E
[email protected]
- Jan 1, 1970
- 0
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
load, the load is connected between each amplifiers output. The power
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?
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
load, the load is connected between each amplifiers output. The power
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?