Formula calculate capacitance for PSU, it's correct?
Say total current is 2x100mA = 200mA. C must be > 200mA*10ms/10mV = 200mF! That's 200 000uF
Where 10mV is max ripple.
TIA
Felipe
Say total current is 2x100mA = 200mA. C must be > 200mA*10ms/10mV = 200mF! That's 200 000uF
Where 10mV is max ripple.
TIA
Felipe
The calculation looks fine to me, if theoretical reservoir cap value is what you are after. Not the way to design a real PSU, though.
If you really need only 10mV ripple, consider using either multiple stages of filtering, like say C-R-C, or C-L-C, or use an active device to produce a regulated power supply.
This level of ripple is too much to ask for, from a single stage filter, so you're getting a nonsense value for the filter capacitor.
This level of ripple is too much to ask for, from a single stage filter, so you're getting a nonsense value for the filter capacitor.
A close enough approximation to do rule of thumb calculations for a full-wave rectifier:
V = I/(2fC)
where
V is the peak to peak ripple voltage
I is the current draw on the power supply
f is the source (line) frequency of the AC power
C is the capacitance
Which for the case you present indicates a capacitance of approx 200,000 uF.
But I also agree that there are better and more cost effective ways at this current draw level to get better performance than using straight capacitance. I’d look at using A C-L-C pi filter. For less money you could get the ripple down in the 1-2 mV range. Another worthwhile approach would be to use a capacitance multiplier, but while very cost effective it is more complicated to implement.
Cheer,
Terry
V = I/(2fC)
where
V is the peak to peak ripple voltage
I is the current draw on the power supply
f is the source (line) frequency of the AC power
C is the capacitance
Which for the case you present indicates a capacitance of approx 200,000 uF.
But I also agree that there are better and more cost effective ways at this current draw level to get better performance than using straight capacitance. I’d look at using A C-L-C pi filter. For less money you could get the ripple down in the 1-2 mV range. Another worthwhile approach would be to use a capacitance multiplier, but while very cost effective it is more complicated to implement.
Cheer,
Terry
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That's an insane amount of filtering for one stage.
I don't have city water. I have a well but this month it could go dry. I have some damp mud. Can I filter it good enough to drink, to wash the baby, in a 1-stage filter? No, I would do it in many stages: log/frog-screen, leaf-screen, gravel screen, sand screen, silt screen, a 50u paper filter and a 5u string filter.
As a pencil-mark, your first filter cap can have 5% ripple. Or considering the low-low cost of e-caps today, maybe 1%. On a 100V supply, 5V to 1V ripple. Each subsequent filter stage can knock-down the ripple by a factor of 10:1 to 30:1. You can't realistically do a lot more than 30:1 in each stage because parasitic resistance typically exceeds calculated capacitive reactance.
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