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I have made a simple model of heat transfer between ambient and a silicon chip (module) from which I can read its internal temperature $T_m$. I do not need fancy equations and an approximate model will do just fine.

I have modelled the system using electrical equivalent model (RC), but determining parameters requires equilibrium point data.

Using the heat transfer slope I could potentially determine the model parameters, only that the data is noisy and even using Rick Chartrand, "Numerical differentiation of noisy, nonsmooth data" algorithm I still couldn't get a good estimate for the actual derivatives.

My solution now is to fit data and extract parameters from empirical data.

After I heat the module I let it cool down, but to measure the internal temp I need some part of the module to run, which should account for a residual power $P$.

The model is $\frac{\delta T_m}{\delta t} = c (T_a -T_m(t)) + P$

I can easily assume that the room will not change temperature because of a very small chip running, such that $T_a$ is constant.

I can determine estimates of $c$ from halftime and $P$ from solving $\frac{\delta T_m}{\delta t} = 0$.

A linear sweep of values around this estimates will give probably sufficiently good estimates.

My problem comes from the fact that if I solve the diferential equation I end up with this solution

$T_m(t) = T_m(0) e^{-ct} + (cT_a+P)t$

which makes no physical sense since it simply states temperature will rise infinetly.

What is the right solution for this diferential equation of heat transfer?

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    $\begingroup$ That's not the correct solution. It would be $T_m(t)=(T_m(0)-P-cT_a)e^{-ct}+(cT_a+P)$. $\endgroup$
    – lemon
    Feb 18, 2016 at 10:54
  • $\begingroup$ Is the constant term actually $T_m(\infty)$ the solution of $\frac{\delta T_m}{\delta t} = 0$ instead of $cT_a + P$ in your solution? $\endgroup$ Feb 18, 2016 at 11:08
  • $\begingroup$ Since you are right that $T_m(t) = (T_m(0)-T_m(\infty))e^{-ct}+T_m(\infty)$ in the absence of P $\endgroup$ Feb 18, 2016 at 11:11
  • $\begingroup$ Yes, they're the same: $T_m(\infty)=cT_a+P$. $\endgroup$
    – lemon
    Feb 18, 2016 at 11:14
  • $\begingroup$ Thank you for your help. Would you like to post this as an answer? I would want to accept it $\endgroup$ Feb 18, 2016 at 11:16

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The correct solution to your differential equation is: $$ T_m(t)=(T_m(0)-\frac{P}{c}-T_a)e^{-ct} + (T_a+\frac{P}{c}) $$ or more succinctly, $$ T_m(t)=(T_m(0)-T_m(\infty))e^{-ct}+T_m(\infty) $$ where $T_m(\infty)=T_a+\frac{P}{c}$.

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