How long for an initial NiMH charge?

I have a standalone charger for some AA (and AAA) cells when AA cells held about 500 to 600mAH.

The charge rate is a steady 150 mA (50 mA for AAA) and there is no cut- off mechanism. The charger leaflet recommends 5 hours for cells between uses and that means for cells which have a reasonable amount of charge already.

So when I get a brand new set of NimH cells (presumably with almost no charge) the initial charge needs to:

(1) fully charge the cells (2) sightly overcharge some cells to equalize their charges

QUESTION: As a rule of thumb what extra percentage of charge time should I add on for this first charge?

I was guessing wildly at 80 extra: 18 hours at 150 mA for 550 mAH cells. What do you reckon?

Reply to
Andy
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18 hours at 150 mA for 550 mAH cells is *way* too much. The "overnight" or 14 hour rate is c/10, which means 55 mA for 14 hours for those cells.

To use your charger for this, modify it. Otherwise, you take the risk of overcharging. The modification is simple - requires three parts: an LM317 and a 22 ohm resistor in series with a 1 ohm resistor to make 23 ohms.

----- Charger + ---Vin|LM317|Vout---+ ----- | Adj [23R] | | +----------+---> To NiCd +

Gnd ------------------------------> To Nicd -

That will give you a charge current of a bit over 54 mA and you can put your batteries in the charge for 14 hours (or more) without damage.

Ed

Reply to
ehsjr

Thanks for the reply.

I may not have expressed myself clearly but I am not looking to recharge within a particular fixed time. I am looking to see what OVERCHARGE I should give the cells in order to fully top them all up.

My charger is 150 mAh and is recommended that it be used for 5 hours to TOP UP "standard" cells. When it was bought, standard meant about 550 mAh.

I take your point that the starting recommendation may need revising although the current assumes a certain type of cell with a certain impedance and I think that as cells have increased in capacity then so has their impedance which in turn will possibly (?) affect the nonimal current. Is this right?

Assuming that the charge current is right (we can assume a change in the

150 mA to whatever value we wish) then ...

... my QUESTION is what amount of extra charge would you guess is needed for the inital charge?

This is where a group of cells are primed to work together and run flat together and which need all topping up (to 100%) together even though some will have started with more or less charge than others.

Reply to
Alex Coleman

The answer already given is the answer to your question.

The "overcharge" you should give your cells is C/10 for 14 hours, but that's not really overcharge as the word is normally used.

It is not really overcharge, per se. Cells do not charge at 100% efficiency. So if you apply a 55 mA charging current to a 550 mAH cell, it takes more than 10 hours of charge time to get 550 mAH of energy into the battery - thus the

14 hour charge period. The "overcharge" is the extra 4 hours over the 10 hour period during which the charger provides a total of 550 mAH. Since the battery does not absorb 100% of the energy supplied by the charge during the 10 hours, you need to "overcharge" for another 4 hours to give the battery enough time to reach full charge.

The value of the "14 hour approach" is that you do *not* overcharge the batteries. It applies equally to fully or partly discharged packs. You provide sufficient energy over time for the pack to fully charge, but it is delivered at a rate that won't damage the cells. Mind you, if you are going to leave the pack on the charger for extended periods well beyond 14 hours, damage will result. For extended period charging, the rate needs to be reduced significantly after the pack reaches full charge.

Overcharging NiCds is not recommended, where you use the word overcharge to indicate raising the cell's voltage higher than it should be. The example you gave of charging a 550 mAH pack at 150 mA for 18 hours will damage the pack. It would be fine for a 1500 mAH pack.

Ed

Reply to
ehsjr

Hi Ed.

Interesting reply. Well, it was too obvious and I missed it. I had completely overlooked efficiency. 700 mAh in does not equal 750 mAh stored. Damn you. Heh! :-)

OK, now I get what you are saying. It is interesting that you mention packing charge in.

Is it a general truism that a slow charge packs in more energy in total than a fast charge? (I.E. the energy density per unit of storage is greater for a slow charge?)

Also, is there perhaps a "knee" on the speed versus density chart which is the sweet spot between fast/slow charging and improved energy density? If so, then what charge rate does it indicate is a good rate for denisty.

Finally, I think I do mean "overcharge". Let me explain. If one cells starts with 30% charge and another with 45% charge then to get both cells fully charged (in a charger which takes the two in series) then the cell starting with a 45% charge will get overcharged.

If the charge rate is gentle then I guess this may not lead to as much damage as a brutally fast rate and maybe there is no harm done but I would still call it "overcharged".

I have the hunch that many modern cells are ok with very rapid charging rates such as C or C/2 and do not suffer as much as the old cells did at a charge rate of C/5. Do you know if this is actually true?

Reply to
Alex Coleman

I don't know. I wasn't trying to compare a fast, automatically terminated charge with a slow charge. I was comparing a fast charge that goes *WELL* beyond where it should terminate to a slow charge. Lets say you need to put in 700 mAH, due to charging inefficiency in the cell, to fully charge your 550 mAH cells. With 550 mAH cells being charged at 150 mA for

18 hours, that's a total of 2700 mAH, or 2000 more mAH than the cells can use for charging. That extra energy is converted 100% to heat starting ~ 4 hours and 20 minutes into the cycle - you are cooking the cells for 13 + hours at 150 mA * (Vcharger - Vbattery). This is likely to damage the battery.

With the C/10 14 hour charger (55 mA) it takes about 12 hours and 45 minutes to fully charge the battery. You cook the battery for about an hour and 15 minutes at a much lower heat 55 mA * (Vcharger - Vbattery). Vcharger will be lower as well, since it needs to be at a level above Vbattery that will drive only 55 mA, not 150 mA. The net effect is the the heat in the cells is much lower, and the cells can dissipate it without venting.

The above is not in any way applicable to an 18 hour charge at ~ C/4 (150 mA with a 550 mAH pack). At issue is getting a full charge, without damaging the pack, with a charger that has no automatic control. Knees of the curve and sweet spots are 100% irrelevant to the issue. If you had a full featured charger that could sense the condition of the pack and adjust and terminate the charge as appropriate, it might be germaine if one stretches logic into some bizzare shape. But for a battery *pack*, knees of curves are irrelevant for the reason you mention below: the cells may differ. And if the cells differ in charge, the knees would have to differ.

Ok, then you need to stop meaning it. It is BAD for NiCds. There is unavoidable overcharging, and designers have used a number of techniques to minimize it - taper charge, float charge, multi-stage charge, terminated charge and so forth. As you used it, you seek to overcharge. Not good. When charging is placed in your hands, as it is with your charger that has no auto shut off and a single (relatively) high charge rate, *you* need to be gentle to the pack.

With an automatic charger designed for the cells you are charging, you plug it in and let the charger do its thing. The *charger* decides what is a charge and what is an overcharge, and adjusts its output accordingly.

It's leaving the pack on the high rate charger

*WAY TOO LONG* that is the problem.

Yes, there are cells that can be charged quickly with a charger designed for that. None of the fast chargers charge the pack for anything remotely resembling 18 hours, and they all operate automatically, either terminating the charge cycle or reducing the charge rate dramatically.

Ed

Reply to
ehsjr

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