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Itsu and Steve,
For example, please review the attached simulation. This involves a three winding transformer assembly whereby L1 is the primary wound on the center leg of an E-core and L2 and L3 are wound on the outer legs of the core. An initial bias current of +/_100ma is applied to L2 and L3 respectively. L2 and L3 are therefore in a buck condition with a net inductance of 3.5 mH. This equates to an initial bias energy of .1^2*.0035/2 = 17.5uJ . There is also a 100ma bias current applied to L4 the constant current inductor.
Normally with C2 out of the circuit, VL2 and VL3 would be at ~90% 65% of VL1 and the currents in L2, L3, and L4 would remain nearly the same as the starting bias currents. L1 would begin to ramp in current and if all the energies were totaled, the net result would be conservative. However, if we now add C2, the voltage on VL2 rises much slower than the voltage on VL3 with the net result being an increase in L4 and this is the source of gain in the overall device. There is also another potential gain source with this device and that is the resultant inductor currents at the end of a cycle but that is for a later date.
We will also change the rise time on the applied voltage to L1 which gives us a better control means for more practical operation. We can see that the total input energy from the ramped source V1 is 1.187uJ . We will cover later how to produce the various signals used here along with the switching requirements but what is important here is to see that OU is possible using classical electrodynamics electromagnetics.
From the lower plot, we see the waveforms for the various components in the circuit. From the plot data, we see that at the end of the 1.64us cycle, the currents in L3 and L4 are -/+110.7ma respectively. We also see the ending currents in L1 and L2 are ~60ma.
The energy gain in L4 is (.1107^2-.100^2)*.025/2=28.2uJ .
We then have the energy recovery waveform analysis in the upper plot from the circuitry at the right side of the schematic which shows the voltage across C1 to be 50.4 volts when the currents in L1a, L2a, and L3a are all equaling ~5ma. The energy recovered in C1 is 50.4^2*.01e-6/2=12.7uJ . We will neglect the recovered energy in the inductors as it is small.
So we have a net input energy of 1.187uJ+17.5uJ=18.687uJ . The recovered and generated energy is 28.2uJ+12.7uJ=40.9uJ for an apparent COP=40.9/18.687=2.18 .
This is just the tip of the iceberg!
Regards,
Pm
Edit: Oops, forgot the proper upper plot!
Edit: Corrected VL2 voltage percentage.
Edit: See post #308 below. These calcs are incorrect!!!
« Last Edit: 2022-08-19, 14:49:44 by partzman »
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Topic: partzmans board ATL (Read 36313 times)
