0
\$\begingroup\$

What are the temporal dynamics for the temperature change in a resistor driven by a current source? I am hoping that there is a simple equation that works well for most resistors so that I don't have to do something like solving the heat equation for a cylindrical conductor.

\$\endgroup\$
2
\$\begingroup\$

To begin, most resistors these days are rectangular surface mount resistors, not cylindrical.

Second, the resistive element is usually not a cylindrical conductor, but a resistive surface layer which is rather thin, both for SMT and through-hole cylindrical components.

Third, for a given dissipated power (P=R*i^2), the body temperature of a resistor will be mostly determined by heat fluxes to the soldered ends/pads, and each pad/mounting hole will have certain thermal impedance in each particular case of board layout (and associated thermal mass/path).

More, solving equations for cylindrical conductor would be not enough, you will need to "solve" for heat transfer by free convection into ambient air, which would depend on component orientation relative to the gravity field and air density. Plus you would need to add an effect of thermal radiation.

In short, temperature of resistive element is defined by a balance between a given Joule dissipation and the rate of heat loss, which is usually unknown. So there is no "simple equation" to determine its temperature.

[A paradoxical example: if a resistor (heating element) is placed in a fully insulated enclosure, its temperature will rise to infinity]

\$\endgroup\$
4
  • \$\begingroup\$ There are all kinds of ways we could complicate this question. I am betting there are useful equations describing the dynamics of T change in a small resistors regardless of the resistor geometry. Please assume that radiation losses are absent. \$\endgroup\$
    – D_J_S
    Apr 6 '18 at 22:16
  • \$\begingroup\$ @D_J_S, if you would be so kind to explain any particular and legitimate reason for your exercise, then maybe the answer could be different. \$\endgroup\$ Apr 6 '18 at 23:03
  • \$\begingroup\$ I am thinking that a combination of the simple Newton's Law of Cooling with a source term that in some way depends upon the resistance and current source is what is needed. I am assuming that this is probably a well known problem to some folks and am looking for a quick answer. \$\endgroup\$
    – D_J_S
    Apr 7 '18 at 0:02
  • \$\begingroup\$ The problem is that there is no quick answer that's generally applicable. Usually you simply don't need to know the temperature of a resistor, only whether the power dissipated in it is within its rating. \$\endgroup\$
    – Chromatix
    Apr 7 '18 at 4:41
0
\$\begingroup\$

These days one would likely just model the system and simulate it with finite element analysis or multiphysics software. Real resistors usually have trimming artifacts which probably limit pulse-handling capability.

Anyway, you can make a crude model using lumped circuit elements - resistors and capacitors, and evaluate using SPICE, easy if heat loss is only through conduction, less easy if convection.

\$\endgroup\$
0
\$\begingroup\$

Assuming the resistor maintains its ohmic value (which should be approximately true if you're operating it within its power rating), the power dissipated in watts is equal to the current through it multiplied by the voltage developed over it, hence P = I^2 * R.

To determine the resulting temperature, you'll need to know the thermal mass of the resistor (probably quite small, unless it's a high-power type) and the heat conductivity away from it (which will in turn depend on lead length, trace geometry at the ends of those leads, and whether the air around it is moving).

It's probably easier to determine these properties by empirical measurement than by calculation. It most certainly is not possible to come up with a general rule of thumb.

There are applications where the temperature of a resistor is actually important in some way. Most obvious is when the heat dissipated in the resistor is the application, such as in a cooker or kettle. In that case, you can mostly neglect the thermal mass of the resistor itself, and assume that the thermal mass of the load it's coupled to is the relevant property. You would then engineer the thermal coupling between resistor and load for efficient heat transfer.

\$\endgroup\$
2
  • \$\begingroup\$ Correction: The power dissipated is equal to the current squared multiplied by by the resistance. \$\endgroup\$
    – Barry
    Apr 7 '18 at 1:36
  • \$\begingroup\$ Yeah, that was a brainfart. Edited. \$\endgroup\$
    – Chromatix
    Apr 7 '18 at 4:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.