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Re: gEDA-user: ESR of 2.2u ceramic capacitor




On Feb 25, 2006, at 5:23 PM, Bill Sloman wrote:

At 23:01 25-2-2006, you wrote:

On Feb 25, 2006, at 1:39 PM, Karel Kulhavy wrote:

What does it mean? That the capacitor loses capacity when switching
between charge and discharge?

Yes. If you start with your 2.2 µf capacitor discharged, charge it up to a volt, it might take the full 2.2 µC it's supposed to. But then discharge it to zero volts, you might only get 2.1 µC back! The charge doesn't disappear (averaged over many cycles the net DC current will be zero), but in general when you push charge in with voltage, the capacitor pushes back with less voltage when it's discharging. That's a net energy loss to the circuit, dissipated as heat. A "Q meter" will register this as resistance, but it's not ohmic, and may not have the same effect as ohmic resistance on circuit operation.

Put on pedant hat.

Charge is conserved

Sure. I said that.

Charge is *always* conserved. Charge in one wire equals displacement charge in the dielectric equals charge out the other wire. But this is a little trickier. If it was ohmic leakage you could keep pushing charge through forever with a fixed voltage, but in dielectric hysteresis you can't.

- if you could convert electrons to heat, you could fry your circuit, and eliminate our dependence on oil.

Remove pedant hat.

What you are talking about here is "charge soak" which is often modelled as extra capacitance-plus-series-resistance in parallel with the ideal capacitor you thought you were buying. I've seen time constants varying from a few seconds to a number of minutes, depending on what I was doing. The charge is still in the capacitor, but it's coming out slowly.

It can take years. That's how "electrets" work. In the vacuum tube era (yes, I'm that old, barely) I remember reading about folks using this effect to make a rechargeable "bias cell" for grid bias!


John Doty              Noqsi Aerospace, Ltd.
jpd@xxxxxxxxxxxxx