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This is an improved version of Post #353 with lower supply voltage, higher efficiency, higher self-capacitance, and with a bi-directional switch for VL2 using Ixys IXTH3N120 high voltage mosfets since I have them on hand.
To maintain any given peak pulse voltage on VL2 for a fixed supply and first phase timing, as the self-capacitance of L1 and L2 decreases so must the bias current in L4. IOW, if the self-capacitance of L1 and L2 is increased, bias current in L4 may also be increased which results in a higher COP. In this example, the Bvdss min is 1200v so VL2 must be held at or below this level.
A typical self-capacitance for the winding area of a 5mh inductor used in a bench device for the core I've chosen is ~56pf. To raise this self-capacitance to 200pf, several options are available such as bifilar or multifilar winding, using a flat magnet wire, sandwiched flat coil sections, etc. Once this is accomplished, we have a go.
The operation of this topology may at first glance appear to be rather inefficient if one is paying attention. This is due to the first part of the VL2 pulse going negative which has the effect of reducing the bias current in L4. However during this time, the current from L4 entering the summing junction at VL2 has the effect of increasing IL1 and reducing IL2 as can been seen. For the following positive half cycle, the current in L4 is restored, the current in L2 falls to near zero, and the current in L1 holds at a reasonably high level due to the constant current effect of L4. This is the gain mechanism of the device.
This may not be apparent unless one studies my previous posts and analyzes the reactive power levels in L1 and L2 during an "aiding" pulse generation.
Looking at the data from the sim, we see that the input energy drawn from the supply V4 is 86.62uJ .
The starting bias in L4 is 450ma and the ending current is 451.2ma for a slight gain of (.4512^2-.45^2) * .025/2 = 13.5uJ .
Both L1 and L2 start at zero bias and end in 437ma and -3.4ma respectively. With recovery at 100%, this results in a buck energy gain of (([.437]+[.0034])*.5)^2 * .001/2 = 24.2uJ and an aid energy gain of
(([.437]-[.0034])*.5)^2 *.018/2 = 423uJ for a total net energy gain of 447.2uJ .
Therefore, the apparent COP = (13.5+447.2)/86.62 = 5.32! In reality, the energy gain in L4 would just be allowed to carry on if positive and topped off if negative.
This will probably be my last post for awhile as I am going to the bench to build this thing.
Regards,
Pm
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Author
Topic: partzmans board ATL (Read 36315 times)
