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Recently I have undergone a project involving a Thermo-Electric Cooler (TEC). I successfully recreated the manufacturer's datasheet curves by simulating the TEC in LTSpice using thermal circuit modelling techniques. In doing so, I discovered a question that, while irrelevant, distracted me for hours.

Consider a big block of aluminum of some mass such that its thermal capacitance is 1000 (J/K). Imagine it is connected to a smaller block with thermal capacitance of 500 (J/K) through a thermal strap with thermal resistance of 0.1 (K/W). The thermal circuit equivalent is below (yes, GND should be zero Kelvin, but shush the end result is the same shifted by 273.15C in this case): LTSpice thermal model

Now, here's where things get knocked into twelfth gear. I set the big block's initial temperature to 20C by doing a ctrl+rmb over the capacitor and typing "IC=20" in the "SpiceLine" field. In a similar fashion I set the small blocks temperature to 80C.

My question is: how does one find an equation (closed-form or otherwise) in the time domain AND the Laplace domain for currents (heat flows in this case) and voltages (temperatures in this case) in circuits in which Ls and Cs have initial conditions? That is, what are the lumped Laplace models for Ls and Cs with initial currents and voltages, respectively? I also humbly request some intuition as to why these equations make sense!!

LTSpice has no problem telling me the system settles to around 40C, as seen in the figure above. To help me answer this conceptual question of modeling initial conditions, I looked for practice problems solving Thévenin Equivalent Circuits. I can find countless examples involving Ls and Cs in textbooks and on the internet. However, in EVERY SINGLE ONE they assume zero initial conditions!

I will attempt to answer this question below using Laplace expressions, however if someone could point me in the direction of a time-domain expression for two capacitors sharing charge, it would be much appreciated! I've been racking my brain at this for days.

Impedance of Capacitor C in Laplace domain--> 1/(sC)

Laplace model of Capacitor C with initial charge --> 1/(sC) + Vi/s (?)

Let's investigate using LTSpice by removing the initial voltage on the caps and instead placing a series step with each: enter image description here suspiciously similar results...

How about Inductors with some initial currents through them? For this we must leave the thermal stuff behind, because as far as I'm aware there are no thermal energy storage phenomena in life that lend themselves to inductors:

Impedance of Inductor L in Laplace domain--> sL

Laplace model of Inductor L with initial current --> sL || Ii/s (?) enter image description here

banjoeschmoe
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  • It could be analogous if R2 shunts both inductors. – Tony Stewart EE75 Dec 21 '20 at 01:56
  • That wouldn’t answer my question about the true way to represent these energy storage components though. – banjoeschmoe Dec 21 '20 at 02:41
  • Energy storage depends greatly on more than one property than heat mass, but also heat velocity and heat loss or emissivity and fluid flow of air, or liquid. So a capacitor is too reductionist in thinking it will be an accurate analog for capacity unless perfectly insulated. https://heatsinkcalculator.com/ – Tony Stewart EE75 Dec 21 '20 at 03:07
  • I appreciate your helpful comments on this - my understanding of thermal representations could use it! To clarify for anyone else reading this comment exchange, the true intent of this post is to understand energy storage components with initial conditions in a purely electrical domain. – banjoeschmoe Dec 21 '20 at 14:44
  • Zero order approximations use R's only for steady state and for dynamic changes, the dielectric with the lowest capacity C*Rth = T has the fastest overshoot of temperature before equilibrium is reached. This often the junction interface to case. Rjc even tho Rjc is fairly low, you are correct in thinking C being so small causes an overshoot temp until flow equalizes all time constants to a steady state after say 10T for the longest T of any layer. – Tony Stewart EE75 Dec 21 '20 at 17:33
  • For the inductor case, doesnt your initial condition itself violate charge continuity? – MAM Dec 24 '20 at 23:00
  • Thank You Adil, I've edited my post. – banjoeschmoe Dec 25 '20 at 02:29
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    @banjoeschmoe I did something not exactly the same, but similar to what I think you are asking about, here. – jonk Dec 25 '20 at 05:10

1 Answers1

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Okay so lets make it a more general setting first. The question you want to ask yourself is, how does one represent the laplace transform of the nth degree derivative operator, including the initial condition.

The first derivative has the following result: $$\mathcal{L}\left[\frac{dx}{dt}\right] = sX(s) - x(0^-)$$

Higher order derivatives can be attained by the repeated application of the above result.

The general result is: $$\mathcal{L}\left[\frac{d^nx}{dt^n}\right] = s^nX(s) - \sum_{k=1}^{n} s^{n-k} x^{(k-1)} (0^-)$$

Note in this notation \$ x^{(r)} (0^-) \$ is the \$ r^{th} \$ derivative evaluated at \$ t = 0^- \$. This is the initial condition.

Now for capacitors: $$ i = C \frac{dv} {dt} $$ In the laplace domain: $$ I(s) = C (sV(s) - v(0^-))$$

For inductors, its just the dual: $$ v = L \frac{di} {dt} $$ In the laplace domain: $$ V(s) = L (sI(s) - i(0^-))$$

You can now solve the above systems using these results.

For example, I have done the capacitor system for you:

$$ I_1 = C_1 (sV_1 - v_1(0^-)) $$ $$ I_2 = C_2(sV_2 - v_2(0^-)) $$

But $$ I_1 = - I_2 = \frac{V_2 - V_1}{R}$$

Now lets solve for \$ V_1 \$ by elimating the rest, we get:

$$ V_1 = \frac{(sRC_1C_2+C_1)v_1(0^-) + C_2v_2(0^-)} {s^2RC_1C_2+s(C_1+C_2)} $$

To get the final values for the voltage, we use the final value theorem (\$ \lim_{s\to0}sV_1(s) \$) and get:

$$ v(t\to\infty) = \frac{C_1v_1(0^-) + C_2v_2(0^-)}{C_1 +C_2} $$ This makes sense because it is essentially saying the new voltage across both capacitors will be the total intial charge redistributed to the total capacitance: $$ v(t\to\infty) = \frac{Q_1(0^-) + Q_2(0^-)}{C_1 +C_2} = \frac{Q_{Total}(0^-)} {C_{Total}} $$ Plug in your values and you get 40V as shown by LTspice: $$ \frac{1000\cdot 20 + 500\cdot 80} {1500} = 40V $$

EDIT 1: So let me make it more clear as to why including the step function works, it follows from this expression:

$$ I(s) = C (sV(s) - v(0^-))$$

Invert the equation in terms of \$ V(s) \$: $$V(s) = \frac{I(s)}{sC} + \frac{v(0^-)}{s}$$

As a circuit, the above equation describes a series connection of a capacitor and a step function!

EDIT 2 : Also note how this implements the initial condition as a series circuit. You can also implement it instead as a capacitor and a shunt connected constant current source, if you stick to the original equation in terms of current.

MAM
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  • I appreciate you breaking down the math. However, this still doesn't answer my original question. Perhaps I wasn't clear - I basically want to isolate an L or a C which has an initial condition, and model it as a pure uncharged L or C with the addition of a separate source. The idea being I don't want to have to solve a large differential equation when taking a Thevenin equivalent. At this point, I'm convinced the step voltage/currents I've included in my OP are correct (based on simulations), but I still need to come to terms with why they are correct. – banjoeschmoe Dec 28 '20 at 15:16
  • It does answer the question you posed exactly in the bounty and it also answers why the step function works. Look at the Laplace representation of the capacitor in my answer and write it out in terms of a voltage. Infact, let me do an edit and spell it out. Hope that makes it clearer. – MAM Dec 28 '20 at 15:42
  • Thank you for your patience with me, this makes sense now! – banjoeschmoe Dec 31 '20 at 01:54