Ceramic capacitor polarity trick

Spehro Pefhany wrote in news: snipped-for-privacy@4ax.com:

Nice tip. Thankyou.

It's not possible for a hysteretic to become unstable. The main issues there are bounding frequency of operation across all possible combinations of input voltage and output loading, and achieving low output ripple at high ratios of output to reference. The first is overcome by using a topology of fixed ON time inversely proportional to input voltage magnitude with some non-linear foldback to handle discontinuous mode, and the second by various little RC feedback networks containing zeroes that pass the ac-component of ripple unattenuated.

I heard left end.

Tim

It was this one:

Hi Walter,

Well that's certainly at the far end of "less" of "more or less" like a short circuit, eh? ;-) Thanks for the clarification. I've seen people suggest the "trick" of using two non-polarized caps back to back for years now, but never any particularly discussion of just how well it does or doesn't work.

---Joel

In message , dated Sat, 5 Aug 2006, Joel Kolstad writes

It works. Packaged 'reversible' elcos are sometimes made that way. The wrongly-biased section never gets the chance to pass damaging leakage current, because that is stopped by the correctly-biased section.

But the behavior will be complex. Very low-frequency capacitance will effectively be about 2x the high-frequency value, with an obvious nonlinear/distortion zone somewhere in the middle. They're OK as long as the AC swing across them is low.

John

In message , dated Sat, 5 Aug 2006, John Larkin writes

Where does non-linearity come in?

Of course; that's what capacitors in series with the signal path are designed to have. If there is any appreciable voltage across such a cap, it's value is too small.

The back-biased cap leaks with a current that's nonlinearly dependent on voltage. Any cap whose C is a function of V (even a linear function) generates harmonic currents, and the DC capacitance here is twice the small-signal AC capacitance.

Right. So they aren't ideal in filters or crossovers.

John

In message , dated Sat, 5 Aug 2006, John Larkin writes

The current is controlled by the insulation resistance of the correctly-oriented capacitor in series.

That's for DC. Regards,

Mike Monett

Antiviral, Antibacterial Silver Solution:

In message , dated Sun, 6 Aug

Yes; the alternating current doesn't damage the dielectric. What comes off on one half-cycle goes back on during the next. If that wasn't so, power supply filter capacitors in compact fluorescent lamps wouldn't work: they work with a large ripple voltage so as to reduce mains harmonic current emissions.

Certainly. But apply a DC voltage to a discharged capacitor of this sort. Eventually, most of the voltage will wind up across the conventionally-biased section, with much less across the back-biased one. So the effective capacitance (computed from integrated current and final voltage) is twice the small-signal AC capacitance, and the path to "twice" is complex... the charging curve won't be anything like that of an ideal capacitor.

John

In message , dated Sun, 6 Aug 2006, John Larkin writes

I've measured some (discharge time through a known resistor), and it doesn't happen with my samples. I'd forgotten that I'd done that, but I've just checked my notes.

"It is value is too small"?

Come on, John. You're supposed to be one of the founts of wisdom around here.

I hope that's enough references. If not, there's

That depends on if you're using conventional rotation or electron rotation. ;-)

Cheers! Rich

In message , dated Sun, 6 Aug 2006, Apostrophe Police writes

OK, I did a typo. I do check such things, but since I'm outputting about

only for the first half-cycle after that you have the normal capacitance.

to put it another way. if you take the DC voltage away only half the charge will come out.

Bye. Jasen

that effect only occurs during charging. after they've been charged they behave normally unless the inital charge is exceeded.