# Bridged Amplifier Capacitor Ripple Current

Discussion in 'Electronic Design' started by [email protected], Jun 26, 2007.

1. ### Guest

In the same way that we have easy conversions for dealing with sine
waves such as RMS Current = I peak / SQRT(2) and I ave = 2 * Ipeak /
PI, I would like to know if anyone can help with this question;

Consider two bridged class B (single supply) amplifiers driving a
supply is fed through a large inductor into a large reservoir
capacitor.

For the sake of argument lets say the peak output current is 4 Amps.
This makes the average current 4 x 2 / PI Amps, and this is the
[constant] current in the inductor. The ripple current in the
capacitor is the difference between this average current and the
instantaneous current being supplied into the load. I get the RMS
ripple current as 0.31 Ipeak (using a spreadsheet) but am not sure how
to do the maths to get this answer. Can anyone help?

2. ### EeyoreGuest

You're using a switchmode power supply ?

Graham

3. ### EliotGuest

I'm not sure what you are getting at - the inductor is "large" and so
carries a constant current.

4. ### EeyoreGuest

I'm asking a question is what I'm getting at, because I haven't seen such a
power supply ever for an audio amplifier.

It'll have to be very large indeed for that.

Is this some hypothetical scenario ? It doesn't sound like a real world one.

Graham

5. ### EliotGuest

It is real world. It's an RF amplifier. By "large" I mean the
inductor has a value so the current barely changes over 1 cycle. It's
µH's.

6. ### EeyoreGuest

If it's that large then surely there'll be next to no ripple current in the caps.

Graham

7. ### EliotGuest

I think you have misunderstood this somewhat;
The inductor supplies the average current,