Real OU-Effect to Share with everyone!!!

[PaulLowrance]

I'm wondering what the maximum voltage spike is. I've seen transistors that are good for at least 1500V. IMO that would get the best results with the proper timing, but then again such experiments are almost entirely only good for a learning experience. NRG still needs to know the capacitance just before and after the experiment. Also, IMO he should try to replicate his own coil.

PL

[duff]

I replicated the basic experiment last night using only the battery and capacitor.

The cap was a .85uF 2300V and was charged to 340V from a single pulse.

It appears to me that at the moment of the pulse the cap is charge to 24v and then inductor acquires its energy/charge after which the circuit is disconnected and the bemf of the inductor is released through the single wire connection to the capacitor.

Clever!

I'll see if I can implement a commutator on a low rpm motor as a reliably switch.

-Duff

[robbie47]

Wouldn't the use of relay's work better?
Relay's normally would have very reliable contact plates.
You could always use the rotor to power the relay coils.

[PaulLowrance]

Even better yet a 400+ volt transistor.

PL

[wattsup]

@guys

Sorry for regressing back to post #1 since I did not have time to read the whole thread given I spend more of my time on building and testing things, but I do have a concern of the uTube video showing the two capacitors as OU.

There is one medium DC capacitor of high uF and low voltage transfering a charge to the big non-polarized capacitor of low uF and high voltage. You then measure the voltage in the big cap and do some math to get the comparative joules of energy. But this is not right in my view since the big capacitor will certainly not have a linear specification relative to voltage level and percentage on the dielectric occupied by the charge.

Meaning if the big cap is a 1200volts, and you read 120 volts as a given charge, it does not mean there is 10% of the dielectric being used, hence using the uF value of the cap to do the math would not be right. You would require the real capacitor specifications to understand how a voltage reading realates to the total charge in the capacitor in joules.

Otherwise you would have to then use the charge in the big cap to do some work and compare it to the work being done by the smaller cap.

Another example would be if the big cap was fully charged to 1200 volts and you put a motor load via a resistor and find that the capacitor can discharge all its energy in time and make the motor turn one complete turn. It does not mean if the capacitor is charged to only 120 volts it will make the motor turn 1/10th of a turn. Hence you should not rely on the capacitor value to do your math especially if you are using such a big capacitor.