# resistor heat dissipation

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

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?

2. ### MrChips Moderator

Oct 2, 2009
12,623
3,451
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.

3. ### mcgyvr AAC Fanatic!

Oct 15, 2009
4,784
973
in general your resistor wattage should be sized at least 2x the calculated value.

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
5. ### crutschow Expert

Mar 14, 2008
13,469
3,356
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.

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
4,686
805
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.