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Half bridge or Royer ciruit?

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Colin 99

Mon Oct 23 2006, 05:05AM

Registered Member #192 Joined: Fri Feb 17 2006, 03:08AM
Location: Canada
Posts: 44

Hello,

I was wondering if a Half Bridge circuit works any better than a Royer Circuit for driving a flyback transformer? I've been looking at Steve wards circuit and I might try it.

I've built the royer circuit and it works well!

thanks,

Shaun

Steve Ward

Mon Oct 23 2006, 05:26AM

Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055

The half-bridge gives more flexibilty in my opinion. Ive played with the royer circuit too, im not totally sold on it personally, but lots of people seem to like it, probably for its simplicity, and can still run enough power to melt your flyback. The royer circuit requires trying different capacitors to tune it, where a half-bridge is a simple turn of the potentiometer.

Sulaiman

Mon Oct 23 2006, 06:53AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

To work properly a 'Royer' invertor needs a loaded 'Q' of at least 2.4
(I can't remember the reference for this though)
typical loaded/operating 'Q' would be 5 to 10
which means a half-bridge (or full-bridge etc.) can output at least
2.4 times more power than a 'Royer' configuration (maybe 5 to 10 times)
for a given magnetic core/windings.
PLUS a 'Royer' needs a significant series inductor as well.

A 'Royer' is excellent for fixed-voltage transformation
and gives a nice fixed-frequency sinewave output
but needs extra circuitry for variable output voltage
and isn't suited to the common task of charging high voltage capacitors.

For charging capacitors to high voltage I think flyback mode is best.

cbfull

Mon Oct 23 2006, 01:28PM

Registered Member #187 Joined: Thu Feb 16 2006, 02:54PM
Location: Central Ohio
Posts: 140

Another thing with the half-bridge, is the option of upgrading it to a full-brige with the addition of two more transistors, (and gate drives). In a full-bridge, you get the bonus of having 1/2 the voltage stress on the C-E path of the transistors because you end up with two transistors in series (the load is between them) for each cycle. A small downside is that you end up doubling the "on" resistance, so your supply voltage has to be doubled to get power levels up. An alternative to doubling supply voltage would be to stack your MOSFETs in pairs to cut the resistance in half, resulting in a circuit with 8 MOSFETS.

Ultimately it sounds like a tradeoff. The Royer gives great performance and reduces the complexity, while the half bridge has versatility. Don't be afraid to try both.

Steve Ward

Mon Oct 23 2006, 03:58PM

Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055

In a full-bridge, you get the bonus of having 1/2 the voltage stress on the C-E path of the transistors because you end up with two transistors in series (the load is between them) for each cycle.

Im sorry, but that is incorrect. When 2 fets are ON, and the other 2 OFF, the entire supply is across the fets that are OFF. The fets need to be rated for at least the DC supply voltage, and really, a factor of 50% more is typical id say.

Colin 99

Tue Oct 24 2006, 02:32AM

Registered Member #192 Joined: Fri Feb 17 2006, 03:08AM
Location: Canada
Posts: 44

Well I guess that settles it, I'll have to build a half bridge!

I've built a full bridge before using one of Dan McCulleys Plasmasonic boards. It was just a matters of following the instructions and solder.

I'm going to use some IGBT's that I recently sampled from Fairchild -- 20N60s in a TO-247 case, they are suppose to be good to 70 amps at low frequencies.

Question: Some half bridge diagrams show the reverse collector - emitter diode and some don't. Does that mean I can get away without it? The IGBTs I have have the extra diode in them anyway so it doesn't matter.

2nd question: Do I need to put protection zeners and resistors on the gates?

thanks,

Shaun

Steve Ward

Tue Oct 24 2006, 07:12PM

Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055

Yes, you need the diodes across the CE junction. They conduct any current going back to the power supply. Since your load is reactive, there *will* be some current going back from the load to the power supply. Some schemes probably dont show the diodes because MOSFETs already have an integrated (its really more of a parasitic thing) diode. If you use IGBTs, either get the ones with internal diodes (highly recommended), or attach your own ultrafast diode. For a flyback driver, it shouldnt need to be more than maybe a 5A diode on each IGBT. It should have the same (or higher) voltage rating of the IGBT.

I dont bother with protective zeners on the gates for lower power things like this. You are usually only switching a few amps, so things are fairly tame.

cbfull

Wed Oct 25 2006, 03:29PM

Registered Member #187 Joined: Thu Feb 16 2006, 02:54PM
Location: Central Ohio
Posts: 140

Steve Ward wrote ...

Im sorry, but that is incorrect. When 2 fets are ON, and the other 2 OFF, the entire supply is across the fets that are OFF. The fets need to be rated for at least the DC supply voltage, and really, a factor of 50% more is typical id say.

I don't see how that makes my statement incorrect, you need to support your statement with more than just a couple of sentences. If you are going to make a very direct statement (and a bold one I might add) like that please be considerate and explain it so I can learn something.

When the E-C paths of FETs are connected in series (and off) the blocking voltage is the sum of the two transistors breakdown voltage. This is a FACT. The problem lies in making sure each transistor switches on at exactly the same time, or you will be putting (worst case) full supply voltage across the E-C of the FET that is still off. It's similar to connecting diodes in series.

I promise you I did not make this up, it came from a bridge design manual. If I still had it I would quote it, and there is always the possibility that I am remembering it incorrectly, which I don't think is the case.

Now unless the inductive load (flyback) being placed between the two FETs has some effect that I have not considered, then by all means please tell me!

Please excuse me if I really am way off on this one.

Steve Ward

Wed Oct 25 2006, 05:03PM

Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055

Ok, i will defend my statement.

When the E-C paths of FETs are connected in series (and off) the blocking voltage is the sum of the two transistors breakdown voltage.

Ok, fine, but this is not the case to be concerned with, and designing for this case is pointless, since at some point, the FETs *are* ON.

Looking at only a half-bridge (the voltage requirements/stresses are exactly the same, mind you). Say the bottom switch turns ON. What voltage is across the TOP fet that is OFF? The entire supply voltage.

The problem lies in making sure each transistor switches on at exactly the same time, or you will be putting (worst case) full supply voltage across the E-C of the FET that is still off. It's similar to connecting diodes in series.

That is true for directly seriesing switches. An H-bridge is not the case you just described.

I promise you I did not make this up, it came from a bridge design manual. If I still had it I would quote it, and there is always the possibility that I am remembering it incorrectly, which I don't think is the case.

From what you said, it sounds like you are confusing a simple series switch device and an H-bridge. After years of designing H-bridges, i *know* that i am not wrong. This doesnt mean that we simply arent misunderstanding eachother still. The bottom line is, for either half-bridge, or full H-bridge, the switches must withstand the entire DC supply voltage, plus transients from switching.

If that doesnt settle it, id probably have to draw a diagram or something. It should be simple to prove to yourself by drawing the 2 conduction states of the bridge.

Steve Conner

Thu Oct 26 2006, 12:13AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Steve Ward is right, the switches must be rated for the whole supply voltage. If you don't believe it, try it and watch them "A splode".

You're not the only one that doesn't get it: I've seen Douglas Self say in print that the devices in an audio power amp (basically the same circuit as a half bridge) have to be rated at _twice_ the total supply voltage. On the other hand C-Audio, probably thinking the same as you, brought out a power amp that ran 160V FETs off a total of 200V DC. As far as I know, it "A splode" and they went out of business.

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