A robot-arm kit I am playing with uses 4xD-cell batteries wired in series to provide 6V, with an additional connection on the cross-bar at the end of the pack to provide 3V. This seems reasonable enough, but I wonder if it might have any effect on cell lifetimes?
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asked Dec 11, 2011 at 18:16
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\$\begingroup\$ How do you wire 4 cells in series-parallel to give 6V? 2 cells in series will give you 3V. Two more in series, placed in parallel across the first two will still be 3V. Please give us a circuit diagram - as it stands you're making no sense. \$\endgroup\$– MajenkoCommented Dec 11, 2011 at 19:04
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\$\begingroup\$ My mistake… They are arranged in the pack as described but they are wired in series, with an "extra" lead coming from the middle of the series. \$\endgroup\$– Kaelin ColclasureCommented Dec 12, 2011 at 1:14
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3 Answers
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The cells being used for both loads will be depleted faster than those being used for only one, at least if the lower voltage load is substantial in comparison to the higher voltage one. If you then re-charge them in a charger not suitable for unbalanced starting conditions, this could be somewhat (to even seriously) detrimental, yes.
For chemistries with simple charging methods such as NiCd, NiMH, etc, simply using a low rate (1/10C) charger would probably not result in much of an overcharging problem for the less depleted cells - such chargers often have no timer or cutoff mechanism anyway (though they shouldn't be left connected constantly). But a higher rate or peak-detect charger would not be appropriate.
For chemistries with more precise charging requirements, especially the various lithium cells, you would require a charger with a "balancing harness" - ie, an additional connector tapping out all of the internal cell terminals so that they can be individually charged and monitored to a safe level of charge. It doesn't sound like you are talking about these types of cells, though.
If you have loose cells in a battery holder instead of a soldered pack, you could swap the doubly and singly used cells partway through the discharge cycle to partially equalize the usage.
answered Dec 11, 2011 at 19:14
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A cell has a finite amount of power it can provide. This is measured in "Ah" or "mAh" - the Ampere-hours, or Miliampere-hours. It's the number of amps (or miliamps) a battery can provide over the course of an hour before the battery runs flat.
The more current you draw from a battery the sooner it will run flat. For example, a 2200mAh battery will be able to provide, for period of one hour, 2200mA (or 2.2A).
If you draw 220mA from the battery it will last for about 10 hours before running flat.
If you are running your batteries as I think you are (not as you say you are), which is 4 cells in series with a centre tap, then the 2 batteries "below" the centre tap will be providing current for both voltages. The cells above the centre tap will only be providing current for the 6V supply.
Therefore, the cells "below" the tap will run out before the ones above.
If we say the current being drawn by the 6V portion is 300mA, and we call this \$I_H\$, and the 3v portion draws 140mA - call this \$I_L\$ then we can see that:
\$I_{BH} = I_H\$
and
\$I_{BL} = I_H + I_L\$
Where \$I_{BH}\$ and \$I_{BL}\$ are the currents being drawn from the "high" and the "low" pairs of batteries.
So, the "upper" batteries , if they are 2200mAh, will be asked for 300mA, which means they will last for \$\frac{2200}{300} = 7\frac{1}{3}\$ hours. The "lower" batteries, will be 300mA + 140mA, which equals 440mA - these would last for \$\frac{2200}{440} = 5\$ hours.
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The Robot arm in question probably uses one pair of batteries to run the motors forward, and the other pair to run them backward; it may in fact never use both pairs in series. You could check this by observing the behavior of the unit with each pair of batteries removed. If that is indeed the way the unit behaves, you should regard it as having two entirely separate battery packs. While for most motors the total distance moved forward would roughly match the total distance moved backward, the torque required for the two directions may easily differ; hence battery wear may or may not end up being roughly balanced. In any case, though, I would suggest that for that particular unit the batteries be regarded as two separate pairs. If operation in one direction gets sluggish, replace just the batteries associated with that direction.
In a more general scenario (e.g. if a device needs +3 and +6 supplies, so one pair of betteries will consistently have to supply more current than the other), it may be beneficial to periodically swap the high-side and low-side batteries so as to promote even wear. This may be especially important if it would be possible for the high-side batteries to demand current even when the low-side batteries were completely flat. Rechargeable batteries can be severely damaged if current is forced through them in the wrong direction, and non-rechargeable batteries treated in such fashion may emit caustic goo (were it not for the risk of equipment damage, it wouldn't matter if reverse-driving depleted non-rechargeable batteries would damage them, since the depleted batteries would be useless anyway; nonetheless, there's a very real possibility the damage wouldn't be limited to batteries, but could (via caustic goo) harm other equipment.
