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From: "Fritz Schlunder"
Subject: Re: MOSFET losses,CUK converter !
Date: Mon, 30 Sep 2002 19:48:42 -0700
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"Peter Larsson" wrote in message
> Hello I have tried to build an CUK converter.
> The converter has an IRFP 250 transistor and three ETD44 (overkill, but
> is only for my own test) cores with C85 material.There is a small
> airgap(outer legs too) ~0.4mm on the first coil and it is wound with 10
> Electrolytic capacitors 2x680uF low ERS made for up to 100kHz on primary
> side parallelled with 100nF polyprop cap.
> The transformer is wound with 3 turns on primary side and 10 on
> secondary.(Litz wire 90 strands in one bundle).
> Electrolytic capacitors 1x680uF low ERS made for up to 100kHz on secondary
> A fast diode is used on the secondary side.
> Output core has no aigap and is wound with 10 turns.
> Drive circuitry for the MOSFET is an elantec driver with rise and fall
> about 30 ns (measured).
> Used frequency: 100kHz
> The converter is made to do this:
> 12 to 30 volt conversion with max 5 amps at 30 volt from a 12V car
> Problem: transistor IRFP 250 is running too hot when i load with 5 ohm,
> i load with 10 ohm it is not too hot.
> This converter only have about 50W as output but should easy handle 150W.
Reconsider your choice of MOSFET. The IRFP250
(http://www.fairchildsemi.com/ds/IR/IRFP250.pdf) is a 200V 0.085 ohm part.
Lets do a couple napkin (what if, ultra rough) calculations shall we?
Assume 180W input power (for 150W output) after effeciency loss. So
180W/12V = 15A average DC current draw from source. To handle this alone at
85mohms we are at I^2R heating of: 225x0.085= 19.1W. I'm not too familiar
with your topology or your specific circuit, but presumably the MOSFET won't
conduct continuous DC to handle that power. Instead it will be pulsed (as
seen by the MOSFET) at higher currents at lower duty ratios. Lets just
imagine it handles all that power at twice the current at half duty ratio.
Thus 30Amps^2*0.085ohms*0.50 (duty ratio) = 38.3W. Now imagine the cooling
is insufficient to keep the die at 25 deg. C like the 0.085ohm Rds(on) is
measured at. The temperature vs. on resistance curves of the IRFP250 will
vary by manufacturer so I don't know what the actual result would be. But
imagine a situation where the die temperaure is sufficiently high to double
the on resistance from 0.085ohms at 25deg. C up to 0.17ohms. Now the power
dissipation becomes 38.3W * 2 = 76.4W
Add on top of that your anticipated switching losses. Is your heatsink
capable of handling 76W++ without getting too hot? Likely not... You need
to select a better MOSFET. Reevaluate your circuit/topology to see if you
can reduce the required MOSFET blocking voltage. Then select an appropriate
lower voltage MOSFET with much lower on resistance.
> I have measured the current into my application with a small resistor and
> here I can see that the first core is not in saturation.
> Is there an easy way to measure where the problem(s) are ??
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