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zvs induction heating, water cooling my transistors, still friggin' hot

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Dr. Dark Current

Thu Jun 26 2008, 06:24AM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

Marko wrote ...

Be careful with voltage as the peak voltage devices see is 3.14*supply voltage
...

I have no idea where this number comes from. The voltage "seen" on the winding center tap is a waveform identical to fullwave rectified sine wave (if your circuit is working right), so it should be sqrt(2) times supply voltage. From this the peak voltage the switches see is sqrt(2)*2 times supply voltage so approx. 2.82 x supply voltage.

Steve Conner

Thu Jun 26 2008, 12:17PM

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

No, it's pi. The average voltage across the DC link inductor is zero, hence the peaks seen at the centre tap are higher than the DC link voltage.

Marko

Thu Jun 26 2008, 12:18PM

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

Dr. Kilovolt wrote ...

Marko wrote ...

Be careful with voltage as the peak voltage devices see is 3.14*supply voltage
...

It is amazing where can the magic number appear. I have confirmed by measurement that it is indeed very closely pi.

Apparently I lost the Steve Conner's reply about this, but I never understood too well either.

First thing that needs to be considered is that drain voltage doesn't really depend on supply voltage at all, since it's decoupled from the supply by the impedance of the link inductor.

The voltage only 'happens' to be constant because feedback always keeps perfect zero voltage switching.

Conner implied that value of voltage will be linked to mean magnitude of current which is 2/pi because current in the link inductor resembles fullwave rectified current, and that final voltage is given by doubling effect done by push-pull configuration of devices.

None of this makes sense to me anymore as I can apparently only get 1/pi value out of it, and I still have no clue how does it mathematically link to the voltage.

So it's still sort of a great mystery to me, apparently God just wanted it to be so.

Marko

GeordieBoy

Thu Jun 26 2008, 03:04PM

Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881

The peak switch voltage for this arrangement is indeed pi times the dc supply voltage (Vdd) as Steve Conner said earlier. Whether you beleive it is divine intervention or just maths depends on your viewpoint!

It is relatively easy to see why this should be so. The drain of each MOSFET is connected to the DC supply + rail via inductors. Since ideal inductors have no DC resistance the average voltage at each MOSFET drain must be equal to the DC supply voltage Vdd. We also know that each MOSFET is switched on for half of the total switching period and has approximately zero volts across it when conducting. We also know that when it's not conducting the supply voltage is a half-sinusoid (...because the these two-half cycles add up to produce the voltage sinewave we get out of the circuit.) Now it just becomes simple maths.

If the drain voltage spends half of its time at 0 volts, and the other half of it's time tracing out a half-sine pulse, and the overall average voltage is Vdd, then how high does the half-sine pulse need to be? (This criteria comes from the necessity to have VoltSecond balance across any inductor, otherwise the current through the inductor would increase without limit.) If you work through the calculus it turns out that the drain voltage needs to peak at pi x Vdd to make the overall average drain voltage equal Vdd. In the real world DC resistance and AC resistance of the inductors, and finite Rds(on) act to reduce this voltage slightly.

-Richie,

Tiberius

Thu Jun 26 2008, 08:58PM

Registered Member #1484 Joined: Wed May 14 2008, 03:24PM
Location: Cary, NC, USA
Posts: 27

You said you are using mercury for thermal interface, isn't that a bit worrying? I bet even cheap thermal grease would work up to full dissipation for the package as long as it is properly applied, surfaces are clean and mounting is proper. Grease probably plays just a minor role after that I think.

Cheap thermal grease would indeed be adequate -- I wish I had some handy the other day as initial wetting of the block with mercury is messy and quite laborious in comparison. Other than the aggravation of initial wetting, though, I have no real problem working with it infrequently and additionally in the spirit of not letting switching losses keep my inverter down a few percent lower thermal resistance was not unwelcome.

What are you using for power supply? You should better have some large iron transformers around for power supply.

Power supply is indeed large and iron, 7.8kVA 0-280V Variac limited by 15A 120V breaker currently.

How are you mounting your devices? A single screw is usually poor since it lifts the die from the surface, a clamp from top should be used to provide proper pressure.

Mounting is a #6 machine screw, washer, and nut currently. I'm not familiar with the phenomenon you refer to -- can you clarify?

You should better get somewhat larger MOSFET's if you intend serious induction heating, like IRFP260 or higher current.

Regarding IRFP260, it appears to have the same Vds as the ST MOSFETs I'm already apparently overvolting. The ST MOSFETs are rated 34A continuous and have proven quite capable of switching the full current I can supply without relocating this project outside, so I'm gonna stick with them for the time being.

I glanced at the data sheet for the 20N60 earlier and found a gate resistance / switching loss graph which implied my switching losses could be on the order of 200W per device depending on my unknown frequency. I did some similar guesstimation for 40N60 and if I were using those in the same circuit I could see switching losses on the order of 1200W per device. There's a good chance I was still in the safe operating area of the 20N60, but boy am I glad I only found one of the 40N60s in my parts bin.

---

Update: I decided to measure current instead of voltage for a few runs and I can't say I'm entirely sure what's going on here. I only took it up to 20A on the 120V supply for my Variac and encountered this in the lower third of the approximate range of voltages I've been using. Measuring the variac output current I found currents in excess of 20A nearly as soon as I raised the voltage off of zero. At this range, my efficiency must be terrible, or parasitic oscillations dominant, because I have not been able to heat noticeably above ambient. It's only in the upper range that any heating seems to occur at all.

Tiberius

Mon Jun 30 2008, 02:53AM

Registered Member #1484 Joined: Wed May 14 2008, 03:24PM
Location: Cary, NC, USA
Posts: 27

This evening I noticed that the rectifier section of my inverter was still bypassed from early testing, apparently I've been switching unrectified mains this entire time. I put the rectifier back into the circuit and performance has increased drastically. The DC link inductor no longer noticeably heats and current drawn is down significantly at any input voltage, an order of magnitude or more at lower voltages.

I was able to bring a small Allen-head bit to red heat in a couple minutes at 13.5A measured on the 120V input to variac. During this run the only significant heating occurred in my rectifier and filter capacitors, which appear to be due for an upgrade. I am still using the ST MOSFETs mentioned earlier and they remained lukewarm throughout the run.

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