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# Theory. Charged capacitor used to charge an uncharged capacitor. Many contradictions. Any lights welcome.

A relatively old problem since there are "docs" on Internet about the subject, but they suffer major contradictions.

The problem is about a charged capacitor, 1uF, at 5V, used to charge another uncharged capacitor, also 1uF. What are the final voltages? (That it the simplified problem, that is).

First, there is a need to have a resistor, else, the close loop cannot be equilibrated, or the potential cannot exist, choose the one you prefer. Indeed, without resistor, one of the possible paths passes through the 5V and a parallel path, through the same ending points, passes through a total of 0V. So, we need a not-zero resistor in the model to patch that first inconsistency.

Next, the question about if the caps are in series or in parallel. With the necessity of the presence of the resistor, we should cancel any model based on the caps being in parallel.

Finally, there is an argument about keeping the initial Q Coulomb   (  Q = C V  of the initially charged cap) up to the end  ( Q/2  =  C  V_ending, since it is distributed to the 2 caps.) but that leads to loss in energy ( initial 0.5*^2 *C different than  ending:  2*0.5*C*2.5^2 ). Furthermore, with the obliged equation:  Vc1 = Vr + Vc2  (voltage for the caps and the required resistor), the derivative through time leads to  d(Vr) = d(V1) - d(V2).  So, if the total Q must be kept, d(V1)=d(V2)  and thus, the hypothesis lead to Vr = constant. Which is incoherent.

Thanks to still be here, so my question is about if you are aware of a (theorical)  solution which could, among other things, be applicable to back check by hand the solutions of most of the simulators? The simulators do not, in general, consider the energy, since those implies second degree polynomial equations, not as friendly as linear equations, but it would be fine if the solution can also satisfy the conservation of energy,... if possible :-)

## Top Replies

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• Hi Vanderghast,

I don't think there's any contradiction when considering a basic charge sharing situation. There are some simulations in this exercise:

https://wiki.analog.com/university/courses/electronics/switched-cap-power-supplies

that start out with theoretically (almost) perfect capacitors. The situation with zero resistance is also noted - LTspice doesn't like that at all so we gave the switches a nonzero resistance.

With a nonzero resistance, indeed half the energy is dissipated in the resistor, so energy is conserved.

What if the resistance was truly zero? This is just a thought experiment, but if you had perfect capacitors that had no resistance, but did have parasitic inductance, you'd end up with a tank circuit, and the energy would ring back and forth between 1/2*C*V^2 and 1/2*L*I^2. (This should be pretty easy to model in LTspice, and should approach theoretical behavior.)

-Mark

• A tank circuit may indeed reduce the loss with a resistive circuit (50% efficiency, independently of the value of R) and I will have to investigate this avenue.
I have found my fundamental error, dR/dt  is not 0, but -2 dVc1/dt since the loss of charge in C1 is a gain of charge for C2, so taking in count the sign, which solves the apparent contradiction and I now have a set of valid 3 equations.