I mentioned in a previous post that a 'U' type core with a shunt could be use instead of a small/large core combo. An equivalent to this would be a modified 'E' core with equal core area on all legs. The attached simulation includes the model of such a transformer of a bench version that uses modified Magnetics P42515 'E' cores. The sim and bench results match within ~5% or so after carefully modelling the nonlinearity of each leg of the core.
The attached sim shows the results. The magnetic and electric models of the transformer utilize capacitor/gyrators and the arbitrary behavioral voltage sources B1, B3, and B4 model the nonlinear core characteristics. P1 and P3 represent the low level permeance (reciprocal of reluctance) of the outside legs while P2 is the center leg with gap.
Also shown on the schematic is a table of the what I call "recovery inductance values" which are determined from the actual discharge energy of L2 from the various peak currents shown. Using these inductance values will guarantee 100% accurate recovery energies in the final calculations even though the inductances are nonlinear. The actual inductance levels of L2 are higher than stated in the table because the losses are taken into account. For example, let's say that the power recovered from an inductor back to a 30vdc supply over 66us from a 200ma peak was 2.7w or 178.2uJ. To convert this to an inductance value that will yield this same energy level we use L = 2*UL/I^2 = 2*(178.2e6)/.2^2 = 8.91mh. We now have an inductance value that when used in the recovery calculations of the secondary L2 will give accurate results.
L2 is the 267mH constant current inductor and switches S2, S4, and S8 form a pseudo bridge that drives the primary. During the charging phase of the primary, L2 is connected to the secondary thus supplying a constant current of 200ma to said secondary. During the second phase for the primary it's energy is returned to the supply Vs. At the same time, L2 is disconnected the secondary and is clamped to freeze it's current while the hot end of the secondary is grounded to allow an increase in it's bias current. The cycle is complete at the end of a 10us period and the energies are calculated.
The mean input energy consumed from the supply Vs is shown in the plot math as V(Vs)*I(V1) = 1.715uJ.
Cursor #1 shows the ending secondary current I(V8) = 231.92ma peak for an energy gain of (.23192^2.2^2)*.0088/2 = 60.66uJ. The 8.8mH inductance for the secondary was interpolated from the table.
Cursor #2 shows the ending current in L2 that is I(L2) = 199.54ma for an energy loss of (.2^2.19954^2)*.267/2 = 24.54uJ.
The net COP = 60.66/(1.715+24.54) = 2.31. This is ~5% higher than the actual bench device tested.
Regards, Pm
