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pwm power transistor — Parallax Forums

pwm power transistor

TobiasTobias Posts: 95
edited 2007-09-29 17:09 in BASIC Stamp
I am working with a 2amp 12 volt coil that I am trying to control with a pwm, I am using a tip41 power transistor it works good for a moment but after a bit it gets really hot and does not perform like it should I think the gain on it is 15, I have been lead to using a 1k gain transistor would that solve my problem? I've got the diode in the circuit for that back emf. Your advise will be greatly appreciated.
Toby
Thanks

Comments

  • Mike GreenMike Green Posts: 23,101
    edited 2007-09-21 00:05
    You'll need to post your circuit. It's impossible to tell what's going on without the voltages and resistor values and knowing what's connected to what.· Post your program too.
  • TobiasTobias Posts: 95
    edited 2007-09-21 06:53
    I have got the power transistor's base connected directly to the bs2, the emitter to the coil's neg·and the collector to ground, the positive wire to positive,··I just want to run it at a single valued pwm, but turns hot after a moment.

    thanks

    toby
  • Mike GreenMike Green Posts: 23,101
    edited 2007-09-21 15:05
    You need a current limiting resistor between the BS2 and the base ... on the order of 220 ohms. You may have damaged the Stamp's I/O pin although they're pretty robust.

    If you're using an NPN power transistor, you've got the transistor connected backwards too. Even if you're using a PNP transistor, the coil shouldn't be in the emitter lead. Before you go any further, please look at the examples in www.parallax.com/dl/docs/prod/sic/Web-PC-v1.0.pdf, chapter 5. Any time you have a solenoid or motor or relay of any size, you will need a diode (like a 1N4001) connected in reverse across the solenoid or relay (so it doesn't normally conduct). This prevents the "back EMF" (generated when the transistor turns off and the coil's magnetic field collapses) from destroying the transistor.
  • TobiasTobias Posts: 95
    edited 2007-09-26 04:50
    I put in a current limiting resistor, and I also have the diode in the circuit for the coil, It works good when its cool but warms up after a bit, the transistor is a tip41 with 15 hfe then I also used a 1k hfe tip125 pnp transistor and also got hot.
    What I am trying to control a 2 amp coil with pwm.
    Toby
    Thank-You
  • Beau SchwabeBeau Schwabe Posts: 6,568
    edited 2007-09-26 05:10
    If your current that you are trying to control is 2 Amps, and your hfe is only 15, then you need at least 133mA driving the base of your transistor for proper transistor saturation.
    Since the Stamp can only source or sink about 20mA of current per pin, then I would guess that your transistor is not adequately being driven into saturation.
    ·
    One suggestion is to find a transistor with a higher hfe of at least·100 or more.· Another is to use an intermediary transistor in a Darlington configuration.








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    Beau Schwabe

    IC Layout Engineer
    Parallax, Inc.
  • TobiasTobias Posts: 95
    edited 2007-09-26 05:18
    thats exactly what I just did is hook it up as a darlington configuration and it would still get hot, I hooked the base of the first transistor to the bs2 and then the darlington configuration to the hfe 15 power transistor.
    Your reply will be helpful
    Thank You
    Toby
  • Beau SchwabeBeau Schwabe Posts: 6,568
    edited 2007-09-26 05:40
    You still need a current limiting resistor on the base of the first transistor....

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    Beau Schwabe

    IC Layout Engineer
    Parallax, Inc.
    HFE.JPG
    1141 x 765 - 385K
  • TobiasTobias Posts: 95
    edited 2007-09-26 13:37
    I don't know for some reason I am hooking up my circuit exactly the same way that youre explained and the tip41 still gets really hot, how hot should it get????? I know that the coil max amps is only 2amps.
    Thank you
    Toby
  • allanlane5allanlane5 Posts: 3,815
    edited 2007-09-26 14:09
    What is the voltage across the TIP41 transistor?

    Power = Current * Voltage (P = IV). A 'saturated' transistor (fully on) should have 0.1 volt between Collector and Emitter.

    2A * 0.1 Volt == 0.2 watts, or 200 mWatts. 200 milli-watts is a VERY small amount of power to dissapate -- a TO-220 package should be able to disappate 5 watts or so (though it would get quite warm -- 150 degrees or so -- doing it).

    Since the TIP41 IS getting hot, it sounds like you haven't 'saturated' it, but instead are operating it in some 'linear' region. Worst case would be it's dropping the entire 5 volts at 2 amps (not sure that's possible, but that WOULD be worst case) which is then 10 watts -- which is a LOT of power to try to dissapate.

    So, measure that voltage when it's on, then you'll know how much power it's trying to dissapate. You may have to lower the base resistor on your FIRST transistor, to insure it's puttting enough current into the TIP41 to saturate it.
  • Beau SchwabeBeau Schwabe Posts: 6,568
    edited 2007-09-26 16:02
    Also, what is the voltage across the coil ... AND ... what is the resistance of the coil not connected to your circuit?
    What is your power supply rating and source to the coil?· If it cannot sustain the required 2A it may allow the TIP41 to fall outside of saturation resulting in excessive heat.
    It's also possible that your TIP41 may be damaged from your previous trials.


    ·



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    Beau Schwabe

    IC Layout Engineer
    Parallax, Inc.
  • pwillardpwillard Posts: 321
    edited 2007-09-26 17:52
    If it *is* damaged. Try replacing it with a TIP111. If I were doing this, I also would not be doing it without using a heatsink of some kind initially.

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    There's nothing like a new idea and a warm soldering iron.
  • Phil Pilgrim (PhiPi)Phil Pilgrim (PhiPi) Posts: 23,514
    edited 2007-09-26 18:58
    One problem with a Darlington connection is that it switches rather slowly. If your PWM output is too fast, the Darlington won't be able to keep up and will spend most of its time in its linear region. Combined with the Darlington's rather high saturation voltage, this will cause overheating.

    My personal preference in a situation like yours would be to use a MOSFET and include a MOSFET driver, if necessary, to increase the switching speed.

    -Phil

    Addendum: If you're using the PBASIC PWM command to generate your PWM output, be aware that at 50% duty cycle, you're looking a 113KHz switching frequency. This is much too fast for your switching circuit to keep up with. In general, the PWM command is more suited for producing an analog output though a low-pass filter than for driving inductive loads through saturated switching. Parallax's PWMPAL is one option to consider for a more realistic output frequency. Another would be the MoBoStamp-pe, whose onboard AVRs support PWM output natively.

    Post Edited (Phil Pilgrim (PhiPi)) : 9/26/2007 7:37:40 PM GMT
  • Beau SchwabeBeau Schwabe Posts: 6,568
    edited 2007-09-27 04:15
    Phil makes a good point.

    Tobias,

    Can you tell us what you are using the PWM'd coil for?

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    Beau Schwabe

    IC Layout Engineer
    Parallax, Inc.
  • TobiasTobias Posts: 95
    edited 2007-09-27 06:56
    I check the resistance across the coil, and discovered its 5.2 ohms which makes sence, another thing I discovered is that when Ive got the circuit produceing 1 volt across the coil there are 10 volts across emitter and collector (something wrong, tells me the transistor has to dissapate lots of heat). What causes that? And what could I've done wrong?
    Thank-You much
    Toby
  • TobiasTobias Posts: 95
    edited 2007-09-28 06:06
    How come I get 12 volts across the emitter and the collector? that don't even sound right...
    Thank-You
    Toby
  • Beau SchwabeBeau Schwabe Posts: 6,568
    edited 2007-09-28 15:11
    Tobias,

    What you are observing is an average of the PWM, and most of the voltage observed across the transistor should be during the OFF state of the transistor.

    Let's back up a minute, and forget about the PWM for a moment to make sure that the transistor is saturating correctly when it is ON.

    Your resistance across the Coil is 5.2 Ohms (out of circuit?) and your supply is 12V. Because you are using a bipolar transistor, there is an inherent voltage
    drop of about 0.6V ... There is no way around this, that's just how bipolars work, so for the remainder of the bipolar transistor calculations, I will use 11.4V as my supply voltage
    instead of 12V. (12V - 0.6V = 11.4V)

    The maximum current if the transistor is turned ON should be about 2.2Amps. (I = V/R = 11.4V / 5.2 Ohms = 2.192 Amps)


    Now the question is, can your voltage supply handle the demand of 2.192 Amps?

    This is also where you should measure the voltage across the transistor, and the Coil.

    If the voltage across the transistor is close to 0.6V then we are ok... if it's not then the transistor is not saturated, and you will develop a LARGE amount of heat.

    To see this, consider the 0.6V that should be across the transistor. Remember, the same amount of current is going through the transistor as is the coil.

    Power or Watts = I * V =·2.192 * 0.6V·= 1.3 Watts ... That's a fair amount, and that's an ideal circumstance.


    Now, suppose the transistor was reading 1.1V ... The same formula would apply...

    Power or Watts = I * V =·2.192 * 1.1V··= 2.4 Watts ... That's HUGE amount of difference for only a half of a volt change!



    For a MOSFET solution, what you look for is a low RdsON (<- which means ... A low Resistance across the Source and Drain while the transistor is ON)

    With a MOSFET, you don't have a 0.6V drop.

    Suppose you have a MOSFET with an RdsON of .055 Ohms ... To calculate maximum current, the equation would look like...

    I = V / (Rcoil + Rmosfet) = 12V / 5.255 Ohms = 2.284 Amps

    Now, this seems like it's going the wrong way because you now have a higher current, but look what is happening across the MOSFET.

    Power or Wattage = I^2 * R = 5.217 * 0.055 Ohms = 287mW .... Compare that to the bipolar transistor at 1300mW

    Less energy is being wasted as heat, and more energy is available across·your coil.




    Bottom line is....

    1) Make sure that your power supply can handle the current demand regardless of what transistor type you are using.
    2) If you are using a bipolar transistor, make sure it is properly saturated

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    Beau Schwabe

    IC Layout Engineer
    Parallax, Inc.

    Post Edited (Beau Schwabe (Parallax)) : 9/28/2007 3:18:09 PM GMT
  • LawsonLawson Posts: 870
    edited 2007-09-29 17:09
    another problem with the darlington transistor configuration is that it has a high minimum voltage drop. The power transistor needs at least 0.7v on it's base and the only way to get that with a darlington is to have a collector-emitter voltage greater than 0.7v on the power transistor. I've observed voltage drops of up to 1.2v for fully on, when I played with this stuff YEARS ago. So with a 1.2v drop and 2A current, the transistor has to dissipate 2.4 watts. Try driving the power transistor with a PNP wired up as an inverter. This will give the extra current needed to saturate the power transistor without the voltage drop of a darlington configuration.

    My 2 cents
    Marty

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