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Overunity transformer effect

Started by tinman, March 02, 2015, 06:39:14 AM

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MileHigh

There is a decent sim compliments of EMdevices.  I assume more tweaking and adjusting will be forthcoming.

http://www.overunityresearch.com/index.php?topic=2804.msg46835#msg46835

tinman

Quote from: MileHigh on March 04, 2015, 04:01:04 AM
There is a decent sim compliments of EMdevices.  I assume more tweaking and adjusting will be forthcoming.

http://www.overunityresearch.com/index.php?topic=2804.msg46835#msg46835
Looks very close indeed MH. Lets see if he can get the voltage polarity to reverse on C1 using his sim.

tinman

So a question for you all.
Lets say we have 2 DC cap's-63 volts,and a capacity of 10 000uF.
Now we know if we place these two caps in parallel,we will retaind the voltage rating of 63 volt's,and the capacitance will double to 20 000uF. We also know if we put these caps in series,we will half the capacitance to 5000uF,but double the voltage to 126 volt's. But what if we want to use these two caps on an AC current. For this we would join the two negatives in series,and we would have the two positive terminals conected to our AC source-->but what would be the capacitance of the two cap's now?.

MarkE

Quote from: tinman on March 04, 2015, 09:47:36 AM
So a question for you all.
Lets say we have 2 DC cap's-63 volts,and a capacity of 10 000uF.
Now we know if we place these two caps in parallel,we will retaind the voltage rating of 63 volt's,and the capacitance will double to 20 000uF. We also know if we put these caps in series,we will half the capacitance to 5000uF,but double the voltage to 126 volt's. But what if we want to use these two caps on an AC current. For this we would join the two negatives in series,and we would have the two positive terminals conected to our AC source-->but what would be the capacitance of the two cap's now?.
Two capacitors in series exhibit a total capacitance of:

CSERIES = C1*C2/(C1 + C2)

gyulasun

Quote from: tinman on March 04, 2015, 09:47:36 AM
...
But what if we want to use these two caps on an AC current. For this we would join the two negatives in series,and we would have the two positive terminals conected to our AC source-->but what would be the capacitance of the two cap's now?.

Hi Brad,

It is advisable to use parallel diodes across the capacitors to prevent unwanted voltage polarity across the electrolytic capacitors what the AC voltage would impose on them, I attached a drawing (taken from the web) how the diodes should be connected.
The diodes should have appropiate voltage and current ratings as per the expected peak charge or discharge current values for the capacitors.
And the resultant capacitance of the two series caps would be as MarkE wrote (or from your example, half of the 10000 uF i.e. 5000 uF as you wrote). 

By the way, you can find non polar (bi-polar) electrolytic capacitors, they are manufactured too, see a random link here:
http://tinyurl.com/nzxb3yu      or here: http://www.erseaudio.com/s.nl/sc.7/category.833/.f

Gyula