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Can someone help me find the voltage and charge across each of the capacitors? This is not just a simple/ parallel circuit so I am having a problem in formulating the right equation on how to find the voltage and charge across each capacitor.This is actually a charge pump wherein capacitors pump charges to the load capacitor, CL.enter image description here

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  • \$\begingroup\$ I see that you changed the circuit, invalidating the previous answers. I'd suggest that you either ask a new question, or edit the current one but keeping the original schemating and describing the changes. \$\endgroup\$ – clabacchio Nov 24 '14 at 11:02
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It is a simple series/parallel circuit, V1 does nothing if \$V_{D}\$ is ideal, \$C_{1}\$ can be ignored.

\$C_{X}\$ parallel \$C_{2}=C_{n}=C_{X}+C_{2}\$

\$C_{N}\$ series \$C_{l} = (C_{N} *C_{l})/(C_{N} +C_{L})=C_{tot}\$

\$Q_{tot}=C*U=C_{tot} *V_{d}\$

\$V_{l}=C_{N}/C_{tot} *V_{d}\$

\$V_{N}=C_{L}/C_{tot} *V_{D}\$

Rest:

\$Q=C*U\$

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  • \$\begingroup\$ There's a switch between Vd and C1 and between C2 and CL @P.Koch . How would you do that? \$\endgroup\$ – djambalong Nov 24 '14 at 7:53
  • \$\begingroup\$ Can you edit the schematic? I'm not sure if I unterstand your switching correctly. \$\endgroup\$ – P. Koch Nov 24 '14 at 8:31
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Actually this is a simple series parallel circuit. It is not a charge pump. Whatever AC voltage you put in at Vd will appear at Vl with some attenuation. A charge pump would have two diodes in there somewhere.

Part of the confusion seems to come from the way the circuit is drawn. Perhaps it was drawn this way deliberately to obfuscate it and see if you can interpret it as part of the assignment. The circuit may be more obvious from this schematic:

We can immediately see that C1 is irrelevant in determining the input to output relationship since it is in parallel with a voltage source. It will load the source voltage, but for this analisys the source voltage is what it is. We can therefore simplify the circuit to:

C2 and C3 in parallel are effectively one capacitor with value C2+C3. What is then left is a simple voltage divider implemented with capacitors.

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  • \$\begingroup\$ The OP mentioned there are two switches in the circuit (in the comments: There's a switch between Vd and C1 and between C2 and CL @P.Koch . How would you do that?), one at the input, another one between C2 and C4, i don't know if it's at the C2 branch, or after both of C2 and C3. If it just after C2 and C3, then it has some similar function with charge pump. \$\endgroup\$ – diverger Nov 24 '14 at 15:28
  • \$\begingroup\$ @diverger: I didn't look at the edit history, but what you say is not the case now, nor was it when I answered the question. Besides, anything in comments isn't really part of the question, although I see no such comment anyway. I often skip reading the comments to a question, especially if the comment chain is long. Any relevant information added by the OP must be added to the question itself, not just mentioned in a comment. \$\endgroup\$ – Olin Lathrop Nov 24 '14 at 15:37
  • \$\begingroup\$ Agree with you, but the OP sometimes may change something, even they may change the title, and the answer seems useless. Buy i can't figure out how to get the original question or history. \$\endgroup\$ – diverger Nov 24 '14 at 15:43
  • \$\begingroup\$ @diverger: When you have enough rep, you can click on "Edited xx hours ago" and it will show you the edit history. Go to your profile and click "privileges". That should tell you if you can see the edit history now, and at what rep it kicks in. I just checked, and I think it's tied to being able to edit, which kicks in at 2000 rep. \$\endgroup\$ – Olin Lathrop Nov 24 '14 at 16:42

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