Resistor Power dissipation help

Discussion in 'General Electronics Chat' started by systemcrashdump, Jun 1, 2011.

  1. systemcrashdump

    Thread Starter New Member

    Jun 1, 2011
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    Hello,

    I have battery pack for which I have to perform fuse verification testing by discharging process, I have to connect 100mΩ resistor across battery pack and relay which starts discharge process.

    Max voltage of battery pack can be 48V and can be discharged until voltage reaches down to 12V.

    For this simple circuit, what wattage resistor should I select:confused:. As discharging happens voltage reduces and hence current too,

    THanks in advance.
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Take a look here to calculate the power of the resistor:
    Calculating electric power

    It will be very bad to discharge a battery from 48 Volts to a minor 12 Volts.
    The battery will be unusable afther the experiment.

    Bertus
     
  3. tom66

    Senior Member

    May 9, 2009
    2,613
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    Power dissipated in a resistor is P = V^2 / R.

    Current through a resistor is given by I = V/R.

    So a 10mOhm resistor will draw 4,800A (a lot!) and dissipate 230.4kW!!!! In other words, not only will your battery be ruined, but your resistor will be completely incinerated (and if not, it will make for a nice room heater.)

    A 100mOhm resistor will draw 480A and dissipate 23.04kW, which is still a massive amount.
     
    Last edited: Jun 1, 2011
  4. systemcrashdump

    Thread Starter New Member

    Jun 1, 2011
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    product is rechargeable battery system, it can be recharged, the resistor what I'm supposed to use is 100 mili OHM, so when the battery is at 48V the current would be 480 Amps and power dissipated is 23040W.

    When voltage is 12V current would be 120A so power rating will be 1/4 of max power dissipated.

    Since I have variable power dissipation I'm confused to select resistor.

    I have 300W resistor with me right now, can I do the testing with that resistor ?
     
  5. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    @Kingsparks
    There is a BIG difference between 100 mOhm wich is 0.1 Ohm and 100 M Ohm wich is 100000000 Ohms.
    The power in the 0.1 Ohm will be (48X48)/0.1 = 23040 Watts
    The power in the 100000000 will be 23 μ Watts.

    The battery or the resistor will probably explode with the 0.1 Ohm resistor is connected.

    Bertus
     
  6. systemcrashdump

    Thread Starter New Member

    Jun 1, 2011
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    Basically you can say I'm trying to do short circuit testing.
     
  7. #12

    Expert

    Nov 30, 2010
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    crashdump: I suggest you step out of the room and close the door before turning the test on. That way the battery can paint the walls without hurting you.
     
  8. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    If you put 23040 watts on a 300 watt resistor it will explode.
    If you want the resistor to stay OK the power should be at least 2 X the power to be dissipated And then the battery will explode.

    Bertus
     
  9. beenthere

    Retired Moderator

    Apr 20, 2004
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    Ah, the fuse part helps. Obviously, you need to draw enough current to verify the fusing. One way is to drop a crowbar across the terminals and see if the sparks quit. Another is to devise a load that should just not pop the fuse, and then switch a parallel resistance in to actually exceed the blow point.
     
  10. systemcrashdump

    Thread Starter New Member

    Jun 1, 2011
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    Yeah I'm worried of that so thought of taking guidance from experts here, Basically I m trying to test fuse functionality when there is a short circuit, do you have any other method. At discharging section I have FET which has max handling capacity of 180A, and I have 3 of those, so it would make total of 540A.
     
  11. beenthere

    Retired Moderator

    Apr 20, 2004
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    What is the fusing level? That makes an enormous difference.
     
  12. Adjuster

    Well-Known Member

    Dec 26, 2010
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    300
    I note that the purpose of this system is said to be fuse verification. Does this mean that you are trying to blow a fuse, with a battery pack in various different states of discharge?

    If this is so, a short-term rated resistor may be acceptable, but in this case you may need to include a safety system to terminate the discharge in the case that the fuse does not fail. Can you provide more information?
     
  13. systemcrashdump

    Thread Starter New Member

    Jun 1, 2011
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    product is intelligent battery management system, each cell is connected with fuse links which will break to protect cells when very high current is detected. I have 3 Mosfets at discharging section and each can handle 180A so totally or ideally system can go upto 540 A,

    What I'm trying to test is if there is any short because of any reason my product should not fail, so this can be best tested by short circuting the terminals with a small resistor. The software running in microcontroller will detect levels and opens up FET circuit stopping the discharge, so I want to see this action when I do the test.
     
  14. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Your MOSFETs will have an Rds(on) that you will need to include in the load resistor equation.

    You will likely need a BANK of MOSFETs that are water cooled in order to keep them from melting.
    You will also need very large and heavy copper buses to carry the current, and very heavy-duty copper cables to connect everything to the load.
     
  15. Kingsparks

    Member

    May 17, 2011
    118
    5
    Hey Bertus.

    Thanks for the heads up. My mistake, I caught it and removed the post. Should have read that post more carefully. Only excuse is I am not accustom to dealing with millohms, but that's not really an excuse.

    Yep, I would not want to be around when that is tried. I can't imagine that size resistor. I have some 150W, .128 ohm resistors that are ten inches long. How big would a .1 ohm have to be to carry that load? Rhetorical question.

    Thanks again.
     
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