Hello,
I would like to use a 9V battery at 580mAH to power my Arduino which consumes 620mWh/day.
Does anyone have experience and if this is possible?
I would like to know before I order.
Thanks guys,
Cookie
It will work it will only drain in less than a day
At first glance I'd say yes. You'd have roughly 5220mWh in the 9V battery which should run your 620mWh/day load for several days.
When you calculated 620mWh/day, however, did you include the losses in the linear regulator? I don't know which Arduino you're using, but in the Uno it'd be IC1 -> https://www.arduino.cc/en/uploads/Main/arduino-uno-schematic.pdf
Linear regulators are cheap and simple, but they are not efficient. It will consume (9V - 5V) * the current drawn from the 5V supply.
Yes i didn't account for the linear losses of the regulator
I'm using an Arduino leneardo.
I'm not sure how to calculate the losses due to the regulator, if someone can provide some insight, it would be much appreciated.
I don't know if all of your 620mWh/day are powered by the +5V. If they are, however, then the losses in the regulator would be (9V - 5V) * 620mWh/day / 5V = ~500mWh/day. If you are interested in where that comes from, you can read about it on the Wikipedia page -> Low-dropout regulator - Wikipedia
If it bothers you that you're wasting half your energy in your regulator, you can use a buck converter instead. Something like this ->
MC34063-DC-DC-DC12V-DC-9V-to-DC-5V-Buck-Step-Down-Power-Module-Replace-LM2596
It's not going to be 100% efficient, but it's going to be much better than the linear regulator.
Thanks for your quick response.
I have considered using a buck converter module, however I want to limit the size of my project.
If this is the case, would using less voltage (ie. 4 series AA batteries) have less losses because of the lower voltage?
However, the size is also a main issue. The size of a 9V battery is smaller than 4 series AA battery holder.
I have one component that is using 3.3V, but the three other components are using 5V.
I will look indepth into the linear regulation loss calculations because I really need to consider that issue in my design.
Thanks.
Cookie.
Zkryvix:
It will work it will only drain in less than a day
Seems you didn't notice one number is Amps, the other is Watts.
580mA current @ 9v is 5220mW power, so his battery energy capacity is 5220mWh. He uses energy of 620mWh a day, so there's 5220/620 ~ 8 days' worth of energy in there, as BigBobby points out.
(Always assuming the OP didn't confuse his A with his W 
and that the battery can provide enough current at any instant to drive the device.)
A basic Arduino powered by a 9 volt battery has a current consumption of 50mA. Here are suggestions of what you can do to optimise that:
The power consumption quoted in the OP is 620mWh/day which translates to 25.8 mW, which at 9 volts is 2.8 mA (if my arithmetic is correct) which is remarkably low if the standard voltage regulator is in place.
cookie89:
If this is the case, would using less voltage (ie. 4 series AA batteries) have less losses because of the lower voltage?However, the size is also a main issue. The size of a 9V battery is smaller than 4 series AA battery holder.
I have one component that is using 3.3V, but the three other components are using 5V.
In the case of 4 AA batteries, you would actually have even fewer losses because you wouldn't need to use the voltage regulator at all. See how they connected them here -> http://www.instructables.com/id/Power-your-Arduino-Uno-with-four-rechargeable-AA-b/
They used NiMH cells which are a good idea, as the 4 cells in series will always be between 4.5V and 5.5V. According to Figure 29.6 of the ATmega32U4 datasheet, you need to stay in this voltage range in order to run your Arduino at its standard frequency of 16MHz.
Technically, section 29.1 of the datasheet lists the Maximum Operating Voltage of the ATmega32U4 as 6V so you could use Alkaline cells. Personally I'd prefer to not do that since the Typical Characteristics in section 30 only shows test data up to 5.5V. Also, 4 Alkaline cells in series could be slightly over 6V. You can see discharge curves for Alkaline and NiMH cells here -> Discharge tests of Alkaline AA batteries 100mA to 2A
It's true that 4AA will be roughly twice as big as a 9V, but they will also contain roughly 2x the energy (4 * 1.2V * 2000mAh = 9600mWh). As a general rule with batteries, when you have twice the chemicals, you're going to get twice the energy. Because you wouldn't be using the linear regulator, you'd actually get 3x the runtime too (5220mWh/(620mWh/day + 500mWh/day) = 4.7days for the 9V vs 9600mWh/(620mWh/day) = 15.5days for the 4AA).
If the 9V is the size you want and 4.7days is enough runtime for you, however, then feel free to use it. In this Green world we live in today wasted energy gives me a bad feeling but in reality your project working is all that matters.
Also, yes...you do need to account for the energy required for the 3.3V regulator (U3 on the Leonardo schematic. If you read the wikipedia link, you should be able to calculate that now.
6v6gt:
A basic Arduino powered by a 9 volt battery has a current consumption of 50mA. Here are suggestions of what you can do to optimise that:
Gammon Forum : Electronics : Microprocessors : Power saving techniques for microprocessorsThe power consumption quoted in the OP is 620mWh/day which translates to 25.8 mW, which at 9 volts is 2.8 mA (if my arithmetic is correct) which is remarkably low if the standard voltage regulator is in place.
You calculated the average current over the day. I believe that the OP's loads are only being used for a short period of time each day. He didn't provide information about his peak currents (and as JimboZA pointed out he would need to ensure that his battery can supply the peak currents required).
Also, the 50mA number is for the Arduino Uno, which has two ATmega ICs on it to allow USB communication. The Leonardo only has one, and Table 29-1 of the ATmega32U4 datasheet says the maximum Power Supply Current when Active @ 5V/16MHz is 27mA.
It seems the OP must be planning to use power reduction techniques to achieve the energy/day numbers in his post. The link you provided seems to be an excellent guide for using those power reduction techniques.
If size is that big of a concern, use a single AA (or two) and an efficient 5V boost converter from www.pololu.com.
Feed 5V into the 5V/power header. Still get 2500mAH capacity vs 300-400 from 9V battery.