Help me stop Frying my Circuit!

Discussion in 'The Projects Forum' started by alfredeneuman, May 6, 2011.

  1. alfredeneuman

    Thread Starter New Member

    May 6, 2011
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    1
    I have a DC motor pulled from a cordless hand vacuum, powered by an AC adapter pulled from a laptop computer, and controlled (on/off) by a control board connected to my computer (USB). I will turn it on and off by pressing a key on my computer. Please see the below diagram. I can get the circuit working for a like a minute. But, after being on for about a minute (that is, the DC motor running for a minute), it fails.

    [​IMG]

    I have fried this circuit about 3 times already. The first time I fried the transistor because I did not attach a heatsink to it (was not informed that this is necessary). But, the second and third times I have fried this circuit, I am not sure what I am doing wrong. I am a total beginner and the components in my diagram are all based on someone's recommendation on another message board a few months ago.

    Right now, when you turn on the circuit, the motor stutters. What I mean is, the motor moves in pulses like on for a millisecond, then off for half a second, then on again for a millisecond, then off for half a second, etc., etc.. At first, I thought I messed up the AC adapter, but I plugged it into the laptop (the AC adapter was from a laptop computer) and it is able to power up the laptop. But, if I take some wires and run them directly from the AC Adapter's plug straight to the DC motor (bypass the whole circuit), the motor still does that stutter like I described above. But, if I attach the original rechargeable battery pack that comes with the motor (remember, this motor was taken out of a cordless hand vacuum cleaner), the motor runs fine.

    I forgot to also mention, the reason I thought the AC Adapter might be messed up is because took another AC adapter from a printer that is 16V and like 500mA and I connected that directly to the DC motor as a test. This AC adapter is "smart" in that it has this little green light on it. It tells you when power is going through it (I guess). Anyways, I think it is also has some kind of fuse or circuit breaker in it because when I plugged the AC adapter into the wall, the green light is on. Then when I run wires from the adapter's plug straight to the DC motor, the motor jumps to action for like a split second - and then the green light on the AC adapter turns off. If I unplug and then plug in the AC adapter again, the green light turns on again.

    Does anyone know how to fix this?

    I have a feeling - though not sure that I may need to use a better diode? I am thinking maybe I should use a 1A diode (Model 1N4004) called 1N4004? http://www.radioshack.com/product/index.jsp?productId=2036270

    But, I also see that Radio Shack sells another diode that they sell a 3A diode as well (Model 276-1141). Would that be better? I guess more amps the merrier maybe? But they call it a "barrel diode" - not sure what "barrel" means here, so not sure it can be used. http://www.radioshack.com/product/index.jsp?productId=2062577

    Also, do I need to change the resistor? I see a lot of people mention a 1K resistor on message boards. Should I use that? I see Radio Shack sells a 1K Ohm 1/4 Watt resistor (Model 271-1321). Should I use that instead?

    Help!
     
    Last edited: May 6, 2011
  2. bertus

    Administrator

    Apr 5, 2008
    15,648
    2,347
    Hello,

    What is thee current rating of the used motor?
    The TIP 120 can handle an absultu maximum of 5 A.
    You could use a logic gate fet in stead like a IRL 540 of the TIP 120.

    I have attached both datasheets.

    Also use fast recovery diodes that are rated for about 2 X the motor current.

    Bertus
     
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  3. retched

    AAC Fanatic!

    Dec 5, 2009
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    Have you scoped the signal from the computer to the base pin?
     
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  4. R!f@@

    AAC Fanatic!

    Apr 2, 2009
    8,754
    760
    Sheeeesh!!

    U would use a SS diode as a spike suppressor.
    Use a high speed, high current diode instead of 4148
     
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  5. wayneh

    Expert

    Sep 9, 2010
    12,145
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    Is the solution really so hard to see? Your AC adapter is under-powered, especially when you include the inefficiency of using a transistor to control it (a MOSFET as suggested would help). At less than 90W, there are plenty of laptops that your adapter wouldn't be able to power either. Your hand-vac probably needs at least 100W or so, which you should be able to verify by measuring its current draw when running from its battery pack, or reading its label.
     
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  6. alfredeneuman

    Thread Starter New Member

    May 6, 2011
    22
    1
    Bertus,

    I think the current rating for the motor us 1.8A at no load and 9.16A at maximum efficiency. Is that what you were asking? I have attached the spec sheet for motor. It is model number LS-545S-60938 (the last one listed).

    Thanks for the advice to use a IRL 540. Radio Shack (my local store) has a MOSFET IRF510 Transistor (Model IRF510) http://www.radioshack.com/product/index.jsp?productId=2062618&numProdsPerPage=60.

    Is that good enough? Or is this TIP3055 NPN Transistor (Model TIP3055) better http://www.radioshack.com/product/index.jsp?productId=2062611&numProdsPerPage=60?

    Based on the above. Do you have an recommendations as to what diodes might be appropriate?
     
  7. alfredeneuman

    Thread Starter New Member

    May 6, 2011
    22
    1
    When you ask "if I have scoped the signal from the computer to the base pin?" if you mean have I tested if the computer signal is reaching the control board/rest of the circuit, then yes, I did check that and it is working correctly.

    I also connected another device to the control board later and was able to control it correctly from my computer.
     
  8. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    The difference between the IRL540 and the IRF510 is that
    the IRL540 needs about 5 Volts gate voltage to drive it full on
    and the IRF510 needs about 10 Volts gate voltage to drive it full on (wich the controller can not deliver).

    The 2N3055 needs a lot of current on the base to drive it full (wich the controller can not deliver).

    Bertus
     
  9. alfredeneuman

    Thread Starter New Member

    May 6, 2011
    22
    1
  10. alfredeneuman

    Thread Starter New Member

    May 6, 2011
    22
    1
    wayneh,

    But, even though the AC adapter is able to power my circuit sucessfully for like a minute (that is, the motor will run with the laptop ac adapter for like a minute before it fails), the AC adapter might still be underpowered?
     
  11. bertus

    Administrator

    Apr 5, 2008
    15,648
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    Hello,

    The motor will be to heavy for the used power supply.
    With NO load the current is 1.9 Amp.
    At ideal load the current is 9.16 Amp.
    At stall the current is even more than 46 Amp.

    [​IMG]

    Bertus
     
  12. alfredeneuman

    Thread Starter New Member

    May 6, 2011
    22
    1
    Bertus,

    Thanks. But, what I don't understand is that, using the laptop AC adapter, I was able to run the motor in the beginning for like a minute before the circuit failed. So, if my AC adapter is under powered, then if I just unplugged everything, and replugged everything back, shouldn't it work again (for at least a few seconds again)? Aside from not running correctly, will using an underpowered power source damage the transistor, diode, or resistor?
     
  13. alfredeneuman

    Thread Starter New Member

    May 6, 2011
    22
    1
    Bertus, wayneh,

    Just to be clear - I very much appreciate you letting me know that my AC adapter is underpowered. I just want to make sure if I can use it all or not because I don't have any more powerful power sources at my house and wanted to finish this project this weekend (too late to order a power source on the internet in time) and also, don't have much money to spend on this project. Thanks.
     
  14. wayneh

    Expert

    Sep 9, 2010
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    Using the MOSFET instead of the transistor will help, and maybe just put you over the ragged edge. And maybe not. But it appears to be your only cheap option unless you can find a bigger power source or lower power motor.

    I think you could use the MOSFET from the Shack if you use a small transistor to allow the controller's output to drive that transistor and deliver your power supply voltage to the gate of the MOSFET. Come on back if you need help with designing that.
     
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  15. tom66

    Senior Member

    May 9, 2009
    2,613
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    It's quite possible the AC adapter is shutting down because it is getting too hot. It is getting too hot because you are drawing too much current.
     
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  16. Wendy

    Moderator

    Mar 24, 2008
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    While you have a high gain transistor, I suspect that base resistance is too high. I assumed 5V since you didn't specify. This puts the base emitter current at 1ma. Any transistor at saturation has a lot less gain, for a single transistor the rule of thumb is a gain of 10, for a darlington you will not get the listed beta.

    Try dropping that base resistor to 470Ω. 10ma X100 (gain) is 1A, so you may need to drop it even further.
     
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  17. alfredeneuman

    Thread Starter New Member

    May 6, 2011
    22
    1
    wayneh,

    Thanks for the idea. I might go ahead and do that.
     
  18. alfredeneuman

    Thread Starter New Member

    May 6, 2011
    22
    1
    tom66,

    Thanks for letting me know that AC adapters can shut down. I was thinking that it could possibly shut down, but did not know for sure or not if they did.
     
  19. alfredeneuman

    Thread Starter New Member

    May 6, 2011
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  20. Wendy

    Moderator

    Mar 24, 2008
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    Not really. It works like this...

    If you have 5V being fed to the base, and the transistor is a darlington, then the resistor is dropping around 3.8V (Darlingtons drop 1.2V BE, conventional transistors drop 0.6V BE). Power = Voltage X Amps, or Voltage² / Resistance (in ohms).


    So, 3.8V X 3.8V / 470Ω = 0.031W

    A ¼W resistor is 0.25W, so the quarter watt is the best choice. Generally you want at least the double the wattage the resistor actually dissipates, a form of life extension and redundancy.
     
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