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I have a 2.7V 50F* capacitor and a 4.2V high-power Li-ion slave battery. Based on an ideal circuit that only has the internal battery resistance of approximately 1 m \omega\$ , the RC time constant would be 5ms. However, since charging it above its nominal voltage gets messy, I want to lower the voltage discharged by the battery in a way that offers negligible resistance. How can I do this?

Thanks,

BH

*EDIT: 100F, not 50

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  • \$\begingroup\$ Depending on your requirements, it could be as simple as two diodes in series. There will be power loss due to the voltage drop. If you want to reduce the loss, or to have tighter control over the current and voltage, then a buck regulator may be the answer. An up-to-25A low voltage buck regulator is common enough that it is well supported, but if you want to charge at say 100A+, you probably would need to do more design work. \$\endgroup\$
    – rioraxe
    Commented Dec 5, 2015 at 8:41
  • \$\begingroup\$ I looked up a 50F 2.7V capacitor, its maximum absolute current rating is 27A. \$\endgroup\$
    – rioraxe
    Commented Dec 5, 2015 at 8:50

1 Answer 1

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If you want to charge the capacitor quickly, but not overcharge it, then just use a large power FET and a comparator that turns it off if the capacitor is > 2.7 V.

However, if you don't limit the current, it will be many (say 10) A, and the power in the FET will be ~ 1.3V*10=13 W. It will take 13 s to charge the capacitor, and the FET would need a heatsink. Therefore I suggest that you do add a series resistance.

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  • \$\begingroup\$ Note that, assuming 1 milliohm battery resistance, and a 2-volt difference (4.2 battery to 2.2 cap), charging current is 2,000 amps. \$\endgroup\$
    – WhatRoughBeast
    Commented Dec 5, 2015 at 3:41
  • \$\begingroup\$ Actually the supercap will generally be the limiter -- perhaps between 25 and 100 mohm. The FET RDSON would likely be the next most significant contributor, especially as it may only have a few volts VGS \$\endgroup\$
    – jp314
    Commented Dec 5, 2015 at 4:13
  • \$\begingroup\$ You won't get 2700 A from any common Li-Ion cell, or any common super cap. Charging caps in series doesn't properly balance them, and one will develop a higher V than the other, and eventually fail. Finding a single FET that can switch 2700 A would also be a challenge -- likely you'd need about 20x 2 mohm FETs in parallel, and a sophisticated physical interconnect to keep series resistance low. \$\endgroup\$
    – jp314
    Commented Dec 5, 2015 at 20:00
  • \$\begingroup\$ I don't know how Li-Ion cells would fuss, although connecting discharged 100 F directly across any Li-Ion cell with built-in short protection would trip that protection circuit. \$\endgroup\$
    – jp314
    Commented Dec 6, 2015 at 4:23
  • \$\begingroup\$ OK. You'll know when it is fussy when it catches fire and adding water will make it worse. Be very cautious. \$\endgroup\$
    – jp314
    Commented Dec 6, 2015 at 18:33

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