Back 2 back electros bipolar?

When you say, "positive to negative", it sounds like all you're doing is putting caps in series:

• + ------||-----||-------

What that gives you is a polarized cap that you'd use the series cap formula on.

For a non-polar, you'd put them like this:

• + ------||------||------

Then, the total capacitance is equal to the capacitance of one cap; but it works on AC.

Some people like to bias them:

+Vcc | [~1M] + | + ------||---+---||------

Cheers! Rich

No.

Try series connection, or a true bipolar.

Sort of. the midpoint charges up (or down) by reverse leakage when the "wrong" capacitor is reverse biased, once it's charged you have a bipolar device.

a high resistance to a suitable supply could be added to ensure the charge remains otherwise the pair will go non-linear near the absolute peaks in whatever current signal the capacitors pass .

same as any other two capacitors in series.

probably not enough to charge the forward one either, include an 1meg (etc. - scale aprropriate to supply voltage and leakage current) pull up (or down) on the midpoint to ensure they are correctly biased.

Bye. Jasen

no, it's same capacitance as the series combination above.

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Bye.
   Jasen

Hmm - apparently I'd assumed that the backwards one acts like a short for its half-cycle.

Anybody got any docs on this, or done the experiment? I haven't, but I have no problem with being corrected when I'm wrong.

Thanks, Rich

Small-signal AC capacitance is C/2, and DC capacitance is C. So it can create harmonic distortion.

John

only the first time, the second time it's got a charge on it and is working like a capacitor.

it's fairly simple to model, IIRC it was done in SEB some time last year.

C1 1F C2 1F ||+ +|| 1V - --- --- --- -o--||--o--||----. / \\ / \\ / \\ a || b || | 0V - + + + + + ---+-- \\ / \\ / ////// -1V - --- --- 0 1 2 3 4 5 6 7 0) start with both caps discharged, a at 0V b at 0V

1) ramp the voltage on a upto 1V , C1 is reverse biased and leaks current , c2 charges up with 1C b is at 1V a is at 1V 2) reduce the voltage on a to 0V again and 0.5C charge flows out of C2 into C1, a is a 0V b is at 0.5V , 3) now take A negative, the other 0.5C flows out of C2 and into C1, b is at 0v a is at -1 points 4,5,6,7 are left as an exercose for the reader should he need more convincing,

for a swing of 2V 1C flows... capacitance is 0.5F a real reverse biased electrolytic is going to be some non-zero voltage drop, so you may need to use more than 1V for the signal :)

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Bye.
   Jasen

Thanks for this - it really does clear it up for me.

Thanks! Rich