Re: A few questions...

• To: tesla-at-pupman-dot-com
• Subject: Re: A few questions...
• From: Ed Phillips <evp-at-pacbell-dot-net> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)
• Date: Sun, 25 Jul 1999 17:49:06 -0600
• Approved: twftesla-at-uswest-dot-net
• Delivered-To: fixup-tesla-at-pupman-dot-com-at-fixme

"A Tesla coil is pulsed with giant spikes that can reach 1000 amps many
times every second.  Even though the spikes may get to 1000 amps, the
RMS
or equivalent DC current level is around 15 amps.  Thus, the cap has to
dissipate the heat internally that a 15 amp current would create.  Since
a
MMC has an internal resistance of say 0.06 ohms, the power dissipated
would
be 15^2 x 0.06 = 13.5 watts.  That may not sound like much power but to
a
plastic cap with 20kV across it can be a lot.  If you change the cap to
a
paper dielectric, the internal resistance may go up to 2 ohms.  15^2 x 2
=
"BAD" :-))"

	The peak current in amps can be calculated as follows:

	Let f = frequency in kilohertz, C = capacitance in microfarads, V is
the peak voltage across the capacitor.

	Then Ipeak = (f x C x V)/159.15

	Examples:

	f = 250 kHz,  C = 0.01 ufd, V = 15000

	Ipeak = (250 x 0.01 x 15000)/159.15 = 235.6 amperes

	f = 100 kHz,  C = 0.1 ufd, V = 20000

	Ipeak = (100 x 0.1 x 20000)/159.15 = 1256.7 amperes

Ed

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