Re: CAPACITOR CHARGE RATE
Subject: Re: CAPACITOR CHARGE RATE
From: mark.graalman-at-mediccom.norden1-dot-com (Mark Graalman)
Date: Thu, 23 Mar 1995 17:00:00 -0500
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Resent-Date: Thu, 23 Mar 1995 19:05:51 -0500
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Te> * Original msg to: Esondrmn-at-aol-dot-com
Te> Quoting Ed Sonderman:
ES> In a recent communication, you said a 12kv 120ma 60 cycle n
ES> power supply will effectively charge a .026mfd capacitor.
ES> I've been curious about this and was doing some math and I
ES> came up with a different answer. Maybe others will check
ES> this out and comment.
Te> Oh lord Shaitan help me!
Te> You guys know that math and I don't get along. Just the thought
Te> of meandering through this makes my head start to hurt. Who is
Te> the resident MATH expert anyway?
ES> I calculate as follows: The available time to charge the
ES> capacitor is one quarter of the 60 cycle waveform (peak
Te> My understanding was that current was delivered from zero voltage
Te> point to zero voltage point on the 60 cycle waveform. That covers
Te> one-half of the waveform as opposed to 1/4 of the waveform that
Te> you calculated from. Again, I could be wrong, but I believe that
Te> this is another transmission line problem that Mark Graalman
Te> would be the resident expert consultant on. I figured that the
Te> voltage peak lagged behind the current peak in the line charging
Te> the capacitor.
MG> Ed, you are correct. actually, a 1 milliamp source will charge any size
capacitor, just as a small battery charger will charge a large
automotive battery. It will just take longer, if the source can't
charge in the required time frame you simply wind up with a average
voltage level that is much less than would be possible had the source
charged at a faster rate.
ES> This is .0042 seconds. To achieve 97% charge on a capacitor
ES> we need 3 time constants. The thevenan equivalent of a 12kv
ES> 120 ma transformer must look like an infinite current 12kv
ES> supply with a 100k ohm resister in series. We know we must
ES> achieve full charge (97%) in .0042 sec and it takes 3 time
ES> constants to do this so we have .0042 / 3 = .0014 sec for
ES> one time constant. We know one time constant = RC and C =
ES> one time constant / R. Solving for C we have .0014 sec /
ES> 100k ohms = .014mfd. Thus, a 12kv 120ma transformer will
ES> fully (97%) charge up to a .014 mfd capacitor.
Te> Oh boy. You lost me after "This is .0042 seconds."
Te> My tried and true (in practical applications and circuits) Tesla
Te> coil calculator states simply that at 60 cycles a 12kv 120ma
Te> power supply will fully charge a .0265 mfd capacitor. This is not
Te> to say that this is a 100% correct answer, but it works OK.
MG> The formula works by accident, the imped. of a transformer is a
CONSTANTLY CHANGING attribute, the current level and voltage level are
different at every point on the 60 hz. sinewave. At the begining of the
capacitor charging curve (0 degrees) the current is maximum and the
voltage level is minimum, at the top of the sinewave (90 degrees) the
voltage is maximum and the charging current is minimum. So for the sake
of exhample, at the bottom of the curve you may have a neon delivering
60 ma. at 1 volt and at the creast of the curve it may be delivering
15000 volts at 1 ma. So what do call a match!?!? The reactance of the
capacitor never changed, the transformer voltage and current was all
over the place!!
ES> If this math is correct, I calculate that we need a 12kv
ES> 171ma transformer to fully charge a .026mfd capacitor.
Te> My calculator says this power supply will charge a .0377 mfd
Te> capacitor at 60 cycles. I do not have the latest version of Mark
Te> Graalman's calculator program, so if you do not get a response
Te> from him, forward your post directly to his attention.
MG> In TESLAC I use the same accidental formula that every one else does
because is gives a workable answer. But, to insure there is a safety
margin in the calculation I convert the RMS voltage level of the
source to the peak value, this results in a capacitor calculation that
appears too small, but it insures the source has ample reserve to
charge the capacitor. It is better to have a capacitor thats a bit too
small, than a bit too big.
ES> What happens when you have a power supply large enough to
ES> charge the tank capacitor in say 1/8 of a cycle? Then we
ES> need a rotary spark gap to take advantage of the extra power
ES> -- do I have this correct? Ed Sonderman
Te> This is correct.
MG> Absolutely! POWER POWER POWER!
Te> Richard Quick
MG> Mark Graalman
... Alias, Mark the spark
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