Mosfet heating Up in DC DC converter

Hello, I have a problem in power supply which i have designed. I have designed a dual supply of 3.3V and 5.4V from 30 V using semtech SC2442H DC DC converter and Infineon BS0604NS2 Mosfet. The problem is that mosfet is getting heated up a lot, it goes to about 100 C. I have calculated the power, it turn to be 0.72 W and the Mosfet can handle upto 2 W. Can some one point why is the mosfet getting heated???Because of this the inductor and capacitor near it is also getting heated up. DC DC converter operating frequency is 250 KHz.

Thanks and regards Praveen

Reply to
praveen
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And the connection with Satellite's and GPS is ?

Reply to
Nospam

This is a nice misunderstanding. The fact that your mosfet can handle 2 W does not mean that it does not heat up at 0,7W. Check the thermal resistance from the datasheet, multiply that with your power (0,72W) and this gives you the temperature rise above ambient. If that gives you 100C, the circuit is ok and you just need to cool the mosfet. If the calculated rise is lower, you either do not have sufficient drive voltage to the gate (mosfet not completely on) or you have an unwanted oscillation at a high frequence which can also produce a lot of dissipation.

Meindert

Reply to
Meindert Sprang

Did your calculations allow for the switching losses?

I've just been simulating a simple 1300 Hz PWM switch where the simulation shows that the MOSFETs dissipate about 2.5W each. But only 1.5W of that is the "dc" loss, another 1W comes from the switching losses. In fact the simulation shows a peak of 160 V*I during each on/off edge.

--
Tony Williams.
Reply to
Tony Williams

for

Yes I agree with you, but I still ask the question, what is the relevance to Satellite's and GPS.? All the replies are also coming to sci.geo.satellite-nav group, just because the original person could not be bothered to check what groups he (she) was sending to, and all the people replying just reply without checking where their replies are going to.

Reply to
Nospam

To the original idiot crossposter:

Get a 13.2V power source such as a car battery. Get a 240 ohm, 1 watt resistor. Connect the resistor across the power source (0.72 W) and hold the resistor in your hand for one hour. Come back and report your findings.

--
The state religion of the USA is atheism, as established by the courts.
Reply to
clifto

It may have escaped your attention that sometime things are crossposted, for example I am reading your post in the NG 'sci.electronics.design', which would seem to be appropriate ??

-- Regards ..... Rheilly Phoull

Reply to
Rheilly Phoull

Multiposted to far too many groups. Reply the SED only.

The new SO8 FETS are indeed specified for 2W. It is your problem to get rid of the heat though. To give you an idea, a double layer copper on a pcb of 1 inch in square gets pretty hot with 1 W.

Rene

--
Ing.Buero R.Tschaggelar - http://www.ibrtses.com
& commercial newsgroups - http://www.talkto.net
Reply to
Rene Tschaggelar

to

because

was

Ahh, such is human nature. Better to 'bend with the flow' you are never going to change this sort of thing (worst luck), theres always the old killfile thingy :-)

Cheers.... R.P

Reply to
Rheilly Phoull

Sounds like inadequate heat sinking. This is an SO-8 dual MOSFET, and it doesn't have real good thermal properties to begin with. See Infineon's application note on thermal considerations with their SO-8 devices:

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If you are using the device without any provision for heat sinking, you're going to see it run something close to 80K over ambient at 0.72 W. You need a low-thermal-resistance path to several square centimeters of copper just to get the thermal resistance down to 60K/W or so. If this isn't feasible, you need something in a different package for which you can more readily arrange a proper heat sink.

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Chris Green
Reply to
Christopher Green

Hello Mr. Meindert,

Thanks for the reply.

About maximum temperature rise: Thermal resistance =150 K/w (50+100) So temperature rise= 114 K so it means 50 C increase in temperature. So junction temperature is around 80 c which very much below the juction temperature (150 c). About the drive voltage higher FET has around 35-40 V and lower 8-10 V.

Please correct me if i am wrong. Is there some thing which i am missing??? Is have not provided any heat sink. My boss is not agreeing with any PCB changes. If it is very much require i have to convince him. It need your feedback from you people.

Thanks and regards Praveen

Reply to
praveen

Quite so. Just that it is not strictly necessary to explain his situation. His 100C temperature is consistent with his calculated 0.72W dissipation and something around 100K/W thermal resistance. An SO-8 with no heat sink will do that without any further design flaws.

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Chris Green
Reply to
Christopher Green

Hello Mr. Tony,

Thanks for the reply.

Yes i did. 0.72 W include the total power losses.

Total conduction loss:0.27W switching loss: Upper FET: 0.45W Lower FET: 0.03 W Gate loss: 0.03 W(upper)+.0075 W(lower) Switching frequency 250 kHz. Average current =2A rise time/fall time=30 nsec Thanks and regards Praveen

Reply to
praveen

Hello Chris, Thanks for the reply.

I donot have any heat sink, but 80K over ambient temperature looks ok. AS the junction temperature is around 150 C.

If i am wrong please correct me.

Thanks and regards Praveen

Reply to
praveen

Another one that will cause it to overheat is if the driver circuit for the MOSFET either has too slow a rise/fall time or does not drive it completely on or off. Too much time in the linear region causes lots of heat problems too.

mikey

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Reply to
Mike F

Either get a new boss or go into biz with him.

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Best Regards,
Mike
Reply to
Active8

What you can do to improve the thermal resistance is to glue a heat-sink onto the package. There are special glues for that so they do not get loose with the high temperatures. This is of course only a bad substitute for a better layout with large free copper areas as recommended in the manufacturers data sheets.

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ciao Ban
Bordighera, Italy
Reply to
Ban

I think it should be RMS current for the I-squared*R power dissipation calculation.

Any chance that your lack of heat sink puts the device into the dRds(on)/dT thermal runaway ambush?

ie, The increase in Tj increases Rds(on), which increases the power dissipation, and if the heatsink is inadequate the resultant increase in Tj increases Rds(on) further, increasing Pdiss..... and so on.

This phenomenon is nicely explained in International Rectifier App Note No. 942.

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Tony Williams.
Reply to
Tony Williams

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