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Consider the following situation:

  • There is a capacitor C1 (polarity does not matter). The 'top plate' of the capacitor is the one at the top of the diagram and has two ideal switches connected, one connecting to ground during φ1 and one floating at φ2.

  • The bottom plate also has two ideal switches, one connecting to Vin during φ1 and one connecting to ground during φ2.

  • In phase 1, the capacitor is charged to Vin with polarity as shown in the diagram. The bottom plate has +Q charge and the top plate has -Q charge.

  • In phase 2, the bottom plate of the capacitor is switched to ground and the top plate is switched to the floating node, thus a voltage of -Vout will appear at the floating node as the capacitor will maintain the charge.

At the end, the capacitor no longer has +Q and -Q on each of it's plates, it kind of has -Q and -2Q. The positive charges that were at the bottom plate should all go to GND and thus to maintain the same charge across the capacitor (Q=CV), the top plate charge should go more negative.

Where does the top plate receive these extra electrons from? Is it just from the metal of the wire and hence no current actually flows?

My question is really about the polarity of the charges on the capacitor as we go into phase 2. I've always struggled with this.

enter image description here

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  • \$\begingroup\$ Please note that such a switched capacitor can mimic a resistor only when its right side is connected to ground, to "virtual ground" (opamp input) or to a sufficiently large capacitor. \$\endgroup\$
    – LvW
    Nov 9, 2022 at 8:10
  • \$\begingroup\$ I’m not trying to mimic a resistor here. It’s just a thought experiment. \$\endgroup\$ Nov 9, 2022 at 12:15
  • \$\begingroup\$ OK - but with an open end at the right side the whole circuit makes no sense at all. \$\endgroup\$
    – LvW
    Nov 9, 2022 at 13:56

1 Answer 1

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You're correct up to and including what happens during Phi1. The capacitor's plates are charged to +Q and -Q respectively.

During Phi2, the capacitor gets connected "the other way around" to ground and a floating node, producing -Vin at the floating output. Note that this node is floating, and therefore no current flows, so the charge on the capacitor plates doesn't change. (Current is just charge going somewhere, so if there's no current, the charge is not going anywhere.)

The charge on the plates therefore remains at +Q and -Q.

The only charges that are transferred in the circuit when switching between Phi1 and Phi2 are those needed to recharge parasitic capacitances. However, in an ideal circuit there are no parasitic capacitances, so the change in charge on the capacitor plates really is zero. In practice, the parasitic capacitances are much smaller than the actual capacitor's capacitance, so the change in charge is still approximately zero even in a real circuit.

The charge on a plate (+Q in your case) can still exist even if that plate is connected to ground. The charge on one plate (-Q) attracts the charge on the opposing plate (+Q) and keeps it physically in place. (And vice-versa.) This prevents the charge from flowing to ground and also ensures that the charges on both capacitor plates are always equal in magnitude (and of opposite sign).

Alternative explanation: If you view the capacitor as a whole, the sum of all charges on it is zero (+Q and -Q cancel each other). Therefore, there is no charge to equalize if you only connect one terminal of the capacitor to any potential. You have to force the potentials on the capacitor's plates apart to add or remove charge.

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  • \$\begingroup\$ What confuses me is that in Phi2, the bottom plate of the capacitor is connected to ground, yet that bottom plate still has +Q positive charge on it? \$\endgroup\$ Nov 8, 2022 at 23:20
  • \$\begingroup\$ I suppose that’s due to the electric field holding the charges as they were and since no current can flow, there is no way for those +Q charges to flow to ground. The +Q and -Q designators always confuse me. \$\endgroup\$ Nov 8, 2022 at 23:22
  • \$\begingroup\$ That is correct. I've edited my answer to explain the physical phenomenon. \$\endgroup\$ Nov 8, 2022 at 23:23
  • \$\begingroup\$ Excellent answer. Thank you so much. Cleared up a doubt that I always had with switched cap circuits. \$\endgroup\$ Nov 9, 2022 at 0:24

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