Today I made a negative DC offset Module. After correcting a diode it now works.
it also immediately made a problem clear, which might be a solution!
attached scope shot shows it.
ignore the orange.
in yellow (DC coupled probe) shows the positive impulse, and the negative DC offset.
impulse = + 1500V
while DC offset is -1300V dc
problem, or solution! is that the impulse completely (even more) discharges the DC offset.
What I had planned, is to short the DC to ground, and then slowly bring it back up by letting the impulses charge the DC back up.
And I wanted to do it at the time of the impulse.
So a sharp discahrge like the impulse first half (quarter wave).
keep it zero, then slowly charge back up.
resulting in a single direction displacement current.
So, now I'm still thinking I could use the impulse, to discahrge. And then use the parallel mosfets, to keep the voltage of L2 at zero volt. for, lets say until the next impulse triggers again.
now... what happens then?
impulse voltage is MAX positive. So this L1 coil dielectric energy is MAX.
Then the L2/C2 tuning cap junction is switched to ground by the B MOSFET.
Thus, L1 dielectric field will fully discharge. thus the second quarter period (second half of the impulse) is cancelled.
Beautiful, so far so good.
But now comes the deal with the DC offset. C2 and C6 at both ends of L2 are charged with negative voltage.
Lets first deal with C2.
C2 is lets say 45nF. and charged to lets say -2kV
it now has the B mosfet connecting it to ground.
B mosfet switched on when the impulse was max
L1 has discharged.
But C2 charge? how ever it will play out, it will be discharged through B mosfet to ground.
So... C2 is empty. zero volts.
Now C6 comes into play.
C6 has L2 between itself and the B mosfet to ground.
So C6 discharge will be somewhat delayed, due to the L2 resistance (assuming impedance is "zero" due to series resonance)
C6 has a MUCH bigger charge (tension) then C2.
C6 will want to be discharged through L2 through the B MOSFET.
but we first need to look at L2
L2 will have a massive voltage difference over it, due to C2 being zero, while C6= charged to -2kV
That happens every time, when the impulse occurs, so nothing extra special.
So back to C6. I would like to think that the L2 resistance will limit the discharge current of C6.
from here on there are 2 scenarios.
1 is we let C6 discharge (heating up the B mosfets)
2 we turn the B mosfet off. and don't discharge C6 (maybe only partly)
1 is simple. B mosfets burn up, get hot from the massive charge dump of C6.
L2 will be provided by max current, limited by voltage of c6 and L2 resistance.
giving a discharge curve, that slowly slows down.
Meanwhile B mosfet stays grounded, C2 stays grounded...
This is just wasting energy away.
So scenario 2.
B MOSFET has discharged L1 capacity, and C2 capacity, after this short discharge period (however long it takes I dont know)
We turn off B mosfets.
Then, C6 is still charged, and will charge C2 back up with negative voltage.
this charging up of c2 is slowed down, by the L2 resistance.
A slow charge curve is present, until C2 is charged again.
Hej!
Thats what we need! fast discahrge, slow charge. perfect.
Now the C6 tension detirmines the charge time, and we can change that by changing the capacity of C6.
bigger C6 capacity, is bigger tension, more current, faster charge.
smaller C6 capacity, is smaller tension (for smaller voltage) less charge current, slower charge.
If matched properly, we could even do a discharge like this, every impulse.
But I do not think that L4 will like that.
L4 needs the impulses to get swinging. So I would prefer a current impulse once every. 10? voltage impulse?
This also means, since C6 will barely discharge, I dont need to analyse the charge time of C6.
but... the B MOSFET should be delayed, until the C6 and C2 caps are fully charged.
and... lets not forget, after C6 has charged c2 up, it will need to be recharged by some impulses again
ok. nice!
Very curious how this will work out.