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This is a new concept of an RLE current pump that utilizes a constant current/voltage load in a unique configuration. Let me first say however that this simulation is by no means a final solution but rather a potential concept that needs to be engineered in the real world. The caveat for this sim is that extremely short high voltage transients occur during the complementary current transitions of P1 and P2. This is due to the fact that there are no capacitance's included in the models. This can not be achieved with the various capacitance's involved in normal inductors and switches so this is basically a learning tool to attempt to achieve these results in a bench version.
The basic idea is to provide at least a two primary, single secondary transformer wherein the primaries are alternately switched out-of-phase to each other while producing a constant DC emf in the secondary. This secondary DC emf is matched to a DC voltage of 5.5v as a load through a constant current inductor L2 that maintains a relatively constant current to V15. This is one source of output energy.
To provide a constant current in L2, L3 is added to form a constant current transformer with a coupling k=.5 for simplicity as other k factors may be used . L3 is supplied with a ramped current source I1 during the 200us of operation that starts at 200ma and finishes at 300ma. With the currents in a bucking mode in this current transformer, the current in L2 is maintained relatively constant. As will be shown, this current transformer provides another means of output energy.
In operation, P1 is first connected to the 12V supply V10 via S4 for 10us with a 20us duty cycle. Alternately, P1 is switched off and P2 is connected to V10 via S1 for the same time period. This switching in effect shuttles the flux in the outer legs of the core between P1 and P2 while producing an emf in the S1 of ~5.5v DC. This EMF is the result of P1/P2 coupling to S1 of k=.455 so EMF = k*12 = 5.46v . The constant EMF of the secondary is possible due to the constant current load on the secondary and the alternate switching of the primaries.
Another important aspect of this flux switching is the eventual settling of the peak currents reached in P1 and P2 to plus and minus values. In this core, the coupling between P1 and P2 is k=.45 . During the first switching cycle of P1, the peak current reached is 99ma starting from 0ma. After four cycles, the peak currents in P1 and P2 are ~+68ma and ~-31ma alternately. This can be calculated by +Ipkrun = Ipk*(1/(1+k)) = +.0682 . The negative peak is simply the difference between the initial Ipk and the +Ipkrun value. What is important about this is the fact that energy is returned to the input supply V10 during the negative portions of the input current from both P1 an P2. This effectively lowers the input power when creating the same S1 EMF. As the k factor is increased, the more input energy is saved until theoretically, when K=1 is reached, the input power would be zero.
The first pix is the waveform plot, the second is the schematic, and the third is the power/energy data.
When viewing the data, we see that the input energy from V(vs)*I(V1) is 46.336uJ . Also, the output generated across V(Vv15)*I(V15) is 222.41uJ . The energy across the current ramp I1 is seen to be -V(VL3)*I(I1) which is 148.72uJ which is an actual gain. However, depending on how this current generator is implemented, it could be a loss or gain so we'll consider both options.
The other source of gain is the current transformer L2/L3 which will require explanation. We first need to calculate the aid and buck inductance values for L2/L3. Since M=k*(L2*L3)^.5 therefore M=.5*.025 = .0125h.
Then, Laid=L2+L3+2M=75mh and Lbuck=L2+L3-2M=25mh.
The initial currents in L2 and L3 are +200ma and -200ma respectively noting the dot convention. So we are in a buck mode therefore, the starting energy in L2/L3=.2^2*.025/2=500uJ . The ending currents in L2 and L3 are +205ma and -300ma respectively. Since L2 and L3 are identical in inductance, we can now take the average sum of the current magnitudes to calculate the buck energy. So, Ubuck=(([.205]+[.300])/2)^2*.025/2=797uJ . However, we also have a differential in the + and - currents which allow us to apply the average difference of the current magnitudes to calculate the aid energy. So, Ugain=(([.300]-[.205])/2)^2*.075/2=84.6uJ .
Therefore, we have an ending energy in L2/L3 of 797uJ+84.6uJ=881.6uJ . With the starting energy considered, we have an energy gain of 881.6uJ-500uJ=331.6uJ .
So, the COP with I1 used as a gain source is (222.41e-6+331.6e-6+148.72e-6)/46.336e-6 = 15.16 . With I1 considered a loss, COP = (222.41e-6+331.6e-6)/(46.336e-6+148.72e-6) = 2.84 .
Research is ongoing.
Regards,
Pm
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Topic: partzmans board ATL (Read 36109 times)
