resistor heat dissipation

Discussion in 'General Electronics Chat' started by roadey_carl, Feb 1, 2012.

  1. roadey_carl

    Thread Starter Active Member

    Jun 5, 2009
    116
    5
    Hello everyone,
    I just have a couple of questions about resistor heat dissipation. For example, if you put a 50k resistor parallel with 230vac you would get 1.060 watts. In good practise what watt resistor would you select? 3-5watt? Also is there any way to work out how hot that resistor will actually get?

    thanks for any reply's!
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,449
    3,365
    The higher the wattage the better.

    No, it is not easy to determine the final equilibrium temperature. It depends on too many parameters: length and size of resistor and leads, pcb style, mounting, case, air flow, ambient temperature.
     
    roadey_carl likes this.
  3. mcgyvr

    AAC Fanatic!

    Oct 15, 2009
    4,771
    971
    in general your resistor wattage should be sized at least 2x the calculated value.
     
    roadey_carl likes this.
  4. jimkeith

    Active Member

    Oct 26, 2011
    539
    99
    A good rule of thumb is to run the resistor at 50% of its power rating--otherwise they run 'stinking' hot! It does not hurt to add more safety factor either if your space /cost constraints permit.

    There is a simple /crude rule of thumb that approximates temperature rise as a function of power and surface area--this works for convection cooled surfaces such as heatsinks, but does not account for radiation /conduction losses as in hotter devices such as power resistors.
    T = 400 * P / A Where t is in °C, P is in Watts and A is in cm²
    So if you dissipate 1W on a 10cm² surface, the temp rise would be roughly 40°C.
    Anyone out there, correct me if I am wrong

    Years ago an engineering mentor told me to use the relationship °C * cm² /mW
    Subsequent checking indicated that it was roughly 0.4°C * cm² /mW
     
    Last edited: Feb 1, 2012
    roadey_carl likes this.
  5. crutschow

    Expert

    Mar 14, 2008
    13,052
    3,244
    This is a nit, but if you say ΔT = 400 * P / A, it makes it clearer that you are solving for a temperature rise (delta) rather than actual temperature. :)
     
    roadey_carl likes this.
  6. roadey_carl

    Thread Starter Active Member

    Jun 5, 2009
    116
    5
    Hmm I think I see......
    how about an example? if the resistor is 17mm Long and 6mm Wide. 1.7*0.6= 1.02cm2

    so T= 400 * P / A 400*1/1.02 = 392°C

    Is that right? No? what am I doing wrong?

    thanks for your patience. it great stuff by the way, really useful to me!
     
  7. kubeek

    AAC Fanatic!

    Sep 20, 2005
    4,670
    804
    Resistor is not a plate, it is a cylinder, therefore the area should be 1.7*0.6*3.14=3.2cm2
    so 400/3.2 is a more believable 125°C rise
     
  8. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    Good practice is to always use a resistor rated for at least TWICE the power dissipation in operation. Don't worry about device temp.
     
  9. roadey_carl

    Thread Starter Active Member

    Jun 5, 2009
    116
    5
    I don't really worry, its just good to know! You never know when you need to know! Thanks everybody this is great stuff!
     
  10. bountyhunter

    Well-Known Member

    Sep 7, 2009
    2,498
    507
    The operating temp is different for every type: wirewound resistors run much hotter than carbon comp or metal oxide type resistors.
     
Loading...