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Hi Robert,
Well, one of the differences I did not consider in the placement of the switch is that the direct connection of the puffer cap to the diode bridge would make the cap to be seen by the input current draw all the time (the cap would always be in parallel with the half coils via the diode bridge) while in your schematic the cap is present only when you activate the switching at the diode bridge output and the two input MOSFET switches are also off.
Now I wonder what may cause the "effect"? You wrote that it disappears when the cap is not fully disharged: this means that for the effect to occur, a nearly zero capacitive reactance should suddenly appear across the coils (one half of the coils in this case) and the reactance of a cap is the closest to a short circuit when the cap is fully discharged. And it must be the current (what this cap starts to take from all the half coils from the moment of the cap switch-on) which may make magnetic fields in the half coils that may add / help positively to the normal fields the input current creates. And of course the capacitive current is driven by the instanteneous AC voltage difference across the series half coils (when the input current is off and the AC voltage across the half coils comes from the induction and from the remains of the collapsing fields spikes).
When you connect the resistor, the effect is less pronounced, probably because there isn't an instanteneous short across the coil like in case of the empty cap, and when both the cap and the resistor are present across the diode bridge output, then the instanteneous load is even more like a short (paralell RC impedance is at the smallest value). Obviously, the rise time depends on the load impedance (RC time constant of the charging cap) as you found.
Would like to ask whether you have the possibility to change the duty cycle for the input pulses? Their duty cycle now is less than 50% I guess. Also, you drive the two input MOSFET switches in push pull? when one of them is on, the other is off and vice versa, right?
You mentioned a possible similarity with Jim Murray reactive power experiment: while it is not known yet how Jim does 'his tricks' I think what I wrote above can fairly well explain the effect? You have a DC voltage source at the input and the input current is reduced from it when the load is switched across the half coils via the diode bridge, and it is at the output where the current is instantly high and the voltage is lagging, right? While in Jim's case the input voltage source is AC and (I assume) the lag-lead situation exists for the input voltage and current I guess.
Greetings, Gyula
PS: When you wrote "I put a bridge rectifier on just one half of the coils for this test and only capture the output between the drive pulses. If I try with the two halves, I get the usual drag with a 50% rise in amps." did you mean with "the two halves" that you placed the red coloured AC input of the bridge to the drain of the red MOSFET?
I ask because if the answer is yes, then what if you try a second diode bridge across the red half coils too? and also switch the diode output like for the first bridge (but perhaps with different timing)?
Well, one of the differences I did not consider in the placement of the switch is that the direct connection of the puffer cap to the diode bridge would make the cap to be seen by the input current draw all the time (the cap would always be in parallel with the half coils via the diode bridge) while in your schematic the cap is present only when you activate the switching at the diode bridge output and the two input MOSFET switches are also off.
Now I wonder what may cause the "effect"? You wrote that it disappears when the cap is not fully disharged: this means that for the effect to occur, a nearly zero capacitive reactance should suddenly appear across the coils (one half of the coils in this case) and the reactance of a cap is the closest to a short circuit when the cap is fully discharged. And it must be the current (what this cap starts to take from all the half coils from the moment of the cap switch-on) which may make magnetic fields in the half coils that may add / help positively to the normal fields the input current creates. And of course the capacitive current is driven by the instanteneous AC voltage difference across the series half coils (when the input current is off and the AC voltage across the half coils comes from the induction and from the remains of the collapsing fields spikes).
When you connect the resistor, the effect is less pronounced, probably because there isn't an instanteneous short across the coil like in case of the empty cap, and when both the cap and the resistor are present across the diode bridge output, then the instanteneous load is even more like a short (paralell RC impedance is at the smallest value). Obviously, the rise time depends on the load impedance (RC time constant of the charging cap) as you found.
Would like to ask whether you have the possibility to change the duty cycle for the input pulses? Their duty cycle now is less than 50% I guess. Also, you drive the two input MOSFET switches in push pull? when one of them is on, the other is off and vice versa, right?
You mentioned a possible similarity with Jim Murray reactive power experiment: while it is not known yet how Jim does 'his tricks' I think what I wrote above can fairly well explain the effect? You have a DC voltage source at the input and the input current is reduced from it when the load is switched across the half coils via the diode bridge, and it is at the output where the current is instantly high and the voltage is lagging, right? While in Jim's case the input voltage source is AC and (I assume) the lag-lead situation exists for the input voltage and current I guess.
Greetings, Gyula
PS: When you wrote "I put a bridge rectifier on just one half of the coils for this test and only capture the output between the drive pulses. If I try with the two halves, I get the usual drag with a 50% rise in amps." did you mean with "the two halves" that you placed the red coloured AC input of the bridge to the drain of the red MOSFET?
I ask because if the answer is yes, then what if you try a second diode bridge across the red half coils too? and also switch the diode output like for the first bridge (but perhaps with different timing)?
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