3 ms·
The other side of the 'input' capacitor is pulled down to a much lower voltage, which means the charge prefers to flow into it. (And then this alternates so the
by rcxdude 2y ago
The other side of the 'input' capacitor is pulled down to a much lower voltage, which means the charge prefers to flow into it. (And then this alternates so the next time the one it is transferring charge to is pulled lower)
- amelius 2y agoI still wonder if all the charge moves from left to right, or if some of the charge stays behind. And how would that work in a CCD chip where thousands of positions are shifted through?
- rcxdude 2y agoFrom what I can find, it's something like 99.999% of charge moves, but it varies from CCD to CCD and can cause artifacts in bigger ones if it's too low. I couldn't find a great indication of what affects it, other than defects in the crystal lattice causing charge trapping, which apparently tends to be an issue in aging telescopes, especially space ones like Hubble.
- amelius 2y agoOK but what I mean is that if you tie two capacitors together, and one of them has a terminal at a much lower potential, then unless that voltage is minus infinity, then the other one will not be completely discharged.