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The basic understanding of this that I have is that: Assume clock low is 0V, clock high is 3V and Vin is at a fixed 3V. Ignore diode drops

  • Initially Clock is low (0V) and Vin is at 3V (always). Voltage at the first capacitor node is thus 3V (ignoring diode drop). The left capacitor has got positive charges on it's top plate and attracts an equal number of negative charges on it's bottom plate thus giving 3V across it.
  • Clock goes high (3V). Now since we already had 3V across the first capacitor with respect to ground, it somehow doubles to 6V now? Can someone explain to me with charge carries what happens to the first capacitor? How does it magically become 6V (I realise the reference node has changed, but surely charges on that capacitor must change too).
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  • \$\begingroup\$ Initially, Clock is low and the first capacitor is going to charges to Vin = +3V. And when the clock is high the voltage at the top is equal to Vclock + Vcapacitor= 3V + 3V. So the second capacitor can now be charged to 6V. \$\endgroup\$
    – G36
    Jun 10, 2019 at 14:16
  • \$\begingroup\$ youtube.com/watch?v=I4ED_8cuVTU \$\endgroup\$
    – G36
    Jun 10, 2019 at 14:19
  • \$\begingroup\$ electronics.stackexchange.com/questions/424666/… \$\endgroup\$
    – G36
    Jun 10, 2019 at 14:22

1 Answer 1

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Now since we already had 3V across the first capacitor with respect to ground, it somehow doubles to 6V now?

No, the voltage across the first capacitor remains at 3 volts in that instant but, due to the clock signal rising to +3 volts, it lifts the capacitor up (both plates) by 3 volts so, there's now 6 volts on the upper plate with respect to ground and 3 volts on the lower plate with respect to ground. There is no net charge change on the capacitor in that instant - it still has 3 volts across it.

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  • \$\begingroup\$ Ah yes. A simple KVL. As a matter of interest, when you deal with these capacitor circuits, do you not think of charges on a plate? It seems to just make things complciated for me when I do. \$\endgroup\$ Jun 10, 2019 at 14:27
  • \$\begingroup\$ I think of \$I = C\dfrac{dV}{dt}\$. that tells me what I want to know. So providing I is small there can be little change in the voltage in a tiny amount of time. \$\endgroup\$
    – Andy aka
    Jun 10, 2019 at 14:28

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