High Current switch-will this work?

Discussion in 'General Electronics Chat' started by RodneyIwan, Jun 14, 2008.

  1. RodneyIwan

    Thread Starter New Member

    Jun 14, 2008
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    I am new at using MOSFETs but have an application and some spare IRFZ44's so am thinking of using them as shown in the attachment. If I need to parallel two to get the current capaciy (the motor will draw about 35 to 40 amps) I'll connect the second one at points x,y and z. Will this work, any suggestions appreciated along with any precautions. Thanks Rodney
     
  2. hgmjr

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    Jan 28, 2005
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    The parallel scheme you are considering is probably workable. You may need to increase the size of the 22 Ohm resistor a bit. Are you going to use PWM to control the motor?

    hgmjr
     
  3. hgmjr

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    When paralleling mosfet devices, it is crucial to make sure that they remain at the same temperature. Heatsinking in applications such as the one you are proposing is very important. Any time one of the parallel devices gets hotter than its companion it starts to take a bit more of the overall current being delivered to the load. This in turn causes it to heat up more. This is the thermal equivalent of the "spiral of death". The result is usually the loss of all of the devices. Proceed with caution.

    hgmjr
     
  4. RodneyIwan

    Thread Starter New Member

    Jun 14, 2008
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    I am using a common heat sink (will hold up to 3 of them on one heat sink), the one the FET's were mounted on in the UPS I scavanged them from. How big should the 22 ohm resistor be? Isn't it just to help the FET's share the drive? My biggest problem--I think--will be to have an adequate size bus wire connecting the grounds (sources) and the drains.
    In answer to "will I be using pulse width"; no; just a push button switch to turn the motor on when it is engaged with the props spinner. One more question, how important is it to have the diode accross the motor? Would I risk blowing the FET's if I left it out? Thanks for the replys and suggestions.
     
  5. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    I don't think this is quite right, hgmjr. The IRFZ44, like most vertical FET's, has a positive temperature coefficient of on-resistance. This means that if a FET gets hotter it will take less, not more, of the current. This is one of the great benefits of power FET's.

    See the attached graph from the IRFZ44 data sheet.
     
  6. hgmjr

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    That is a good measure to take.

    Since you are not using PWM to control the motors then switching speed is less of an issue. You can go to 100 Ohms or so. The gate current is not that significant so there will be no degradation to the gate drive.
    Yep. When it comes to high current, size really does matter.

    The inclusion of diodes is well worth doing. The flyback voltage from the motor definitely needs to be suppressed.

    hgmjr
     
  7. hgmjr

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    According to the graph you provided, the rds on does increase with temperature.

    That means that instead of taking more current with the increased temperature it takes less. There would still seem to be a potential for a thermal runaway situation to ensue. Wouldn't there be? If one of the devices in parallel stops carrying its share of the load due to local heating then the others must take up the slack. Eventually, the lion's share of the current would end up being shoulder by fewer and fewer devices. That is until the last mosfet exceeds its current rating and is damaged.

    hgmjr
     
  8. The Electrician

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    If one of the FET's is not securely bolted to the heatsink, then its temperature will increase rapidly and it will burn out, but this isn't thermal runaway in the sense the term is usually used to describe the effect that happens with a negative tempco.

    The FET's should all be bolted securely to a common heatsink to get good current sharing.

    Perversely, if the on-resistances of the FET's aren't well matched from the beginning, a counter-intuitive effect occurs. The FET with the lowest on resistance will carry more current and get hotter. You'd think the FET with the lowest on resistance would run cooler, but this is not the case when you have several FET's in parallel.

    You said:"If one of the devices in parallel stops carrying its share of the load due to local heating then the others must take up the slack."

    When the others begin to take up the slack, they get hotter, their on resistance goes up, and the one that had stopped carrying its share once again carries more current.

    The positive tempco of the FET's tends to equalize the current sharing.

    But, the FET with the lowest initial on resistance ends up carrying somewhat more of the current, and gets hotter. This is contrary to what you would think at first, but a quick analysis shows that it's true.

    Since FET's in parallel have the same voltage, E, across them (neglecting wiring resistance), the power dissipated in each is E^2/Rds_on. E is the same for all the FET's, but each FET has its own Rds_on. If a FET has a lower Rds_on, it will dissipate more power, since that parameter is in the denominator of the expression for power.
     
  9. hgmjr

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    That is very good info. I will keep it in mind the next time I tackle a parallel mosfet driver design.

    Thanks,
    hgmjr
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    I suggest that the 4.7k resistor needs to be decreased to 2k.

    You want 10v on the gate(s) to ensure the MOSFETs are fully turned on. Otherwise, they will be operating in the linear region, and will have a higher resistance than they would otherwise.
     
  11. The Electrician

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    The IRFZ44 has a gate voltage rating of +- 20 volts.

    He should move the top of the 10K over to the left of the 4.7K (and get rid of the 22 ohms), then he'll get 12 volts of gate drive, but still have 14.7K pulling the gate down to zero volts when the switch is open.

    Not knowing where the 12 volts is coming from, but just in case it's a battery in a vehicle, or some such situation, he probably ought to put an 18 volt zener from gate to source (cathode up) to make sure no spikes greater than 20 volts find their way to the gate.
     
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