Battery test - amazing

I have two alkaline batteries in series in my test product. Two days ago the current was about 12mA at about 0.9V. It is now about 4mA at about

0.8V and the current is about 3.8mA.

How can these batteries continue to supply that much current for so long?

Reply to
John S
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We know how. It's normal to get something out down to 0.8v.

NT

Reply to
tabbypurr

It is reading .875V while supplying 3.9mA. This has been going on for at least 2 days. What gives?

Reply to
John S

AA, AAA, or AAAA alkaline battery? An alkaline battery should be around 1.4V, not 0.9V Two batteries in series should be about twice that. I consider an alkaline battery dead at about 1.0V. For a AA alkaline battery, the capacity at around 10 ma is 2000 ma-hr. So, your two unspecified maker, model, and size alkaline batteries should run at 10 ma for: 2000 ma-hr / 10 ma = 200 hrs or about 8.3 days, depending on what your unspecified load can handle for a minimum voltage.

How? Well, you wind them up until the spring that runs the little generator inside the battery runs down. If that's unacceptable, perhaps you could fill in some of the blanks in your information?

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Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
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Reply to
Jeff Liebermann

AA - about 2000 mAHr AAA - about 1000 mAHr

Math happens...

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Rick
Reply to
rickman

Some days we are more jaded than other days. :-)

Mikek

Reply to
amdx

What size?

4mA * 50 hrs is 200 mA-hr. It's chemistry, you are at the tail end of the reaction, record numbers and make a plot (several plots.) (I'm not a battery expert...) George H.
Reply to
George Herold

In the marketplace, the difference between unsellable mediocrity and the next big thing is called "product differentiation", even if the real difference between the two is negligible. It's the same with answering questions on Usenet. Anyone with an engineering background could have produced a polite, accurate, detailed, and informative answer to the question. Only a few can produce an answer that is entertaining and thought provoking. Even fewer will spend the time to do so. I'll called it "Usenet reply differentiation". Again, the difference is negligible, but which would you prefer to read?

Full disclosure: I was looking at a shopping bag full of dead batteries that will eventually be dragged to the recycler. What a waste. There must be a better way. At one point, I thought of a tiny spring wound generator inside a AA package. There was even a sketch on my notepad. It's not practical and won't work, but it does redirect my overly fertile imagination in the right direction.

Half-full disclosure: I've attended many brain storming and problem solving sessions. In some cases, the participants all seem stuck in a rut, following the same fixed ruts in the roadway as their predecessors. To get them back on track and stimulate some genuine creativity, I'm usually the one to produce a succession of strange, absurd, or ridiculous suggestions. As with the tiny generator, these ideas won't work, but they do serve to drag people out of their personal ruts, and get them thinking again. It's like pressing the reboot button. If you can't think of a good solution, think of a ridiculous one. It helps.

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Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
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Reply to
Jeff Liebermann

If you follow ridiculous ones many times you get some winners among the junk. Sometimes it's worth it. Sometimes it's really worth it.

NT

Reply to
tabbypurr

What is the total integral under the I/t curve?

Should be around 3Ah for an AA if the current is suitably low and you can tolerate low voltages - although they may well leak at some point.

I have had genuine Duracells go bad like that after a while :(

Modern batteries are much better about not leaking than in the old days.

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Regards, 
Martin Brown
Reply to
Martin Brown

depends on your definition of the "old days." I've had several Duracells leak so badly that they completely ruined the device. All were WITHIN the expiration date on the battery. Don't remember ever having an alkaline cell other than Duracell fail so quickly and catastrophically in that mode. I'm not buying any more of them!

To their credit, they did reimburse for the devices with no hassle. Would have been different had the devices been no longer replaceable.

Reply to
mike

I have two AAA Duracells which came out of a used FM modulator I bought on ebay. One measures 29.1mV and the other 0.0mV but it isn't open circuit. I have never seen such flatness (flatitude? flatulence?) before.

Cheers

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Syd
Reply to
Syd Rumpo

Sorry I omitted that important info. They are AA cells.

Reply to
John S

Current this morning is 2.9mA at 0.84V.

Reply to
John S

On Fri, 4 Dec 2015 10:19:40 +0000, Syd Rumpo Gave us:

What are these posts? Flatitude platitudes?

Reply to
DecadentLinuxUserNumeroUno

That's happened much more lately - I have several "ancient" devices that never had battery leaks over 30+ years that have had to be opened, washed, resoldered etc. only a year or so after battery replacement. The crap that leaks out of a battery will wick itself through stranded wire all the way to where it's soldered on the other end - eats solder joints and board traces when it gets there.

I blame "industrial engineers" - optimize the process but ignore the consequences for the end user.

Not limited to Duracell - unless they're the end supplier for CVS, Radioshack etc.

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Grizzly H.
Reply to
mixed nuts

That is sometimes true but that's not how it works. The problem is that most people will look for a solution within their areas of expertise and experience, no matter what the problem. I've seen it in this newsgroup a few times, and probably commented on the effect. It's a natural thing to do, but rather limited when the best solution is something out of everyones areas of expertise.

I once attended a meeting where almost all of the participants were electronic engineers or managers with electronics backgrounds. Of course, they offered only electronic solutions to the problems presented, even though it was obviously a mechanical problem, that required a mechanical solution. By offering an absurd mechanical solution to the problem, I managed to get them redirected and thinking in the right direction. Had I argued that it was really a mechanical problem, most everyone would have gone on the defensive, justifying their favorite electronic solution to the death. In the end, they dragged in a mechanical engineer, who quickly offered a good solution. As a major side benefit, it was suggested that I really didn't need to attend such meetings in the future, which was fine with me.

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Jeff Liebermann     jeffl@cruzio.com 
150 Felker St #D    http://www.LearnByDestroying.com 
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Reply to
Jeff Liebermann

Yes, I understand that. These were well-used batteries but I decided to watch them for a while when they ran out of enough juice to keep the boost regulator going. So, maybe you did not notice that the starting point was 0.9V (series connected) and 12mA. The 0.9V is the voltage of both cells in series. Maybe you missed that point?

Reply to
John S

What is your point? You can get the last bit of energy from otherwise dead cells. Ok. Have you taken enough measurements to see just how much (or more accurately how little) energy you have extracted?

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Rick
Reply to
rickman

AHr is not so relevant as WHr. At these voltages the curve tails off to very low figures.

"Better" is not leak proof. I have had alkaline cells leak on me several times in low (but not zero) duty cycle devices. The great thing about Duracell is that they actually stand behind their guarantee. I called them about a radio controlled clock the batteries leaked in and they sent me a check for the replacement. I read the guarantee on Ray-o-Vac cells and they want you to mail the device to them. I haven't bothered with that so I don't know how hard it is to get a check. I think that is the point.

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Rick
Reply to
rickman

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