# Calculating Battery Usage

Discussion in 'General Electronics Chat' started by b1g5l1ck, Feb 13, 2011.

1. ### b1g5l1ck Thread Starter New Member

Feb 12, 2011
1
0
Hi There

Please could someone tell me the estimate battery life with the for the following:

2 x cr2032 batteries - 220mAh each

1 x 2mm led - 5mA usage

Approximately how long would the batteries last for if the led was turned on and ran continuously?

Thanks guys

2. ### cjdelphi New Member

Mar 26, 2009
272
2
if the batteries could deliver 220ma, i doubt it, these batteries lithium or lipo? 4.2v or 3.7v? so realistically speaking, if it's rated 220mah it's probably more like 180mah or if you're lucky 200mah.

So let's just assume it's what it states.

and these 2 batteries? you hooking them up in Parallel or in series? Series would mean power dissipation waste, you're kinda vague really sorry. 2*220 440mah (in parallel) CR2032 im sure are 3v i have a couple of my Fenix these are really quite bad for energy density.. anyway

2*220 (In Parallel) simple answer would be, 220 or 440ma / 5ma but add a resistor and we get something more like this

What's the LED and voltage drop? pluck a number here, 2.6v depends on the bandgap and the LED used, infra red/ red require a lot lower voltage than say the blue or green (something to do with the bandgap it went over my head)

3-2.6 = 0.4v.
3/0.005(voltage will drop after use decreasing brightness) = 600ohm resistor

So hook your LED to a 600ohm resistor for 5ma any wattage resistor will do for such a low amount.

Before anyone shouts my math is wrong, it could be but i have a feeling 600ohm will give you around 5ma maybe a few more or a few less milliamps depending on the battery voltage.

3. ### cjdelphi New Member

Mar 26, 2009
272
2
yeah the answer to the question... 440/5 = runtime in hours (88 roughly hours depending on the how good the battery really is.)

4. ### JMac3108 Active Member

Aug 16, 2010
349
66
Here is the general procedure to estimate battery life.

(1) Convert your battery rating to Watt-hours by multiplying the battery amp-hours by the nominal battery voltage.

(3) Divide your battery rating in watt-hours by the laod in watts and you will have an estimate of your battery life in hours.

Hope this helps

Jul 7, 2009
1,585
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The "easy" way is to do what cjdelphi did in his/her second post: divide the total charge by the charge per unit time and get the time. The battery's rating is usually given in mA*hr, which is a charge; the charge per unit time is a current, so if the two batteries of 220 mA*hr were in parallel, then the life at 5 mA should be (440 mA*hr)/(5 mA) or 88 hr. See how the units cancel?

Now, that's an estimate; in fact, it's better viewed as a guess. A better estimate is gotten by knowing the discharge curve for the battery. The attached graph shows the output voltage of an AA battery (vertical axis) as a function of time (horizontal axis) when a constant current of 500 mA is drawn from the battery. The ideal battery would be a flat horizontal line. But real batteries behave similar to what's shown in the picture. Now, what's important in your application is how long the battery will put out a voltage high enough to keep the LED turned on. If you know this value, then you can find that value on the vertical axis and find the corresponding value on the time axis -- that's your estimate of the life. Alas, such discharge curves don't just typically fall in your lap -- you have to measure them or have a battery manufacturer give you the one you need. The attached graph was measured with a B&K 8500 DC Load.

Of course, you must realize that these things are stochastic variables, so an engineering design would want to know something about the distributions of the values. Then you could make a better estimate of what you were looking for. I'll assume that's not the case here and you're just looking for an estimate that's within, say, a factor of two or so. And there are other things to take into account, such as temperature, storage time, battery chemistry, etc.

An AA alkaline battery has a mA*hr "rating" of around 2800. Thus, using the estimation method given in the first paragraph, you'd estimate the battery could supply this current for 2800/500 hours, or over 5 hours. So the estimate's too high by about a factor of 2, as the battery reached 0.5 volts in 2.4 hours. This is almost certainly due to the fact that the rating of 2800 mA*hr was stated for a lower current draw.

6. ### Audioguru New Member

Dec 20, 2007
9,411
896
Energizer spec's a CR2032 lithium disposeable battery cell at 240mAh when the load is 15k ohms (0.19mA at 2.9V) and the battery voltage ends at only 2.0V.
They show a waveform starting at 3.25V then the voltage dropping like a rock immediately then flattening out to 2.8V in 2 seconds with a 30mA load. When the load is removed the voltage gradually increases to 3.1V in another 1.5 seconds. The test is then discontinued.

So they do not show how long it will last with a constant load current higher than 0.19mA.
You also must find out at which dropping battery voltage the LED becomes too dim.

7. ### JMac3108 Active Member

Aug 16, 2010
349
66

The procedure you describe only works if the load is specified at the battery voltage. This is not always the case and is in general not the case unless your system is designed with linear power supplies.

As an exdample you might have a load (in mA) that runs at 3.3V from a switching regulator. The battery is at a different voltage, for example two LiIon batteries in series, 7.4V nominal. This is why the normal engineering procedure is to convert the battery capcity and the load to Power (in watts) first, then perform the calculation.

8. ### Audioguru New Member

Dec 20, 2007
9,411
896
The capacity of the tiny battery cell is spec'd at a very low current. The current in this LED will be 26 times higher so it is very difficult to calculate how long the battery will last because the current vs duration is not linear.
Also the voltage drops as the battery runs down which causes the current to drop and the LED get dimmer. When is the LED too dim?
When the current decreases then the battery lasts longer. It is a nightmare of mathematics and might be worth speding a dollar for a battery and simply time its survival.