2
\$\begingroup\$

I am trying to see if it would make sense to use a LC network to store small amount of electrical energy (500 mWHr or less) as an alternate of a 3.3 V battery. What I understand is that L and C are reactive components and they do not dissipate energy at least in their ideal case.

Specially if I have to recharge by AC mains outlet or by wireless-charging then recharging an LC would be easier compared to a DC battery.

Any expert thoughts about it?

Can someone throw out some mathematics to prove or dis-approve the above idea?

\$\endgroup\$
  • 1
    \$\begingroup\$ LC probably not. But perhaps with a super-capacitor alone. You might google "battery vs supercapacitor" and read some of the articles. \$\endgroup\$ – scorpdaddy Apr 30 at 15:31
8
\$\begingroup\$

...use a LC network to store small amount of electrical energy (500 mWHr or less)

You might think that 0.5 WHr is "small" but for an LC tank, it is HUGE.

So what is 0.5 WHr?

0.5 Watt = 0.5 Joule / second

One hour is 3600 seconds so that gives a total energy of 0.5 J/s * 3600 s = 1800 Joule

You want to use an LC resonator. In an LC resonator the energy resonates between a capacitor and an inductor.

Let's keep it simple and assume that at a certain time all energy is stored in the capacitor and the inductor is completely discharged.

What value does our capacitor need to be and to what voltage will it be charged to store 1800 J ?

I played around here and here are some solutions:

  • a 90 mF (milli Farad so 90000 uF) capacitor charged to 200 V

  • a 1 Farad capacitor charged to 60 V

  • a 400 Farad capacitor charged to 3 V

Of these the most practical solution is the last one, 100 F 3 V supercapacitors are available. However these capacitors only work with positive voltages. We could solve that by using placing capacitors in anti-series with protection diodes. Then we need 16 capacitors already !

Charging them will take time, you cannot "quickly" (few seconds) charge a super capacitor. Charging will not be much faster than charging a conventional battery. Source: Julian Ilett's Youtube channel, he sometimes messes around with super capacitors.

You could try to build a "better" capacitor and go for the 90 mF, 200 V option. I fear that the capacitor will need to be as large as a washing machine or a truck. Is that practical?

I'm not even going into the inductor as these are even less effective in storing energy in a given volume. You might need an inductor of the same size as a house.

I really think a battery is much more practical.

You could store less energy, less than the 0.5 WHr but then you get in the range where using a battery becomes even more efficient.

\$\endgroup\$
  • 1
    \$\begingroup\$ A 1000 µF, 250 V DC electrolytic capacitor is available in a 35 (dia.) x 50 mm case, so 90 of them in parallel would only occupy a volume of a few litres. Bigger than a battery, sure, but not quite a washing machine. \$\endgroup\$ – nekomatic Apr 30 at 12:09
  • 2
    \$\begingroup\$ @nekomatic But those are polarized, for use in an LC resonator we'd need unpolarized caps. so then we would need 4x more of these (2x for each polarity and then 2x because the value halves as we connect the caps in anti-series). Still not washing machine size I admit but large nonetheless. \$\endgroup\$ – Bimpelrekkie Apr 30 at 12:14
  • 1
    \$\begingroup\$ Fair point, I guess I skipped over 'capacitors alone would be more efficient, but…' \$\endgroup\$ – nekomatic Apr 30 at 12:29
  • 1
    \$\begingroup\$ Please keep in mind that storing that amount of energy in a capacitor is very dangerous. If you short such a capacitor bank it will provide massive amounts of instantaneous current and power. The capacitor bank of parallel resistors has nearly no resistance - as compared to a battery of the same energy. You can vaporize or weld metal if you short it. \$\endgroup\$ – scorpdaddy Apr 30 at 15:36
6
\$\begingroup\$

they do not dissipate energy at least in their ideal case

They do in reality.
Capacitors have a parasitic series resisance called ESR.
Inductors have winding resistance.

Both resistances will dissipate.

\$\endgroup\$
  • \$\begingroup\$ True, but as the OP said "...in their ideal case" they do not dissipate energy. \$\endgroup\$ – Elliot Alderson Apr 30 at 14:04
6
\$\begingroup\$

Your 0.5mWh amounts to 1800 J of energy.

To store that amount of energy in a capacitor, you would need a 56 millifarad capacitor at 230V.

It would take a 1 henry inductor at about 60A to hold that amount of energy.

Each part (L and C) have to be capable of storing the total amount since they are shuttling it back and forth all the time.

You are talking about putting an absolutely scary voltage and a frightening current in your pocket.

If you look up the required components, you will find that they won't fit in your pocket.

You will need like a (well insulated) backpack to carry the stuff, and a bunch of electronics to charge whatever device you intended to use it with because it will certainly want DC instead of the AC your resonating contraption will provide. And, it will self discharge in fairly short order while heating up your backpack.

Stick with the battery.

Smaller, lighter, safer, and more efficient.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.