Schematics - Diagrams

[Romero]

@Wings
The circuit you posted is incomplete.one side of the FWBR is not connected properly, please have a look.
Romero

[Romero]

This driver circuit I tested myself and I find it the best so far.

[darkwanderer]

This driver circuit I tested myself and I find it the best so far.

Romero I have a question about your circuit diagram. Why you put 2 mosfet instead of 1?

[Romero]

This driver circuit I tested myself and I find it the best so far.

Romero I have a question about your circuit diagram. Why you put 2 mosfet instead of 1?

the switching is much better if both ends of the coil have a transistor also the recovery is superior, closer to a mechanical switch.This way we can send the recovery back to source directly.
The circuit is copied from the Neogen Project.

[keykhin]

This driver circuit I tested myself and I find it the best so far.

Romero I have a question about your circuit diagram. Why you put 2 mosfet instead of 1?

When you switch off both rails at the same time, the coil remains free to ring and you create "electron vacuum" - negentropy. You put the electrons in a different charged state (cold electricity) and that may lead to overunity  ;)

[darkwanderer]

This driver circuit I tested myself and I find it the best so far.

Romero I have a question about your circuit diagram. Why you put 2 mosfet instead of 1?

When you switch off both rails at the same time, the coil remains free to ring and you create "electron vacuum" - negentropy. You put the electrons in a different charged state (cold electricity) and that may lead to overunity  ;)

For example, you connect your circuit to your battery and you want to cut his power what will you do? Will you disconnect (+) (-) together? I think disconnecting only 1 is pretty enough. Kirchoff laws rule  ;). Last things about this circuit it's not the best. There are lots of useless things and it's not very efficient.

Regards...

[darkwanderer]

Here you go the better one.

[Romero]

@darkwanderer
it works with your modified circuit too but better try it my way and then see the difference.We are not looking just to drive the coil but to capture the best BEMF.

Romero

[keykhin]

@ darkwanderer, here we're not dealing with laws of classic physics, so maybe the law of Kirchhoff will not apply in this case.

[darkwanderer]

@ darkwanderer, here we're not dealing with laws of classic physics, so maybe the law of Kirchhoff will not apply in this case.

Come on  ;D you make me laugh here...

All right i think i have to explain everything...

The first diagram below shows the current way to drive high-side mosfet as i draw red.

The circuit has 2 problems here:

1-)When you try to charge mosfet's gate you are reducing the coils voltage.(second uploaded image)

2-)When you try to discharge mosfet's gate you cannot discharge the gate capacitance.(third uploaded image) So you just can't turn off the mosfet. It stays open. If the mosfet going OFF state then what can I say, god likes you :)

I don't see any other law here? If you can please tell us...

@romero

If your drive circuit works too fast you'll never get maximum BEMF. Because the coil voltage will be lower so Vcoil/Rcoil=Icoil
the amp on your coil will be lower. So the energy stored in your coil L*I*I/2 will be lower.According to this equation your BEMF will be less.I don't know what's your frequency or anything else. And if your mosfet's gate stay at low voltage you'll have heat problem.That's all I can say by looking your circuit...

[citfta]

@ darkwanderer,

           The MC3415 is a mosfet driver chip.  It is designed to be used just exactly like it is shown in the circuit.  You are only showing the on side of the chip.  There is also an off side that will discharge the gate.  The chip also has a built in hysteresis circuit to ensure quick on and off times.


           My own testing has also proven that turning off both sides of the coil will yield much greater output than only switching one side.  John Bedini has also proven this with his testing.  Please test it for yourself and you will see the difference.


Respectfully, Carroll

[Romero]

@ darkwanderer, here we're not dealing with laws of classic physics, so maybe the law of Kirchhoff will not apply in this case.

Come on  ;D you make me laugh here...

All right i think i have to explain everything...

The first diagram below shows the current way to drive high-side mosfet as i draw red.

The circuit has 2 problems here:

1-)When you try to charge mosfet's gate you are reducing the coils voltage.(second uploaded image)

2-)When you try to discharge mosfet's gate you cannot discharge the gate capacitance.(third uploaded image) So you just can't turn off the mosfet. It stays open. If the mosfet going OFF state then what can I say, god likes you :)

I don't see any other law here? If you can please tell us...

@romero

If your drive circuit works too fast you'll never get maximum BEMF. Because the coil voltage will be lower so Vcoil/Rcoil=Icoil
the amp on your coil will be lower. So the energy stored in your coil L*I*I/2 will be lower.According to this equation your BEMF will be less.I don't know what's your frequency or anything else. And if your mosfet's gate stay at low voltage you'll have heat problem.That's all I can say by looking your circuit...

you'll  see that theory is many times proved wrong in practice.
What I am saying is that my personal testing showed better results using this circuit and the mosfets are not even getting warm, hot, no way... I will use a simpler circuit just with a mosfet if that will be better, I will prefer the most simple circuit and no IC if possible.The voltage drop via diodes should not be acounted that way, I have a different view about that but difficult to explain it.
I don't like my systems to run too fast, so many times lower speed was much better than any high speed and I see that most of the replicators are looking to get high speed - that is wrong in my opinion.
This should not create any kind of dispute here, we should apreciate any info comming from anyone.

Best regards,
Romero

[darkwanderer]

@ darkwanderer,

           The MC3415 is a mosfet driver chip.  It is designed to be used just exactly like it is shown in the circuit.  You are only showing the on side of the chip.  There is also an off side that will discharge the gate.  The chip also has a built in hysteresis circuit to ensure quick on and off times.

Respectfully, Carroll

I'm saying that chip cannot discharge the upper mosfet. See for yourself Carroll i'm not talking with the ideas coming out of nowhere I'm talking with the datasheet information. As you can see with the image below that there's no such circuit like that.

The second image show two independent drive circuit.One is for high-side one is for low-side. If you don't want to see drive circuit effects on your load, you must use a drive circuit like that. Please don't say that wrong circuits work better. I'm saying this because I'm working professionally on power electronics It's my job.If you don't care It's ok go ahead like that.

If you don't believe in me maybe you will believe another persons circuit. Look for the third attachement.

[citfta]

Ok darkwanderer I will concede the IR2111 is a better choice for a high side and low side driver chip.  However I still do not agree the upper mosfet will not turn off.  When the MC3415 output switches it will ground the gate.  Since the other side of the gate will now be floating if the bottom mosfet has already turned off then the gate will actually be reversed biased and will thus turn off. If the bottom mosfet is a little slower to turn off then the gate will be discharged through the coil and lower mosfet.  Either way the upper mosfet will turn off. 

Respectfully, Carroll

[keykhin]

Quote from darkwanderer: "I'm saying that chip cannot discharge the upper mosfet"

That's the reason I always use complementary mosfets like in the schematic below. It works perfect, the transistors don't have heat sinks and I use it for ages without any problems. I know, Romero will say he doesn't like to use so many transistors, but sometime they do the job :D