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How do I change Low current DC to high current DC? Answered

Suppose I have a high voltage source of DC with fractional milliampere current.  How can I increase the current?

I think Watts is fixed with respect to source. So, if there is 12000 Vdc at 1 milliampere (12 Watts??) I should, in theory, be able to get 12 Vdc @ 1 amp.

Practicality aside, is this correct?

Practicality engaged:  Is 12 Vdc at 1 amp possible, and how?  4 amps??? 



turning low a DC current into a high DC current is impossible.Because you always need an AC current to boost the it to a higher voltage.

AC, not always necessary. As I said earlier, older internal combustion engines ran on high voltage spark supplied by a coil in line with a point and capacitor system. (Actually, the same is true today, but no points) Base power is 12 Vdc on primary side, and up to 25,000 volts on the secondary (high performance coils). There is no inverter within the old system.

At 100% efficiency, yes, you could, however in the real world its never that easy.

12000v is a tricky beast, it wants to jump through most insulation, so you need a very shielded design.

DC doesn't directly convert to other DC without being either inverted or pulsed - because transformers only work with 'changing' electrical current.

...The question is, where on EARTH are you finding voltage potential of 12kv for 'free' (per your keywords)? Even at that rate, anything you built wouldn't be worth 12 watts of power.

Back in the old days of points and condensers automotive coils would produce 25000 Vdc, at the plug. Not that that has anything to do with where I'm going, but i have experienced HIGH voltage in the 10K to 20K range.

The voltages are theoretical. I am waiting to see how much voltage is involved in an upcoming experiment. The conversion will be part of the harvest process.


well, by all means, best of luck.

Why (and from where) - do you want to do this?
-tapping power-lines?


Numbers theoretical. I'

Thanks for the reply.

Your math is correct, assuming 100% efficiency and no losses. If you want 4 A, then you have to step down to 3 V (same math).

In reality, I suspect you'll be lucky to get 50% efficiency. You either need a pulsed power supply, or you need to invert the 12 kV DC to AC, run that through a transformer (with losses), then rectify the AC back to DC (more losses).