VIC-PICK UP SECTION — IonizationX

#1 · 8 Mar 2011

Hey guys has anyone figured out how the pick up coil works with the over all circut?

#2 · date not recorded

If by pick up coils you mean the pulse indicator coil, o tell you how it can work. The VDD 12v forms with the resistors a voltage divider that maintain a predefined voltage in the + and - comparator inputs, The diodes limit the voltage difference between the pins to 0,7 volts, and the other resistor determine the gain and the other the load.

I think the resistors should form a 1:5 voltage divider so two resistors of 500 on each side and a 100 ohm resistor going to ground seems ok... The current would be 20ma and the power dissipated 24milliwatts if using 12v as vdd and there would be 2v in the inputs of the comparator (relative to ground). I think that the gain transistor could be something like 5k ohms...

Well this are some values i'm working on to test here...

The pulse indicator coil i would put in a separated coil in series with the resonant coil, so it can get a purer resonant signal... I think a current transformer could also be used. But voltage is ok too. just need to limit the current...

I'm still trying to find witch is the best signal that the pll wants, i mean the voltage of the signal cause the kind of signal is a 50/50% square wave. I just don't know if must be 0-5v or maybe from 2 to 3v square wave-.

If anyone know please... I all ear.

Kali_ma_Amar

#3 · date not recorded

Quote

I just don't know if must be 0-5v or maybe from 2 to 3v square wave-.

Lol, well, it just has to get to the voltage which will be interpreted as high from your PLL-Chip (4046). The exact value, you can get from a datasheet.
E.g. if you let the 4046 run on 12volts, then you best put out 0/12V signals from your pickup-OpAmp. Sure the OpAmp is not able to put out a 12V signal if feed with 12V, but e.g. 10V would surely still be OK.
About the resistors. I think 20mA at 12V is a bit much. That much current is IMHO not needed. E.g. Stan used two 10kOhm resistors, which is about the size I would have taken too. I would have gone a bit smaller, not to be influenced too much by induced currents (noise).
You also have to remember, that if there's a resonance in the VIC, the voltage of the pickup-coil will be much larger than 12volts.
And if you wanna use a current transformer, then the PLL gets a problem. For the current has 90° offset. Then you would have to take the first comparator of the PLL, instead of the 2nd. But the first has the problem, that it also locks on harmonics, as it doesn't use a flip-flop.
BTW: Just for informational purposes, he used a 1MOhm in series with a 100KOhm for the feedback. The feedback is responsible for the hysteresis.
It does looks, like he additionally added, a capacitor. IMHO I already thought this is strange before, as I designed additional filter capacitors to my Pickup-circuit. You just don't wanna have some high frequency noise on the pickup-voltage.

sebosfato

#4 · date not recorded

Thanks for the input man. I'm having many problems here, i just blowed three mosfets...

I only have 1 now.

I'm going to wind more turns on the primary cause i think i wounded not many turns.. I found that when the duty cycle is 0 even being no pulse there the mosfet stay on so i think that when the inibit pulse leaves maybe a high logic.. i need to solve that. Don't know what is happening, my oscilloscope don't measure dc. Seems some kind of avalanche effect. I think that i'm going to add a diode in parallel with the mosfet and with the coil too. maybe the mosfet is not able to turn off correctly because of the impedance of the driver, i'm going to take out the 100 ohms resistor and connect the driver to the source, and not to the ground...

the problem is that my money finished...also the credit.. hehehe

Kali_ma_Amar

#5 · date not recorded

Well, what kind of protection do you have? IMHO a varistor or transil-diode is an absolute must have in such a prototype circuit, where you never know, what you will get (the integrated Z-diode of the FET cannot handle much energy). And, as always surely some z-diode(s) for the protection of the gate (the gate is quite sensitive to overvoltage). And last but not least you certainly need a fuse, to protect the FET, if anything goes wrong in the driver circuit, so that not too much current can flow, if the FET stays too long ON.

This is how I made my driving circuit. And then you don't so easily kill it. If something is going wrong, you will usually first see this, by the fact, that one of your protection devices are getting hot. then you know. Immediately switch off, and look why.

What FETs do you use? What driver?

TonyWoodside

#6 · date not recorded

Yes I have figured out the Pulse Indicator circuit. When the circuit is out of resonance the output of the Op Amp will be high and when this High signal is sent to pin 14 of the PLL circuit and this lets the PLL circuit know that resonance hasnt been found. When the feedback coil senses resonance, the output of the Op Amp will produce 50% square waves at the same frequency as sensed by the feedback. Very simple I might say :-) I will let you guys know more about this when i finished up building the circuit and testing.

sebosfato

#7 · date not recorded

Thanks for the description.

I'm using the ir4427 driver and was using irfp460 mosfets, all died including a driver and one section of the other driver...

Don't know what to do..

I need something that rocks solid up to 250v 3 amps...

Could you help me design it? maybe you can make a schematic?

I tried all that i could but could not keep the mosfets from blowing.

Tomorow i'm going to get this mosfets and some resistors

irf740 x3
irf840 x2
Irfp460LC x1
Irfpg50 x1

If you know any part numbers you recommend for working at this voltages...

I still have some IGBTS modules here so i will still working until i have components to blow...

The igbts are more solid... i hope...

sebosfato

#8 · date not recorded

Quote from: TonyWoodside on March 10, 2011, 20:59:56 pm

Yes I have figured out the Pulse Indicator circuit. When the circuit is out of resonance the output of the Op Amp will be high and when this High signal is sent to pin 14 of the PLL circuit and this lets the PLL circuit know that resonance hasnt been found. When the feedback coil senses resonance, the output of the Op Amp will produce 50% square waves at the same frequency as sensed by the feedback. Very simple I might say :-) I will let you guys know more about this when i finished up building the circuit and testing.

Nice one thanks very much this explain manythings... but if so then the resonant scaning circuit is a must have not an option. when i tried it two years ago i didn't figured out to it work the way you describe. would you have a diagram with the resistors values?

TonyWoodside

#9 · date not recorded

no problem, it looks like, if im reading the data sheet correct, that pin 14 has a built in inverter which will convert the High signal to Low. The ONLY time the indicator circuit will output the square wave signal is when its in resonance

sebosfato

#10 · date not recorded

I thought it was only the input of the phase comparator. But is clear if the phase remain aways high the filter capacitor will be dicharged and thus the freqency will change going down so not necessarily it will search for the resonance ..

Stan was very genius...

sebosfato

#11 · date not recorded

The igbt solved pretty much all the problems, it runs solid with nothing connected to it seem that it already have the protection.. look the pic... I worked till the fuse blew. I was puting like 175 volts...

I found that it can limit the current if i reduce the voltage in the gate. AMp restriction... hehe

Now i'm going to add more turns on the primary.. .and try adding a diode in parallel with the primary..

igbt.png

TonyWoodside

#12 · date not recorded

here is a simulation scope shot of the output of the "Pulse Indicator Circuit" when its in resonance and when its out of resonance. The image on the left is the resonance signal from the feedback coil. The top line is the output signal of the OpAmp and the bottom signal is the resonant signal. The image on the right is the non-resonant signal, top line is the output of the OpAmp and the bottom is just basic square waves from the feedback coil. Basically resonance is detected when the input signal to the OpAmp is BIPOLAR and when the signal is UNIPOLAR it is considered out of resonance. Very simple huh?
(http://www.globalkast.com/images/tonywoodside/Pulse_Indecator_Circuit_Resonance_Detected.PNG)

sebosfato

#13 · date not recorded

I understood, thanks for that. I'm chilling a bit later i'm going to make more experiments. i'm going to add another coil in series in the same core, to try reducing the resonant frequency further.. i'm also going to wind more coils in the primary to have more impedance so my voltage to amp ratio is greater as load for the variac..

could you provide the circuit schematic with the resistor values..?

what is this program? simulator?

TonyWoodside

#14 · date not recorded

I use MultiSim 11 Pro Edition for my simulations, its a really good program. Heres the schematic with values for the Pulse Indicator circuit.
(http://www.globalkast.com/images/stanmeyer/Fig.9.PNG)

Kali_ma_Amar

#15 · 11 Mar 2011

@Tony:
Just for informational purposes: The resistor values of the LED part are not like the ones Stan used in the original. Just to mention. Sure it's not important, as it's for just driving the LED anyway...
The capacitor,as you placed it, wouldn't IMHO make too much sense, and is not the way Stan put it. And the symbol of the capacitor shouldn't be a normal electrolyte cap, but a bipolar one.
Sure, these are just details, but I like to be quite straight in these things...I know, I'm a little picky...

Edit:
BTW: Why did you remove the VDD connection in the schematic? You need to tie the circuit to a voltage which lies in the OpAmp input range. Otherwise the circuit will not work all the time. You don't need to make 2 coils and tie them in the middle, as in Stans patent circuit. It is enough, when you just tie one side. I would tie the neg side to 5V. Like that you can make sure, that the inputs to the OpAmp are always in a valid range. I think Stan did the same, for in the 5-coil-pic you can clearly see one 5V wire going to the windings.

Edit2:
Hmm, it seems like the server was having some real difficulties. E.g. the few posts after this one (especially the one by Don) were deleted.
Don stated the real values for the LED-circuit, as can be seen from the original VIC-PCB-picture. And as it seems, Tony now corrected these resistor values.

webmug

#16 · date not recorded

In the patent "WO1992007861A1 CONTROL AND DRIVER CIRCUITS FOR A HYDROGEN GAS FUEL PRODUCING CELL"
TX3 has also a center tap, connected to Vdd Positive supply voltage.

br,
Webmug

Newguy

#17 · 15 Mar 2011

is there anything from the negative side of the overall circuit part of the pick up section?

Newguy

#18 · 19 Mar 2011

i was really hoping i guess for something along the lines of picking up FID signal.
http://www.magritek.com/videos.html

stanmeyerlives

#19 · 6 Apr 2011

Wow, you guyz are really smokin'!
This is a complex puzzle, but still a puzzle.
The more we work on it, the closer we will get.

Meyer LIVES!

sebosfato

#20 · date not recorded

Quote from: newguy on March 19, 2011, 12:17:17 pm

i was really hoping i guess for something along the lines of picking up FID signal.
http://www.magritek.com/videos.html

2,5khz hun
cool

stanmeyerlives

#21 · date not recorded

Tony,

Can you verify the ECG918M is the correct IC for the pulse indicator ckt?
It is a hard-to-find part if it exists!
I do not see the two 1N4003's on your board either, do you use them?