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I have a toy that works with 3 AA Batteries in series. I switched to NiMH batteries but my voltage is less now.

Do you think that the middle Battery in the series can be swapped to a 14500 Li-ion?

That way it runs in between 2 NiMH batteries and the total voltage would be 5.1 in a perfect world, with more mAh.

What would happen? The toy is a GPi Case running a Raspberry Pi Zero, or upgraded to a CM3 module if I wanted.

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    \$\begingroup\$ Definitely not. And when recently charged your voltage would be 4.2 + 1.3 + 1.3 = 6.8 volts, way out of spec (actually the NiMH could well initially be more). \$\endgroup\$ Sep 30 '20 at 22:40
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As a general rule, it is not adviseable to mix batteries' or cells' chemistries.

(edited after a comment)

And any kind of mixing should be avoided: don't mix cells of same chemistry, but different capacities; don't mix cells of different brands; don't mix cells of different age; don't mix cells of different charge levels...

In short: if there is any difference between cells or batteries (capacity, age, chemistry, brand, charge level), don't mix them.

The different internal resistance of the different kind of batteries may increase the discharging of one type in relation to another, and you can end with some of the batteries completely depleted before the others.

When using Li-Ion, is adviseable to run them with battery managers IC's, which assures that the battery is not depleted below the voltage levels that could damage the battery in an irreversibly way. Of course, if you mix different chemistries, you won't be able to use a battery manager IC.

By the way, besides these reasons for not doing this, I should add that the position of the Li-Ion relatively to the NiMH ones is irrelevant. You could put Li-Ion at the start of the series, in the middle of two NiMH, and at the end of the series, and the final result would be the same: bad things.

As a final warning: although I wouldn't recommend this solution that you describe, I still have to say that, if you want to try this solution, NEVER EVER recharge this trio of different batteries together: when charging, Li-Ion would go to a Li-Ion charger, and NiMH should go to a NiMH charger. But the best, rational and safe option is really DO NOT DO this.

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  • \$\begingroup\$ excellent answer. we also should mention one should not only avoid to mix batteries' or cells' chemistry's but also cells of different capacities, cells of different brands, cells of different age, cells of different charge level,... \$\endgroup\$
    – schnedan
    Sep 30 '20 at 22:38
  • \$\begingroup\$ So since I have your attention, what if I Mod the case to run in parallel instead of in series and use 3 14500. The voltage would then be less than 3 AA's but more mAh. What would happen? \$\endgroup\$ Sep 30 '20 at 22:54
  • \$\begingroup\$ Don't use Li-Ion in parallel without an apropriate controller, either. Besides that, a 14500 Li-ion has nominal voltage of 3.7 V and RPi Zero nominally needs 5 V. There are some online blogs that tells that they got RPi Zero working at 3.7V, but I never tried (I have 4 Pi's, any of them is a Zero). It may be possible, but, as a general advice, it's the kind of thing that is done by a skilled tinkerer. With 3.7 V Li-Ion, you would need a step-up DC-DC converter. But, tell me, why don't you get rid of those Li-Ions batteries (remember, they need controlers ICs) and go for an USB powerbank? \$\endgroup\$
    – mguima
    Sep 30 '20 at 23:12
  • \$\begingroup\$ An USB powerbank has proper chosen and adjusted controller IC for its batteries, the controller manages the charging and the discharging (you could even keep using the device while charging), and it supplies exactly the regulated 5V that RPi Zero needs. \$\endgroup\$
    – mguima
    Sep 30 '20 at 23:18
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    \$\begingroup\$ @mguima Thank you for the support in fighting the crowd here. For you it's easy in here maybe because you are used to life in a zoo. Thank you also for visiting my website and liking the scope project. \$\endgroup\$
    – Moty
    Oct 16 '20 at 21:27
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Put 4 NiMH cells = 4.8V or fully charged 5.6V

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    \$\begingroup\$ A warm welcome to the site. Your Answer needs to contain some explanations rather than just firing off one line. It should be written clearly enough to teach and future readers. Please can you edit it and greatly improve it, making it informative rather than just factual. Thanks. \$\endgroup\$
    – TonyM
    Oct 1 '20 at 6:26

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