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Hi, I have some confusion about this (It's a stupid question).

  • Initially, the capacitor has no charge stored with S1 and S2 open.
  • S1 is closed and it now has charge C1Vin (after a while).
  • S1 is opened and S2 is closed. Now, shouldn't the C1Vin charge already on the capacitor get discharged to Vout? Why would the capacitor change it's charge stored to C1Vout? Or should I think of it as the charge passed to the output will depend on the Vout level? So it's possible that the capacitor isn't fully discharged and only C1Vout get's pushed out.
  • \$\begingroup\$ In the text, they speak about charge transfer....however, where? To which unit for further processing? \$\endgroup\$
    – LvW
    Jan 20, 2019 at 11:02
  • \$\begingroup\$ @LvW It isn't stated to where. \$\endgroup\$ Jan 20, 2019 at 22:08

3 Answers 3


Yes that's a particularly confusing piece of text. Your analysis is basically correct:

1) With S1 closed and S2 open, the capacitor charges or discharges until the voltage across it is Vin, and the charge is C1 * Vin. (they've called this qin)

2) With S1 open and S2 open, and an ideal capacitor the capacitor's charge does not disipate and the voltage remains the same.

3) With S1 open and S2 closed, the capacitor charges or discharges until the voltage acorss it is Vout, and the charge is C1 * Vout (they've called this qout).

If you do the sequence of 1 then 2 then 3, then 1 again.

During step 3, the charge on the capacitor starts at qin and ends up as qout so during that step a charge of qin-qout is transferred to the output

During step 1, the charge on the capacitor starts at qout and ends up as qin so during that step a charge of qout-qin is transferred from the capacitor to the input, or more sanely a charge of qin-qout is transferred from the input to the capacitor.

So after one complete cycle a charge of (qin-qout) is transferred from the input to the output.

plugging in the values you get qin-qout = C1*(Vin-Vout).

As an aside the problem with this sort of design is that you'll end up dissipating a lot of energy in resistance of the wires and switches, due to equipartition as the charge flows in and out to/from the decoupling capacitors on Vin and Vout. Most uses of this sort of capacitor switching power supply are used where Vin is equal to Vout, either to turn a +ve voltage into a -ve voltage (the switches turn the capacitor upside down), or as a doubler (the switches take the capacitor and stack it ontop of the supply voltage). Both types can be seen in the popular MAX232 series of line drivers.

  • \$\begingroup\$ Awesome. That last part of point 3 in your answer really cleared it up. The capacitor charges/discharges only until the voltage across it is at Vout. Now, it's still a bit confusing since Vout technically would be something we're measuring? And in this case it's kinda of assuming theres another voltage source at the output. \$\endgroup\$ Jan 19, 2019 at 18:39
  • \$\begingroup\$ yes, the sort of thing the question is supposing to be connected to vout is a large capacitor and, if you want vout to remain constant, something to stop the voltage going up like a zener diode. Either way, the idea being that that the quantity of charge transferred by your switching capacitor is small compared to the change it makes to Vout so you can analyse it as if Vout were fixed. Again the original text could have made this a lot simpler to understand. \$\endgroup\$
    – james
    Jan 19, 2019 at 18:42
  • \$\begingroup\$ Your point 3 is incorrect. With S2 open, how can C1 acquire a voltage equal to Vout? \$\endgroup\$
    – Andy aka
    Jan 19, 2019 at 18:47
  • \$\begingroup\$ @Andy - how is simple, there's now a wire joining them. I suspect the motiviation for your question was: how is a voltage Vout maintained that is different to Vin, given that if there was nothing else connected to Vout then the voltage would be exactly the same as Vin and no current or charge would flow into or out of the capacitor? Given that the text is asking about that charge, you have to assume the the text is assuming something connected to Vout capable of sourcing or sinking that charge or current, be it an ideal voltage source or something else. As I said the text is somewhat unhelpful \$\endgroup\$
    – james
    Jan 19, 2019 at 19:02
  • \$\begingroup\$ @james I think he meant that part 3 should be "With S1 open and S2 closed" \$\endgroup\$ Jan 19, 2019 at 20:03

When S1 opens and S2 closes AND the voltage on C1 does not equal Vout (set by external processes), then either: -

  • Charge flows out of C1 to slightly try and raise Vout (per switching cycle) OR
  • Charge flows back into C1, having the effect of lowering Vout (each switching cycle)

This means that when S2 opens and S1 closes: -

  • Charge flows into C1 from Vin
  • Charge flows from C1 into Vin

I understand your problem....Surprisingly, in many books and other contributions you can find a figure similar to the one shown by you. But one important part is missing: The theory of such an S/C block requires that the output is connected either to a large capacitor (much larger than the switched one) or to ground. In most applications, however, the output is connected to the virtual ground provided by the inverting input node of an opamp (charge transfer to the feedback capacitor).

Explanation: The principle of the S/C techniques is based on a CHARGE TRANSFER from one capacitor to another one or to a current-processing unit. The switched capacitor must be discharged during one clock cycle. Hence, in nearly all cases, the Vout node is connected to the inverting input of an opamp) .

It is NOT sufficient to explain the S/C principle with an open output node.


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