28
\$\begingroup\$

I am currently using AA batteries in my projects but would like to look at cell batteries for my next project, due to their compactness.

Apart from the physical appearance, how do cell batteries differ from the more chunky AA batteries? (I am guessing the mAH would suffer for example?)

\$\endgroup\$
26
\$\begingroup\$

Let me sum up your limitations of a CR2032:

  1. 10mA is about the max current you want to pull from a single, it is easy to put them in parallel, but a large amount of testing(more than 2000 batteries worth) has confirmed this.*
  2. They can be purchased to have 400mAh, the less current you pull the closer to this it will be, pulling more then 1mA decreases this a decent bit.**
  3. Under a 1mA load they will decay all the way to 1.5V before they fail, they will be at 2.7V almost right away.
  4. You can measure an almost full voltage on them with a multimeter when they are dead. This is solved by placing a load on them.***
  5. If you are lazy, it is very easy to tell how much charge they have left by how much they make your tongue tingle. Your tongue acts as the load and measures. This is probably by far the easiest way to test them, although it does pull a decent bit of current.

I think Thomas wrote a good answer, I just thought it might be helpful to give some details of the coin cells since it seems you have used AA quite a bit.

*Wikipedia says up to 15mA pulsed, but we confirmed that up to 1mA shows a nearly consistent capacity.

**Wikipedia shows a standard that is a bit lower, but my company would always purchase 400mAh or 450mAh CR2032. When you buy a "standard battery" you can expect 200mAh it seems.

*** People often will measure batteries without load, when someone tells you on a project that ran out of power early, ensure their original battery measurement was under load, very easy mistake.

\$\endgroup\$
  • 1
    \$\begingroup\$ Love the tongue test. Just tried it :D \$\endgroup\$ – Mr. Hedgehog Jan 7 '11 at 13:18
  • 4
    \$\begingroup\$ @Mr. Hedgehog, you can touch a new one, touch a dead one, and you are fully calibrated for determining the age of a battery. It is very easy, very effective. \$\endgroup\$ – Kortuk Jan 7 '11 at 14:16
16
\$\begingroup\$

I'll compare against alkaline AA's and lithium CR2032's as these are the most likely candidates to replace each other. However, they have several differences:

  1. Coin cells are usually 3V, instead of 1.5V.
  2. Coin cells can only deliver a few mA before the voltage drops too low - you can power an LED with them at most.
  3. Coin cells have significantly less capacity, you'd be lucky to get 100mAh out of a coin cell whereas 2000mAh isn't unusual for a good AA and even the cheapies will do >1000mAh.
  4. Coin cells are usually more expensive per cell compared to an AA cell.
  5. Their terminal voltage characteristic is to keep a stable voltage (>2.7V) for a long time before dropping very quickly, whereas alkalines generally fall to ~0.9V before dropping out completely. This can create difficulties for low battery warning circuits. Here's an image that shows what I mean: alt text
\$\endgroup\$
  • 1
    \$\begingroup\$ I can get the chinese made CR2032 at 3USD per strip of 8 (2USD each, weird economics, I know). But the capacity and discharge are very tight constraints in your design. \$\endgroup\$ – J.P.Wack Oct 26 '10 at 17:12
  • \$\begingroup\$ I picked up a blister pack of 12 cells for £3.50, or about $5.50, which is 45 cents each, but I can pick up AA alkalines for about half of that. \$\endgroup\$ – Thomas O Oct 26 '10 at 17:26
  • 1
    \$\begingroup\$ We have done a large amount of testing and I have to disagree with 3 and 5. I do not know AA costs well enough. \$\endgroup\$ – Kortuk Oct 26 '10 at 19:47
  • 1
    \$\begingroup\$ These were standard. It has a curve close to the one you have the lower the load. as the load increases it goes closer and closer to linear. \$\endgroup\$ – Kortuk Oct 26 '10 at 20:07
  • 1
    \$\begingroup\$ Yes, the way that a battery fails is not by a decrease in voltage, but an increase of internal resistance. \$\endgroup\$ – Kortuk Oct 27 '10 at 15:19
1
\$\begingroup\$

This isn't a direct answer to your question, but it's related. I assume you're using multiple AA batteries in series (in order to achieve a higher voltage for your circuit to operate). In that case, you might consider using a buck/boost circuit to boost the voltage from one AA battery. You'll pay a penalty in terms of how much current you can supply, but compared to a coin cell, it will be better. Sparkfun sells a kit that will take in down to 0.3v and output 5v at 500 mA, for example.

\$\endgroup\$
  • \$\begingroup\$ If even one AA is too big, you can use an AAA cell. An AAA cell is both shorter and slimmer than an AA. However, an AAA has less than half the capacity of an AA. Consider using a rechargeable AAA to save money. \$\endgroup\$ – unforgettableid Jan 29 '13 at 7:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.