No Load
L1 L2 close coupled, L3 loose coupled 15mm to L2
L1 unifilar, L3 unifilar L2 bifilar
C2=5nF
C3=2x 0.68=1.36uF
yellow= L2 C2 voltage
orange= L3 C3 voltage (ZERO!?)
green= L3 current=2.19A pp

F=268kc/s
PSU: 2*12.1V @ 0.36A =8.7W dc input power

impulse still present, around -300V
L2 sine wave around 420Vpp

Frequency is to high, will make L3 bifilar.
How can L3 measure amperes if there is no voltage???
I remeasured with a normal voltage probe (10:1) and that is scr447
then again with a 1:1 which is scr448

So around 1 V pp on L3... very low. very very low.
but... at a relative high frequency, so lets lower it by making L3 bifilar, and increase the L2 C2 capacity

so I should recalculate the frequency, for th3 cap that I now am using, to get the same impedance

Looks like what you are seeying is a transient comming from the switches.
A mere split second were the sourcing current is "running up against a wall"

As it seem to happen every time the switch opens(after the sharp impulse) (edit: i see the switch is still in its high state after the impulse)

In this split second the current has no place to go too till the diodes open up guiding it away to your bus capacitor(assuming your still reusing the "bemf" from the previous pulse)


So then if this is not caused by the switch.
I would try and look at the phase relation of the gatedrive in comparison to the actual primary LC.
In the scope shot it looks as if the pulse happen on the voltage phase reversal.
But the ringing is actually a little late.
With DRSSTC teslacoil switching we applied a little thing called phase lead.
Where we take feedback from the primary LC and we lead the phase with a inductor (to switch a little early)
We do this to take the switching delays caused by the various components into the mix.

So what i think it works like is basicly like with a swing.
Push to early and you get the load on yourself.
Push to late and you have to be fast to put in energy.
Push at the right time and there is a high energy transfer.

With pushing to late comes that the current is allready reversed and to keep up with it you have to go faster (wich the mosfet switching a little later has to keep up with)
Going faster could mean wiring starts to induct(parasitic inductance).
But then again i cant be sure without measuring what you expirience.

If I connect my external load to the resonant capacitor, it will change the impedance
of the resonant system.

so... what if I charge the resonant capacitor
and only connect the load when it is fully charged (from resonance)

and, what If I also at the same time disconnect the cap from the resonant coil?

and what if I charge the resonance with a pulse of power that happens when the load is switched off?

So I power the resonance with a pulse when the load is off
and I power the load when the pulse is off

and then Do the same trick with feedback.
power up the capacitor parallel to the first mosfet switch. while the pulse is off and load is off.

hmm, that would change all the impedances all the time. and the power source would shift through the circuit.

the cap, feeding the L1 coil,
the L1  coil feeding the resonant coil/cap
the resonant cap c3 feeding the load
but... what Would I use to feed the first cap?
maybe... the dis connected resonant coil. hmm no as it is depleted when the cap is full.
maybe time it better.
disconnect the cap and coil when both are still charged. then I would have amps from the cap for the load
and an impulse from the coil for the first cap.

so between the max current and Max voltage points of the resonant sine wave,
there is the maximum power point (which happens twice per period).

If I disconnect the resonant coil and capacitor at the maximum power point,
then the cap can power the load,
while the coil recharges the dc capacitor at the beginning of the circuit, providing feedback.

I still don't know how to do this, if It can be done using diodes and mosfets.

but very interesting.

So a 240V ac 42W halogen bulb (assuming it is fully resistive) has a resistance of:

first, turn ac to DC, 240/1.4142=169.7V dc
P=I*I*R
P=U*I, I=42W/169.7ohm=0.2475A
R=P/I*I=42/0.06125=685.7 Ohms

To get a capacitive reactance of 685.7 Ohms, with a capacitor of 1463nF I need a frequency of:
Xc=1/(2pi f C)
685.7ohm=1/(2pi f C)
f=1/(2pi C 685.7)= 158.7 c/s
wow... That is way to slow.
I made some errors in my previous calculations. lets do it again...


So a 240V ac 42W halogen bulb (assuming it is fully resistive) has a resistance of 685.7 ohm

To get a capacitive reactance 685.7 Ohms at 100kc/s, I need a capacitor of
Xc=1/(2pi f C)
685.7=1/2pi 100000 C
C=1/(2pi 100000*685.7)=2,231 nF

Hmm... that appears to be very small.

so how does it work?

The load is in parallel with the cap, so... if they are equal then... I need
ahrg I dont get it

Vrms=240V ac
Vpeak=sqrt2x240=339.4V ac

psu 0.55A 2x 13.1V=14,4W
f=100kc/s
only L1 coil with 615pF
measuring source.

so a 370V DC squarewave at 100kc/s
it has a large overshoot of 150V why is that? diode seems to have trouble to block or is it due to the parallel caps?


Rings around 20Mc/s after discharge. probably, because it is not a discharge to ground.
So I need to ground positive again, ALSO, I still have a diode between V+ and drain, which should be removed

I measured in AC coupling, should have been DC

L2 connected to the square wave capacitor, with the inside rim. L2 is unifilar (parallel connected).
outside rim is open ended (probed by yellow probe)
orange probe is on the source of the mosfet.

The larger ring=4.73Mc/, is to be expected as the coil has some interwinding capacitance.
the at the source it rings 4x faster around 4x 4,87329=19.5Mc/s

PSU=2*11.9V @ 0.53A=12.6W @ 103.0kc/s

I now added the full coil capacitor, by close coupling L3 to L2.
L3 is bifilar, and grounded on the outside rim.
Since L2 is connected to the square wave cap by the inside rim, this makes L2 and L3 180 degrees out of phase, adding to the voltage difference, while still having Ideal coupling. (I could use a single speaker wire coil, with its 2 windings disconnected, to act as the coil cap, Less inductance that way.

green= Cap (inside rim L2)
yellow = L2 outside rim
orange= L3 inside rim

NOTE that the green channel has 200V/div, while the others have 100V/div
The orange shows the bifilar L3 not being able to handle the high dV/dt from the discharge.
To do so, I will next, add the parallel capacitor to L3

drive frequency: 106.2kc/s
PSU: 2*11.9V 0.49A=11.7W

I added 9nF in parallel to C3 making it parallel resonant.
f=100.6kc/s
PSU 0.53A @ 2X11,9V= 12.6W

Now the L2 follows the voltage of L3, which has a larger impedance (frequency depended resistance to voltage change)

Current is Low, only 1.6A pp but these are still low settings.
Also, I will need a larger capacitor to get higher currents.

My resistive lamp impedance calculations failed,
so instead I will use the 100 Ohm of my oil radiator (529VA) lowest setting.

Right now Xc=1/(2pi x 100600 c/s x 0,000000009 F)=175.8 ohm.
I want less impedance to get 100ohms, so I can match it to the load.
Still doesnt make full sense. I think it is better to disconnect C3 from L3 and connect C3 to the Load.
L3 will then also change in impedance during the charge of L2, by being disconnected from C3
would that matter? hmm???

glad i am not being distracted anymore

scr462 shows L3 current in green

the ringing of L2 is too high infrequency, to harmonically match L3.
So I will instead, of unifilar, I will make it bifilar. while still connect the inside rim of L2 to the c2

I did some tests to make a impedance graph.
only changing the value of L3:

6nf      f=122.7 kc/s   1.4847App     Xc=216 ohm
7nf      f=113.7kc/s    1.6709App     Xc=200 ohm
8nf      f=106.5kc/s    1.5337App     Xc=187 ohm
9nf      f=100.6kc/s    1.519App       Xc=176 ohm
10nf    f= 95.2kc/s     1.47App         Xc=167 ohm
11nf    f= 90.9kc/s      1.6611App!   Xc=159 ohm
16nf    f= 75.1kc/s     1.3083App     Xc=132 ohm
21nf    f= 65.7kc/s      1.2887App    Xc=115 ohm

The current probe didn't move, so why is 11nF suddenly much higher in amps? Due to high Q tuning error?

the cap values are approximations.

L2 is ringing mad, so I will make it bifilar.
I am also wondering if it can be made series resonant, with c2
 I think not, as it is open ended.
but... I can add a parallel capacitor to L2.
Then, I would have a very interesting series parallel setup in L2.

L2 now is bifilar, with a parallel capacitor C4 over it of 18nF equal in size to C3(of L3)
Power in: 2*5.5V @0.38A=4.18W
f=51.34kc/s
C3=C4=18nF
C2=615pF

now L2 is also resonant with C4,
mirroring the L3 with C3 parallel resonance

L1 is not coupled! only L2 and L3

Xc=172 ohm huh... no, this is now a dual resonant system, so there are 2 inductances and 2 capacities. from the mutual capacitance and inductance. So I can't just use that formula, as it is just for a single coil.

yellow is C4 outside rim L2
orange=C3 L3 inside rim
green is C2 source L2 inside rim

I could flip L2 over.

But... what I wonder about, is what happens if L2 and C2 become resonant also.
forming true series parallel resonance,
between c2 L2 (series) and L2 and C4 (parallel).

Since there are two diodes at the source, conducting positive, the L2 c2 resonance, whould be clipped when it enters the positive half wave since L1 has a higher impedance, the energy would return to V+ ground.
So a diode and a cap would be needed to capture it and recycle it, as before.

right now, green shows C2 is not resonant, altough it still does ring.

I can raise the negative voltage By reducing the size of L4.

I also wonder at what frequency the lmd mode is with being close coupled.

Since I now work with the TEM mode, which is the first lowest resonant mode.

close coupled means the frequency is much higher.

With a smaller C4 I will need a larger C3 to get the voltage down.
a larger C3 will also give more power.

since the lmd mode is out of phase, I can again connect L2 the same as L3, so outside rim grounded on C2.

I should swap names of C4 and C2, so it connects to L2. that is easier.

the beauty in this induction is still that PSU power is not influenced by the load or is it?

I like it that C4 adds to the negative voltage discharge. but... does it really work and help?

I need power amplification, I don't have a load yet.
But since I now use a single signal generator,
I can again use a second switch, to disconnect the load. but is that needed? worth seeing if it can be done

the dielectric induction should also work better in LMD mode. but will be tricky to find, as it is much higher in Q and frequency

I relabelled the circuit,
C1 is the square wave cap
C2 is parallel with L2
and parallel cap to supply is removed.

I reduced C2 to 9nF
and raised C3 31nF

TEM is 48 kc/s
LMD is 204 kc/s, very high.
With LMD it is ringing madly, but also inphase with L2 as predicted.
So I should now reverse connections on L2 again.
power input at LMD mode was around 0,5A at 2x 17V.

After that I went back to TEM mode, with the high power supply on.
Luckily I turned it off quickly. no damages so far I can see. the wave shape was high, and sagging in amplitude.

Interesting to see Lr (orange) ring in LMD mode. Wonder if those harmonics can be tamed by proper tuning.

So if L4 extra coil is high enough in voltage, the discharge of L2 could work, as long as they are out of phase.
only wonder what the effect is of L1 L4 coupling.
I dont count the charge, as it is much slower as the discharge so the displacement is less

to get the highest capacity, I would need the closest coil to L2, have the highest voltage.
So that would be the extra coil.

Then the parallel resonant L3 could be coupled to L1, as it will have much lower voltage but high amps.
L3 should then be induced by L4.
L3 would need to feedback into L1, reducing power supply.

Other thing I am again thinking of is that energy flows. and is set into motion by discplacment current.
When the discharge is quick enough, The energy flow is longitudinal. in or out of the conductor with the high dV/dt.

But the energy flows into from all sides of the plate coil, bakc and front.
So if it is coupled to a coil with a constant voltage (high impedance parallel resonant)
then the other side will have the the energy flowing in.

So that side needs some room/space, to let the energy flow in.
Then we can direct the energy in, by placing another constant voltage coil loose coupled to the back side.
The energy will now flow in from the gap.
This inflow, is at 90 degrees angled, and should feed into  the magnetic field of the loose coupled coil.

So ... lets try working with 3 coils.
center coil is the L2. and one side has a close coupled coil, and the other side a loose coupled coil. aka, extra coil and secondary.
lets see how that works.

Video shows true radiant effects, like wireless capacitor charging, and dc offset from radiant effects.

video is in russion, but you can turn on auto translated subtitles and choose any language (from pc).

https://youtu.be/uhNECKoqR1o

the video shows a positive capacity discharge,
which confused me.
but the primary exciter coil is. grounded and disconnected, by the open spark gap.

when the spark triggers, the coil suddenly jumps up in voltage to the capacitor level (positive).

this is the same as my negative current impulse.

the coil needs some inductance to have the voltage difference, but it will also make it ring.
as long as the spark continues, which isn't good. a resistor could also work,

as the effect is. from the high dV/dt when the coil is first connected to the cap.

the secondary would still recieve the displacement current and start oscilating.

Fascinating.

I wonder If not better can use a foil primary of 2 windings

inspired by the video,
I connected the coil capacitor, with outside rims to ground, which where facing each other.
Inside rims, open ended, orange and yellow probes for L3 and L2
C3 tuned to 18nF
F= 70 kc/s
power supply: 0.5A 2x 8V=8W

L3 only gave 150mA pp and 110Vpp Very low.
Much less then the previous tests which used the increase voltage difference of the facing resonant coils.
So What I was doing was correct.

The video just didn't have the resonant frequency of the primary exciter coil tuned to the secondary.
A crude setup, as can be expected of a spark gap setup.

I had no parallel cap over L2, so I used circuit 4

ok, lets see If I can get better high voltage out of phase difference between l2 l3, by grounding the L3 inside rim, making the outside rim resonant.
L2 is connected to the square wave cap by its outside rim, so inside resonant, which makes it out of phase with l3

I tuned both L2 and L3 with the same capacity, 9nF each for c2 and c3.
L2 grounded on outside rim, inside rim resonant
L3 grounded on inside rim, outside rim resonant
F=72.17kc/s
psu=2x 7.3V @0.42A
L3 is 600V peak to peak (low current)

scr478 shows resonant L2 inside rim in yellow, with the nice vertical voltage jumps. orange is L3 outside rim resonant 600Vpp
scr479 orange= L3 voltage, green L3= current 260mA (low)
scr480 is actual voltage difference for displacement between L2(yellow outside rim) and L3 orange resonant outside rim.

So during the charge (purple squarewave goes low) there is no voltage difference.
but during the discharge (square wave goes high) there is 500V diffenence, which is discharged to 300V.
I want to have this on both inside rims of L2 and L3, so this again means flipping L3 over, to keep it out of phase with L2.

L4 would then be series connected with L3, and L4 would have a large parallel capacitor, to produce high resonant currents.
L4 to be coupled to L1?  Lets keep L4 out of it. get this right first.

flip L3 over. ground both coils on outside rim.

 L2 would not need to face the L3 coil with it resonance so:
L2 needs to face L3, with its outside rim.
L3 needs to face L2 with its inside rim.


C3 and C2  could be unbalanced. one large one small.
large is more current less voltage
small is more voltage less current.
but I already know from practice this isn't fully true.

L3 in previous experiment with voltage impulses had both current and voltage large.
best was to make c3 small, so lets do that and make C2 big.

we can make the caps even very big, so the LMD mode can be used at a lower resonant frequency.

voltage will be low, but current high.

ok, I flipped L3 over. grounded L2 and L3 on outside rims. and made the inside rims resonance face eachother.
This works great. very happy with this result.
voltages are much higher now. for smae power at the same frequency.

PSU=2x7.3V 0,41A
F=70.98
L2 (yellow) is 747V pp (including DC charge and discharge voltages)
delta V of L2 and L3 (inside rim displacement current) is now max 730V,
which is twice discharged, one time from the c1 cap charge, and the other with the c1 discharge.
so truly pumping energy.

but now the other side of L2 is still open. but thats not for now.

this was with 2x 9nF, so c2=c3=9nF

changing the balance of C2 C3 capacitance should not make a difference in the voltage level between L2 L3.
but if it does. then it points to wards an energy gain.

C2 bigger, C3 smaller.

Then if it works. treat C3 as extra coil, and connect L4 in series, with a much larger capacitor in parallel.
or, see If I can make this work in LMD mode (which also requiers larger caps to get the frequency down