-5
\$\begingroup\$

After observing battery voltage gauge on my car, I can see my lead acid battery operate voltage is from 11.8v to 14.8v.

The car is about 5 years old, and while start/stop, the car drains the lead acid battery, and voltage drops from 14.8v immediately to 12v and when it drops to ~12.2v, the engine ignites again. the stop won't last more than 30 seconds, which is annoying.

Now I would like to have a new battery that prolongs the start/stop time.

I intend to use Panasonic NCR18650B Batteries. 3 in series many in parallel. lithium ion packs with BMS battery management system.

But charging will easily pass 4.2v for each cell. 14.8/3=4.93v, when discharging, batteries is ok: 11.8/3=3.93v.

If I use 4 in series, 14.8/4 = 3.7v and 11.8/4 = 2.95v. And batteries will only be partially charged to 3.6~3.7v, which is another problem. and might get over discharged.

Any suggestion for me to use these NCR18650B to replace the traditional car batteries?

\$\endgroup\$

closed as too broad by Leon Heller, ThreePhaseEel, dim, uint128_t, laptop2d Dec 2 '16 at 19:25

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    \$\begingroup\$ Planning on setting fire to your car ... ? \$\endgroup\$ – brhans Dec 2 '16 at 1:51
  • 1
    \$\begingroup\$ I think you have some shorted diodes on 3 phases , there are 6 diodes and if Vbat after full charge drops below 12.5 there is a bad cell. Loose contacts and fan belt prevent correct operation too. A 650A CCA LiPo array and regulator is expensive \$\endgroup\$ – Sunnyskyguy EE75 Dec 2 '16 at 2:00
  • 1
    \$\begingroup\$ @brhans not fire, but innovations \$\endgroup\$ – c2h2 Dec 2 '16 at 2:14
  • 1
    \$\begingroup\$ The basic premise of this "question"(?) borders on the insane. If there is a problem with the charging circuit, then it is likely a simple matter to fix quite economically. Engineering a $1000+ exotic solution to a common vehicle electrics problem is just silly. \$\endgroup\$ – Richard Crowley Dec 2 '16 at 4:35
4
\$\begingroup\$

Your car provides a amp draw of 100+ while on, headlights, fan. Those 18650 cells are not even going to last a minute.

If your car is dropping to 12V at a stop light, then you have alternator system issues. A loose cable, alternator belt, bad battery, something. A five year old car should not be experiencing such issues yet. Fix the underlying problem instead of trying to bandaid a hole in the titanic.

Edit: Of course if you are trying to build a proper Lithium equivalent battery, 600+ Cold Cranking Amps capable, it still won't work properly if you try to run the car on it without a working alternator. The Alternator or Generator powers the car while on, the Battery is mainly for starting the car.

\$\endgroup\$
  • \$\begingroup\$ 250F super cap + lithium batteries for handling 600A issue and 100x 18650 batteries, makes 3.4Ah * 3.7v * 100 = 1258Wh or 12v 100Ah \$\endgroup\$ – c2h2 Dec 2 '16 at 2:16
  • \$\begingroup\$ please read: jalopnik.com/5527441/porsches-new-1700-car-battery-what-you-get \$\endgroup\$ – c2h2 Dec 2 '16 at 2:20
  • 1
    \$\begingroup\$ I too, think this is a band-aid to the real problem - probably a faulty alternator. But all those lithium-ion batteries, in the summer, in a hot engine compartment, with the vibration, are going to be a fire/explosion waiting to happen. In the winter they will die below freezing. Try it if you must, but I'd suggest carrying a fire extinguisher in the car. \$\endgroup\$ – rdtsc Dec 2 '16 at 2:55
  • 2
    \$\begingroup\$ @c2h2 No offense, but do you think your home made battery will be equivalent to a Porsche battery? They most likely have a team of people working on this (that do this kind of thing for a living). \$\endgroup\$ – Tyler Dec 2 '16 at 3:50

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