Self running coil?

[Lakes]

The video is a bit confusing, he says there is no current being drawn, but there must be *some* current (mA) being drawn to drive the pulse circuit.

How can this be self running when he has that very large battery connected?

The pickup coil with the rectifier charging the other cap needs to be removed (for now) to have a clearer picture of this setup.

I would hope the next step is to run the pulse circuit from his Cap bank...

[wings]

Thanks nice work Luc.

[mscoffman]

Hi Folks,

I see this problem differently.

I think the problem is not 'overdriving' the MOSFET gate with DC and AC which then leaks through. 
The problem is the MOSFET itself that has interelectrode voltage dependent capacitances, so it is very far from an ideal switch in this respect.

If you accept Luc used the MOSFET's output capacitance as the C member of a resonant LC tank circuit, then it will be the same whether you use opto coupling at its gate input.  And the MOSFET's gate-drain capacitance will also remain the same value as the data sheet says, where the input pulses can leak through regardless of the input opto.

@Mark, I do not get this:

"Also, decrease the drive signal until it is 2.5+0.7 = 3.2*vpp* using a resistive voltage divider on the IC side."

What is the 2.5V and what is the 0.7V?

Thanks,  Gyula

OK...Well, he seems to have picked a better solution...A little bit risky one
but I guess he won with it. Circuits may not be that efficient and may require
lots of extra components.

The interelectrode capacitance of that particular Mosfet... was the problem.
High power transistors have more capacitance...This is going to let AC through
no matter even if the DC is blocked. Rectified AC = power. Also operating fets
on low voltage is tricky..they like 10Vdc or more.

2.5V = approximate maximum voltage on the bulk capacitors actually 2.85Vdc
0.7V = one silicon voltage (pedestal) drop that mosfets like to see on their Gate
         when Source is operated <8Vdc. This will minimize fet drive AC.

With an opto, that same thing *would* happen except input AC power would be
coming from the bulk capacitor side...If you could make it work based in the
efficiency.

---

gotoluc;

Good work so far, show us the circuit and your current set up.
A you-tube video would be nice. Voltages are now getting a
little bit on high side. Don't let free run-time cause the capacitors
to overcharge and exceed their WVDC working voltage. This
part of your circuit is really going to need some more work.

Don't blow it, now.    Overunity Yea!    Sell Oil, buy Ferrites 

:S:MarkCoffman

[gyulasun]

...
The interelectrode capacitance of that particular Mosfet... was the problem.
High power transistors have more capacitance...This is going to let AC through no matter even if the DC is blocked. Rectified AC = power. Also operating fets on low voltage is tricky..they like 10Vdc or more.

2.5V = approximate maximum voltage on the bulk capacitors actually 2.85Vdc
0.7V = one silicon voltage (pedestal) drop that mosfets like to see on their Gate when Source is operated <8Vdc. This will minimize fet drive AC.

With an opto, that same thing *would* happen except input AC power would be coming from the bulk capacitor side...If you could make it work based in the efficiency.
...

Hi Mark,

Ok, now I understand what you meant with the 2.5V and I agree with that, and agree with all the rest I quoted from you above, however, I do not think we can speak about the 0.7V silicon voltage in the gate-source relation of a MOSFET. These FETs are manufactured as insulated-gate MOSFETs, meaning the gate is almost a pure capacitance with respect to the source or drain electrodes. Maximum gate source voltage rating normally +/-20V on the gate wrt source, within this voltage range the gate is an open circuit for DC and a capacitance for AC.

I am aware of DC pre-biasing the gate of a MOSFET to help ease AC driving requirements but then it is the minimum gate treshold voltage (V_TH) which should be considered as starting voltage level, (2V if I recall correctly in case of IRF640 and DC bias the gate to 1.7-1.8V or so).

Re on Luc latest youtube video: I think he still utilizes the MOSFET output capacitance to reach resonance with his still magnetically tuned toroidal coil but now the output capacitance is much lower (about 250pF) because the DC voltage across the drain source is about 17.6V and he is now at the 36.7kHz range instead of the 18-19kHz used earlier.

It would be good to know what was the drawback in using the 555 and why he found the SG IC is better?  Maybe operating frequency range was wider for the SG IC?

rgds,  Gyula

[mscoffman]

@gotoluc,

Just saw Video Test #4. 

Quote: "this can't possibly be leakage through the mosfet gate
charging the bulk capacitors"

me: Oh yes it can be... not the DC but a rectified AC pumping
action through the gate-to-source interelectrode capacitance of
a power mosfet. The same way a disposable camera flash unit
has a 3volt battery charging up a 300 volt capacitor.

gotoluc find a large bandwidth optoisolator and drive the gate
of the mosfet through that. If you have any opto in your parts
supply try that. Use your scope to make sure the input and
output signals end up the same shape eventually.

You may need a small or large mosfet to drive the led of the
opto. 10ma @ 12Vdc through a 470ohm resistor. When you
are looking at the opto spec sheet try to find the led to output
line capacitance of the opto...it should be a few pf's, very
small. The Opto should have NPN ouput.

On the output side of the opto, they bring the base of an
output transistor out, put a 100K ohm resistor or whatever the
spec sheet on the opto says to the opto collector. Connect
opto collector to the caps. Connect the emitter to the mosfet
gate. You will need a pull down resistor to pull the gate to
ground. Now use your o'scope to make sure all your signals look
correct. That the gate of the mosfet goes to ground and that it
is turning on fully. (The mosfet gate could also be pulled down
by the opto but you would need to invert the opto drive signal
and gate pulled up by a resistor).

That will do it...LF AC can not flow through a few pf's so any
energy driving the gate comes through the bulk capacitors. If
you adjust the resistors higher and use your scope to keep the
signals clean, that should work. The mosfet drive circuit will
then be properly decoupled from the signal source. Then your
above quote will be true.

---

By the way I like your using of a external probe pickup coil to
look at signals external to toroid. You may be able pick up these
b curl fields or saturation fields to help determan what is going
on, outside the toroid.

---

@ gyulasun

these details shouldn't matter...the mosfet should at least start
working, if it stops working at higher voltages it could then be
debugged. The higher that AC gate voltage the more drive
energy there is to leak through.

The SG IC is optimised for mosfets. It probably uses the bipolar
transistor to terminate stuff. They spend a whole lot of time
optimizing those circuit but only for particular apps.

:S:MarkSCoffman