Re: Suggestion on Power Supply?

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Jeremy Scott by way of Terry Fritz <teslalist-at-qwest-dot-net>" <supertux1-at-yahoo-dot-com>

I'm sorry, I hit send and while driving home I
realized an error. It's not Vrms^2, is Vpeak^2,
so instead of 10,000V^2 it's probably more
like 14,140 squared. Ultimately it gives you
3J instead of 1.5J :)


--- Tesla list <tesla-at-pupman-dot-com> wrote:
 > Original poster: "Jeremy Scott by way of Terry Fritz
 > <teslalist-at-qwest-dot-net>" <supertux1-at-yahoo-dot-com>
 >
 >
 > Hi,
 >
 > I'll take it from your email address that
 > you're using 50Hz, so your AC half-cycle
 > length is 5ms. And at 200bps, your capacitor
 > would have to charge and be discharged at
 > least twice within that 5ms. That is,
 > you need to charge and discharge in 2.5ms.
 >
 > I'm assuming your power supply is rated 10KV,
 > but I'm not sure of the amperage, but I can
 > tell you what it has to be to accomplish
 > the above:
 >
 > Charge Time = Z * Cp * 5
 >
 > Z = V / I
 >
 > .0025s = (10,000V / I) * .00000003F * 5
 >
 > Solve for I: I = 600mA
 >
 > 10,000V * .6A = 6KVA
 >
 > Energy = 0.5 x C x Vē
 >         = 0.5 x .00000003 x 10,000^2
 >
 > Energy = 1.5J
 >
 > Now, you'll notice I left out * BPS out of that
 > formula because I don't think this is an entirely
 > accurate way to measure the output 'power' of a
 > coil. (It is theoretically, but misleading
 > because we like to measure things 'per bang',
 > not usually per second.)
 >
 > The problem is when you multiply it by BPS,
 > what you are getting is an average output over
 > the course of one second. If you've ever seen
 > a waveform of the voltage off a secondary's toroid,
 > you'll see that it's not continuous over a full
 > second,
 > rather it's only for a few hundred microseconds
 > during
 > every AC cycle. (Check out Terry's pspice model of a
 > coil.)
 >
 > So we have this figure, 1.5J which is the amount of
 > energy stored in your capacitor. If we multiply
 > it by BPS, we get the average power over one second,
 > which is of course VERY low due to the minimal duty
 > cycle of the tank circuit. (Charge for a relatively
 > long time, discharge fast, charge some more etc...)
 >
 > 1.5J * 200BPS = 300W
 >
 > Not very useful... good if you want to know how
 > much it costs to run your coil for an hour :)
 >
 > But the instantaneous 'bang' power is what we're
 > interested in, and for that we need to divide the
 > stored energy in the capacitor by the time it takes
 > to
 > discharge it:
 >
 > Power = Energy / Time
 >
 > Power = 1.5J / t
 >
 > Power = depends on t.
 >
 > If you make t smaller, power per bang goes up.
 > There is really only one way to make t smaller
 > and that's to lower the resistance of the primary
 > tank circuit. Big conductors and properly quenched
 > spark gaps are how we do this. Higher voltage also
 > facilitates the quick discharge, so not only is
 > it an important factor in the amount of energy
 > stored
 > in the capacitor, but it is also a factor in quick
 > discharge. (That is, it pushes more amps)
 >
 >
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 >
 > --- Tesla list <tesla-at-pupman-dot-com> wrote:
 >  > Original poster: "Christoph Bohr by way of Terry
 >  > Fritz <teslalist-at-qwest-dot-net>" <cb-at-luebke-lands.de>
 >  >
 >  > Hi!
 >  >
 >  > this thread really brought up new insight into
 > the
 >  > topic to me, but araised
 >  > one further question.
 >  >
 >  > I'm using a 30nF tank cap with a 200BPS async
 > rotary
 >  > ( BTW for those who
 >  > remember, I just cant get that darn thing to
 >  > sync....probably I removed too
 >  > much material from the rotor but works not too
 > bad
 >  > this way )
 >  >
 >  > referring to that formula from richie burnetts
 >  > homepage:
 >  > http://www.richieburnett.co.uk/rotary.html
 >  >
 >  > P = 0.5 x BPS x C x Vē
 >  >
 >  > Is it right that I can only get a power
 > throughput
 >  > of ca 450 Watt at 10KV
 >  > that way? Or am I messing up the units and it
 > should
 >  > be 4500Watt?
 >  >
 >  > from the reading of my amp-meter it should be
 >  > something like 450Watt.
 >  > My power supply should be able to deliver around
 > 10
 >  > to 20 times more, but I
 >  > think with that cap and 200 BPS I can't put more
 >  > power through the system.
 >  > Just wanted to be sure before messing around with
 > my
 >  > rotary......
 >  >
 >  > I know my lack of math knowledge must really
 > hurt,
 >  > but please help me anyway
 >  > ;-)
 >  >
 >  > Thanks in advance.
 >  >
 >  > Christoph
 >  >
 >  >
 >  >
 >  > ----- Original Message -----
 >  > From: "Tesla list" <tesla-at-pupman-dot-com>
 >  > To: <tesla-at-pupman-dot-com>
 >  > Sent: Tuesday, May 06, 2003 4:02 PM
 >  > Subject: Re: Suggestion on Power Supply?
 >  >
 >  >
 >  >  > Original poster: "by way of Terry Fritz
 >  > <teslalist-at-qwest-dot-net>"
 >  > <FutureT-at-aol-dot-com>
 >  >  >
 >  >  > In a message dated 5/5/03 9:44:49 PM Eastern
 >  > Daylight Time,
 >  >  > tesla-at-pupman-dot-com writes:
 >  >  >
 >  >  > >John, thanks for clarifying... I didn't think
 >  > your
 >  >  > >formula had much to do with BPS at all,
 > simply
 >  >  > >what happens in ONE shot.
 >  >  >
 >  >  >
 >  >  > Jeremy,
 >  >  >
 >  >  > You bring up an interesting point.  In my
 > tests,
 >  > I used
 >  >  > break rates from 30 bps to 1000 bps or so.  At
 > 30
 >  > bps,
 >  >  > the output spark length was quite short and
 >  > didn't come
 >  >  > anywhere near the predicted lengths.  At 60
 > bps
 >  > however
 >  >  > the coil performed close to the predicted
 >  > lengths.  I concluded
 >  >  > that such low breakrates of 30 bps or lower do
 >  > not keep the
 >  >  > arc streamer channels hot enough to maintain
 >
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