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>


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)





















--- 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
 > spark growth,
 >  > at least in the coil setup I was using.  I know
 > that the spark
 >  > length can be maintained at much lower breakrates
 > however in
 >  > some tube coil designs.  My formula was simply
 > based on
 >  > observed spark lengths of various coils.  At the
 > relatively
 >  > low power levels I worked with, 120 bps gave
 > better "efficiency"
 >  > than higher or lower breakrates did.
 >  >
 >  > John
 >  >
 >  >
 >  >
 >  >
 >
 >


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