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How to estimate losses when hard-switching?

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Dago

Wed Aug 29 2007, 01:58PM

Registered Member #538 Joined: Sun Feb 18 2007, 08:33PM
Location: Finland
Posts: 181

Is there a way somehow to estimate losses when hard-switching?

I'm making an induction heater which adjusts the VCO frequency based on the desired LCLR tank voltage (to limit the power, kinda) so this thing will be hard-switching. Is there any simple way for me to estimate how much power the switches will dissipate? The load current seen by the H-bridge will be sine-waveish (at perfect resonance it is a sine-wave and goes toward a square wave depending how much the inverter is off-tune of the resonance frequency). Lets assume I will use IRFP460s as the switches and I will be putting out about 2kW and the bridge is powered from rectified mains (~340VDC @ 6A). Is there any way to even give a rough estimate how much power the switches will be dissipating?

Steve Conner

Wed Aug 29 2007, 02:06PM

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

If you know the turn-on and turn-off times you are achieving with your gate driver, and you know the current flowing at the (beginning of the) switching transition, you can roughly estimate the energy loss in joules per transition as:

0.5 * switching time * current * DC bus volts

and multiply by the switching frequency to get the power loss in watts. Double it if the turn-on is hard too.

You may think that speeding up the gate drive will make things better, but only up to a point, since the energy 0.5*L*I^2 stored in stray inductances must also be dissipated at each transition. If you try and turn it off too fast, it will slam the drain voltage up as high as it needs to go to dissipate itself in the time available. That includes avalanching the device and killing it if necessary. (This is how bad layout kills devices.)

Dago

Wed Aug 29 2007, 02:14PM

Registered Member #538 Joined: Sun Feb 18 2007, 08:33PM
Location: Finland
Posts: 181

Steve Conner wrote ...

If you know the turn-on and turn-off times you are achieving with your gate driver, and you know the current flowing at the (beginning of the) switching transition, you can roughly estimate the energy loss in joules per transition as:

0.5 * switching time * current * DC bus volts

and multiply by the switching frequency to get the power loss in watts.

Speeding up your gate drive may not work beyond a certain point, since the energy 0.5*L*I^2 stored in stray inductances must also be dissipated at each transition, and if you try and turn it off too fast, it will slam the drain voltage up as high as it needs to go to dissipate itself in the time available. That includes avalanching the device and killing it if necessary. (This is how bad layout kills devices.)

I think the rise- and fall-times are in the order of bit under 100ns is under soft-switching if I recall correctly so it will be bigger when hard-switching.

Since I cant know the current flowing when the transition begins (since its kind of random how it "lines up" with the load current) I assume I have to put somekind of an average value here? How do I come up with the average value?

Steve Conner

Wed Aug 29 2007, 02:15PM

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

I would just assume the full rated load current, since that's the worst-case scenario. If you worked it out, you might find that it's actually sqrt(2) times less bad than you thought, or whatever.

Marko

Wed Aug 29 2007, 02:18PM

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

I think the rise- and fall-times are in the order of bit under 100ns is under soft-switching if I recall correctly so it will be bigger when hard-switching.

For ZCS, probably not too much.
Miller effect is present with or without load, so difference will generally be small.
Switching speed will in that case depend mainly on your drain voltage.

Dago

Wed Aug 29 2007, 02:19PM

Registered Member #538 Joined: Sun Feb 18 2007, 08:33PM
Location: Finland
Posts: 181

Soo 0,5 * 0,0000002s * 6A * 340VDC gives out 2,04e-4J of loss per transition so at 100kHz its about 20W per switch, correct?

And this is assuming 200ns switching times and full load current at every transition so its a REAL worst case scenario. 80W dissipated in the switches when hard-switching 2kW as the worst case scenario sounds fairly correct and fairly manageable with proper aircooling to me I guess. In a real situation I guess its way less than half of this, says my intuition (gotta measure the switching times)

Conner: yeah I though about sqrt(2) as the "magic" value too

Marko

Wed Aug 29 2007, 02:47PM

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

And this is assuming 200ns switching times and full load current at every transition so its a REAL worst case scenario. 80W dissipated in the switches when hard-switching 2kW

Your rise times of 200ns are pretty pessimistic for IRFP450's - even with crudest drives like direct-SG3525 you should be well under 100ns.
Other than that, you are unlikely to be switching that much power when you are out of tune.

Although it may just be more efficient to use some kind of simple dimmer to regulate the input voltage. Since you don't need bus filtering you could run with high power factor and use simple off-shelf dimmer.

Marko

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