P-Channel MOSFET overheating issue

Discussion in 'The Projects Forum' started by russpatterson, Nov 1, 2010.

  1. russpatterson

    Thread Starter Member

    Feb 1, 2010
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    Hello,

    I've got a design here that does a good job of cooking P channel MOSFETs. Even if I don't switch the gate, just leave the thing on so it conducts, it heats up. Which leads me to believe it's not all the way switched on. The way I understand the P-Channel MOSFETs work, which may be wrong, is that with a Gate to Source voltage differential of 0, they are off, or they don't conduct.

    So in this circuit, when I drive the GP5 pin low, there's no current on the base of T1 so the voltage at the gate of Q2 is the same as that on the source so it should not conduct.

    When I drive the GP5 pin high then T1 conducts and the voltage at the gate of Q2 drops to ~0V and Q2 conduts.

    I'm using the IRF5305 http://www.irf.com/product-info/datasheets/data/irf5305pbf.pdf

    I attached an LTSpice sim as well.

    I'm not sure why the MOSFET heats up. Any ideas? Thanks.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    I guess you're still not paying attention to Vgs.

    You do realize that the absolute maximum for Vgs is +20v/-20v, right? And if you exceed those limits, you will destroy the MOSFET?

    Describe the signal that is on GP5_panels in your real-life situation.
     
  3. russpatterson

    Thread Starter Member

    Feb 1, 2010
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    The signal on GP5_Panels is from an I/O pin on a PIC. It's 0 to 5V. I just use it like a relay. Switching frequency is less than 1Hz.

    VGS is Voltage on the gate minus voltage on the Source correct?

    I see what you're saying. I'm generating a 20V differential, VGS when I turn on that BJT. How does that relate to VGS(th)? That's the threshold that causes the MOSFET to turn on correct? Should I be concerned about not exceeding the max VGS(th)?

    I put in a voltage divider to hold the voltage swing on the gate from 16 to 20V. (in attached .asc file)

    For some reason I've messed up the LTSpice sim so that current is not flowing through the MOSFET like it was yesterday. I'm not sure what is wrong there.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    I don't see an inductor, nor a diode. The idea of the inductor is to keep the current relatively constant; to store the power that's coming out of the MOSFET. The diode is to provide a current path for when the MOSFET is turned off.

    Since you do not have a current limiter (inductor) in the circuit, the MOSFET is dumping as much power as it can into the battery.

    Vgs is the voltage on the gate, using the source terminal as the reference point.

    You're pulling the gate to nearly 0v potential. Since there is 20v on the source terminal, Vgs is nearly -20v, which is near the breaking point of the MOSFET.

    The threshold is where the MOSFET conducts at about 250uA. You only look at the threshold to determine how low you need to get Vgs in order to turn off the MOSFET. Basically, if a power MOSFET is passing less than 250uA (0.25mA) it's considered off.

    You need to start looking at Rds(on) instead of the threshold.

    The absolute maximum limits for Vgs are +20v/-20v. If you exceed those limits, you will destroy the MOSFET. There really isn't any purpose in exceeding ~+12v/-12v, unless you simply have too many MOSFETs lying around, and you like changing them.

    If you use a voltage divider, you'll increase the gate charge/discharge times, resulting in more power dissipation in the MOSFET. Use a voltage follower that has it's input excursions limited instead.

    I haven't looked at your simulation yet.
     
  5. nickelflipper

    Active Member

    Jun 2, 2010
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    I'm interested in the outcome on this topic. Having personally smoked a couple of IRFR5305's in a solar panel buck converter, I can appreciate Wookie's Vgs concerns. Using a SFP9540 (with Vgs +/-30V silicon RevC) cured the Vgs problem at the expense of higher Rdson. The reason I mentioned the RevC silicon, is that DatasheetCatalog.com is listing RevB silicon which is only +/-20V Vgs.
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    What are you actually trying to do, i.e., what does the series RC circuit connected to the drain represent?
    In your sim, no current will flow through the drain once the capacitor is charged.
     
  7. SgtWookie

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    Jul 17, 2007
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    I'm fiddling around trying to modify Helmut Sennewald's NiMH AA battery model into a general purpose lead-acid battery model, and adding a few more features to it.

    Helmut's models are available on Yahoo!'s LTSpice User Group.
     
  8. russpatterson

    Thread Starter Member

    Feb 1, 2010
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    The RC on the drain of the MOSFET is supposed to represent a 12V battery. Is there a better way to do this?

    I've attached an updated .asc file with the voltage follower to hold the gate voltage closer to the source voltage. So VGS never gets higher than about 11.5V.

    I'm going to try it out, hopefully later today, and see what happens.
     
  9. Ron H

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    You're connecting a 60 milliohm MOSFET between 30V and 12V. Use Ohms law to calculate why your MOSFET is getting hot.
     
  10. Ron H

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    If your 30V supply really has a 4 ohm output resistance, then it will get hot instead of, or in addition to, the MOSFET.
     
  11. russpatterson

    Thread Starter Member

    Feb 1, 2010
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    I had a bogus (poorly chosen) part in the sim for the MOSFET that's why it didn't run. Thanks for looking at it with me.

    The 30V power source represents a 7 Amp solar panel. In real life, not the sim, the panel's voltage collapses to that of the battery. I'm working on another project to DC/DC that down to the battery's level. In the mean time I need to walk before running and be able to successfully switch the MOSFET at low frequency without it heating up.

    I think the VGS probably is not the issue since the panel voltage (connected to the MOSFET source pin) collapses to 12V to 14V. So even grounding the gate, 0V, would only produce a VGS of -14V which is within range (+-20V) of the IRF5305 I'm using.

    So the mystery continues as to why the thing cooks when 4-7 amps flow through it. This is just leaving it switched on. In <5 seconds it's too hot to touch.

    Out of sun today but tomorrow I'll try it with a voltage divider of a 5.1K / 1K (on the gate) to get the VGS right about -10 and see what happens.
     
  12. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Four amps through 0.06Ω is almost a watt. This will probably get too hot to touch. Seven amps will cause about 3 watts dissipation. You need a good heat sink. Maybe you should try IRF4905 (Rds(on)≈.02Ω), which will cut your dissipation to about 1/3 that of IRF5305.
    If your switching frequency is less than 1Hz, gate drive rise and fall times are not too critical. The average dissipation decreases as the transition times get faster, but if you have, say, 1uS transition times, the peak dissipation during that time will be high, but the average dissipation is pretty low.
    That voltage follower you drew was not connected to power, so it did nothing. If you are going to use a buffer, that op amp is not the right part. It will only work with 12V or less, and the inputs and output ranges are inadequate. If I were to use a buffer, I would probably just make a push-pull emitter follower from a PN2222 and a PN2907.
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    Finally making some decent progress on the batt_SLA model.

    Have a look at the attached schematic; it shows the simulated battery starting off at 100% charged, being discharged with a 2A load, then being recharged with a constant 2A (this is too much for a 7AH battery, but this is just a simulation.) at three different temperatures; green is -20°C, blue is 5°C, red 30°C. Note that the 100% charged voltages are different over temperatures, and so is capacity.

    Note that it takes longer to charge at 2A than it does to discharge at 2A, which is due to the charge efficiency; currently it's set to 75%.

    It's not a perfect model, but sure beats a resistor and a cap in series.

    batt_SLA.zip contains:
    batt_sla.asy - goes in Program Files\LTC\SwitcherCad\lib\sym\misc
    batt_SLA.sub - goes in Program Files\LTC\SwitcherCad\lib\sub
    SLA Batt Test.ASC - simulation file; goes in Program Files\LTC\SwitcherCad
     
    Last edited: Nov 5, 2010
  14. russpatterson

    Thread Starter Member

    Feb 1, 2010
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    Thanks for the replies.

    So I can get to your .96 watt at 4 amps calculation but I'm not understanding the 3 watts for 7 amps. I get 1.68 watts for 7 amps. Thanks for the info that 1 watt would generate enough heat to make the part too hot to touch. Here's how I got to .96 watts

    voltage drop across resistor V=IR so Vdrop = 4*0.06 = 0.24 V
    power dissipated P=VI so P= 0.24 * 4 = 0.96 watts

    Did I do that right?

    I put on a good sized heat sink by bolting the TO-220 package to my board and putting some folded tin on the top and that seemed to keep things cool enough so the IRF5305 would not self destruct. (Although my motor controller MOSFET burned up on me for some reason. Sparks, smoke, complete meltdown. I could see it igniting nearby items had I not been there right next to it. I might have shorted something or it might have been damaged when the IRF5305 burned up right next to it last weekend. I've replaced it and tested it on the bench and it works ok there at this point.)

    I'll get some IRF4905's next time I put in a digikey order.

    I did some searching on the push-pull emmiter follower but I'm not sure where to set the bias voltage and what node is the output. Given this schematic would I just make a voltage divider to set the voltage at Vin then drive the gate where Rload is shown?
     
  15. russpatterson

    Thread Starter Member

    Feb 1, 2010
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    Thanks a lot for the battery model. I was not able to get the sim working however. I had to rename the batt_sla.asy file to battery_sla.asy because it complained that it could not open "battery_sla.asy". Now it opens up the schematic ok but when I run the sim I get a Spice Error: "Unknown subcircuit called in: xu1 vcc 0 battery_sla". Maybe I don't have the intended files. Really interesting stuff though.
     
  16. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    You did the first calculation right, but you screwed up the 7 amp calculation. You apparently said 7A*.06Ω=.42, but then you multiplied that by 4 instead of 7 to get the power dissipation. A simpler way to calculate it is P=I^2*R, i.e., P=7*7*0.06.
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    OK, I goofed! Uploaded the wrong test.asc file; sorry about that.

    Rename battery_sla.asy back to batt_sla.asy, and try the attached batt_sla test.asc
     
  18. russpatterson

    Thread Starter Member

    Feb 1, 2010
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    Sorry for the delay, got distracted with a pay LED architectual light project and a pay iPhone app gig. So thanks again for the great info. I finally swapped out the P Channel Mosfet (IRF5305) for the IRF4905 this morning. The RDSon certainly makes a difference. I can feel the heat generated still but it's much cooler than the 5305 even at 7 amps. I still need to layout a schematic with the dual BJT voltage follower so I can quickly switch the 4905 off and on. The things been running fine for weeks as is though, (at the on/off no faster than .5 Hz).

    I did get the battery simulation running correctly. Thanks for that! I need to research how to make a part in LTSpice IV and take a look at what you're doing. I'll add that battery model to my simulations next time I open that up.

    So the outcome of this thread was that I underestimated the heat that would be generated by 0.06 Ohms (the RDSon of the FET). Seems like nothing when you're thinking about 100 Ohm resistors and such but it's there and you pay for it.
     
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