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External recovery diode

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nrhoades

Wed Nov 14 2007, 03:15PM

Registered Member #610 Joined: Wed Mar 28 2007, 09:44PM
Location: Middletown, RI
Posts: 110

unless the current in the recovery diode just happens to stop flowing for some other reason

What you are refering to here is the free-wheeling current in the opposing switch?

I may also be using cross-conduction and shoot-through as synonyms which may be incorrect, though I don't know if that changes what you said.

Steve Ward

Thu Nov 15 2007, 05:35AM

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

What you are refering to here is the free-wheeling current in the opposing switch?

I may also be using cross-conduction and shoot-through as synonyms which may be incorrect, though I don't know if that changes what you said.

Say you have an inductive load and after driving current into it, you switch your fets off. The current cant suddenly stop, so it starts flowing back to the power supply through the diodes. Now, suddenly the opposing fets turn on (before this "recycling" current stops) and basically you have to force the diode recovery (since the MOSFET is taking over). Deadtime doesnt help here, and thats all i was getting at. You *will* force a diode recovery, and this time for recovery is basically a shoot-through condition (except its through diode and opposing fet, instead of 2 opposing fets).

AFAIK, cross conduction and shoot through are the same thing.

Steve Conner

Thu Nov 15 2007, 10:31AM

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

I don't think the other Steve has it quite right.

With an H-bridge driving an inductive load, as soon as you turn your MOSFETs off, the current will commutate to the diodes of the *other* two MOSFETs that you are about to turn on. Some time after these new MOSFETs have turned on, the current will reverse so it's flowing in the channels of the new MOSFETs, and when these turn off the process will repeat. There is no forced recovery, but the devices have to hard switch current. So this mode doesn't hurt MOSFETs, but IGBTs don't like it. With suitable deadtime and snubbing caps, it leads to zero voltage switching, similar to Class-E.

With a capacitive load, by the time you go to turn your MOSFETs off, the current already passed through zero and commutated to the body-drain diodes of the very same MOSFETs you're about to turn off. When you turn them off, nothing happens: the load current still keeps flowing in the diodes. Now turning the next FETs on causes forced recovery of the diodes. This is zero-current turn-off because the devices have stopped carrying current before the gate drivers turn them off. It is nice for IGBTs, since they don't like turning off huge currents and also have fast co-packaged diodes that can take a little forced recovery.

If you tune it just right for a sinusoidal load current at unity power factor, you have perfect zero current switching.

So in a nutshell these are the concepts of ZVS and ZCS as applied to H-bridges.

Marko

Thu Nov 15 2007, 02:00PM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

Steve: does this surely mean that with inductive load, and 50% duty cycle, diodes won't conduct at all?

That's somehow intuitive, but how can then I in case of a SSTC, have primary current which is many x bigger than supply current and still sine, like if I'm driving a resistive load? That should formally be a series of inductor and resistor...

...so if there is difference in currents like that power should be returning to the supply? So do the diodes conduct then?

And what wen current sops being sine due to loading?

Something just doesn't fit there.

With a capacitive load, by the time you go to turn your MOSFETs off, the current already passed through zero and commutated to the body-drain diodes of the very same MOSFETs you're about to turn off. When you turn them off, nothing happens: the load current still keeps flowing in the diodes. Now turning the next FETs on causes forced recovery of the diodes. This is zero-current turn-off because the devices have stopped carrying current before the gate drivers turn them off. It is nice for IGBTs, since they don't like turning off huge currents and also have fast co-packaged diodes that can take a little forced recovery.

Ok... but how can voltage on a fully capacitive load ever get higher than supply voltage? I guess there must be inductance in the circuit?

I may have caused confusion by my statement that mosfet bodies may conduct current instead diodes - in bridges we use it never happens since there is half of supply voltage working against it, as I think.

But from the other side, one must consider that mosfet *works* exactly as well with drain-source and source-drain polarization - the diode is what limits it's use in latter;

Synchronous buck is actually nothing but halfbridge with load put down onto GND and uneven duty cycle;
Lower mosfet conducts only in reverse, in direction of it's diode, but the diode never actually conducts because it never gets enough forward voltage, as drop on the mosfet is always lower (if it wasn't our sync rectifier would lose the point).
Not to mix this with upper stuff though.

I'm frustrated to see how much I *don't know* - I hope no trouble will come when two Steves are disagreeing about something...

Marko

nrhoades

Thu Nov 15 2007, 03:57PM

Registered Member #610 Joined: Wed Mar 28 2007, 09:44PM
Location: Middletown, RI
Posts: 110

Marko:

I'm frustrated to se how much I *don't know*

Even after a Masters in Electrical Engineering, it *still* takes most the better part of a career to really understand this stuff. It's a difficult subject to learn about on your own, which is why it is impressive to see how far some can get without formal education. I definately still have a long way to go and I've been in school for 5 years. Don't feel discouraged, your well ahead of the game, and you already have enough skills to make you valuable if you decide to pursue EE as a career.

I think both Steves are talking about the same thing. I also discussed the external recovery diode with a resident hardware engineer here and he basically said the same thing as Conner.

Steve Conner

Fri Nov 16 2007, 11:33AM

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

Hey Marko,

An inductive load can't dissipate power, so all the energy that passes into it from the DC bus via the MOSFETs has to pass back again via the diodes. So if you try driving a pure inductive load with a 50% duty cycle, you'll find that the duty cycle increases to 100% of its own accord! When the current commutates to the diodes, the load voltage looks as if the other MOSFETs have turned on, and the diode current has to last just as long as the MOSFET current did, for conservation of energy to be satisfied.

On a low voltage circuit like a gate driver, though, you can see the diode drops on a scope, so you know what's really going on...

BTW, I am a professional EE, though I somehow ended up with a PhD in mechanical engineering. But I recently got a job that involves way too much programming! I'm currently in the middle of a nightmare world of C++ and Win32, and hoping I can get my half-assed soup of classes and objects to behave before the project deadline! :[

Marko

Fri Nov 16 2007, 11:46AM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

So if you try driving a pure inductive load with a 50% duty cycle, you'll find that the duty cycle increases to 100% of its own accord!

Yes, but that wasn't my point; I refer 50% duty cycle to what you refer as 100% - I mean, full duty cycle. I read both of these on various places and I don't know what is correct to say.

So do the diodes conduct at all with inductive load if there is no deadtime, full duty cycle or whatever so one of mosfets is always on?

And why you say there is forced recovery with capacitive and not with inductive load? I thought it's vice versa. I can't figure out your post.

nrhoades

Fri Nov 16 2007, 04:48PM

Registered Member #610 Joined: Wed Mar 28 2007, 09:44PM
Location: Middletown, RI
Posts: 110

But I recently got a job that involves way too much programming! I'm currently in the middle of a nightmare world of C++ and Win32, and hoping I can get my half-assed soup of classes and objects to behave before the project deadline! :[

Join the club!

Steve Ward

Fri Nov 16 2007, 06:51PM

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

Conner is right, I was wrong (i was stuck thinking about my DRSSTCs where the current REVERSES direction before the IGBTs transition, this is definitely not the case with an inductive load). I think if i ever disagree with someone like Conner, its because ive made a mistake somewhere in my "mental simulation" of whats going on.

Marko

Fri Nov 23 2007, 12:07PM

Registered Member #89 Joined: Thu Feb 09 2006, 02:40PM
Location: Zadar, Croatia
Posts: 3145

OK...

Can someone... for me, put some headwords when diodes *do* and *don't* conduct in weakly-coupled-series RLC (SSTC);

-when coupling is high
-when coupling is low
-when primary inductance is large
-when primary inductance is small
-when resonator Q is high/low
-anything else
-etc.

?
Marko

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