60W potentiometer extremly heating

Discussion in 'General Electronics Chat' started by coco243, Feb 23, 2013.

  1. coco243

    Thread Starter Active Member

    Sep 2, 2010
    62
    0
    Hello,

    I have a 47 ohm 60W potentiometer that is heating extremly and I don't know why. I power this circuit with an 24VDC and 3.4 ampers transformer.

    I connect it in series with an electric break for a motor.

    It is heating when I power the circuit, it is no necesary to start the motor.

    If it has 60W and 47 ohm, and the voltage is raised at 27VDC so through the potentometer shoud pas aprox 0.6 ohm.

    But the potentiometer has 60W and at 27VDC can suport 60W/27V=2.2Ampers,
    so, why the potentiometer is heating?

    Please help me to figure it out.

    I atached a schematic.

    Thank you verry much, and please help me.
     
  2. #12

    Expert

    Nov 30, 2010
    16,266
    6,777
    The only thing I can think is that you are turning the adjustment to a low ohm setting and a lot of current is going through a small part of the 60 watt potentiometer.

    The 60 watt rating is only for the whole potentiometer, not a small part of it.
     
  3. coco243

    Thread Starter Active Member

    Sep 2, 2010
    62
    0
    The same thing is happening when the cursor is at maximum resistence.
    My big wondering is even if the electrobrake is in short circuit, and the potentiometer has all voltage 27 VDC on it, at his 60W trough it will pas i=60/27=2.2Ampers so the potentiometer shouldn't overheat.
     
  4. tubeguy

    Well-Known Member

    Nov 3, 2012
    1,157
    197
    What is the resistance of the motor brake ?
     
  5. #12

    Expert

    Nov 30, 2010
    16,266
    6,777
    What is the REAL resistance of the potentiometer!

    When you follow all the rules and still have smoke, something very basic is wrong.
     
    PackratKing likes this.
  6. coco243

    Thread Starter Active Member

    Sep 2, 2010
    62
    0
    And the second thing, if the power of the potentiometer is 60W at 47ohm, and P=r*i*i it results that i*i=P/r, let's say we put the potentiometer at 2 ohms resistence then i*i=60/2=30 and I=square root from 60=4.47 ohm on the other hand if we set the potentiometer at 47ohms will have a current i=1.27, smaller.
     
  7. coco243

    Thread Starter Active Member

    Sep 2, 2010
    62
    0
    I don't know, it's an old motor and I don't see it's characteristics. But I think it is bigger than 47 ohms because I have to
    go with the potentiometer close to the 0 ohms in that way the motor brake to do the job, the most of the voltage it's on the motor brake.
     
    Last edited: Feb 23, 2013
  8. coco243

    Thread Starter Active Member

    Sep 2, 2010
    62
    0
    47ohms, measured

    I am affraid that I don't know how to measure the motor brake to find if it is in short circuit. And if it would be I don't think that the potentiometer should overheat because it has 60W at 27VDC, it can support direct connection from source voltage to the grounding.
     
  9. tindel

    Active Member

    Sep 16, 2012
    568
    193
    The image is hard to see, but I'll give it a shot anyway.

    The power the pot is dissipating is P = I^2 * R.pot. Based on you drawing, the amount of current will be I = V / (R.motor + R.pot)

    Example 1
    Let's assume your motor has 1 ohm of equivalent DC resistance. Just a guess.
    Also assume your pot is adjusted to be 47ohms

    I=27/(1+47) = .56A
    and
    P=.47^2*47=14.8W
    In this scenario, you should be just fine, but that's a bit of power and the pot will probably still get warm.

    Example 2
    But, the trim pot can be anywhere from 0 to 47 ohms. So, let's assume you've trimmed your pot to be 1 ohm

    I=27/(1+1)=13.5A
    and
    P=13.5^2*1=182W!
    This is too much power for the pot, obviously, and will cause damage to the pot, and may potentially start a fire.

    Anyway, as your resistance goes down, your current and power go up. In many applications like this you will see a 10ohm resistor in series with the pot to ensure the pot doesn't dissipate too much power.
     
  10. tindel

    Active Member

    Sep 16, 2012
    568
    193
    Put an ohm meter across the leads of the motorbrake with it out of circuit, that will tell you what the resistance of the motor break is.
     
  11. coco243

    Thread Starter Active Member

    Sep 2, 2010
    62
    0
    I did that, it was aprox 3,1 3,2 ohms. How I can figure if it in short circuit? i supposed that is not a short circuit.
     
  12. #12

    Expert

    Nov 30, 2010
    16,266
    6,777
    This statement is correct. Therefore, I think you must have a mistake in the wiring.
     
  13. coco243

    Thread Starter Active Member

    Sep 2, 2010
    62
    0
    From what I remember, I have a source thar outputs 27VDC at 3.4 Ampers,
    so I don't kow if the current can increase at 13,5 Ampers but indeed I can rise at 3.3 Ampers that is more that i=squareroot(60W/47ohm)=1.13A.

    Begining from your 10ohm resistence I wil have
    I=27/(1+1+10) so I=2.25 Ampers
    and
    I*1+I*1+I*10=27
    and the voltage on the motor brake will be 2.25*1=2.25V and the voltage on the series resistence will be 22.5 but the motor brake I think it's works at 24VDC, and the voltage will be not enough.
     
  14. tindel

    Active Member

    Sep 16, 2012
    568
    193
    if the resistance was 3.1 and 3.2 ohm, then that is a 'short-circuit'. But when dealing with power circuits then 3.1 ohm might be your load.

    I uploaded a Excel file (in zip format) that shows how the rated pot power is diminished over pot resistance (assuming a linear distribution), and the power in the pot over pot resistance given a 3.1ohm load. It should come as no surprise that the maximum pot power is at 3.1 ohms, as it is matched evenly with the load. The pot power is less than 60W all of the time, but the rated pot power at 3.1 ohms is much lower than 60W. Basically, you shouldn't let the pot go below 20ohms if you want to keep your pot 'safe'.

    Another thing that may be causing the heating - some resistors require adequate heat-sinking when dissipating large amounts of power to keep the power rating. Check your datasheet and ensure you have adequate heat sinking. Can you post a link to the datasheet?

    Another thing I thought about - are you really powering using a DC transformer? I've never heard of a DC transformer.
     
  15. crutschow

    Expert

    Mar 14, 2008
    13,000
    3,229
    What do you mean "heating extremely"? Is the pot starting to emit smoke?
     
  16. gerty

    AAC Fanatic!

    Aug 30, 2007
    1,153
    304
    What is the load on this motor?
    Do you get the same results with the load disconnected?
    Is this something that has been running ok , but started having problems ?
     
  17. coco243

    Thread Starter Active Member

    Sep 2, 2010
    62
    0
    So, it is or it si not not a short-circuit on the motor brake?

    Interesting but where is from that formula P disipated accordung the pot cursor =R according pot cursor / ( 60W * 47 ohms ) ?
    And thanx for the excel is verry pro :)

    Here are the descripsions of the potentiometer:
    http://www.tme.eu/en/details/pot60w-47r/wire-single-turn-axial-potentiometers/tt-electronics/

    [/QUOTE]

    No, I have an 230/24 VAC transformer and after a bridge rectifier, and at the terminals of the motor brake a condenser. I measured the condenser with the buzzer and it didn't had continuity.

    However, this was the system configuration until I come to see where is the problem. The old potentiometer with the same characteristics had the wire that was supply the motor brake disconected, I conected back, I assumed that the problem was solve and I was leaving, but they tell me to come back the day after, I have seen the potentiometer broken and I buyed another to change it, and surprise, the new one is overheating!! Why?
     
  18. coco243

    Thread Starter Active Member

    Sep 2, 2010
    62
    0
    No, my luck was that I sticked the terminals with tin, and when I powered the
    circuit the wire was unstiking, anyway then I fixed well the wire on the potentiometer's teminal and start to make tests, when I put my finger on the potentiometer I was hurt, it was very warm. I don't let it until to emit smoke I was scared that it can be damaged.

    I am afraid to let the potentiometer at full resitence ( ohms ) because i think it may be damaged.

    I have a series of contacts before the pot. supply, I don't think to be problems
    in circuit before potentiometer because, he voltage on the power supply is the same with the voltage supply on the potentiometer.

    Another question is, if my transormer has 3.3 ampers, at were I have a current that heats in same way the potentiometer?
     
  19. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    The maximum current through the potentiometer is √(60W/47Ω)=1.13A. If you exceed that current with any viper setting, you are asking for trouble.
     
  20. tindel

    Active Member

    Sep 16, 2012
    568
    193
    It is the load. The term 'short-circuit' is relative. In this case, I would say that it is not short-circuit, since you are trying to do something with this resistance.

    I just assumed that the rated power dissipation is linearly derated with resistance. For example, at 47 ohms you can dissipate 60W, and at 23.5 ohms you can dissipate 30W, etc. Maybe someone can correct me on this if I'm wrong.

    This is not a datasheet. A datasheet will give part dimensions and specifications about the part. The link you provided just give basics. I did try to go to the ttelectronics website and find the datasheet but could not find it

    When you say condenser, do you mean capacitor? What is the size of the capacitor? A capacitor should not have continuity through it.


    Things only overheat when they dissipate too much power. The lower the resistance, the higher the power (assuming the supplied voltage remains the same).

    Some suggestions:

    Double check all of your wiring. Then check it again one more time.

    Have someone else check your wiring, often times someone else can see a wiring error better than you can.

    Provide adequate heat removal (heat sinking) from the part. 60W is a lot of power to dissipate out of a device this size. It could be that all the heat that is being generated is trying to get out of the part through the terminals and melting the solder joints. You have to have proper heat sinking.

    Providing some pictures of your application may help as well.

    Provide a full schematic that is easy to read.
     
    Last edited: Feb 24, 2013
Loading...