I use eneloops (AAA and AA) in my primary lights. However, if I wanted to recharge the cells very quickly, say at 2 amps with AAA or 5 amps with AA eneloops, I'm sure most of you would not recommend it due to heat build up. But can't I use cold air when charging the cells to ameliorate this overheating problem? I've heard that this is a bad thing to do, but I would like to know why. Any thoughts?
Fast chargining and heat
- Thread starter h2xblive
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JWP_EE
Enlightened
Thanks. I came across that thread, but I don't think they directly addressed my question. Still a good read, though.
I would say at those rates don't expect your cells to last nearly as long as they would, you may lose half the recharge cycles or more due to the excessive heat in charging and the faster chemical reactions causing essentially almost boiling in the cells themselves. I would say you could cool the outside of the cells but it would be like microwaving a chicken from the inside first the heat has to build up inside where it is starting to damage things before it gets to the outside of the cell where you can cool it off.
JWP_EE
Enlightened
From everything I have read in this forum the problem with cooling the cells is with smart chargers missing termination. Most smart chargers use the temperature of the cell as a safety cutoff on a missed termination. With cooling this cutoff may not work. The negative delta V is a result of heating in the cell. Cooling may make the voltage drop smaller and again missed termination. This is my understanding of how it works based on my reading of this excellent forum.
LukeA
Flashlight Enthusiast
I would just buy some more Eneloops.
Hmmm, that's the best explanation I've heard so far. It certainly makes sense.
I understand that. I wouldn't expect external cooling of the cell to completely counter all the heat generated from extreme charging rates. However, assuming a fast charge had to be done, would external cooling be better than nothing? Perhaps JWP_EE has answered my question?
Mr Happy
Flashlight Enthusiast
The 15 minute chargers like the Energizer one use a charging current of 7.5 A for AA cells, and the cells do not get particularly warm for the first 10 minutes or so.
So the answer is to charge very fast at the beginning, and slow the charge down near the end. A really sophisticated charger could control the charge current automatically to achieve this, but you won't normally find one of those. A simpler option is to use a 15 minute charger for about 10 minutes, and then move the batteries to a more normal charger to finish off.
So the answer is to charge very fast at the beginning, and slow the charge down near the end. A really sophisticated charger could control the charge current automatically to achieve this, but you won't normally find one of those. A simpler option is to use a 15 minute charger for about 10 minutes, and then move the batteries to a more normal charger to finish off.
TakeTheActive
Enlightened
CLICK on my Sig Line LINK and read the THEORY article by jtr1962.
+1 :thumbsup:
But be aware of the potential 'Dead Zone' many chargers may have at the beginning of a Charge Cycle where they IGNORE any termination signals for xx minutes to avoid false triggers.
Some 'SMART' chargers may charge a FULLY CHARGED cell for 45 minutes before giving up, while others may recognize the SOC and stop after 10.
SilverFox
Flashaholic
Hello H2xblive,
You can get away with it for awhile, but you should exercise extra caution when doing so.
The RC people used to charge at 2C in order to warm the packs up. After several explosions resulting in injuries, the charger manufacturers have now cut the chargers back to charge at 1C. When you go to charge a cell you are asked for the capacity of the cell (or pack). This triggers a maximum charge rate.
A warm cell seems to be able to spit its internals with quite a bit of force when the vent is unable to release the pressure.
Tom
You can get away with it for awhile, but you should exercise extra caution when doing so.
The RC people used to charge at 2C in order to warm the packs up. After several explosions resulting in injuries, the charger manufacturers have now cut the chargers back to charge at 1C. When you go to charge a cell you are asked for the capacity of the cell (or pack). This triggers a maximum charge rate.
A warm cell seems to be able to spit its internals with quite a bit of force when the vent is unable to release the pressure.
Tom
Battery Guy
Enlightened
In an ideal world, that brilliant comment from Mr. H would end the thread.
However, this is not an ideal world, so I will do my best to fill in the gaps.
To the OP: it is very hard to fool mother nature. All batteries eventually degrade in performance because of parasitic or side chemical reactions. Temperature is your enemy. As temperature increases, the rate of these side/parasitic reactions also increases.
With a NiMH cell, you also need to deal with gas generation. The reason that NiMH cells show an increase in temperature towards the end of charge is because oxygen generated at the positive electrode reacts exothermically with hydrogen in the negative electrode to make water. The faster the charge rate, the more gas is generated. If you generate oxygen too fast (i.e. charge at too high of a current) you will literally tear apart the cathode due to gas generation.
So, for these reasons, as well as those given by other contributed to this thread, charging NiMH cells at extremely high currents (e.g. 4C) will result in faster degradation in cell performance.
Cheers,
Battery Guy
But if temperature could be externally lowered, would oxygen generation be lowered as well?
the problem is the heat will still have to soak through the cell from the inside to be cooled. It is like blowing on a red hot heat gun you may remove the heat around the element but it will still be hot itself and the more you heat it the hotter it will still get. You could also take a hot light bulb that takes 6v and put it on ice and then run it at 8v instead of 6v and the ice will have little effect on the bulbs longevity.
Mr Happy
Flashlight Enthusiast
No, that is not so. The oxygen is generated when the positive electrode is fully charged and can absorb no more charge. After this the oxygen is generated by electrolysis, splitting water into hydrogen and oxygen. The amount of oxygen produced is in direct proportion to the amount of current flowing.
In another post it was said that the heat is generated by the exothermic recombination of oxygen and hydrogen. Another way to understand this is that the heat is generated by resistive heating from the current flowing through the cell. When the cell can absorb no more charge the energy being supplied by the charging current has to go somewhere, and it gets transformed into heat. In a fully charged cell the electric power equation applies,
Power = Current x Voltage Difference
At high charging currents the voltage difference across the cell terminals is higher as well as the current, so the amount of heat generated is even greater.
I dare to say that the lower the cell temperature, the more visible the -dV will be.
This is so for NiCD.
http://www.americantoppower.com/battery_cd.htm
and according to uniross for niMH:
links show plots...
http://www.uniross.com/images_industrial/Curve-1_charge-nicd-nimh.gif
and at faster charge rate:
http://www.uniross.com/images_industrial/curve-2_charge-fast-nicd-ni.gif
This is so for NiCD.
http://www.americantoppower.com/battery_cd.htm
and according to uniross for niMH:
links show plots...
http://www.uniross.com/images_industrial/Curve-1_charge-nicd-nimh.gif
and at faster charge rate:
http://www.uniross.com/images_industrial/curve-2_charge-fast-nicd-ni.gif
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Battery Guy
Enlightened
No. In fact, it might actually be generated faster. Oxygen generation begins when the positive electrode potential exceeds the oxidation potential of the aqueous electrolyte. While oxygen is being generated, two electrochemical reactions are occurring in the cell: the normal charge reaction and the generation of oxygen. If you are charging fast and cooling the cell, the internal resistance of the cell will be higher, which will result in more polarization of the electrodes and oxygen generation will start sooner in the charge cycle. You can actually see this effect in the plots posted by s0lar in the post above this one. At lower temperatures, the cell voltage is higher and the dV/dt=0 occurs earlier in the charge cycle. The dV/dt=0 point occurs because the cell temperature is rising due to oxygen recombination, therefore the internal resistance of the cell decreases. Nice plots s0lar!:thumbsup:
In addition, the heat generated is beneficial with regards to oxygen recombination at the negative electrode. As the cell temperature rises, the oxygen recombination rate also rises, so you can charge at a higher current. If the rate of oxygen generation (determined by the current and positive electrode voltage) exceeds the oxygen recombination rate, the pressure in the cell will rise and the cell may vent oxygen.
So while heat is the enemy of batteries with respect to longevity, in the short run, you rely upon the NiMH cell to heat up in order to charge it at extremely high currents.
Hope this helps.
Cheers,
Battery Guy
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Battery Guy
Enlightened
I'll nit pick here. Oxygen generation occurs when the potential of the positive electrode exceeds the oxidation potential of the electrolyte. If you charge a NiMH at a very slow rate, you can reach 100% SOC without generating oxygen, and charging beyond that point will result in oxygen generation as you stated. But at higher rates, the cathode potential exceeds the electrolyte oxidation potential before it is at 100% SOC. When this occurs, the charge reaction is occurring simultaneously with the oxygen generation. Initially, most of the current go into the charge reactions, but as the cell potential rises, an ever increasing proportion goes towards generating oxygen.
Agreed, except at that OH-, not hydrogen gas, is generated along with oxygen due to the high pH of the electrolyte.
We may have different ways of talking about the same thing. This is how I think of it. There are two sources of heat generation: internal resistance and oxygen recombination. During the first part of the charge cycle where no oxygen is being generated, the only heat is due to internal resistance. This is the "P=I x dV" portion that you mention above. Another way to think of it is "P = (I^2) x R". Both are equivalent.
When oxygen is generated at the positive electrode, the oxygen gas diffuses to the negative electrode and reacts chemically with hydrogen in the negative electrode to produce water. This reaction is exothermic. So in the later stages of the charge process, heat is being generated by two mechanisms: internal resistance and oxygen recombination. This is why the temperature rises more in the later stages of the charge cycle.
Cheers,
Battery Guy
45/70
Flashlight Enthusiast
Boy! You guys are really into it! Very interesting, although perhaps a bit over my head, at the moment anyway.
I think a simple "It's best to not charge NiMH cells at a rate higher than 1C on a regular basis due to internal reactions that occur within the cell when subjected to higher rates of charge" would suffice, however.
s0lar, yes they are nice graphs. Unfortunately, they are hotlinked from the Uniross site, which is against CPF rules. You need to replace the graphs with links instead. Don't say I didn't tell you so!
Dave
I think a simple "It's best to not charge NiMH cells at a rate higher than 1C on a regular basis due to internal reactions that occur within the cell when subjected to higher rates of charge" would suffice, however.
s0lar, yes they are nice graphs. Unfortunately, they are hotlinked from the Uniross site, which is against CPF rules. You need to replace the graphs with links instead. Don't say I didn't tell you so!
Dave
Mr Happy
Flashlight Enthusiast
Heh heh. Thank you for supplying the extra detail.
However, if we look at the whole cell, and look at both half reactions together, there has to be charge balance. Since the oxygen gas is neutral and comes from the water in the electrolyte, the other reaction product must be neutral too. If we add together the two electrode half reactions we have in total:
4M + 2H2O + [energy] => 4MH + O2
The hydrogen is produced and then absorbed at the negative metal hydride electrode. The energy in this equation comes from the supplied electric current.
Later on, the oxygen diffuses across to the negative electrode and is catalytically recombined with the hydrogen, reversing the above reaction to form water again:
4MH + O2 => 4M + 2H2O + [energy]
In this reaction the energy is liberated as heat and the cell warms up. But the heat originally came from the electrical energy used to split the water into hydrogen and oxygen.
Quite. In science there are always different ways of looking at and understanding what is going on. That is the beauty of it.
In terms of physics we have an energy balance:
[energy supplied] = (I^2) x R = [energy stored] + [energy dissipated]
So in our cell we find the electrical energy supplied is either stored internally as chemical energy or dissipated as heat. When the cell can accept no more charge then all energy supplied is being converted into heat. Internally the oxygen recombination cycle may be taking place, but when looking at the cell from the outside we don't need to know about that. Chemistry may happen, but physics cannot be denied.
