Re: Suggestion on Power Supply?

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

Original poster: "Malcolm Watts by way of Terry Fritz <teslalist-at-qwest-dot-net>" <m.j.watts-at-massey.ac.nz>

Hi Jeremy,

On 5 May 2003, at 19:40, Tesla list wrote:

 > Original poster: "Jeremy Scott by way of Terry Fritz 
<teslalist-at-qwest-dot-net>" <supertux1-at-yahoo-dot-com>
 >
 > John, thanks for clarifying... I didn't think your
 > formula had much to do with BPS at all, simply
 > what happens in ONE shot.
 >
 > Let's see if I got this right, one shot, no BPS rate
 > involved:
 >
 > To charge a .03uF capacitor ONCE to 20362V in under
 > 8.3333ms requires 14400V delivering .270A of
 > charge:
 >
 > Charge Time = Z * Cp * 5
 >
 > Charge Time = (14400V / .270A) * .03uF * 5
 >
 > Charge Time = .008ms
 >
 > Power = V * I = 14400V * .270A
 >
 > Power = 3888W
 >
 > That's 3888W of power input from the
 > secondary side of the transformer.
 > The law of conservation of energy says
 > that the output cannot be more or less.
 > (we'll assume no energy is lost as heat
 > due to conductor resistance)

More. There can always be losses resulting in less.

 > Energy at this point in the capacitor is:
 >
 > Energy = 0.5 * Cp * Vp^2
 > Energy = 0.5 * .03u * (20362V^2)
 > Energy = 6.22J
 >
 > The capacitor has 6.22J of stored potential energy.
 >
 > The required gap conduction time is about 1.6ms
 > in which all the energy in the capacitor is
 > released. (probably less time in the real world
 > because some of that energy will be lost as heat
 > etc...)
 >
 > Power = Energy / Second
 >
 > Power = 6.22 J / .0016s
 >
 > Power per Bang = 3888W
 >
 > 1.7 * sqrt(3888) = 106 inches
 >
 > 106 / 12 = 8.8 feet
 >
 > This is regardless of how many breaks per second,
 > but being able to charge every 8 ms affords 120BPS
 > exactly.

True, but the output voltage soars to get there as the energy in the
secondary is now in the kJ range. Your secondary coil is going to
have to be pretty tall to withstand voltages in the MV range.

Malcolm

.............msnip