Pulse power supply - how quickly does fusion happen?

It may be difficult to separate "theory" from "application," but let''s see if this helps facilitate the discussion.
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AllenWallace
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Pulse power supply - how quickly does fusion happen?

Post by AllenWallace »

I am considering a low duty cycle pulsed power supply. It would deliver 17 KV pulses at 18 amps! that's 300 KW! The problem is that the pulse width is only 1 us. Does anyone have an idea how long it takes fusion to happen? What might be the minimum pulse width? I read that Gas Discharge tubes trigger in nanoseconds.... I guess the question might be how long does it take a deuteron to travel the distance from the ionization area to the posior?
Richard Hester
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Re: Pulse power supply - how quickly does fusion happen?

Post by Richard Hester »

I would use high voltage and lower current get the average ion energy into a region where fusion is more likely.This way you are more likely to get something for your trouble.
3l
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Re: Pulse power supply - how quickly does fusion happen?

Post by 3l »

The crossection at 17 kev is poor at best.
I started pulsing at 10kev but the crossection sucked!
the most I ever got was 6 x 10^4 neutrons @ 1024 J's.
34 kev is where I would start....things move readily there.
I know I have not posted straight pulse in a while ...
Here's where the work is now.
I use my 10 kv @33uf to drive a single turn transformer that has 12 turns on the secondary.
My pulse is under 2 microseconds with a trigatron style gap.
1024 j/2 x10-6 = 512 x 10 ^6 or 5.12 x 10^ 8 Watts on the primary. The secondary has 4.267 x 10^7 watts. or 120 kv @ 355 A.
Plenty for fusion.
I will post my pulse box pictures soonest.
The problems begin with how to apply that monstrous current w/o arcing. The Xrays are very harsh on my first trial. I pegged my meter >10 rem. I shelved the work until outdoor shielding becomes availiable. My drive for the linac began at the realization that a spherical fusor would be prohitative at that voltage. It's not the time but crossection that is critical.

Fusion is fun!
Larry Leins
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jlheidecker
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Re: Pulse power supply - how quickly does fusion happen?

Post by jlheidecker »

Why used a "pulsed" power supply? Why not DC?
DaveC
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Re: Pulse power supply - how quickly does fusion happen?

Post by DaveC »

Question and an answer.

Q. What will you use for a cathode? 18 amps is a pretty big current.

A. To determine the transit time, the formal solution involves determining the ion acceleration at each point along the path, taking a small length (delta S(x,y,z)) of the path and assuming the acceleration is constant in that region, thus getting a small increment of transit time over the region (delta S ). The total transit time is then the sum of these increments of time across the whole distance from cathode to anode (or poissor). This usually requires some form of numerical integration, and a lot of guesswork, since the actual path is not really known and more importantly, the electric field which creates the ion's acceleration, is not known at all if there is space charge involved..

So..... a simple approximation is to assume the electric field is uniform and thus the acceleration of the ion is uniform and then the time is gotten by using the simple expression,

S = 1/2 a t^2 or t = sqrt(2aS),

where a is acceleration in m/s^2, t is time in seconds, and S is distance in meters.

The acceleration (a) is determined by the electric field , volts per meter, mass (m) of the ion in kg and the charge in coulombs according to the formula:
a = f/m = qE/m = qV/(Sm)

Putting it together gives:

t = sqrt(2aS) = sqrt(2(qV/Sm)(S)) = sqrt(2qV/m).

Note that distance (S) drops out of the final expression.

At first glance this might seem strange, but we remember that "voltage" is energy gained (or lost) per coulomb. When distances are small the energy is gained rapidly (high acceleration) over a short distance. When distances are large, the energy is gained slowly (low acceleration) over a long distance. In either case, the transit time is the same. It is governed by the ion charge, mass and applied voltage.

For your data of 17 kV, and deuterons, T ~ 0.784 usec . just under one microsecond, irrespective of how big your device is.

But I think a focussed steady beam could be more effective.

Dave Cooper
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