Hey Goat

Well that is an issue, getting it back to the source. Especially all of it and or more. ;]  At least with what we have so far. Thats the way we have to think about it. Use the kick in the pants to at least know its real and we are not wasting out time.  ;]

Ive read that a cop of at least more than 1.8 - 2 would be needed. 
So once we get just above cop>1 we still have a ways to go.

But we are at the least getting more to a load than what the battery could put out  by itself over time.  Like an electric bike could go 50% further just on a regular charge, etc. Its not useless.  These are just beginning things, but at least we are here.  ;]

I gotta get some sleep.

Mags

Mags

Just ground both capacitor and battery negative and touch positive of capacitor to positive of battery, you will see a back current flow to battery if capacitor if of correct value and enough charged (of cource voltage above battery).
The ultimate source for better HV devices is Earth ground ! Don Smith

OK magluvin

Good idea to transfer here.

I made a circuit with what i understood from your recent post

The result is impressive, but i wonder if i missed something

Thank's for your apreciation

good luck at all

Laurent

http://www.youtube.com/watch?v=Auv_66Ke-iw

Quote from: woopy on March 25, 2011, 03:02:37 PM
OK magluvin

...

I made a circuit with what i understood from your recent post

The result is impressive, but i wonder if i missed something

...

Laurent

hi Laurent

your calculation of energy supplied by the input cap has a mistake

you must find the energy on the input cap for its start voltage and then subtract the energy remaining on the same cap for its final voltage - this will give you the correct input energy

if you need to be more precise about energy transfers between capacitors then you must measure the actual value of the capacitors and not rely on the case marking (which can be around  +/- 20% of the actual value!)


here is a good online capacitor energy calculator to save time with cap energy calculations:
  http://www.electronics2000.co.uk/calc/capacitor-charge-calculator.php

the page at that link can also be saved to your computer to use offline


hope this helps!

cordialement
np

http://docsfreelunch.blogspot.com

Ok  this is the general idea.
There are 4 pics below that show where Im going with this. It is not an idea that I just think works. It just does.

The first pic is the same circuit Woopy made the first video of. With the addition of the diode across the batt and switch.

1  The first pic shows the circuit idle no current flowing and no charge in the cap. The scope shots are Source, Inductor and Cap in order from left to right. Source is 5v dc.  The resistor is added to the inductor to give its real world values in the circuit. Woopys inductor.

2  The next pic shows the switch closed and a charge building in the cap, through the inductor and diode at upper left with current flowing clockwise. The diode on the right is not conducting anything yet.

3 This pic shows me open the switch when the cap reaches 5v, equal to source. Now the diode on the right is conducting, current is still going clockwise, just not through the battery, but now the diode.

4 Now the currents have finally stopped. The cap is charged to 6.32v.
The flywheel crammed that extra 1.32v into that cap after the battery was disconnected.

Is that extra 1.32 not free?  When we charge a cap from a source, the cap when fully charged by the source, will contain and equal amount of energy that was taken from the source to charge it.

So when we cut off the battery here, when the cap reaches 5v, and we know the source is 5v, then we have to know that the 1.32v worth of energy in the cap came from where?   ;]