0
\$\begingroup\$

I know that capacitors discharge quickly, but it seems like there has to be a way to replace deep cycle batteries. A capacitor with a constant current source that is rated to drain the cap slowly, should allow the cap to act like a battery, yes? Isn't a boost converter a constant current source (or variable but not so high as to drain the cap bank)?

To be clear, I am wanting to use caps for a datacenter 8 hour backup, but I imagine solar and other enthusiasts want to know.

\$\endgroup\$
  • 3
    \$\begingroup\$ Fast discharge is synonymous with low energy density = bad for 8 hours with or without a boost converter. You're missing the forest for the trees by focusing on boosting the voltage. \$\endgroup\$ – DKNguyen Jul 29 '19 at 14:36
  • 5
    \$\begingroup\$ Irrelevant because the low energy density problem is MUCH more severe than you are thinking. \$\endgroup\$ – DKNguyen Jul 29 '19 at 14:40
  • 4
    \$\begingroup\$ 1) the energy stored in a capacitor is E = CV^2 (see: hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html). 2) Determine how much energy is needed for "datacenter 8 hour backup" 3) Play with the calculator from the hyperphysics link and find a solution to get the amount of energy you need. 4) compare to a battery based solution. 5) conclude that super capacitor technology simply isn't there yet, you'd need far more volume (for all the caps) and money compared to a traditional battery based solution. \$\endgroup\$ – Bimpelrekkie Jul 29 '19 at 14:40
  • 3
    \$\begingroup\$ What about a typical home? Don't change the question! My procedure is universal as it simply compares energy stored (in a cap) to energy needed. So for a "typical home" or a supercapacitor powered gadget, the equations are the same. If you want to know: do the calculation. Also see where supercaps are actually used in products because that's where there is an advantage (if not the manufacturer would have used a battery). I'll give one example: a real-time clock inside a measurement product. Even for that solution: I simple coin cell is hard to beat. \$\endgroup\$ – Bimpelrekkie Jul 29 '19 at 14:47
  • 5
    \$\begingroup\$ "... but it seems like there has to be a way to replace deep cycle batteries." - Why does it seem like there has to be? \$\endgroup\$ – marcelm Jul 29 '19 at 15:27
7
\$\begingroup\$

Yes, you can use switching power supply topologies to slowly and safely discharge a large capacitor bank. These can be controlled to create exactly the load current you need. Basically every switching power supply has some DC intermediate circuit with a capacitor bank, exactly as you described above.

Still, the remark from the comments seems to be limiting factor here. First you talked about datacenter usage, then you changed to a typical home.
Let's just assume an average consumption of 2kW during the 8h. This seems legit for home usage, but is way to low for a datacenter.

\$ 2 kW * 8h = 16 kWh = 57.6 MJ.\$
This is the energy we somehow need to store in the capacitors.

\$ 57.6 MJ = E_{Cap} = \frac{1}{2} \cdot CU^2\$

I just checked at a big electronics distributor and the "best" solution I could find would be an 8.4 V supercap with 15 F.

\$ C = \frac{ 2\cdot E}{U^2} = \frac{2 \cdot 57.6 MJ}{(8.4 V)^2} = 1.6 MF \$
Megafarad. Mega. Farad.
With the above mentioned capacitor with 15 F each this would just require... 100,000 peaces! With a price of ~15$/piece you can calculate the cost of this. And this calculation assumes that the switching power supply can discharge the caps with 100% effiency over the complete voltage range down to 0 V. Not likely.

And now let's not talk about a supply for a datacenter, right?

\$\endgroup\$
  • 1
    \$\begingroup\$ I think your first statement is misleading. If the switching converter has some specified input voltage and you need some specified amount of power out, you can not arbitrarily control the current drawn from the capacitor. The "exactly what you need" is really "exactly what the load needs", plus a bit more because the power supply is not 100% efficient. \$\endgroup\$ – Elliot Alderson Jul 29 '19 at 15:40
  • \$\begingroup\$ Well yeah, that is true, when the caps are discharging and voltage decreases, the discharge current has to increase (for U->0 the current would theoratically go to infinity. But because voltage squared law, the energy left in the cap is negelctable for low voltages). But the current on the output of the supply is stable and of the desired value the whole time. \$\endgroup\$ – jusaca Jul 30 '19 at 4:56
  • \$\begingroup\$ Looking at this again, I think the issue is where you placed the word with in your first sentence. It looks to me as though you meant "discharge....with exactly the current" but maybe you meant "use switching...with exactly the current [to the load]" In any event, it is best to avoid complex sentence structure when explaining engineering concepts. \$\endgroup\$ – Elliot Alderson Jul 30 '19 at 10:53
  • \$\begingroup\$ I changed the first sentance, now it should be clearer. \$\endgroup\$ – jusaca Jul 30 '19 at 10:58

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.