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Induction heater (featuring my new PLL driver :-O)

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MRacerxdl

Sun Dec 14 2008, 02:43PM

Registered Member #989 Joined: Sat Sept 08 2007, 02:15AM
Location: SÃ£o Paulo, Brazil
Posts: 476

No, its not. 18A for the IRFP460 is only 87W. (It have Rds(on) about 0.27R)
If you get ohm law: P = R * IÂ²
P = 0.27 * 18Â² => 324 * 0,27 = 87W

Considering that IRFP460 can dissipate about 300W, that is far from the more that it can support.

uzzors2k

Sun Dec 14 2008, 05:28PM

Registered Member #95 Joined: Thu Feb 09 2006, 04:57PM
Location: Norway
Posts: 1308

Actually I used IRFP450s, each doubled up. SO, 18A peak or 13A RMS shared (hopefully) between two mosfets with an rds of 0,44R. That's about 20W of power dissipation per device, but since they only conduct for half the time we'll say 15W. (Switching losses play in as well.) Giving 60W of power lost in the fets. It's important to note that:

1) This is power LOST in the MOSFETs, the only way it affects output power is whether it approaches the maximum of what the IRFP450s can handle.
2) The 18A circulate in the tank/inverter loop and is largely reactive current, the actual current drawn from mains is probably 3-5A.

Experimentonomen

Sun Dec 14 2008, 05:47PM

Registered Member #941 Joined: Sun Aug 05 2007, 10:09AM
Location: in a swedish junk pile
Posts: 497

paralelling fets meand half the rdson, not twice as you imply.

Raka

Sun Dec 14 2008, 09:36PM

Registered Member #1838 Joined: Tue Dec 02 2008, 06:01PM
Location:
Posts: 38

I obviously have a huge gap in my understanding of the FET specs. If I look at the save operating area graph of the IRFP460, at 300V (drain to source) a dc value of about 0.7A can be drawn continuously? Thus for a 50% duty cycle it must be about 1.7A per FET. Can someone please refer my to a webpage that can help to explain.

Steve Conner

Sun Dec 14 2008, 10:05PM

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

The beauty of switched-mode power is that the FET is either off or on. (Anything in between - part on or part off - is not switched mode, but linear operation, which went out of fashion in the 1970s.)

When it's off, it has a high voltage across it, but no current flowing, therefore the power dissipated is zero. (Power is voltage times current, and anything multiplied by zero is zero.)

When it's on, it has a high current flowing, but no voltage across it. (We assume Rds(on) is so small that the IR drop is negligible.) So the power dissipated is again zero.

So this means a switched-mode circuit can pass infinite power without getting hot

Well not exactly, but it means that it can handle far more than what you said, because the 300V and 1.7A are never present at the same time. I would expect a fullbridge of IRFP460s to deliver a couple of kilowatts at hobbyist levels of reliability, while dissipating roughly the 50W or so of heat that Uzzors suggested.

uzzors2k

Sun Dec 14 2008, 10:11PM

Registered Member #95 Joined: Thu Feb 09 2006, 04:57PM
Location: Norway
Posts: 1308

For the love of god, my inverter uses IRFP4 -> 50 <- mosfets. Not IRFP460, so everyone stop bringing them up.

Now Raka, the maximum drain current specified in mosfet datasheets is DC current. Currents beyond this heat the junction (inside the mosfet package) faster than heat can be removed from the package. This has to do with thermal resistance. For all practical use the junction temperature is what limits the mosfet's current rating. You seem to have Ohm's law down, but not generel circuit analysis.

The following example applies to a single IRFP450, and isn't related to the inverter. It's just food for thought.

When mosfets are used in switching applications like this, they can be thought of as switches with some resistance, in this case 0,4 ohms. Now apply Ohm's Law to the switch. 0,4R * 14A = 5,6V So the voltage across the mosfet is only 5,6V when it's on and actually passing current. That means the dissipation in the conducting mosfet has to be 78W. The rest of the 300 - 5,6 = 294,4V are across the load, which I'll call LOAD. The power dissipated in LOAD is 294,4V * 14A = 4kW. When the mosfet is off the resistance is astronomic, and the full 300V will be across it. However, since no current flows through the mosfet the power dissipated in it has to be 0W. Is it any clearer?

Edit: Steve beat me to it.

Raka

Mon Dec 15 2008, 03:08PM

Registered Member #1838 Joined: Tue Dec 02 2008, 06:01PM
Location:
Posts: 38

Its clear now, thanks.

Raka

Thu Dec 18 2008, 12:18PM

Registered Member #1838 Joined: Tue Dec 02 2008, 06:01PM
Location:
Posts: 38

Ok, another question: In a setup like yours, with FETâ€™s in paralleled. What is the pros and cons of a winding for each FET on the gate drive transformer or just one winding for the high side and one winding for the low side, driving more than one FET with a winding?

uzzors2k

Thu Dec 18 2008, 12:54PM

Registered Member #95 Joined: Thu Feb 09 2006, 04:57PM
Location: Norway
Posts: 1308

The simplest approach is to simply parallel the mosfets and then consider them a single larger mosfet. The gate capacitance will be larger, the current handling roughly the same, but the power dissipation will be halved. I found that the gate driver had no problem driving twice the capacitance fast enough. Unless the switching waveforms become unreasonably sloped, there is little point in using a second GDT with separate gate drivers to power the other set of mosfets. If you're thinking of using four secondary windings and putting pairs in parallel, that will only reduce the resistance of the secondary, which is often negligible to start with.

Raka

Sat Dec 20 2008, 05:55PM

Registered Member #1838 Joined: Tue Dec 02 2008, 06:01PM
Location:
Posts: 38

I know you donâ€™t like the mention of the name

But what made you decide on the irfp 450? It seems to me as if the irfp 460 is a superior IC

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