I'm building a launch controller for a model rocket, and I've worked out that I need to draw 2 A for no more than 5 seconds. Given most commercial 9 V batteries can give me ~500 mAh, would a 9 V alkaline smoke alarm battery give me what I need?
Electronic – Drawing 2.0 A from 9 V battery for 5 seconds
batteries
Related Solutions
You are very close. The average power is a very accurate way to do this given that you are not pulling such a high current that the effective capacity of the battery fluctuates.
Batteries, Batteries, and More Batteries
There is one very important term, and that is the self discharge rate of the battery. This is dependent on chemistry, but lets say you get a nickel-metal hydride. The self discharge rate is "20% or more in first 24 hours, plus 4% per day thereafter" if it is not a low self discharge rate NiMH, which still discharges around 25 or so % a year.Lithium batteries have some of the best characteristics for self discharge rate and my experience supports this fact. I think battery university has a great site to discuss many different battery characteristics and I often point people there to learn about batteries when they are starting to work with them. If you want to compare battery discharge rates they have an entire article discussing the phenomena.
This is a bit around the point, but I always try to make this point, when you measure battery voltage you need to have it under load. This varies with chemistry, but it is paramount in lithiums. I had a coworker placing bad coin cells in our devices and using them because the coin cells showed almost full voltage with no load. Under a load of any amount(10kohm aprox .2mA) they were flat dead.
Your Microcontroller and You
As you are dealing with using the manufacturer sheet on leakage current there are also many different issues you will have to deal with to keep to those specs that are probably also work thinking about. The biggest I have seen is a floating input. Many engineers will leave unused pins as inputs thinking, "Hey, what harm can this do?" Quite a bit if you are talking microamps. A floating input will have its transistors changing state constantly and the fluctuations cause a power draw difference. We once had a reduced lifetime in a product because we had an error that left 2 pins floating causing our standby current to more then double on our MSP430. You need to drive all of your pins to output and let them hold a state.
It is easy to miss when doing these calculations things like wakeup time. I seem to remember our MSP430 had a non-negligible wakeup time if you were doing it very often. It also had a larger power pulse for just a moment as it came online. Our little homespun RTOS had to try to take this into account and if the shutdown was less then X milliseconds we skipped it with NOPs and saved some power.
If you are looking at a very long life product, you are going to need conformal coating. The oils in your skin are not an issue immediately, but with time they form a lightly conductive material on your board. Conformal coating protects your board from this little current sucking side affect.
Read any app notes they have about low power operation, it probably covers issues like the pins need to be held as output and many other important and useful facts.
Last but not least, Dont let yourself get relaxed just because you have read the app notes and everything seems okay after a week of running your product, you have to do as clabacchio says, you must measure and make sure. You debug your code normally, this is part of it, you need to find out if you made a mistake that is causing your idle current to be mAs instead of uA or even just if you did what we did and a pin is floating on accident. Make sure you use buffered measurements when you do this, if you have a large leakage on your device taking the data you can make a mountain out of a molehill when testing. Also, never forget about pullups, they are little power hogs if you are not careful.
0.7 means 0.7 amps! (i.e. 700mA). I would be very worried if it was drawing 7 amps :)
You are nearly there with your calculation though: with two AA batteries in series, the total capacity is still 1800mAh. If you had it in parallel, then it would be 3600mAh (but batteries in parallel is not a good idea in general, the exception being lead acid batteris and similar, under certain conditions). So 0.7 amps with 1800mAh capacity means it would last 1.8/0.7 = 2.57 hours (2 hours and 34 minutes)
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Best Answer
An Estes rocket engine squib looks about like this:
(The above picture comes from a document I wrote on the topic.)
Estes provides very clear all-fire specifications for their squibs:
This implies an average of \$\sqrt{\frac{\left[\frac{\frac12\:\text{J}}{50\:\text{ms}}\right]}{\frac23\:\Omega}}\approx 3.9\:\text{A}\$.
As a \$9\:\text{V}\$ alkaline battery will have about 3 times that resistance, internal to the battery. This means that the supply voltage must be \$V \approx 10.4\:\text{V}\$. If you go review the charts at this site, Discharge tests of 9 Volt transistor radio style batteries, I think you'll find that even at just \$1\:\text{A}\$, a fresh \$9\:\text{V}\$ battery might deliver about \$6.5\:\text{V}\$ for a short time. (That chart on that web page is consistent with the figure of about \$2\:\Omega\$ for the internal battery resistance, fresh.) If you were to attempt to draw four times that much, you can be pretty sure that it won't meet the all-fire specification. Not even close.
If you do the calcs, you can perhaps consider adding a \$22\:\text{mF}\$ capacitor with a voltage rating of over \$10\:\text{V}\$, placed in parallel to the \$9\:\text{V}\$ battery. That may work. But they do cost some money and they aren't small.