I am making a LiFePo4/supercapacitor-hybrid.

Spike current draw can reach 300A for a few seconds. The LiFePo4 batteries I am using is rated for 100A max discharge rate. In order to make sure the supercapacitors do the heavy lifting, I need to limit the max current allowed to be drawn from the LiFePo4's.

In my circuit, the polyswitch will "switch off" in milliseconds at that kind of current draw. The power resistor will allow the supercapacitor to be recharged slowly while the polyswitch cools down. An obvious problem here is a waste of energy.

Is my idea completely senseless, or could it work?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ To me, it would be much more sensible to switch the recharging branch instead of the high-current branch. But: Why switch at all? \$\endgroup\$ – Marcus Müller May 15 '17 at 9:19
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    \$\begingroup\$ "300 A for a few seconds" is WOAH. Let's say "a few" is 3. thus, C1 needs to supply Q = 900 As = 900 C. That would mean its capacity needs to be Q/U = 900 C / 12 V = 75 F ; your 70F might be slightly underdimensioned. Also, I'd very much like to mention that at these currents, you need a very proper way of designing your power rails. \$\endgroup\$ – Marcus Müller May 15 '17 at 9:23
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    \$\begingroup\$ Please link to a polyswitch that you think might do the job. \$\endgroup\$ – Andy aka May 15 '17 at 9:26
  • \$\begingroup\$ To my understanding, polyswitches also degrade with use (slower switching). This could be a problem. \$\endgroup\$ – Joren Vaes May 15 '17 at 10:03
  • \$\begingroup\$ This should do the trick: no.mouser.com/ProductDetail/Littelfuse/AHEF1000/… \$\endgroup\$ – skogs May 15 '17 at 12:19

Theoretically, an RLC circuit should work:


simulate this circuit – Schematic created using CircuitLab

L1 is supposed to limit the current during the spike, while providing a lossless current transfer on DC, and making use of energy stored in the B field to charge the capacitor after the spike is over. R1 may be needed to keep overshoots limited. You may still want to include some sort of fuse to protect the battery in case your current spike is longer than you have expected.

However, my feeling is that for your parameters you will end up with a prohibitively large L value for the inductor.

  • \$\begingroup\$ Thank your for this! I will try to do the math and see if i can find the correct inductance needed. \$\endgroup\$ – skogs May 15 '17 at 12:53

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