# Can I get the internal resistance of a capacitor from the voltage, capacity, and construction type?

I have a model of a system and I'm trying to use gradient decent to find a locally optimal solution. One of the variables is the voltage of the capacitor. I think I'll be using a film capacitor, because it is non polar, but I might choose an aluminum electrolytic capacitor, depending on other design decisions.

I am choosing to hold the mass constant, and since mass is proportional to energy I choose to think of it as constant energy. (capacitors have a range of energy densities, I'll choose the mean value and use that to convert between mass and energy). So to derive the capacitance I can use the equation below to figure out what capacitance to input to my system model. $$E=\frac{1}{2}CV^2$$.

But my system model also needs to know the internal resistance of the capacitor. Before, I was just looking at specs and plugging in values of specific capacitors, but I'm trying to stop guessing and checking, and start using gradient decent to find an optimal capacitor. When my script finds a solution, I can take the voltage, capacity, and ESR values and go find a capacitor that is close enough. But that means my script must be choosing values for the ESR based on the voltage and capacity that are realistic, that a capacitor really might have.

Is there a function that, given a voltage and capacity, returns the internal resistance or ESR?

I'm guessing there is, because voltage and ESR are proportional, and capacity and ESR are inversely proportional.

Worse comes to worse, I can download a million capacitor spreadsheets and do a regression, but I'd prefer not to. If I knew the average power density of a film capacitor, I could use that, I think. (I understand the power density isn't exactly well defined, because the power decreases as the capacitor discharges). But I couldn't find any numbers on film capacitors anyway.

• You can get a very rough approximation, but other than that no. Commented Jul 4 at 0:54
• I only need a rough approximation. Commented Jul 4 at 1:48
• Why do you say that ESR is inversely proportional to voltage and Capacity and ESR are proportional. In my experience the opposite is true. Commented Jul 4 at 2:08
• @KevinWhite, oops you're right. In my mind I was just thinking Capacity and ESR "get better" together, so they're "proportional". Of course, ESR "gets better" when it goes down, not up. The mistake is corrected. Commented Jul 4 at 3:40
• What makes you think that "mass is proportional to energy"? What mass and what energy are you talking about here? Commented Jul 4 at 7:11

EDIT: most of this answer is wrong when dielectric barrier is small. But I'm leaving it here for history's sake. Skip to the edit below for a better answer

Everything I know about capacitors leads me to conclude this answer is correct. However, a few of the comments on my question contradict some of this solution. If this is wrong, I'd love to know why. But here's my reasoning

Suppose you have a few capacitors in parallel. The following equation describes the total capacitance.

$$C_{total}=\sum C_n=N*C_n$$

Wired in series, the capacitance is

$$C_{total}=\frac 1{\sum\frac 1{C_n}}=\frac 1{\frac N{C_n}}=\frac{C_n}{N}$$

assuming $C_n$ is equal for all n. Of course this works in reverse too, if you're breaking a capacitor apart.

Imagine that your capacitor is an infinite number of itty bitty capacitors wired up in parallel. (That is in fact what a plate is; Any section of a plate is itself a plate with some capacitance it is contributing to the whole)

Then suppose you removed some ratio r from that capacitor (for example, suppose you reduced it by a factor of 2, r=2). Then the following transformations would describe your new capacitor. $$C_{new}=\frac{C_{old}}{r}$$ $$R_{new}=rR_{old}$$ $$V_{new}=V_{old}$$ $$M_{new}=M_{old}/r$$

Where C is Capacitance, R is resistance, V is voltage, and M is mass of the capacitor. Suppose we put each of those infinitesimal capacitors now in series. This would be another set of transformations, described below. $$C_{new}=C_{old}/r$$ $$R_{new}=rR_{old}$$ $$V_{new}=rV_{old}$$ $$M_{new}=rM_{old}$$

Combining these two transformations together (i.e, take half from parallel and put them in series), we get: $$C_{new}=C_{old}/r^2$$ $$R_{new}=r^2R_{old}$$ $$V_{new}=rV_{old}$$ $$M_{new}=M_{old}$$

From these we can derive the change in power and energy:

$$E=\frac12\frac{C_{old}}{r^2}(rV_{old})^2=\frac12C_{old}V_{old}^2$$

The above is in direct contradiction with one of the commenters: "Usually electrolytic capacitors have an approximately constant CV product for a given can size. Since energy increases as the square of the voltage the energy will increase in proportion to voltage."

I don't know what either he or I have done wrong (probably what I've done wrong). But until someone corrects one or both of us, I'm going to continue with this line of reasoning assuming I'm correct.

The equation for power will be $$P=\frac{(rV_{old})^2}{rR_{old}}=\frac{rV_{old}^2}{R_{old}}$$

So if my reasoning is correct, Power will increase by a factor of $r$, And energy will remain constant. (The fact that the power, or at least the peak power, increases by $r$ screams 'erroneous!' by every fiber of my being; but intuition is often wrong, I can't see why my reason is flawed, so I'm going with this)

These equations describe all sorts of hypothetical capacitors relative to one real capacitor. So assuming my hypothetical capacitor has the same mass, I can work out r from the voltage of the hypothetical capacitor and one real capacitor. Once I have r, it's just plug and chug to get the other values of the hypothetical.

I'm fully aware that not all of these assumptions are completely sound; For example, mass might not remain the same after this hypothetical transformation, given that a capacitor with higher voltage is probably using a thicker barrier, not smaller capacitors in series. But it was never my intent to get a perfectly accurate estimate of what capacitors might be out there. I just need a ball park system I can run gradient descent on, without it optimizing by choosing a completely ridiculous non existent capacitor.

EDIT

Thanks to the commenters and This answer about ESR. The solution seems so easy now.

"Usually electrolytic capacitors have an approximately constant CV product for a given can size", can be written as the following equation:

$$kv=CV$$, where k is some constant specific to a can size, v is volume, C is capacitance, and V is voltage. Keep the volume the same, and C*V is (approximately) the same.

But also, changing the dielectric barrier width has almost no effect on the ESR (see the linked QA). So, if my solver wants to change the voltage by $$\\Delta V\$$, we have: $$V_{new}=V + \Delta V$$ $$C_{new}=\frac{kv}{V_{new}}=\frac{C_{old}V_{old}}{V_{new}}$$ $$R_{new}=R_{old}$$

(My own statistical analysis suggest the claim that increasing the voltage has "almost no effect on the ESR" is not correct, but it is close enough for my purposes. Perhaps in practice increasing the dielectric barrier is not the only way voltage is increased, but only the dominant way? I don't know, I'm not the man to ask.)

• A mathematical argument is not a contradiction of physical manifestation! It is a similar fallacy to saying, by the Banach-Tarski theorem, we can duplicate an orange; a real orange is not infinitesimally (and perfectly precisely) divisible, violating assumptions of the theorem, and so we do not have an infinite food glitch IRL. (Heh, by the same argument, since your capacitors are infinitely divisible, the same theorem might apply here; thus taken to the logical extreme, any capacitance and rating might be possible!) Commented Jul 7 at 10:47
• The correct approach to this problem is to survey the market and see what coordinates are available on the CV, CV^2 and C*ESR/V axes. Including price/qty might be another useful axis: for example, C0G of some 100s V are quite dense types, but if you need any meaningful energy storage, you're going to be spending $1k's, versus$1's for electrolytics. Commented Jul 7 at 10:49
• Your argument is based on the construction of the capacitor being the same as the design voltage changes - that is not the case. I just did a quick calculation for capacitors from the VIshay 190 range (mouser.com/datasheet/2/427/190rtl-2953442.pdf) and my statement seems to hold true. The energy storage for a capacitor in a 16*35mm can increases from 717mJ for 5600uF@16V rating to 1875mJ for a 1500uF@50V rating device. Commented Jul 7 at 13:05
• @TimWilliams, No, this does not allow us to conclude that any arrangement of capacitance and voltage are possible for a given mass. It reduces those two degrees of freedom down to one (the one degree of freedom that I think actually exists). Yes, you can in fact split a metal plate in half and get two smaller metal plates, and two metal plates close together is a capacitor, so that doesn't violate any of my assumptions. Commented Jul 7 at 14:37
• @JosephSummerhays - As Tim alludes to the dielectric thickness does not scale well at low voltages. The volume of a capacitor has to include much more than just the active part. In electrolytic capacitors reducing the dielectric thickness (the oxide layer) has very little effect on the total volume. Eg halving the thickness will double the capacitance and halve the voltage rating but have negligible effect in the total volume - hence giving a constant CV product. Other technologies have similar scaling effects but may be less marked. Commented Jul 7 at 15:33