ZVS Mazzilli Driver rectified

[nix85]

I'm considering ZVS driver but i need pulsed DC output, so, will ZVS Mazzilli Driver still oscillate if load inductor is connected through FWBR?

[nix85]

If anyone has ZVS and is willing to try this simple test, would be much grateful.

[ahbushnell]

ZVS (zero voltage switching)
FWBR???  What is that. 

There are many circuits that can do ZVS.  Not enough information. 

[nix85]

ZVS (zero voltage switching)
FWBR???  What is that. 

There are many circuits that can do ZVS.  Not enough information.

FWBR - Full-Wave Bridge Rectifier, what else.
I made clear i am talking about Mazzilli ZVS.
All needed information is given and question
is clear.

[nix85]

Studying the circle it appears it should oscillate, looks like there is absolutelly no difference in comparison to ordinary LC, but i'd like your opinions.

[james_s]

You can certainly rectify the output, I've used the basic ZVS circuit many times to drive ferrite transformers and rectifying the output of those works the same as with any other transformer. The ZVS device is a form of Royer oscillator which has been around for decades.

[nix85]

You can certainly rectify the output, I've used the basic ZVS circuit many times to drive ferrite transformers and rectifying the output of those works the same as with any other transformer. The ZVS device is a form of Royer oscillator which has been around for decades.

Nice to hear that! [edit] Just to repeat i do not plan to use a transformer, just an air inductor, but doesn't really matter as longs as oscillation is preserved. And i did know it is a variation of old Royer saturation oscillator.[edit]

Can you tell me another thing that's puzzling me now, since voltage and current in LC tank are 90° out of phase, how does power transfer happen here?

Best animation of LC tank i seen, you can slow it down to 25%.

[Seekonk]

I wouldn't think it would be too pulsed if FWBR, need damn fast diodes.

[nix85]

I wouldn't think it would be too pulsed if FWBR, need damn fast diodes.

We are not talking some extreme frequencies, less than 100KHz.

[Berni]

Yes it will run.

Not sure why you would want to do that tho unless your goal is to create a compact powerful voltage doubling circuit.

As for the phase shift. Its only 90 degrees out of phase when there is no net power transfer. As soon as you start putting a real load on it the phase shifts start moving back towards zero.

[nix85]

Yes it will run.

Not sure why you would want to do that tho unless your goal is to create a compact powerful voltage doubling circuit.

As for the phase shift. Its only 90 degrees out of phase when there is no net power transfer. As soon as you start putting a real load on it the phase shifts start moving back towards zero.

Like i said goal is to run the 2 Ohm air core electromagnet (~600 turns) and that is the only load.

So same thing as in transformer happens, "real load" making current and voltage more in phase.

[Berni]

Yes transformers simply "transfer" the load across to the other side as if it was connected there. If you put a resistive load on the secondary of the transformer than the primary will also start to act like a resistive load (provided the load is much larger than the transformers own reactance to swamp it out).

But if you are going to put full wave rectified 100KHz AC into that 2 Ohm 600 turn electromagnet load you can expect the current trough your load to not be all that pulsed anymore. The inductance of that coil is likely significant(unless those 600 turns are around a very tiny coil former) so it will attempt to maintain the current flow.

Also placing non linear loads like a full wave rectifier on the output of a ZVS driver will change the waveform to no longer be a nice pure sinewave (If that's important to you)

[nix85]

Yes transformers simply "transfer" the load across to the other side as if it was connected there. If you put a resistive load on the secondary of the transformer than the primary will also start to act like a resistive load (provided the load is much larger than the transformers own reactance to swamp it out).

Already wrote about it, last post. https://www.eevblog.com/forum/rf-microwave/microwave-tl-coax-impedance/25/

But if you are going to put full wave rectified 100KHz AC into that 2 Ohm 600 turn electromagnet load you can expect the current trough your load to not be all that pulsed anymore. The inductance of that coil is likely significant(unless those 600 turns are around a very tiny coil former) so it will attempt to maintain the current flow.

According to calculation, inductance of the coil will be ~0.18H. Are you saying it will act as a choke, smoothing pulses into DC? If so how do i preven it, by no means is it acceptable that output turns to smooth DC.

Also placing non linear loads like a full wave rectifier on the output of a ZVS driver will change the waveform to no longer be a nice pure sinewave (If that's important to you)

I didn't know FWBR is a non linear load, i plan to get one with shortest recovery time, not sure if that helps with how non-linear it is.

[Berni]

Semiconductors usually make for non linear behaviors. But a ZVS circuit won't mind a non linear load, its just that the uneven power draw distorts the sine wave in the tank cirucit.

And yes 0.18H is a significant inductance. Putting pulsed 100KHz into it will pretty much make for a smooth DC current flow trough the coil and make it produce a mostly DC magnetic field. Hence why i don't see much use in using a ZVS circuit versus connecting the coil directly to your existing DC supply.

If you do want to quickly raise and lower the current in your coil you need to apply a very large voltage on the terminals in order to make the current ramp up quickly across such an inductance. The exact amount of voltage depends on the peak to peak current waveform you want to create in your coil.

The problem you have run into is the reason why only video display CRTs use magnetic deflection coils while oscilloscopes always use electrostatic deflection. Its difficult to swap the direction of a large magnetic field at high speed.

[nix85]

Semiconductors usually make for non linear behaviors. But a ZVS circuit won't mind a non linear load, its just that the uneven power draw distorts the sine wave in the tank cirucit.

And yes 0.18H is a significant inductance. Putting pulsed 100KHz into it will pretty much make for a smooth DC current flow trough the coil and make it produce a mostly DC magnetic field. Hence why i don't see much use in using a ZVS circuit versus connecting the coil directly to your existing DC supply.

If you do want to quickly raise and lower the current in your coil you need to apply a very large voltage on the terminals in order to make the current ramp up quickly across such an inductance. The exact amount of voltage depends on the peak to peak current waveform you want to create in your coil.

The problem you have run into is the reason why only video display CRTs use magnetic deflection coils while oscilloscopes always use electrostatic deflection. Its difficult to swap the direction of a large magnetic field at high speed.

I need ~7-8A in that coil. What if i stay in region of let's say 15KHz (so 30KHz pulsed DC), is that low enough for pulsed DC to remain pulsed DC?

Or will i have to replace ZVS with some big amp that can output bigger voltage?

EDIT: If ZVS cap and my air coil form an LC tank (determining the ZVS output frequency), isn't the inductance of the coil irrelevant since there is no reactance??

[ahbushnell]

Semiconductors usually make for non linear behaviors. But a ZVS circuit won't mind a non linear load, its just that the uneven power draw distorts the sine wave in the tank cirucit.

And yes 0.18H is a significant inductance. Putting pulsed 100KHz into it will pretty much make for a smooth DC current flow trough the coil and make it produce a mostly DC magnetic field. Hence why i don't see much use in using a ZVS circuit versus connecting the coil directly to your existing DC supply.

If you do want to quickly raise and lower the current in your coil you need to apply a very large voltage on the terminals in order to make the current ramp up quickly across such an inductance. The exact amount of voltage depends on the peak to peak current waveform you want to create in your coil.

The problem you have run into is the reason why only video display CRTs use magnetic deflection coils while oscilloscopes always use electrostatic deflection. Its difficult to swap the direction of a large magnetic field at high speed.

I need ~7-8A in that coil. What if i stay in region of let's say 15KHz (so 30KHz pulsed DC), is that low enough for pulsed DC to remain pulsed DC?

Or will i have to replace ZVS with some big amp that can output bigger voltage?

EDIT: If ZVS cap and my air coil form an LC tank (determining the ZVS output frequency), isn't the inductance of the coil irrelevant since there is no reactance??

If you want to ramp the current 8 amps in a saw tooth peak to peak at 15 kHz with a 0.18 henry inductor, it will require 43 kV square wave across the inductor.

What are you trying to do? 

[Berni]

Yes ahbushnell is correct, you will need a LOT of voltage to do this.

What are you trying to even do with your coil? Its possible you are going in the wrong direction with your end goal, making the implementation of your idea harder than it should be.

If you really had to do this the more sensible way is to split your coil  into perhaps 60 seperate coils of 10 turns and then push 8A into each one of those coils, resulting in the required supply voltage being only a 700V square wave at 480A. And yes that is not a mistake. you need to push 350kW of power into the magnetic field in order to build such a strong magnetic field so quickly, and then pull 350kW back out in order to make the field disappear again. In a sense you are making a 350kW induction heater (Id sure like to see what happens when you place various objects inside that coil)

[nix85]

Guys, i wish you both just replied to my edit so i repeat:

If ZVS cap and my air coil form an LC tank (determining the ZVS output frequency), isn't the inductance of the coil irrelevant since there is no reactance??

[Berni]

Yes extra inductance connected to the LC tank circuit becomes part of it, but not when its on the other side of a full bridge rectifier. The diodes prevent the energy from freely flowing back into the LC tank.

You could skip the rectifier and connect it directly in parallel to the LC tank circuit. But this adds a lot of extra energy storage to the tank so its resonant frequency will drop below 10KHz. You can partly mitigate this by making the capacitor in the tank circuit very small in value in order to get the resonant frequency back up. But this also reduces the current in the tank circuit at a given voltage level. So to still get your 8A in such a LC circuit you need to drive it at a higher voltage, again needing the 40kV across your coil, so as a result you need the DC supply to your ZVS circuit to be about 20kV and it must have transistors rated for 40kV.

There is no free lunch in physics. This is how much energy it takes to make such a strong magnetic field.

[nix85]

Yes extra inductance connected to the LC tank circuit becomes part of it, but not when its on the other side of a full bridge rectifier. The diodes prevent the energy from freely flowing back into the LC tank.

How do you imagine "diodes prevent the energy from freely flowing back into the LC tank", diodes are part of the LC tank and energy is perfectly free to flow between L and C both ways (only for L it is always in the same direction).

You could skip the rectifier and connect it directly in parallel to the LC tank circuit. But this adds a lot of extra energy storage to the tank so its resonant frequency will drop below 10KHz. You can partly mitigate this by making the capacitor in the tank circuit very small in value in order to get the resonant frequency back up. But this also reduces the current in the tank circuit at a given voltage level. So to still get your 8A in such a LC circuit you need to drive it at a higher voltage, again needing the 40kV across your coil, so as a result you need the DC supply to your ZVS circuit to be about 20kV and it must have transistors rated for 40kV.

"You could skip the rectifier and connect it directly in parallel to the LC tank circuit" i'm not sure what you mean here. I assume you mean the coil but coil is part of LC tank.

In any case there is no need for extreme voltages and power you guys mention since we are talking LC tank, no reactance here. This is why the ZVS heaters are so popular these days, they can push hundreds of amps at up to 200KHz (there are examples on YT). Zero voltage switching and no reactance, pure ohmic resistance and cold FETs.

I just calculated, with 0.18H inductor and 68nF cap resonance would be at 1.44KHz which is quite low. I will probably lower the number of turns/inductance to raise it to some degree.

There is no free lunch in physics. This is how much energy it takes to make such a strong magnetic field.

That is so wrong it's beyond belief. Truth is the exact opposite. I humbly suggest you to start learning about overunity devices, you are already 100+ years behind on actual developments. Here's a good book to start your research. It's only 2555 pages long, so loads of fun and enlightment ahead for you.

http://www.free-energy-info.com/PJKbook.pdf

[capt bullshot]

How do you imagine "diodes prevent the energy from freely flowing back into the LC tank", diodes are part of the LC tank and energy is perfectly free to flow between L and C both ways (only for L it is always in the same direction).

If you connect the inductor this way, the circuit won't work anymore. You cannot insert a rectifier into an LC tank. The LC tank is an important part of the ZVS circuit, there's no other oscillator than the LC tank. So if you break up the connection between C and L, no oscillation will occur. The circuit probably will do anything from nothing to auto destruction.

You can rectify the voltage across the LC tank, say have an inductor connected at the circuit and then rectify the voltage across the inductor and connect another inductor at the DC side of the rectifier.

[nix85]

How do you imagine "diodes prevent the energy from freely flowing back into the LC tank", diodes are part of the LC tank and energy is perfectly free to flow between L and C both ways (only for L it is always in the same direction).

If you connect the inductor this way, the circuit won't work anymore. You cannot insert a rectifier into an LC tank.
You can rectify the voltage across the LC tank.

Any why do you think so, do you see any logical reason why it would not oscillate, energy can freely and cannot but exchange between L and C just as if rectifier was not there.

[capt bullshot]

How do you imagine "diodes prevent the energy from freely flowing back into the LC tank", diodes are part of the LC tank and energy is perfectly free to flow between L and C both ways (only for L it is always in the same direction).

If you connect the inductor this way, the circuit won't work anymore. You cannot insert a rectifier into an LC tank.
You can rectify the voltage across the LC tank.

Any why do you think so, do you see any logical reason why it would not oscillate, energy can freely and cannot but exchange between L and C just as if rectifier was not there.

For a LC tank to oscillate, the current must flow in both directions (AC), from L to C and vice versa. The rectifier doesn't allow that, because it's purpose is to convert AC to (pulsating) DC.

[nix85]

How do you imagine "diodes prevent the energy from freely flowing back into the LC tank", diodes are part of the LC tank and energy is perfectly free to flow between L and C both ways (only for L it is always in the same direction).

If you connect the inductor this way, the circuit won't work anymore. You cannot insert a rectifier into an LC tank.
You can rectify the voltage across the LC tank.

Any why do you think so, do you see any logical reason why it would not oscillate, energy can freely and cannot but exchange between L and C just as if rectifier was not there.

For a LC tank to oscillate, the current must flow in both directions (AC), from L to C and vice versa. The rectifier doesn't allow that, because it's purpose is to convert AC to (pulsating) DC.

Wrong. It does flow in both directions from one side of the cap to another and back just as if rectifier wasnt there, it's just that it always flows in same direction through inductor.

[Berni]

Yes extra inductance connected to the LC tank circuit becomes part of it, but not when its on the other side of a full bridge rectifier. The diodes prevent the energy from freely flowing back into the LC tank.

How do you imagine "diodes prevent the energy from freely flowing back into the LC tank", diodes are part of the LC tank and energy is perfectly free to flow between L and C both ways (only for L it is always in the same direction).

A LC circuit needs power to flow back and forth between the L and C in order to have it resonate. The diodes let power flow into the inductor but the inductor is not able to send the power back trough the diodes into the capacitor.

If you see a way to send power back trough the bridge then please explain how.

You could skip the rectifier and connect it directly in parallel to the LC tank circuit. But this adds a lot of extra energy storage to the tank so its resonant frequency will drop below 10KHz. You can partly mitigate this by making the capacitor in the tank circuit very small in value in order to get the resonant frequency back up. But this also reduces the current in the tank circuit at a given voltage level. So to still get your 8A in such a LC circuit you need to drive it at a higher voltage, again needing the 40kV across your coil, so as a result you need the DC supply to your ZVS circuit to be about 20kV and it must have transistors rated for 40kV.

"You could skip the rectifier and connect it directly in parallel to the LC tank circuit" i'm not sure what you mean here. I assume you mean the coil but coil is part of LC tank.

In any case there is no need for extreme voltages and power you guys mention since we are talking LC tank, no reactance here. This is why the ZVS heaters are so popular these days, they can push hundreds of amps at up to 200KHz (there are examples on YT). Zero voltage switching and no reactance, pure ohmic resistance and cold FETs.

I just calculated, with 0.18H inductor and 68nF cap resonance would be at 1.44KHz which is quite low. I will probably lower the number of turns/inductance to raise it to some degree.

Yes indeed ZVS induction heaters work really well at high frequency. But notice that the coils they use have small cross sections and only a few turns? This is to keep inductance low so it can resonate so high while maintaining a very large current at voltages that MOSFET transistors can handle. The small coil has much less energy storage capability so a lot less power is needed to build and destroy the field around it.

All you need to do is to drastically reduce the number of turns to get to a coil design similar to a induction heater and then it will operate at high frequency without involving the massive power levels.

There is no free lunch in physics. This is how much energy it takes to make such a strong magnetic field.

That is so wrong it's beyond belief. Truth is the exact opposite. I humbly suggest you to start learning about overunity devices, you are already 100+ years behind on actual developments. Here's a good book to start your research. It's only 2555 pages long, so loads of fun and enlightment ahead for you.
http://www.free-energy-info.com/PJKbook.pdf

Magnetic fields store energy just like a capacitors stores energy with electrostatic fields. The inductance in Henries defines how much energy the inductor can hold at a given current (just like Farads define energy for a given voltage on a capacitor). So just like you need to put energy into a capacitor to raise its voltage you need to put energy into an inductor to raise its current. Because the inductor you are talking about is clearly large and you want to put a lot of current trough it this means the coil will produce a large and strong magnetic field, this is why its inductance in Henries is large.

I don't want to start off a free energy discussion because of how such discussions usually end up.