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kevj
- 16th October 2007, 03:41
Question about figuring expected battery life based on MAH.

I plan to run some CR2032 cells (3v, about 240 mah cells). These are usually used for very low current drain applications. My application requires a higher drain but short lifespan is probably acceptable - the applicaiton requires as micro-sized batteries as possible.

We'll be running 2 of them in serries to get 6 volts and approx 480 mah combined capacity.

Lets say my circuit draws 20 milliamps of current - can I do the obvious math here and figure an approximate lifespan of about 24 hours?


Someone told me the current from one cell forced through the other cell will cut the effective MAH in half because they are wired in serries. This didn't sound right but who knows.

Also, does the current draw itself impact the mah? For example, maybe it's 240 MAH under a load of .2 milliaps (the example in their data sheet), but when we crank it up 100 times to 20 milliamps, does that mah rating now get smaller because we're sucking the current much faster?

I would assume this would be the case if we draw enough to start generating lots of heat, but even at 20 mA, that shouldn't start smoking the batteries or anything.



Long story short, we're going to draw 20 to 30 mA, and I'd really like to get 15 hours out of a set of batteries if possible. We need at least 10 hours.

ronsimpson
- 16th October 2007, 04:16
The CR2032 was designed for currents less than 3mA. If you use it at 10X times that, you will not get all of the 240mah.

You assumed that the mah will double when using two batteries in series. That is not true. You will get 6 volts at 240mah. Not 480mah.

Data at
http://www.renata.com/pdf/3vlithium/DBCR2032.05.pdf

T.Jackson
- 16th October 2007, 05:05
We'll be running 2 of them in serries to get 6 volts and approx 480 mah combined capacity.
Lets say my circuit draws 20 milliamps of current - can I do the obvious math here and figure an approximate lifespan of about 24 hours?


Don't think it works like that (Not linear) Probably best to consult the manufacturer's discharge curve.

kevj
- 16th October 2007, 08:51
Hmmm... well that's not good news.

Can anyone recomend other batteries I may consider that are very small?

Also, is there an easy way to jump up voltage without loosing a lot of power in the process? There are other 1.5v and 3v solutions, but I need 5 volts.

I'd be really surprised if in the current advancement of really small electronics that the most I can draw out of something this small is 3 mA.

Suggestions?

mackrackit
- 16th October 2007, 12:08
I do not have a battery suggestion other than using button cells series / parallel to get the voltage and amps needed.

But why 5 volts? If this is PIC based there are many chips that will run at 3 volts and less if a 4 mhz osc is used. The current consumption is lower too at the lower voltages. Supporting circuitry may have a low voltage solution also.

ronsimpson
- 16th October 2007, 15:51
I use boost up converters like the ZXSC100 in high volume. We boost 3 volt batteries up to 5 volts. There are many different types of boost up ICs Some of them work down to 0.8 volts input.
http://www.zetex.com/3.0/pdf/ZXSC100.pdf
http://www.zetex.com/3.0/pdf/zxsc410-420.pdf

tenaja
- 16th October 2007, 17:06
I use boost converters from www.maximic.com . They have some that are pretty small with minimal external parts. (That means less board space.) The one I use is about 87% efficient going from 1.2v to 3v.

When adding batteries, if you combine them in series, you add the voltages. When combining them in parallel, you add the capacity. If you want to add both the voltages and the capacity, then you have to add a set in parallel and a set in series.

You should look at your data sheet for answers to your particular questions. Most batteries are rated at a mimimal discharge current, and increasing the discharge current will decrease the capacity.

Note that if you can figure out a way to run at 3v instead of 5, your batteries will last up to 40% longer--which might allow you to eliminate that second battery. The reason is that your current draw will drop significantly--just look at the PIC data sheet in the Electrical section for current draw at various voltages. When working with small packages, it could be worth the effort. At the worst, you might be able to run the PIC at 3v, and your other peripheral at 5.