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Basic Electronics » Understanding battery life

 December 02, 2009 by mikedoug Let's add another tool to our belt: With the 5V regulator, a 9V battery, and a measured current usage I, what is the expected lifetime of the battery? Thanks, MikeDoug Hi MikeDoug, Let's take an example Duracell Ultra Alkaline 9V battery and grab the technical datasheet. Look at the graph titled "Delivered Capacity vs. Power Drain" on the second page. This has a few different lines, each with a voltage label, which indicates that the line indicates how much capacity is delivered before the battery voltage reaches a certain level. (Depending on your particular 5V regulator, the output will stop being 5V when the input voltage drops below a certain level -- somewhere in the 7 to 7.5V range for the 7805, but can go as low as 5.5V with some "low dropout regulators" you can find mentioned in other forum posts. So we'll take the 7.0V level on this chart). At low discharge levels, this indicates about 0.5 amp-hours. When you discharge faster, the overall capacity decreases a bit too. But say at 10mA, you've got very roughly (500 milliAmp hours) / (10 milliAmps) = 50 hours of run time. Different 9V batteries can vary dramatically, and 500mAh is probably on the high end, down to maybe 50mAh for non-alkalines. Here's some data. If you're trying to optimize for battery life, you can't ignore the "quiescent current" of the voltage regulator itself. For the 7805, that's about 6mA used just by the 7805 itself -- that's with zero current going through the output! There are a few other things to do to increase battery lifetime: Lower operating voltage. This reduces the current, and the power, and for example may let you run directly off two or three AA's in series. (A good alkaline AA battery can be in the 2000-3000mAh range, which is far more than even the best 9V's. Batteries in series do NOT have their capacities add, which is why you are supposed to use the same kind of battery especially in series combinations, and to keep pairs together forever when using rechargables.) Reduce clock frequency. Look at Figure 29-2 on page 317 of the ATmega168 datasheet. Current is basically proportional to clock frequency. (But then again, the same unit of computation will take proportionally more time, so you may not end up saving anything, depending on your application.) You also might have to reduce this because of the lower operating voltage -- see Figure 28-2 on page 306 of the ATmega168 datasheet. Make good use of sleep modes. Of course you've already been looking into this a lot based on your other forum posts, but just wanted to have this on the list in case others come across this post in the future. Hope that helps! Mike Hi Everybody, I am interested in lowering my operating voltage down to 3V for the main reason of extending battery life. As Mike mentioned above, AA batteries hold 2000-3000 mAh. I need my device to operate on batteries so that it is portable. I found a display I think will work (http://www.newhavendisplay.com/index.php?main_page=product_info&cPath=2_87&products_id=1880) and it looks like the controller (SPLC780D‐01) is compatible with the HD44780 controller that comes with the nerdkit. Further, I will address the drain the display has on the system with an on/off switch of some sort so that the display only draws current when I need to see the information on it. My question is what are the ramifications and requirements of lowering my operating voltage? It looks like I will have to lower the frequency to <10 Mhz? I am not sure how to fascilitate this change beyond changing the oscillating crystal. I believe I also have to change my ADC and beyond following the instructions that came with my Nerdkit, I am not really sure how to address that. I am sure there are other issues that have to be addressed when lowering the operating voltage of the MCU beyond what I have mentioned. Thanks, Devin

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