mosfet heat in buck converter

Discussion in 'The Projects Forum' started by mah, Apr 3, 2014.

  1. mah

    Thread Starter Active Member

    Mar 15, 2010
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    i am doing buck converter to transfer 50 watt of power at pwm of 100Khz but the mosfet heat increased with time when it works at 2 Amp however duty cycle is 50% power loses should be small , should i use bigger heat sink . i attached the circuit board with heat sink.
     
  2. crutschow

    Expert

    Mar 14, 2008
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    How hot does the heat-sink get?

    If it's too hot, and since the heat-sink look reasonably substantial, then it appears that the MOSFETs may either have too high an ON resistance or they are not being driven properly so that they rapidly switch on and off. But you certainly can increase the heat-sink size if needed.
     
  3. mah

    Thread Starter Active Member

    Mar 15, 2010
    276
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    may be around 60 C . the mosfet is IRFZ48R it has very low Ron. what do you mean by "are not being driven properly" also the temperature of the toroidal inductor has increased . i drive it using PI controller there is ripple in the control signal it changed very fast up and down around constant value may be this is the reason . i got almost zero error but oscillations around it i attached error signal picture
     
    Last edited: Apr 3, 2014
  4. Papabravo

    Expert

    Feb 24, 2006
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    "Not being driven properly" means that the signal driving the MOSFET gate takes to long to switch between on and off. If you look at the gate waveform you will see the voltage change rapidly at first, flatten out onto a plateau, and then after some time finish the transition to the opposite state. When the gate voltage is "on the plateau" the MOSFET is operating in the linear region and dissipating large amounts of power. The "voltage source" driving the gate is having trouble charging(discharging) the gate capacitance.

    Drive the MOSFET with a current source to alleviate this problem.
     
  5. mah

    Thread Starter Active Member

    Mar 15, 2010
    276
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    this is the driving signal it is very normal
     
  6. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    If your buck circuit is 80% efficient, a typical number for something that is not thoroughly optimized, then with 50 W ouput you are dissipating another 12.5W in the regulator components, mostly in the transistors and the inductor. 60C is about 35C above ambient, and lets say that 80% of the regulator heat is in the transistors. That works out to about 3.5 degrees C per watt, and based on the photo I'd say your heatsink is small enough to support that.

    ak
     
  7. mah

    Thread Starter Active Member

    Mar 15, 2010
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    what about the raised temperature of the inductor is that normal?
     
  8. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    If you show us the 'scope photo of the FET drain voltage, we can see if it is switching fast enough.

    Also, you can reduce the frequency a lot, looking at your inductor type. If you halve the frequency you halve the switching losses and efficiency goes right up. Just watch for inductor saturation.

    A schematic would be nice too! :)
     
  9. mah

    Thread Starter Active Member

    Mar 15, 2010
    276
    2
    what is the need for drain voltage it is just dc voltage of the voltage source?
    here is the schematic. the inductor is T-94 26 and i used 50 turns around the core of two 25 turns connected in parallel
     
  10. RamaD

    Active Member

    Dec 4, 2009
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    He must have meant the Source Waveform.
    The problem appears to be bootstrap capacitor is not getting charged enough after an initial charge through the load. It appears that you have not connected any load or connected a light load, and the inductor current is zero, freewheeling diode not conducting and there is no charging path for the bootstrap capacitor.
     
  11. mah

    Thread Starter Active Member

    Mar 15, 2010
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    you must have not seen the schematic well, the load is connected it is 10R .there is inductor current or how could the coil get hot . freewheeling diode is connected look carefully it is two diode and one common cathode terminals ,
     
  12. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Yes. The wire in an inductor has resistance, and the heat loss comes from a combination of the peak and average current through it. Plus, the core generates heat as magnetic flux runs through it.

    A side note - switching regulators can be very efficient, and way more efficient than linear regulators - but the big efficiencies (90%, 95%, etc.)come only when the regulator is loaded close to its maximum capability. A 100 W circuit delivering only 20 W can be less efficient than a linear regulator in the same application.

    ak
     
  13. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Just a couple of observations, probably not related to your problem.

    First my assumptions. Since you say 2 amps into 10 ohms the output must be about 20 volts and with ~50% duty cycle the input 40 volts.
    So when I look at your scope picture it says the trigger point is 11 volts meaning the signal is only about 12 volts. The drive signal should be on top of the 20 volts on the source so a signal from 20 to 32 volts. Maybe the scope was AC coupled?
    If you get rid of R3 (1k from gate to source) you can use a much smaller boost capacitor than C1.
    Your right it is a pretty good FET. Most of the power should be lost in the diode. Maybe you can tell if the diode is much hotter than the FET. If it is maybe you could just use a bigger heat sink. If not we need to check some more.
    PS. If the diode is the hot one you could use both of them in the package. It might help a little.
     
  14. RamaD

    Active Member

    Dec 4, 2009
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    Oh, I missed the 10 Ohms in the corner.
    It is not the connection to the freewheeling diode and I did see that. You should read

    'not connected any load or connected a light load, and the inductor current is zero, freewheeling diode not conducting and there is no charging path for the bootstrap capacitor.'

    fully, especially 'not connected any load or connected a light load, to understand the rest of it.
     
  15. THE_RB

    AAC Fanatic!

    Feb 11, 2008
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    Normally a buck uses a p-channel FET and the drain is the pin connected to the inductor. You used a NFET and a special high side driver IC which was information you had not supplied at that time. So it's the waveform at the FET pin that joins the inductor. :)

    The inductor current waveform is really handy too, it will show how bad your current ripple is and how close the inductor is to saturation. And how much you can slow the switching freq. To see that waveform you can put a 0.1 ohm resistor between the inductor and C6 output cap.
     
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