# Constant power discharge of a non-ideal capacitor

My employer sells boost converters to hold up motor drives during power loss. These boost converters are fed from capacitor banks. In order to size these banks correctly, we need to take their voltage, capacitance, and ESR into account, to ensure that there is enough available energy from the capacitors to hold up the drives for a spec'd time at a spec'd power. Right now we do this with an approximation method, but it would be nice to have a more exact equation.

We're assuming ESR, capacitance, and load power are constant.

$$I \text{: current}\\ P \text{: power}\\ R_{C} \text{: ESR}\\ C \text{: capacitance}\\ t \text{: time}\\ V \text{: capacitor voltage}\\ \text{Standard capacitor equation:}\\I(t)=CV'(t)\\ \text{Power out of the cap equals power into the ESR plus power into the load:}\\ V(t)I(t) = P + R_{C}I^{2}(t)\\ \text{Substitute:}\\ CV(t)V'(t) = P + R_{C}C^{2}(V'(t))^{2}\\$$

If I'm right, this gives me a non-linear differential equation, which puts me well past my mathematical comfort zone. If I understand correctly, solving a new non-linear differential equation would qualify as a significant contribution to the field of mathematical knowledge. Given that, I'm unlikely to solve this on my own.

Does anyone know any good approaches for solving for V(t)? Does anyone know if this equation has already been solved? Am I possibly misunderstanding the problem? Or should I move this to the math Stack Exchange?

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 How precise do you need to be? The amount of power lost to ESR will vary non-linearly with supply voltage, but one can easily calculate upper and lower bounds for the amount of energy that can be harvested from the cap as it falls from one voltage to some lower voltage; the smaller the drop in question, the closer the bounds will be. So if the cap starts at 50 volts, one could compute upper and lower bounds for how much energy would be recovered as it falls from 50 to 40 volts. If the difference between upper and lower bounds is too great, one could calculate the energy... – supercat Aug 16 '12 at 16:29 ...as it drops from 50 to 45 and then from 45 to 40. If the gap is still to great with those step sizes, subdivide further. If all parameters were known precisely, one probably wouldn't have to subdivide too much to get upper and lower bounds within 20% or so of each other. Given some imprecision in the parameters, it probably wouldn't be useful to go much beyond that. – supercat Aug 16 '12 at 16:32 Really, I suppose we have three questions. Is this solvable? If so, how? If not, what's the next best approach? We're looking for an exact solution to the equation, but if there's not one, what you describe might be a good backup plan. – Remiel Aug 16 '12 at 16:32 You could try modelling your circuit as an ideal capacitor, ESR resistor, and a load impedance all in series. By solving for the nodal voltages & current flow (which should all be linearly independent) you can find ESR losses vs. load power consumption. The only stickler would be estimating Z_L, though I think you should be able to figure it out by back-calculating from what power rating and acceptable voltage drop you expect your design to have. – helloworld922 Aug 16 '12 at 17:21 @Remiel: It's common to model real-world capacitors as a combination of ideal capacitors, resistors, and inductors, and such a model will be closer to reality than one which simply expected a real cap to behave like an ideal one, but the "ideal non-ideal cap" is still just an approximation. In the real world, ESR and capacitance can both vary with voltage in weird non-linear ways. An equation that precisely describes the behavior of a model may be no more accurate than a discrete-time simulation in predicting the actual behavior of a real circuit. – supercat Aug 16 '12 at 17:40
The equations were solved by others here. Unless I've missed a sign somewhere, this formula gives the time it takes for a cap to reach internal voltage V, starting from voltage $V_{0}$, with a given ESR and capacitance, and a fixed power discharge.
$$t(V) = \frac{C}{4P} (V_{0}^2-V^2 + V_{0}\sqrt{V_{0}^2-4PR_C} - V\sqrt{V^2-4PR_C}) + CR_C(\ln\big(V+\sqrt{V^2-4PR_C}\big) - \ln\big(V_{0}+\sqrt{V_{0}^2-4PR_C}\big))$$
Note that since V is the internal, unloaded voltage of the cap, "behind" the ESR, to find the time it takes for the cap to reach a specific terminal voltage while loaded, we must use the substitution: $$V=V_{min}+\frac{PR_{C}}{V_{min}}$$ where $V_{min}$ is the minimum desirable terminal voltage.