Please help me choose a FET

Discussion in 'General Electronics Chat' started by gte, Sep 16, 2010.

  1. gte

    Thread Starter Active Member

    Sep 18, 2009
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    I'm having a hard time finding a FET that will perform the way I need it to without having to use a heat sink.


    Needs:

    14vDC/15A
    Vgth of 1.2 to 3v/N channel
    Thermal characteristics that allow it at least 140 watts at a maximum of 75C
    Frequency of < 1mHz
    Reasonably priced
    Harsh environment rating



    Any thoughts/ideas?
     
  2. Audioguru

    New Member

    Dec 20, 2007
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    If you want to make a heater then why use a Mosfet? Use nichrome wire instead.
    Why are you looking at the threshold voltage where it conducts a whopping 0.00025A?

    Of course it needs a heatsink if it dissipates 140W. Maybe the load dissipates 140W and you are looking for a Mosfet to switch it on. Then you need a Mosfet with a very low on-resistance so it does not heat much. The Mosfet will be completely turned on when its gate-source voltage is 10V.
     
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  3. tom66

    Senior Member

    May 9, 2009
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    If you want to dissipate 140 watts without a heatsink you'll be in for a world of hurt.

    Take the IRF530, I chose this as it was one I have, not because it is suitable for your application. It is only rated for 79W of dissipation. But let's say you put 140W through it with no heatsink, and with no air flow. The thermal resistance, junction to ambient, is 62.5°C/W (this is fairly typical of TO-220 devices.) At 25°C ambient, the FET will be cooking along nicely at nearly 8,775°C junction temperature, and it will be toast in milliseconds if not less.
     
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  4. Ghar

    Active Member

    Mar 8, 2010
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    That 79W rating is an impossible to achieve "Tc = 25C" rating, where the case temperature (Tc) is held at room temperature by an infinite heatsink.

    Anyway, what is harsh environment supposed to mean?
     
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  5. tom66

    Senior Member

    May 9, 2009
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    Yes, of course, that's why I said "rated for 79W of dissipation", not "will handle 79W".

    But of course you could use some good 'ol LN₂.
     
  6. gte

    Thread Starter Active Member

    Sep 18, 2009
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    It's an environment that would be similar to an automotive environment, with changing environment variables and temperatures. Sounds like I'll need a heat sink?

    Here is the transistor I bought

    I wanted to use the TO-220 package

    http://www.infineon.com/dgdl/IPP_B_...a4339&fileId=db3a304412b407950112b42a72ce4342



     
  7. tom66

    Senior Member

    May 9, 2009
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    That has a thermal resistance of 62.5°C/W junction-ambient (same as my example), so will cook in milliseconds if dissipating 140W.
     
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  8. gte

    Thread Starter Active Member

    Sep 18, 2009
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    Thanks and yes I agree

    Now I just need to figure out the layout of the sink and how to fit it to my PCB
     
  9. Audioguru

    New Member

    Dec 20, 2007
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    Why is it dissipating 140W?
    Its on-resistance is typically 0.01 ohms when it is cold and might be 0.015 ohms when hot. For it to dissipate 140W then its current must be 96.6A but its pins melt when the current is more than 60A.
     
  10. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
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    14v, 10A for the solenoid.

    It won't dissipate 140w, but it will conduct it.



     
  11. Audioguru

    New Member

    Dec 20, 2007
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    The Mosfet must conduct only 10A and the load dissipates 140W.
    With a current of only 10A your 0.015 ohm Mosfet dissipates only (10 squared x 0.015)= 1.5W. It will need a little heatsink.
     
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  12. gte

    Thread Starter Active Member

    Sep 18, 2009
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    Oh, I see what I did wrong.

    That would explain why it didn't melt down when I tested it on the board?

    So if I was running 2 amps through it @ 14v, it would be between 200 and 300mW?

    Thanks for the schooling Audioguru! :)




     
  13. gte

    Thread Starter Active Member

    Sep 18, 2009
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    Am I getting this?

    BODY { MARGIN: 8px } .LW-yrriRe { FONT: x-small arial } .MsoNormal { MARGIN: 0px } The Rds(on) for this transistor is is 16 milliohms. I can take my projected max field operating temp of 75C and subtract it from 175C max transistor temp to get 100C. Then 100C/62.5C per degree of unsinked rise in temperature and I get 1.6W of dissipational power for the transistor.

    If Pd = (I^2 * R), then 1.6w = I^2 * 16 milliohms or I^2 = 1.6/.016, I^2 = 100amps and I = 10amps. So I can flow 10amps (max) through this at 75C without the need for a sink?

    But then my calcs up top don't add up, it would actually be about 64mW it will dissipate at 2amps?
     
    Last edited: Sep 16, 2010
  14. Audioguru

    New Member

    Dec 20, 2007
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    No.
    Your Mosfet has a max on-resistance of 0.01 ohms when at room temperature. Then with a current of only 2A it dissipates (2 squared x 0.010= 0.04W which is 40milli-Watts.
    If the ambient temperature is 75 degrees C then the on-resistance might be 0.012 ohms and it would dissipate only 48milli-Watts.
     
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  15. gte

    Thread Starter Active Member

    Sep 18, 2009
    347
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    Ok,

    I was using the value of 16mW, so I believe my calculation formulas were correct, but I guess I should have used 10mW and 12mW.

    Thank you for the explanation. :)

    If it was 10mW I could conduct up to 12.65A 1.6W/.010Ω = I^2
    If it was 12mW I could conduct up to 11.55A 1.6W/.012Ω = I^2
    If it was 16mW I could conduct up to 10A 1.6W/.016Ω = I^2

    And if I doubled them up, 15A would be easy with no sink?
     
  16. Audioguru

    New Member

    Dec 20, 2007
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    The on-resistance of a Mosfet is measured in ohms or milli-ohms, not mW.

    I never operate any semiconductor at its maximum temperature. They are inexpensive so some might fail at that temperature then the manufacturer says, "Sorry" and gives a replacement.

    If you double Mosfets then the dissipation is not shared equally because each Mosfet is different.
     
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  17. marshallf3

    Well-Known Member

    Jul 26, 2010
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    Have you spent some time going through the drill-down filtering process at a place like http://www.mouser.com?

    As you've probably learned by now your 140W doesn't really apply, your main selection criteria depends on finding a MOSFET that can meet or exceed your voltage requirements yet has the lowest on resistance. Once you find the most likely candidates you've got the equations.
     
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  18. SgtWookie

    Expert

    Jul 17, 2007
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    Just a few notes...

    To keep things simple, just look at three things for the moment:
    1) Vdss rating (maximum voltage to be applied across the drain and source terminals)
    2) Rds(on), or the resistance between the drain and source terminals when the MOSFET is ON.
    3) Qg (total gate charge)

    You want to select a MOSFET that is at least rated for the Vdss it will be exposed to in the circuit. If you go higher than necessary, than Rds(on) and Qg will suffer.

    For a given Vdss rating, Rds(on) and Qg are pretty much tied together. In order to get a higher Vdss, the channel region has to be made thicker. In order to get a lower Rds(on), the area of the channel needs to be increased, which increases the Qg.

    The higher Qg is, the more capable gate driver you will need in order to switch the MOSFET on and off quickly. It is important that the MOSFET spend as little time as possible switching from ON to OFF or vice versa, as that is where the MOSFET is in the linear (resistive) state, and dissipates power as heat.

    But, without knowing what you are using as a driver for the MOSFET gate, it's going to be impossible to determine how much power is going to be dissipated during transitions.

    Since you are driving an inductive load (sinking current), turn-ON times are not as critical as turn-OFF times, because at T=0 (beginning to turn the MOSFET on), the current through the inductor will likely be minimal - unless you are doing something like a synchronous switch.

    If you're working with an automotive environment, you may experience transients (voltage spikes) up to 60v when there are load dumps (such as turning off a heavy load like the headlamps, etc.); it takes a few mS for the alternator to adjust/correct for the sudden removal of such heavy loads.
     
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  19. gte

    Thread Starter Active Member

    Sep 18, 2009
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    Audioguru,

    Is the dissipation so unequal that this is not a possibility, or are we talking about a milliamps difference? If I have 8 amps of current go through 1 transistor and 7 amps go through the other to get my total of 15 amps (instead of a split down the middle at 7.5) that will not be a problem.




    Thanks Sgt ... I was considering a flywheel diode like I did with my relay, but I believe I'd have to go from the drain to the supply, and I'm not sure I need it. I believe the inductor in the solenoid will try and keep the voltage level the same in between the - pin on the solenoid and the drain of the transistor as the transistor is shutting off, which is rated up to 100v. Do you think I'll need the flywheel diode?
     
  20. SgtWookie

    Expert

    Jul 17, 2007
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    You have a gross misconception of how inductors perform electrically.

    You need to study up on them.

    Inductors are an electrical equivalent of inertia. They resist any change in current through them. If you do not provide an alternate path for the current when it's cut off, the voltage across the inductor will climb, climb, climb until it blasts it's way through whatever is necessary to find a current path.

    This usually results in smoke.
     
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