Because the dc offset created on L2, by the L1 impulse, is slightly less in voltage due to losses, The peak of the impulse on L2 becomes positive, which in turn opens up the body diodes of the B mosfets.
greater isolation of the high voltage DC might make it better.
I don't expect to much problems from this as it only is the top of the impulse.
The impulse is used to excite L4, which should still work. (needs further testing at higher voltages)
So the positive peak of the impulse is passing though the body diodes of the B mosfets to the negative side of C4, which in turn will be slightly deminished in voltage. But I assume this wont be a problem either.
B mosfets, should be placed between C4 (drain) and C2 (source), with short thick leads.
the DC offset is greatly deminished during these first tests, when the B mosfets are working.
Instead of seeing a DC discharge and recharge, the voltage stays low.
the impulse has less peak voltage, but still should charge the DC up. Where is the energy going?
Why is the DC so low? Maybe 4 steps is simply to little. lets double it to 8 impulses per discahrge