Steve is right on the money here. Many capacitor filters will cause peak
currents about 5 times average but the current only flows 1/5 of the cycle.
Since heating is related to current squared, the total heating in this case
would
approximate 5 times 5 or 25 times dived by 5 for a total of 5 times as much
heat compared to a resistive load.
The high peak current also causes 5 times the voltage drop at the peak of
the cycle. Since we are trying to charge the filter to the peak voltage this
means the voltage sag is also five times more than with a resistive load. More
copper is needed to compensate.
If the power supply uses really large capacitance values the situation is
even worse than this!
The Dahl transformers seem to handle the rated load so they must have the
correct recipe.
73,
Gerald K5GW
In a message dated 2/9/2006 8:45:41 A.M. Central Standard Time,
g8gsq@eltac.co.uk writes:
Partain, Chuck wrote:
> I have a few questions, I'm yet to fully understand all that needs to be
> known and never will for that matter but!
>
> In the design of HV transformers, what's the difference between
> designing a transformer for use
> With a capacitor filter and a choke filter?
Allowances for much higher peak current with capacitor input - the high
peaks result in higher rms current, which is the value you need to use
when looking at the heating effects. Many people use (sqrt2)*dc current,
which is likely predict heating effects around 1/3 of real. You can
play with examples and see the effects with the Duncan Amps PSU freeware
designer prog.
In my experience, transformer winders don't fully appreciate the
consequences of capacitor input peak currents, and the 'standard'
figures they use usually underestimate the heating effects.
Steve
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