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This is a test of the previous asymmetrical xfmr with a 100 ohm 1% precision film resistor load to show the device produces real excess power.
The schematic is shown first and the the load R1 is seen connected between the low side of L2 and ground therefore, it has current conducting through it for the whole cycle. The L2 secondary has been changed to 90T and at 200ma has an inductance of 14.7mH.
Once again, in the following scope pix CH1(yel) is generator input, CH2(blu) is supply voltage, CH3(pnk) is S5, CH4(grn) is the current probe, and the Math(red) is the average of the instantaneous products of voltage and current.
Pix 1,2- The input power is 5.304w over 7.358us for Uin = 5.304*7.358e-6 = 39.03uJ. The power returned to the power supply is 4.769w over 6.442us for Ups = 4.769*6.442e-6 = 30.72uJ. The net input energy loss is Uinloss = 39.03uJ-30.72uJ = 8.31uJ.
Pix 3,4- The L5 start current is 202.2ma and the finish current is 201.7ma for a net energy loss of UL5loss = ((.2022^2-.2017^2))*.267/2 = 26.96uJ
Pix 5,6- The L2 start current is 203.3ma and the finish current is 216.3ma for a net energy gain of ((.2163^2-.2033^2))*.0147/2 = 40.09uJ.
Pix 7- The current through R1 is 206.1ma avg (the rms is nearly identical) for a power of R1pwr = .2061^2*100 = 4.25w over 14.04us for an energy of UR1 = 59.67uJ.
So, the net input energy consumed is 8.31uJ + 26.96uJ = 35.27uJ. Please note that both the energies in R1 or L2 are greater than the input energy. The overall COP = (40.09+59.67)/35.27 = 2.83.
However, if one considers the energy in L2 recycled back to the supply at 100%, then the energy in R1 is totally free.
As a generator, L5 would not be recharged cycle-by-cycle but rather after a number of cycles to keep switching losses down.
Regards, Pm
Note: With this version, there may still be gain but it is difficult to say without re-running the test fixture. Pm
« Last Edit: 2021-12-08, 20:09:08 by partzman »
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