What is the purpose of a H-bridge in a solid state tesla coil driver?

Was thinking a bit about this tonight while I was winding my tesla secondary.

I think I have the normal tesla coil sorted in my minds eye by picturing the anology of pushing your kid on a swing.

The NST or oil burner transformer is a bit like me giving the kid a push when the swing reaches its zenith.

If I push to early or to late then I don't impart the maximum amount of energy to the swing to make it go higher.

The spark gap is much like the pivot point of the swing at the top bar.

So that is essentially what tuning the tesla coil is all about.

You have to match the resonant frequency of the combined spark gap, tank cap and primary coil etc to the NST frequency.
.
.
.
However with a solid state tesla coil they seem to have an arrangement similar to my current flyback transformer driver where a mosfet directly switches the current on and off.

Many of them seem to also have an antenna that picks up RF from the tesla coil and passes it, plus the oscillator signal, through an AND gate of some sort.

I presume this sort of acts like the spark gap restricting the 'push' to the zenith of the current 'swing'

Without the antenna it is a difficult task to precisely match the oscillator frequency to the resonant frequency of the coil.

With the regular tesla coil the match does not have to be quite so precise because the NST is not directly driving the coil - all it has to do in that case is charge the tank cap.

But I can't quite figure out the H-bridge that many people use.

I also not that they are using mains alternating current to drive the coil and have the driver optically or inductively coupled to the mosfet gates for obvious reasons.

My thoughts are that the H-bridge is necessary purely due to their use of alternating current which can only pass through the individual mosfet in one direction.

But they need to let the AC drive current swing both ways through the primary coil.

In which case a H-bridge is not necessary for me because I intend to drive my SSTC with pulsed direct current - I am not at all confident at using direct mains voltage to drive my tesla coil.

Would this, about the h-bridge, be correct?

With a H bridge you can push and pull current through the enegising coil. With DC you can only push current through it so you will get half the power.
Frank

chuckey said:

With a H bridge you can push and pull current through the enegising coil. With DC you can only push current through it so you will get half the power.
Frank

Click to expand...

I suppose I could build an inverter for the car batteries.

I was just thinking.....

If you pass a 555 signal through any form of NOT gate you would invert the duty cycle rather than get the to BJTs to switch on alternative with the same duty cycle.

What sort of circuit would you use to to get one BJT to switch on while the other is off but retain the same duty cycle on both?

I am assuming that you are using full H bridge? Run your 555 at twice the frequency required. Use the output from this into a flip flop. Now you have two square waves exactly out of phase.
Remember, you must take great care that there is some way to stop the wrong transistors conducting at the same time. Imagine, the top Lh. and bottom Rh transistors are both on. If either of the other two transistors conduct then there will be a direct short circuit across the Vcc line, which will burn out a pair of transistors. The easiest way to do this is to connect a diode as a gate from the collector of the bottom transistor to the drive of the "other side" transistor (for both transistors). So if one transistor is still conducting (Vce = .2V?), it sucks the drive from the other side so that transistor cannot conduct at the same time.
Frank
Frank

chuckey said:

I am assuming that you are using full H bridge? Run your 555 at twice the frequency required. Use the output from this into a flip flop. Now you have two square waves exactly out of phase.
Remember, you must take great care that there is some way to stop the wrong transistors conducting at the same time. Imagine, the top Lh. and bottom Rh transistors are both on. If either of the other two transistors conduct then there will be a direct short circuit across the Vcc line, which will burn out a pair of transistors. The easiest way to do this is to connect a diode as a gate from the collector of the bottom transistor to the drive of the "other side" transistor (for both transistors). So if one transistor is still conducting (Vce = .2V?), it sucks the drive from the other side so that transistor cannot conduct at the same time.
Frank
Frank

Click to expand...

I am building a coil at present but I am just trying to learn how the things are driven.

I believe I have gotten my head around this half bridge / full bridge thing.

I guess with resonance the the collapsing magnetic field in the secondary induces a high reverse current in the primary and the bridge configuration provides an alternative path for this reverse current to flow with each forward and backward swing getting larger and larger until you shut the bridge off entirely.

Hence the need for some sort of interuptor, that turns the bridge off after a set period of time, so that the primary current does not get so large as to blow up the mosfets or igbts.

Can you post a very simple spec for this scheme - having a little trouble picturing what you mean?

Heres a simple circuit you could start with :- http://www.kenyabeach.com/kenyabeach-old/askari/sstesla2.gif. Have you used Google to find useful circuits?
Going back to your original point :- The secondary of a tesla coil is a high Q circuit, this means in the absence of a spark when a current pulse is introduced into it, the voltage across the coil will be a sine wave which will decay by 1/Q every cycle. So with a Q of say, 500, it will have decayed to 37% of the original voltage in 500 cycle!!! The primary of the tesla coil is just a very low inductance, so when you pulse it with a current, it drops very little voltage, so conversely, if you switch a 12V supply across it the current will be very high, just limited by external circuit resistance. So to preserve the life of the transistors, you need to limit the current into them. So lets put a figure on the voltage drop of 1V, now on the next cycle if the current pulse is timed correctly, the voltage being fed back might be as high as .998V ( 1- 1/500), tho' probablely much lower, so the next current pulse will pump up the voltage by another 1V. . So until the coil sparks across the voltage across it increases BUT the voltage on the primary cannot exceed the supply to your transistor, because it would be impossible to pump another current pulse into it. At this stage the current consumption would fall to virtually nothing ( Iinitial/500). A diode is normally put across the transistor to stop the collector being reverse biased by excessive volts being fed back.
Frank