[Boatanchors] Using ac motor start capacitors...

[J. Forster]

> On 1 Oct 2009 at 12:00, boatanchors-request at mailman.qth.net wrote:
>
>> In answering a question posted by Greg Mijal, Marvin Match wrote:
>>
>> <snipped>
>>
>> "BUT, on the next half cycle, the cap will see minus 520 volts in
>> addition to the plus 520 volts from the previous half cycle, so the
>> DC breakdown voltage has to be at least 1040 volts for a cap rated
>> 370 volts AC."
>>
>> This would be for a cap used in voltage doubler power supply
>> service.
>
> No It's not. Drew, you missed my meaning completely.
>
> Look at it this way: Take a cap and charge it to 100 volts. Now while it's
> charged, reverse the terminals and connect it again to a 100 volt supply.
> In
> this case, what voltage would the dielectric in the capacitor see?

========

Essentially infinite current. You HAVE to include a charging resistor in
the circuit or you see singularities in the equations.

The thing is modeled as a capacitor in series w/ a resistor. A femtosecond
after you connect the capacitor to the reversed voltage supply, the
capacitor voltage HAS NOT CHANGED. It is still +V and ALL the doubled
voltage appears across the resistor. Eventually, after many time
constants, the voltage across the capacitor will go to -V and the voltage
across the resistor will drop to zero.

The ONLY case where the capacitor could see more than the peak AC supply
is if there is an inductor and the thing is resonant.

-John

==================


I contend that it "sees" 200 volts, therefore the cap must be able to
withstand 200 volts
> DC. A capacitor rated for 200 volts DC would be safe in an AC application
> for only 71 volts RMS.
>
> Or, in rating a cap for AC service, the dielectric must be able to
> withstand the
> peak-to-peak value of the voltage applied, which is 2.8 times the AC RMS
> voltage, not the peak value.
>
> Marvin
> KA7TPH
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