

C = 18A/(6·35Hz·0.5V) = 0.171 F = 171 mF.

Does anyone know if this formula is correct?

Thank you.

\$\endgroup\$
5
  • \$\begingroup\$ You should add a serial inductor just before the capacitors ... also to limit inrush current. \$\endgroup\$
    – Antonio51
    Commented Nov 21, 2022 at 11:47
  • 3
    \$\begingroup\$ I don't understand, you have there LTspice, you've created the schematic, you have calculated the value, what's stopping you from pressing "run" and test for yourself? Here are a few tips: if you mean to have the source with a positive sequence then you need 0,-120,120 degrees for phases, you don't need 3 caps in parallel (just use one with equivalent value), it will help to use either a .model for diodes or select one from the database, and, finally, use a current source as the load with the flag load. \$\endgroup\$ Commented Nov 21, 2022 at 11:48
  • 2
    \$\begingroup\$ Also, the ground on the sources side should not be tied directly to the sources -- use a 1Meg (or so) valued resistor to ground. If you'll plot the current you'll see why. Or use the same approach for the load side, but not with a common ground. \$\endgroup\$ Commented Nov 21, 2022 at 11:52
  • \$\begingroup\$ In other words, with two grounds where you have them, diodes are shorting out the supplies. \$\endgroup\$
    – Andy aka
    Commented Nov 21, 2022 at 11:56
  • \$\begingroup\$ Ok, thank you. But I put the image just to get the idea. \$\endgroup\$
    – Pablo
    Commented Nov 21, 2022 at 11:56

1 Answer 1

2
\$\begingroup\$

In any case, just be aware of inrush current, overvoltage's and some other things.

Here are 3 pictures which show these "problems".

enter image description here

enter image description here

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.