3
$\begingroup$

I'm building a magnetic loop antenna and was wondering if the voltage ratings on the vaccuum variable capacitors I have are:

  • RMS
  • Peak to Peak
  • Peak
  • other?

I have been looking at this wonderful magnetic loop calculator to help choose a capacitor and it gives the required cap voltage in RMS.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Remember, RMS is an invention of people. The capacitor will see the peak to peak voltage ranges of the applied signal. $\endgroup$
    – K7PEH
    Jul 10, 2016 at 15:06

1 Answer 1

Reset to default
1
$\begingroup$

You raise an interesting point; intuitively, what I'd have thought is that the peak voltage matters – because discharge through the dielectric doesn't happen for any reason but an electrical field strength (at least for vacuum as dielectric, which can have no electric polarization of any kind, being not a material) above the Schwinger limit which you will never reach even remotely.

Now, necessarily, your cap doesn't hang in perfect vaccuum, but let's assume it happens in a quality of vacuum (gas pressure) that guarantees a breakdown value at least equivalent to that of air (see Paschen's Law), but with negligible polarization losses; then you'd still have around 2.5 MV/m breakdown voltage (probably more).

Notice how I precluded polarization effects – this is my guess why these caps use vacuum instead of just being filled with some inert gas: Without any material to be subject to changing electrical field strength, you can't have molecules/atoms align their electric momentum with the field, which can, when quickly enough changing that field, reduce the breakdown voltage of a capacitor, and lead to energy losses and thus, heating of the dielectric.

For considerations of how much energy will get lost in a material, the RMS voltage is a good measure – same goes for the breakdown voltage reduction.

However, as a vacuum cap sees neither of these effects under "earthly" conditions, RMS is not what you need to know – you really just want to know how much voltage you might ever see, because, as soon as that voltage crosses breakdown voltage, there will be an arc within your cap – and that might become a self-sustaining arc, considering that its heat will possibly vaporize and splutter some of the conductor plates into the vacuum, making it a box of conductive plasma.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Quantum effects are not really relevant here. We're talking about a capacitor, which has conductive plates in addition to the vacuum. Long before the vacuum itself becomes nonlinear, the field strength will be more than enough to pull electrons directly from one plate to the other. A current will flow, but probably not an arc. X-rays will probably be generated. I would expect this to start happening at about 10^10 V/m -- i.e., a field of a few volts across the width of an atom. $\endgroup$
    – Dave Tweed N3AOA
    Jul 10, 2016 at 22:45
  • $\begingroup$ @DaveTweed exactly, that's why I said you'll never hit Schwinger limit. You're right, I simply forgot about the fact that electrons might simply spontaneously emit from the metal surfaces (and that whilst thinking about my tube radio – 'doh!). What's the mechanism behind that? I'd simply assume it'd be temperature/phonons "kicking" them out? How does one estimate at which rate that happens? $\endgroup$
    – Marcus Müller
    Jul 11, 2016 at 8:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .