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This is a new form of application using a constant current source to accomplish RLE or a Reduced Lenz Effect. This concept exhibits a gain but has one function that would be difficult (but not impossible) to achieve. That function is the requirement of the voltage source for the solid state constant current generator I1 to track the voltage waveform of VL2.
VL2 is a resonance voltage generated primarily by the parallel combination of L2 and it's self capacitance of 300pf. The current in L1 leads the resonance voltage at VL2 as seen but the current in L2 remains constant at the value designated by I1. This condition can only exist with the self capacitance of L2. If external capacitance is added, then the current in L2 will follow the resonant frequency negating this circuit's gain advantage.
Both L1 and L2 start with zero currents and L1 is charged to ~123ma from Vs during the first 20us of the cycle. At the end of the first 20us, I1 is connected to L2 and as is seen, L1 now reaches a higher peak value that is determined by the resonance of L2 and the magnitude of I1. Also at this time, L1 is reverse connected to Vs to allow any accumulated energy in L1 to be fed back to the Vs supply. As can be seen, a much larger amount of current is fed back to Vs as compared to that drawn from Vs.
The only energy consumed by the circuit apart from minor dcr losses, etc, is generated by the current from I1 the voltage at VL2.
The simulation of this circuit is attached below with the plot waveforms.
The sim is stopped at 41.45us and the initial cycle is complete and from this we can analyze the data. Basically at this point in time we see that L2 has 200ma of stored current , L1 has ~ 5.9ma of stored current which is discarded, and VL2 has reached ~0 volts. What we now have is a 12mH inductor containing 200ma which has an energy level = .2^2*.012/2 = 240uJ.
We see that the input energy required from the supply Vs is V(Vs)*I(V5) = -138.2uJ. IOW, we have placed more energy back into Vs than was required to initially charge L1.
We also see that the energy consumed by I1 and the voltage VL2 for the cycle is V(Vcs)*I(I1) = 146.03uJ. VL2 and Vcs are essentially the same voltage minus the small loss in S3 which we neglect.
So, we now have a COP = ([138.2]+240uJ)/146.03 = 2.59 . As I1 is increased, the COP increases but so does VL2.
Again, the difficult part is being able to have the current generator I1 track the compliance voltage VL2 which reaches a positive peak of ~750 volts in this case. One solution would be to wind L2 so the self capacitance is higher which would reduce the peak voltage of VL2 and also reduce the resonant frequency.
Regards,
Pm
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Topic: partzmans board ATL (Read 36328 times)
