Hi Partzman,
I am failing to see the gain you speak of in this circuit. From my calculations, you have a loss of 7.43 mJ in Cx1 and a gain of 3.25 mJ in Cs1 so the gain in the latter is less than what is drawn from the former. I'm also unclear how such a circuit is feeding anything back into the LAB as there doesn't appear to be a current path through the battery. Maybe you can elaborate on the numbers you've crunched or maybe I'm not seeing a crucial detail in circuit operation?
Thanks,
Dave
Hi Dave, I can see that some confusion might exist because of the way that I've presented this concept. My bad! Anyway, take a look at my post #698 with the circuit and scope traces. This post discloses the energy generation phase to the LAB of the whole concept. The circuit is not intuitive and must be studied closely because it's operation is somewhat complex. L1 is initially charged by the differential voltage between Cs1 and Cx1 that is created by V+. The instant that V+ is applied to L1, the voltage on the negative end of Cs1 (which initially is at zero volts) goes to ~+3v due to charge separation which is the volts/turn of the primary circuit. IOW, 12.5/4=3.125v . Also during the charging of L1, the voltage across Cs1 decreases and the voltage across Cx1 increases theoretically but in actual measurement may not appear to be so. This is due to the individual esr's of Cs1 and Cx1. This is evidenced by the sudden rise in the voltage on VL1a at the switching of M1 and M2 and at the end of the completed cycle. After L1 reaches a predetermined current peak, M1 switched off and M2 is switched on and again due to charge separation, the voltage on VL1a is reduced by the volts/turn to ~0v . This allows the current in L1 to collapse again thru Cs1 and Cx1 and also thru V+. During this collapse, energy is generated in V+ plus the voltage across Vs1 is reduced more along with the voltage across Cx1 is increased more. When the current in L1 reaches zero, M2 is switched off resulting in the increase in VL1a due to the esr. At the end of this cycle, we now have new voltage levels from beginning to end. These ending voltage levels are the beginning voltage levels used in the recovery circuit shown in post #713 . So yes, there is energy lost in Cs1 and energy gained in Cs1 as they are restored to their starting voltage levels. This is done at no cost to V+ . The charging of L1 also takes no energy from V+ . So, the gain of 1.928mJ seen in post #698 comes at no cost and is therefore infinite. I might add that the 160uH inductor labeled L1 in the recovery circuit should be labeled Lr as it is separate from L1. Also, L1 is shorted in the recovery mode to prevent unwanted interaction from it's open inductance as the current thru Ls1 creates induction in L1! Regards, Pm
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