resistor heat dissipation

Thread Starter

roadey_carl

Joined Jun 5, 2009
125
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!
 

MrChips

Joined Oct 2, 2009
19,383
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.
 

jimkeith

Joined Oct 26, 2011
540
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:

crutschow

Joined Mar 14, 2008
23,500
.................
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.
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. :)
 

Thread Starter

roadey_carl

Joined Jun 5, 2009
125
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!
 

kubeek

Joined Sep 20, 2005
5,622
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
 

bountyhunter

Joined Sep 7, 2009
2,512
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!
Good practice is to always use a resistor rated for at least TWICE the power dissipation in operation. Don't worry about device temp.
 

Thread Starter

roadey_carl

Joined Jun 5, 2009
125
I don't really worry, its just good to know! You never know when you need to know! Thanks everybody this is great stuff!
 
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