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As promised, here is a special pulse generator with RLE or constant current load applied.
At start up, two bias currents are applied, 200ma to L2 and 100ma to L4. D3 is disconnected allowing the AC waveform at VL2 to oscillate at ~1.5kv peak to peak until the end of the cycle at 16.385us . At the 10us point in time, S4 is turned "off" and VL2 is connected to L4. S5 for this example is never turned off over the full cycle. The skin effect of L1 and L2 would need to be low by using Litz wire for the windings.
The 100ma bias current in L4 is increased by the positive voltage swing on VL2 to a peak of 154ma at the zero crossing of VL2 at the end of the cycle. This represent an energy gain of 171.5uJ .
At the end of the cycle, we see that IL2 is essentially 0ma and IL1 is 154ma. So, IL1 started at 0ma and ends with 154ma and IL2 started with 200ma and ends with 0ma. This current swap represents an energy loss of (.2^2-.154^2)*.005/2 = 40.7uJ .
We also see from the plot math that the input energy consumed from Vs for the complete cycle is 75.54uJ . Therefore, the apparent COP = 171.5/(75.54+40.7) = 1.48 .
Note that the VL2 voltage ends with three complete cycles and that the current in L2 is essentially zero volts. This is critical timing and the self capacitance of L1 and L2 was fudged to 21pf to accomplish this timing. Depending on the self capacitance of L1 and L2, it should be possible to have a single cycle for minimum core and coil loss in the device.
Also note that there can be no effective self capacitance in L4 for this device to work as shown. We all know this is impossible right, but here is where thinking outside the box is fruitful. If on a common core two identical windings are operated in a buck configuration, the individual self capacitance's will effectively cancel. So, all we need to do is to choose the individual winding inductance's along with the appropriate coupling or k factor to reach our final required net inductance.
For example, using the buck equation on the schematic, we could make each winding 25mh with a k=.5 and we end up with a net 25mh inductance with no effective parallel capacitance. Or, we can solve for the inductance of each winding knowing the final net buck inductance verses differing k factors. For example, Lpri = Lbuck/(1-k)*2 or for k=.4, Lpri = .025/(1-.4)*2 = 20.83mh .
How do we make such a transformer? Use an E-core like the EC series where the outside leg area is nearly equal to the center leg area and place the windings on the outside legs. The k factor is then adjusted with the center leg gap such that the smaller the gap the lower the coupling or k factor. There would possibly be gaps needed in the outer legs to linearize the BH curve for the current levels needed so all these gaps would need to be adjusted for the final inductance and k factor required.
Regards,
Pm
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Topic: partzmans board ATL (Read 36303 times)
