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Anyone tell me if this capacitor question is correct please?

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gjpollitt

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as below. I would have thought there would be 0v at c3 as c2 would block the dc ?
 

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Hmm, by what it says, i'd also give that answer, but I agree thats a bit of an odd question. The text makes sense, but the schematic confuses it, I thought what you thought when I saw it. Surely you'd use resistors for this anyhow :lol:
 
I really think it should be 0v across C3, and I think I'm going to set it up in lab and see what I get.
 
Actually now that I think about it, its correct. You have to consider the transient response of the caps, so they will both charge to equal values. It make sense to me now. I'm still going to test it though.
 
Yes, as I suspected the 2nd time. I however, used a 1k resistor to limit the current, but the voltage across C3 is due to the transient effects (charging times of the capacitors).
 
_3iMaJ said:
Yes, as I suspected the 2nd time. I however, used a 1k resistor to limit the current, but the voltage across C3 is due to the transient effects (charging times of the capacitors).

if you were to charge C2 with 20v and connect it, then C3 would indeed have 0v across it.

and it's a common mistake to think that one capacitor would block the other from charging since it's between C3 and the supply, however all the current that enters one terminal of a capacitor must also leave the other terminal, and that current charges the other capacitor as well. (it goes right back to the fundamentals of a series connection... the currents are equal)

ah, the joys of forgetting about transient response :lol:
 
C2 and C3 will each have 10v across their terminals (this means C3 is at 10v versus gnd and C2 is at 20V versus gnd). There are many ways to look at it. Logically there is no way the circuit could favor C2 over C3 or vice versa for one. Charging a cap requires current and that same current must flow through the second cap (Kirchoff's Current Law). Any time current flows through a cap it either charges or discharges it. In this case the same current flows so they both get the same charge.

For example, say the caps are only 100pF and you put a meter on C3. The capacitance of the connection and the load impedance of the meter, though very high, may quickly discharge C3 to 0v and cause C2 to charge to 10v. There are many ways to look at what just happened but for starters Kirchoff's Voltage Law says C2 + C3 = 20v at all times.

The internal leakage current as well as any board leakage can cause the same problem, especially over a significant period of time. It gets worse if Vc2=20V, Vc3=0v and you remove the 20V source and replace it with a load. Current will flow out of C2 but in that direction it will charge C3 backwards, which will kill it if it's polarized.

Thus it is potentially unreliable to try to increase the voltage rating of the cap (the most plausible reason for doing this) this way without using something to stabilize the center voltage like overwhelming the leakage with an external voltage divider. Also zener diodes can be used.
 
Oznog said:
C2 and C3 will each have 10v across their terminals (this means C3 is at 10v versus gnd and C2 is at 20V versus gnd). There are many ways to look at it. Logically there is no way the circuit could favor C2 over C3 or vice versa for one. Charging a cap requires current and that same current must flow through the second cap (Kirchoff's Current Law). Any time current flows through a cap it either charges or discharges it. In this case the same current flows so they both get the same charge.

For example, say the caps are only 100pF and you put a meter on C3. The capacitance of the connection and the load impedance of the meter, though very high, may quickly discharge C3 to 0v and cause C2 to charge to 10v. There are many ways to look at what just happened but for starters Kirchoff's Voltage Law says C2 + C3 = 20v at all times.

The internal leakage current as well as any board leakage can cause the same problem, especially over a significant period of time. It gets worse if Vc2=20V, Vc3=0v and you remove the 20V source and replace it with a load. Current will flow out of C2 but in that direction it will charge C3 backwards, which will kill it if it's polarized.

Thus it is potentially unreliable to try to increase the voltage rating of the cap (the most plausible reason for doing this) this way without using something to stabilize the center voltage like overwhelming the leakage with an external voltage divider. Also zener diodes can be used.

This analysis is correct. In practice though, usually resistors are paralleled across the capacitors that split the voltage so they "balance" or share the charge more equally.
 
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