# energy density comparison between inductors and capacitors

The figure below is about energy density comparison between inductors and capacitors. This thesis concludes that capacitors possess greater energy density than inductors, as seen in the figure.

I would like to know how to get the data as in the figure. For example, say for one cubic millimeter what formula, which value of L, C, current, voltage do you use to get that data?

• There are a few things that can be worked out in the direction of your question, in theory. There are material limitations, in terms of Teslas. These are "volt-seconds per m^2". "Supermendur" might get to 2 Teslas. (A medical MRI is about 3 T.) There is also a limitation in amps/meter (amps divided by magnetic path length) given copper wire and its diameter and your ability to pack it. Multiplying Teslas by amps/meter gives Joules/m^3. The Teslas part is easy (table of values.) The wire limitations you'd need to examine a little more to get the practical limits. But that's not intractable. – jonk Aug 28 '17 at 7:04

Every inductor has a rated current, every capacitor a rated voltage. And of course their rated inductance or capacitance. And they come in a defined package where you can calculate the volume.

With capacitors you have to be careful though. Ceramic capacitors with some dielectrics will change their rated capacitance based on the voltage they are charged, so it's not quite as simple.

As an example, let's take a look at this inductor:

• 10 µH
• 5 A
• 6.65 x 6.45 x 5.8 mm³ = 248.8 mm³

Neglecting the resistance of the inductor:

energy stored: $E_{ind} = \frac{1}{2} L I^2 = \frac{1}{2} 10~\text{µH} (5~\text{A})^2 = 0.125~\text{mJ}$

Energy density: $0.125~\text{mJ} / 248.8~\text{mm}^3 = 0.5~~\text{µJ}/~\text{mm}^3$

And for a capacitor let's take one of this series:

• 47 µF
• 25 V
• 5.8 x 5.2 x 5.2 mm³ = 156.8 mm³

Again neglecting parasitics:

Energy stored: $E_{cap} = \frac{1}{2} C U^2 = \frac{1}{2} 47~\text{µF} (25~\text{V})^2 = 14.7~\text{mJ}$

Energy density: $14.7~\text{mJ} / 156.8~\text{mm}^3 = 93.8~~\text{µJ}/~\text{mm}^3$

So for these two the numbers of the study seem to hold up. I neglected the loss of parasitics, you can model these as well and get smaller numbers.

• Thanks. Excellent answer but I would like to wait for some time. – anhnha Aug 28 '17 at 6:21
• @anhnha I'm in no rush to get accepted answers, on the contrary, it is often interesting to see what others write. I learn quite a few things that way... – Arsenal Aug 28 '17 at 6:55
• @anhnha Yes, there's no need to rush. Despite that, I do not understand you, because Arsenal gave you perfect answer to your question. He wrote equations, samples, that perfectly explain the point. – Chupacabras Aug 29 '17 at 19:12