MOSFET Heating Problem

Discussion in 'General Electronics Chat' started by Engr, Aug 15, 2011.

  1. Engr

    Thread Starter Member

    Mar 17, 2010
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    Hello everyone,

    I have a circuit which uses a MOSFET as a switch. The MOSFET that I am using is the IXTK90P20P which has a rating of -200V, -90A. The problem that I am encountering is that the MOSFET is heating up so much even though the voltage and current applied to the MOSFET is only -150V, -1A. The heat is causing the MOSFET to get damage when being used. I need help on how to solve this kind of problem.
     
  2. t06afre

    AAC Fanatic!

    May 11, 2009
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    I think you forgot to post the schematics. Then your IXTK90P20P is turned fully on. It has a RDS(on) ≤ 44mΩ. I suspect your FET is not turned fully on. So the RDS is higher than the 44mΩ. Check your VGS voltage.
     
  3. Engr

    Thread Starter Member

    Mar 17, 2010
    114
    5
    Thanks for the input. Sorry I cannot post the circuit schematic since it is a confidential document of our office. I will try to post a simplified version of the schematic maybe later. I will also check on the Vgs voltage base on your input. Just for clarification, if the IXTP90P20P MOSFET is not fully turned on then the 1A current could not pass through the MOSFET right? If it is correct then why would the MOSFET heat up so much that it would cause the MOSFET to get damage? Could you explain more further on why it would damage the MOSFET if it is not fully turned on. Thanks
     
  4. praondevou

    AAC Fanatic!

    Jul 9, 2011
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    With "not fully turned ON" means it's still low resistance, so there will still be almost 1A.
    Ok, it's confidential, but since the circuit isn't working, nobody would try to steal it :D. Just kidding.

    1. Is it mounted on a heatsink?
    2. Is it a high frequency switching application and if so, what's the gate resistor value?

    When the MOSFET heats up, it's maximum allowable power dissipation decreases.
     
  5. t06afre

    AAC Fanatic!

    May 11, 2009
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    A few ohms say 2 ohms added to RDS(on) will not affect the total current in the system much it will only lower the total current with 12mA. But your FET will have to dissipate 2 watts of power. And that may be enough to heat up your FET so it reach a critical temperature. And then the temperature increase the RDS(on) will also increase. You will end up in bad circles of events. You should measure the drain-source voltage. If you control the FET by a pulsating voltage. You must use a scope. As most multimeters will give you a wrong reading. And also have a proper cooling fin on the FET. I am sure some more senior people at your office. Will be more than happy to help you on this matter
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    The datasheet for that transistor can be viewed by clicking this link:
    http://ixdev.ixys.com/DataSheet/DS99933B(IXTK-TX90P20P).pdf
    The total gate charge (Qg) is quite large, at 205nC. It will require a capable gate driver to turn the MOSFET on and off quickly. If the gate driver is not optimal, and the circuit is attempting to switch the MOSFET at a moderate to high frequency, the MOSFET will spend much of its' time in the linear state, dissipating power as heat.

    You could post just the MOSFET, the circuit/method used to drive the gate, the frequency or PRT of the gate drive and the duty cycle or time on/time off.
     
  7. Daniela-21

    New Member

    Aug 14, 2011
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    Definitely you need driver for Mosfet and check gate input (10V for N and up to 20 for P) max. Also search in Element14 for better or lower R like 0.05 minimum and check frequency (Higher mean hotter)
     
  8. t06afre

    AAC Fanatic!

    May 11, 2009
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    If you use Google and this term-> mosfet driver circuit. You will on the first page find helpful information.
     
  9. Engr

    Thread Starter Member

    Mar 17, 2010
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    We have already tried mounting the MOSFET on a heatsink but it still gets very hot that the MOSFET is still getting damage. The Drain of the MOSFET is connected to a capacitor while the Source of the MOSFET is connected to a 1ohms resistor. The 1ohm resistor is used as the current limiter. We had a control circuit connected across the 1ohm resistor which will turn off the MOSFET once the current passing through the MOSFET is greater than 1A. The Gate is connected to a parallel combination of a 1k resistor and a MOSFET which is use to quickly turn off the main MOSFET. The MOSFET connected to the capacitor acts as a switch which automatically turn off once the capacitor is already fully charge. The constant charging of the capacitor is causing the MOSFET to heat and eventually causes the MOSFET to get damage.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    You need an inductor between the MOSFET drain and the capacitor. Otherwise, the capacitor "looks like" a dead short to ground to the MOSFET; you will not be able to turn the MOSFET on and off quickly enough to prevent it from heating up.

    Adding an inductor will cause the current flow through the MOSFET to start at zero, and then (relatively) slowly build up. Once the voltage across your 1 Ohm resistor reaches the threshold, you will then need to turn the MOSFET off for a period long enough to completely discharge the inductor. This is known as "discontinuous mode". If you do not allow sufficient time for the inductor to discharge, you will again wind up with a heating problem.

    You will also need a diode across the inductor and the load, as when the MOSFET turns the inductor current off, the inductor will try to keep the current flowing, resulting in the polarity across the MOSFET reversing, and the voltage to reach a very high peak, which will destroy either the load, the MOSFET, or both.
     
    Last edited: Aug 15, 2011
  11. bobcart

    Member

    Jul 7, 2011
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    The electrical rating of the MOSFET is a separate concept than it's operating temp. I suspect your heat sink solution is not adequate. If the cap is constantly charging as you state, the mosfet will continue to heat up without an adequate sink. You can improve this by adding a fan, increasing mass and fins, improving geometry and reducing thermal resistance by using a thermal grease and/or better mating the MOSFET to the sink.
     
  12. Engr

    Thread Starter Member

    Mar 17, 2010
    114
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    I understand. This type of circuit is somehow familiar to me, having an inductor in series with the MOSFET. With this type of technique we need to have a delay to allow the inductor to discharge, the problem with this one is that we don't have any control on the on-off switching of the MOSFET since this is done automatically. Once the current gets higher that 1A the MOSFET is turned off but once the current goes below 1A it will automatically turn on the MOSFET again.
     
  13. Engr

    Thread Starter Member

    Mar 17, 2010
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    can you explain further what you mean with the "increasing mass and fins, improving geometry". I don't fully understand what you mean with those terms you mentioned. Thanks
     
  14. SgtWookie

    Expert

    Jul 17, 2007
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    Here is what I'm talking about - see the attached.

    On the left is basically your existing circuit; but it's dissipating less power than yours is. Duty is 1uS on, 20uS off. Average power dissipation is 195 Watts. 195 Watts is a lot of power to dissipate.

    On the right is the same duty, but an inductor and diode has been added. Average power dissipation is 10.2 Watts, which is quite manageable.

    If you really want to fix the problem, you will add an inductor and a reverse-EMF diode like I've done, and you will add a fixed time delay to disable the MOSFET turn-on until the delay has elapsed.

    The way that they have it set up now is just silly. Turn off the MOSFET when the current across the resistor trips the threshold, and then turn it right back on again? All they are doing is keeping the MOSFET in the linear region, generating maximum heat.
     
  15. Engr

    Thread Starter Member

    Mar 17, 2010
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    I get your point. Is it possible that the MOSFET will not heat up when we add the inductor if we are going to use a heatsink or the delay is really needed? I am just trying to find another option other than adding the delay since the circuit is already built and adding a delay will cause a modification on our circuit. Is their other way without adding a delay to prevent the MOSFET from heating up without adding a delay on this circuit configuration you posted?
     
  16. SgtWookie

    Expert

    Jul 17, 2007
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    I have no clue what size cap you are charging up. All I know is that the cap "looks" electrically like a dead short when it is not charged; and without some viable method of current limiting, you will continue to roast MOSFET after MOSFET.

    If the design is already in production, then you have a big problem. You will need to scrap or retrofit/re-work any of the existing circuits.

    Get rid of the 1 Ohm current sense resistor, and use a Hall-effect sensor that is in the path of the loop, after the reverse EMF diode connects to the drain/inductor junction. That way you will be monitoring the actual current flow in the inductor, rather than current flow into the MOSFETs' source terminal.

    If you're going to keep the scheme with the current sense resistor on the source terminal, then you will be forced to implement a fixed delay to allow the inductor to fully discharge.

    That discharge time will vary depending upon the actual inductance, the actual voltage applied across the inductor, and how long the voltage was applied. If the capacitor is completely discharged, the voltage across the inductor will be at its' maximum, so current flow through the inductor will increase at the highest rate; which will take longer to fully discharge. As the capacitor charges, the voltage across the inductor will decrease correspondingly, so current flow will increase more slowly, so the discharge time will be less. This makes the fixed delay option far less attractive than monitoring the actual current flow through the inductor.

    If you can implement a Hall-effect feedback instead of the sense resistor, you may be able to salvage the existing circuit; and you will reduce the total power dissipation as the sense resistor can be replaced with a zero Ohm resistor.

    You should also consider replacing that P-ch MOSFET with one that has a much lower gate charge. The trade-off is that the MOSFET will spend much less time in the linear region, for an increase in Rds(on), as you really can't reduce the Vdss requirement by much. Since you don't feel that you can share the MOSFET gate drive circuit, I really can't say a whole lot on the capability of it; but with that large a gate charge you will need a very capable gate driver.
     
    eceblr2011 likes this.
  17. Engr

    Thread Starter Member

    Mar 17, 2010
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    Do you know of any P-channel MOSFET that has a high current rating with a Vds of 200V which I can use to replace the IXTK90P20P?
     
  18. Engr

    Thread Starter Member

    Mar 17, 2010
    114
    5
    How about if I use the IXTP44P15T MOSFET, would it somehow solve the problem?
     
  19. Engr

    Thread Starter Member

    Mar 17, 2010
    114
    5
    Another thing is that our supply to the MOSFET is only limited to 1A current. We are using the PB150S DC-DC Converter as a supply to our MOSFET and this component has a maximum output current of only 1A so even if the current limiter is not working the supply will still limit the current to 1A only.
     
  20. Daniela-21

    New Member

    Aug 14, 2011
    2
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    Bla, bla, bla 1k for P mosfet and 150V ha. For example if you have 12V input (Gate to Source) is 300-1K so 150V ? hmm. Now 1ohm and 150V do what. Please drawing some diagram to give us some ideas. P Mosfet is good only for Dimmers with low frequency like 1kHz
     
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