# Complete discharging of Initially charged capacitor through Inductor

I have been trying to simulate a parallel resonant circuit in Pspice . The idea of the circuit is to discharge C18 (Cin) completely (This cap is initially charged to 1.3 V) through the inductor and charge C19 completely. In order to speed up the process, I have used two equal bypass capacitors (C20 charged to 1.3V) so that it does not take infinite times to discharge and charge the input and output caps respectively. I have used voltage controlled switches with a duty cycle of 0.5 to illustrate charging and discharging. The operation is similar to a synchronous inverting buck-boost converter wherein:

• In the first pulse "DT"(PW), S2, the C20 discharges, discharging C18, and the energy gets stored in L8.
• During "(1-D)T", the energy stored in L8 gets transferred to C21 which charges C19. This process ideally continues until the charge in C18 is completely replenished and transferred to C19. This strategy is basically used to solve the infamous two capacitor paradox.

The values of L and C have been calculated for an inverting buck-boost converter design using design equations. The driving frequency = resonant frequency.

Now my question is:

1. Looking at the graphs, Energy transferred to C19 is 0.25 of Initial Energy at C18.
2. Likewise, Charge at C19 is half as that of C18. Why?

In this case, there's absolutely no use of having the inductor L8, assuming everything is ideal (resistances have been added to prevent floating nodes in pspice), and this should only happen when two capacitors are connected in parallel with a switch.

• Does the frequency have to be rightly adjusted to charge C19 to 1.3V (initial voltage of C18)? (I have used resonant frequency, but have also tried to tune them, but no use!)
• This is an ideal case. How would I achieve complete charge transfer?

Please help!

Cheers, S

• "In order to speed up the process, I have used two equal bypass capacitors" - Huh? How does increasing capacitance by parallel capacitors speed up anything? – JimmyB Jun 14 at 13:55
• @JimmyB The capacitors in the beginning and end are, in reality, two actuators (capacitance of 100F). My intention, was more (strictly for simulation purpose) see, in feasible time, if the small caps of 3uF get charged to 1.3 V, which, in this case, is 0.65V, and I failed to understand why this happened despite using an inductor. – electromaniac Jun 14 at 14:10

## 1 Answer

Assuming a perfect inductor and capacitor and ideal swiches, you cannot "speed up" the process. The capacitors in parallel (C18 and C20) will act as a single capacitance, as will C19 and C21. If there were no series resistance, the voltage and current would be exactly 90 degrees out of phase. The voltage would be maximum and current zero when you first close S2. For a complete transfer of energy, you must open S2 and close S3 at the point where the voltage is zero and the current is at its maximum, 90 degrees from the start point at the resonant frequency. Then, the voltage on C19 & C21 would swing to a negative value, and you must open S3 when the current is zero and the voltage is maximum. You would then theoretically have the same energy in C19 & C21 that you had in C18 & C20 to start with. So as you have rightly suspected, timing is important.

In real life, switches are not perfect and capacitors and inductors have losses, as you have pointed out, so the transfer of energy will be imperfect.

• Thanks! That really helped to refresh my fundamentals. – electromaniac Jun 17 at 11:30