Self running coil?

[gotoluc]

Hi Luc,

From the waveform data your scope shot shows we can approximate pretty well how much driving power goes INTO the MOSFET at its gate source input.
The current flowing across the 100 Ohm carbon resistor is I=.3438V/100=3.438mA.  Here I considered the RMS value of the peak to peak waveform of 3.44V (yellow trace).
This is the current actually flowing into the gate source path (and this path of course includes all the loading effects coming from the output side of the FET i.e. from the drain side).

So the input power to the gate source of the MOSFET is P_inp=4.5V*3.438mA=15.47mW
(I used half of the 9.06V peak to peak value of the gate source voltage (green trace) because for a regular square wave the RMS value is half of the peak to peak value and I took your wave form as a regular one to simplify things.)
We have to add to this the power dissipated in the 100 Ohm series resistor, which is about 1.18mW (.3438²/100).

So your signal generator provides about 15.47 + 1.18=16.65mW input to the MOSFET.  Now I cannot tell you how much from this leaks through to the tank circuit, maybe others can help here too. Will think about it.

rgds, Gyula

Thanks Gyula for taking the time to calculate and post this data.

My local supplier has a CMOS version of the 555 timer and I'll pick one up today. I'll go this direction for now, as it seems to be the easiest answer. I'll power it from the capacitor bank so this way we can compare accurately what is drawn to what is produced by the pickup coil.

Luc

[NextGen67]

Luc,

Would this not be a good solution ? Then no need for a seperate opto at all, as it is build in the mosfet.

http://pdf1.alldatasheet.com/datasheet-pdf/view/139498/TELEDYNE/TC4804.html

EDIT: Attached PDF

--
NextGen67

[wattsup]

@gotoluc

Do you have any germanium diodes. You mentioned somewhere you were using a half bridge and I am wondering if you are not better to make one with germanium diodes. They may leak back somewhat but they only need .2 volts and may just add a new effect.

Also, please make sure you will always be able to go back to square one and get the original effect, while you play around these other ideas.

[gyulasun]

Hi Luc,

I attach a circuit schematic for you to test the opto coupler, first with still your signal generator output. Use a series 1.5-2 kOhm resistor between the generator and the opto diode input to confine forward current to 10mA or even below for the input diode in the opto coupler, ok?

Your 4N35 opto coupler has 7us (microsecond) rise and fall times so it should work up to about 70kHz (1/7+7).
If you still find it cannot switch at 40-50 kHz, than you may use CNY17-3 from Active Tech,
http://www.active123.com/eng/storeSection/redirect.cfm?sectionID=b2c/search/productSearchResults.cfm&itemCategoryLevel2=43754&itemCategoryLevel1=43746&number_results=12   it has 5us rise and fall times, hence should go up to 100kHz, see data sheet:
http://skory.gylcomp.hu/alkatresz/CNY17%20-1-4.pdf

The 10kOhm resistor between the gate source of the IRF640 could be increased to reduce current taken from the electrolytic capacitor, and do not let the voltage in this cap higher that 20V because this is the maximum limit for the gate source voltage for this MOSFET.

rgds, Gyula

EDIT: The output transistor in the opto coupler has a floating base pin, do not connect it to anything, leave it open.

If you can reach higher than 20V in the electrolytic cap, then use another 10kOhm in series with the collector of the opto transistor, this way you divide the higher than 20V into just half of it, now within safe voltages for the gate-source.  Now the upper limit is the collector emitter maximum voltage allowed for the opto transistor, this is 30V.

[NextGen67]

@gyula,

Just send you a PM.

--
NextGen67