Cap Dump Circuit

[Dave45]

Look at Mikes cap dump circuit

The bemf from the coils are pumping massive pos energy into the cap and battery but nothing on the neg side.
When all that pos energy hits the cap it creates a neg deficit on the other side of the cap, the battery has to fill this deficit.

Our circuits need to collect bipolar, pos and neg.

[Matthew Jones]

Thats not hard to do. You have complicated too much using a buck and boost circiut.
You can build self resonator to deliver power to both plates. Works alot as you described with a PNP and NPN.

I'll draw it up for you. I have one the garage if I get time I'll film it for you.

Basically all you need is a Joule thief type circiut with trigger in one of the coil. You can branch trigger to drive both a NPN and PNP with 2 diodes in the opposite direction.

I'll draw it up.

I built that about 8 years ago.

Matt

[BroMikey]

Now yer talking my lingo

Thats not hard to do. You have complicated too much using a buck and boost circiut.
You can build self resonator to deliver power to both plates. Works alot as you described with a PNP and NPN.

I'll draw it up for you. I have one the garage if I get time I'll film it for you.

Basically all you need is a Joule thief type circuit with trigger in one of the coils. You can branch trigger to drive both a NPN and PNP with 2 diodes in the opposite direction.

I'll draw it up.

I built that about 8 years ago.

Matt

@Matt

Glad to hear you say that Matt.

That's a great idea because the the joule thief can be self adjusting.

Mike

[BroMikey]

Gotta Git

Thats a duel N channel driver. It will work but IXYS Fets and IGBT's use larger gate capacitance, so a high amp driver is the way to go to get them moving fast.
Those Fets you using don't require that much. And really you not moving that fast, you only have one shot every so often. At most 2-3 times a second so if you want a high turn on rate you can add more capacitance to the driver side for a pretty quik turn on. Search "Bootstrap drivers in Power circuits"

With what your doing you could get away with driving the gate .5 amp at 18 volt. That will turn that fet on faster than you need.
The bigger the driver the more you have to suppress transient response on the gate circuit. So get ready for that if you move up.

The biggest thing you need to do is learn to calculate the need for power at your gate. All gates have a different capacitance value, and thats what dictates the power of the driver along with the speed the circuit is running.

@ Dave45
You lost me with the magical stuff....


Matt

I will be back later but wow this is what I need right here, what you said about gate times and capacitance. I checked that number at my gate and I think it was 2500PF??? I think so using the calculator?

Is that what I should do? Use a capacitance joule calculator?

Then based on THAT number of joules as long as I do not exceed the value I won't fry my gates, right? I think this must be right.

This is the way I found out about "Turn ON"

By the way thanks for pointing out the special FETS required on that last half bridge.

(1) I put a 10k base to emitter and 500ohms base to supply and proceeded to raise the voltage from 8vdc to 24vdc (My homemade supply won't go any higher yet )

(2)Next I changed the 500 ohm to 200 ohm now and "ON" times went from 500uS to 200uS at the higher voltages like 15vdc and for my fet this is good.

(3) Now I use a 100 ohm to base for a 1 uS rise


So experimenting with more gate capacitance sounds pretty KOOL


Mike

[Matthew Jones]

You don't exceed 18 volt on a gate. Most can only handle 24 volt or so but that is real stress. Only feed it 18 volt.
With the calculator you can figure out how many joule in the gate and you can use whatever method or driver to charge it to that point as fast as you need to. Faster turns ons mean less heat loss less wear and tear.

I'll work on that circuit tonight.

Matt

[BroMikey]

With what your doing you could get away with driving the gate .5 amp at 18 volt. That will turn that fet on faster than you need.
The bigger the driver the more you have to suppress transient response on the gate circuit. So get ready for that if you move up.

Matt

Yes thats why I chose this one because it might actually be ABLE to turn on 12 MOSFETS like I use, all at the same time and have enough power to do it.

Or even 6 IRFP460's would make my day. I will be using this same circuit for a large joule thief (JOHN BEDINI SOLID STATE ENERGIZER) in generator mode operating at 9khz as always.

How many is the question. How many IRFP460's can I turn "ON" effectively with that big half bridge?????

That's the question. The circuits I work on require that I drive more than one or two fets so I think using one driver to power as many as possible makes more sense in the long run. Less problematic all of the way around.

Fewer parts all driven by one driver would be awesome. As it stands I am looking at one driver for 2 FET's and that makes for 6 drivers and 12 devices.

Then I wonder how could 6 driver chips be put in sinc??? And the list goes on.

Will they all pull together

Will the drivers all fire perfectly at the same time or will there always be more deviation going this route???

You see what I mean about the 12 fets being driven by a single driver?

There is a 9amp driver also.

Mike

[Matthew Jones]

I wouldn't drive 12 fets with single driver. You'll get heat in some and not in others, cause they are not all going to match and you will have hard time matching them. Some will have more resistance at the gate and that will be enough to stop it from fully charging.

Here is that circuit. Mind you its just the basics, to show you how the power flows, no specific parts.

https://www.matthewcjones.com/powerBackup1/BothPole.jpg

This will fill a cap quicker than just charging the positive side. you will need 2 separate coils one (North) wound beginning to end clockwise with the trigger winding and the other (South) wound counter clockwise. Do not take clock wise wound coil and run it backwards. This causes BEMF in the coil to reject input power and causes an imbalance. The first windings that receive power should always be closet to the iron core. Do not use a toroid or transformer body as they are designed to contain the magnetic field.

The north coil will produce a positive spike the south coil will produce a negative spike.

Now I understand the temptation to drive this circuit with high power. IE fets and drivers but you are not going to get the return you expect. You want large coils with as little input power as you can stand and bunch of them if you want higher results. Its just like a monopole or joule thief the point is to use as little power for as big a spike as you can get.

This circuit if built and tuned correctly can double your available power. I have done it many times.

I'll add a few details tonight and repost it.

Good luck.
Matt

[Dave45]

Outstanding thanks Matt
We'r on the same page
Hope you dont mind I added ionization electrodes to the circuit.

[Matthew Jones]

Add that to this one.

https://www.matthewcjones.com/powerB.../BothPoleM.jpg

I forgot the caps across the diodes for quiker turn off. Bedini showed us that along time ago, really works good.

Matt

[BroMikey]

Attention

I wouldn't drive 12 fets with single driver. You'll get heat in some and not in others, cause they are not all going to match and you will have hard time matching them. Some will have more resistance at the gate and that will be enough to stop it from fully charging.

Here is that circuit. Mind you its just the basics, to show you how the power flows, no specific parts.

https://www.matthewcjones.com/powerBackup1/BothPole.jpg

This will fill a cap quicker than just charging the positive side. you will need 2 separate coils one (North) wound beginning to end clockwise with the trigger winding and the other (South) wound counter clockwise. Do not take clock wise wound coil and run it backwards. This causes BEMF in the coil to reject input power and causes an imbalance. The first windings that receive power should always be closet to the iron core. Do not use a toroid or transformer body as they are designed to contain the magnetic field.

The north coil will produce a positive spike the south coil will produce a negative spike.

Now I understand the temptation to drive this circuit with high power. IE fets and drivers but you are not going to get the return you expect. You want large coils with as little input power as you can stand and bunch of them if you want higher results. Its just like a monopole or joule thief the point is to use as little power for as big a spike as you can get.

This circuit if built and tuned correctly can double your available power. I have done it many times.

I'll add a few details tonight and repost it.

Good luck.
Matt

Nice diagram.

Not to much more complex, yet it is twice as hard to see for a beginner.

That would be me.

I will be back with an update on my 6 fet's charging the entire bank today. Running all night and today breaking them all in.

Those two large batteries in the basement are getting their backs beaten right now and are at about 28vdc.


Mike

[Dave45]

The trigger coil gives a pos pulse which activates Q1(npn) which pulses L1 with neg current, the bemf from L1 is pos it runs through D1 into the pos side of the cap.

The trigger coil gives a neg pulse which activates Q2(pnp) which pulses L2 with pos current, the bemf from L2 is neg it runs through D2 into the neg side of the cap.

[Matthew Jones]

The trigger coil gives a pos pulse which activates Q1(npn) which pulses L1 with neg current, the bemf from L1 is pos it runs through D1 into the pos side of the cap.

The trigger coil gives a neg pulse which activates Q2(pnp) which pulses L2 with pos current, the bemf from L2 is neg it runs through D2 into the neg side of the cap.

And because of the near stress on both plates you get dielectric charging. So with a 20k uf cap rated at 100 volt it takes about 6 pulses on each coil to hit 28 -30 vdc. Each pulse (In my setup) being .4 joule input + transient spike. 20k uf cap at 30 volt is 9 joules. 12 x .4 is about 4.8 joules. So a 2x gain +-.

Matt

[BroMikey]

Video BroMikey's Dump

And because of the near stress on both plates you get dielectric charging. So with a 20k uf cap rated at 100 volt it takes about 6 pulses on each coil to hit 28 -30 vdc. Each pulse (In my setup) being .4 joule input + transient spike. 20k uf cap at 30 volt is 9 joules. 12 x .4 is about 4.8 joules. So a 2x gain +-.

Matt

Here is my offering for the day of my 6 devices up and running.

Capdump 6devices BroMikey Alum Batteries Only - YouTube

Mike

[Dave45]

Looks like your thumpen the battery's pretty good with pos energy. I have not really studied battery's that much, maybe this is what they need.

It would be interesting to see how they would react to a pos and neg dump.

[Dave45]

Folks this little simple circuit Matt has shown is not something you will see everyday on the forums, not that its not known, its just not talked about.

Study it