+ Post New Thread
Results 1 to 4 of 4

6th September 2019, 17:04 #1
 Join Date
 Jul 2006
 Location
 USA
 Posts
 403
 Helped
 7 / 7
 Points
 4,309
 Level
 15
power consumption and temperature rise
I use a 1V voltage to apply to a 200ohm resistor, so the power consumption on the resistor is V^2/R=5mW. I measured the temperature close to the resistor hoping that the ambient temperature can rise.
However, it does not increase the temperature. How can I detect the temperature variation if the resistor consume 5mW energy? Thank you.

Advertisement

6th September 2019, 17:27 #2
 Join Date
 Jan 2019
 Posts
 372
 Helped
 92 / 92
 Points
 1,992
 Level
 10
Re: power consumption and temperature rise
the specific heat of air at 300 K is 0.718 joules/gram Kelvin
your resistor is dissipating 0.005 W, or 0.005 J/second
it will take 144 seconds, (2 minutes and 24 seconds) to dissipate 0.718 joules
assuming the air around the resistor does not change, so the same air is heated by the resistor for the full
2.5 minutes.
you may be able to do this, if you put the resistor and your temperature probe in a small, closed. well insulated box
the tiny amount of energy your resistor dissipates shows why most of the time most resistors do not get hot

Advertisement

7th September 2019, 03:50 #3
 Join Date
 Aug 2015
 Posts
 1,788
 Helped
 666 / 666
 Points
 11,045
 Level
 25
Re: power consumption and temperature rise
you need to pack cotton wool or sinilar around your resistor at these very low dissipation levels to trap the heat and hence get a temp rise that you can measure ...

Advertisement

7th September 2019, 04:16 #4
 Join Date
 Nov 2012
 Posts
 3,147
 Helped
 777 / 777
 Points
 17,223
 Level
 31
Re: power consumption and temperature rise
However, it does not increase the temperature. How can I detect the temperature variation if the resistor consume 5mW energy? Thank you.
If the resistor is well insulated, the heat produced will have nowhere to go and the temp will rise linearly with time. But no insulation is perfect and heat is lost by conduction, convection and radiation.
When the heat lost is exactly balanced by the heat produced, the temp will come to a steady value.
Your resistor is a (assume) small body with some heat capacity. So it will take some heat initially to get hot (just like a soldering iron).
I guess the steady state will come (including all forms of heat conduction) with an increase in temp of the resistor by about 1C (if the resistor has about the same size of a 1/4W typical resistor).
Much of the heat will be conducted away by the leads; rest will be conducted away by convection (surround air) and a rather small amount by radiation.
It will be tough to measure. What kind of thermometer you are using to measure the temp rise? You will need a thermometer with a very low heat capacity (say a thermistor or a Pt100 temp sensor)
Rest assured that it does increase the temp of the resistor. The heat produced has to go somewhere.
+ Post New Thread
Please login