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I have started to learn about LC circuits, Specifically, I am simulating a case where the initial conditions are where one large capacitor (100F) has previously been charged to 1.3V, considering the ESR to be 10 ohms. I add a small bypass capacitor (lets say 100nF, which is tunable) to this large capacitor with an R of 0.1 ohms. This small capacitor has an initial voltage of 1.3V i.e. the same voltage as the large cap connected to it. When we extract some charge from the small cap to an inductor, the charge is replenished from the large storage capacitor, but at a lower rate- thus justifying the use of small bypass cap.

enter image description here

The target of my simulation is to extract at least half the energy from the initially charged storage cap to an uncharged cap (same order of magnitude). In order to illustrate this effect, after we close the switch (i.e. form the current loop) between the bypass cap and the inductor to transfer the energy, I open the switch at the point where the inductor has the highest peak i.e. highest energy (resonance) and form a second current loop between the inductor and an uncharged bypass capacitor (the aim is to transfer the charge to this small uncharged bypass cap) in the next half period, which will transfer the charge to the second uncharged storage capacitor.

I have simulated my first half cycle well. i.e. the current builds up in the inductor from the small charged bypass cap. For the next half cycle, however, the small uncharged bypass capacitor or the large uncharged storage capacitor does not show any buildup of charge.

I have attached the simulation results below for a fair understanding of the problem here.

enter image description here

3rd plot Red line: V(C17:1,C17:2) is the voltage of large initially charged storage cap/supercap Blue line: V(C18:1,C18:2) is the voltage of small initially charged bypass cap Pink line: -I(L8) is current leaving the inductor

2nd plot Green line:(C20:2, C20:1) is the voltage through small uncharged bypass cap

1st plot Pink line: (C19:2, C19:1) is voltage through the large uncharged cap/supercap (that has to be charged)

Could you please tell me where my intuition is going wrong and how could I transfer the charge better?

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    \$\begingroup\$ A couple of things to consider. 1) The inductor shown has no series resistance - which would affect results. 2) from 883ns to 1µs the inductor has current flowing through it but it is in an open circuit as both switches, U9 & U11, are open. What happens to an inductor with flowing current when you open-circuit it? \$\endgroup\$ – scorpdaddy Apr 25 at 12:13
  • \$\begingroup\$ L8=6.8 nH is basically a 1cm short circuit. What units did you want? mH? \$\endgroup\$ – Sunnyskyguy EE75 Apr 25 at 12:19
  • \$\begingroup\$ Polymer Capacitors of this size are usually in xxx milliohms and Li Ion Batteries are around 10kF and 50 mOhms \$\endgroup\$ – Sunnyskyguy EE75 Apr 25 at 12:31
  • \$\begingroup\$ @scorpdaddy Thanks for the answer. For 1) from 160ns to 883ns when current loop is formed between L8 and the charged capacitors, there is a resistance of 0.1ohms. However. I have increased the inductance value to 1mH, considering 6.8nH is short circuit. Now my question is when should the switches be opened /closed to get C19 charged? 2)I agree that was a mistake, U11 has to be opened at 883ns in this case \$\endgroup\$ – electromaniac Apr 25 at 13:14
  • \$\begingroup\$ @SunnyskyguyEE75 You're right. I have altered the magnitude of L8 in mH. In order to maximize the charge transfer from C17 to C19, I am missing out some basic ideas of timings for opening and closing the switches I believe. \$\endgroup\$ – electromaniac Apr 25 at 13:16

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