Pages: [1]   Go Down
 Author Topic: Battery Requirement Calculations - First Arduino Project  (Read 796 times) 0 Members and 1 Guest are viewing this topic.
Offline
Newbie
Karma: 0
Posts: 8
 « on: June 08, 2011, 02:01:47 am » Bigger Smaller Reset

Hi,

I have my first arduino project running! Very exciting.

With my arduino nano powered by USB and my multimeter between +5v and Ground on the Arduino I get readings of:
- 4.7v
- 3.9mA

1 ) If I want my device to run for 4 hours how do I calculate the required battery? (is it just 3.9mA x 4 hours = 15.6mAh???)
2) I need to find a flat battery like a cell phone battery or coin battery to run this device. Do you have any recommendations?
3) If I combine two of these '3v / 2mA / 17mAh' batteries will that work? http://australia.rs-online.com/web/7020761.html or will I need to limit the 6v back to 5v some how?

Thanks for any guidance.
Cruzah

PS. Device is a Arduino Nano, 16 x 2 LCD (no backlight), and a couple of buttons.
 Logged

Leighton Buzzard, UK
Offline
Edison Member
Karma: 21
Posts: 1339
 « Reply #1 on: June 08, 2011, 02:16:10 am » Bigger Smaller Reset

it may be worth checking the current when your project is doing whatever it does
you mention a display device - can you measure the current while the display is changing?

battery capacity is amps * hours, but you should always allow some slack as not all batteries are equal, and the voltage may drop towards the end of its life

two batteries in series will give double the voltage - probably not what you want unless added together they give 5 volts
two batteries in parallel gives problems as they may not discharge evenly - and end up fighting

better to go for one battery that meets your needs

the battery in your link - hmm hard to see how it could be used!

maximum continuous current is just 2 mA
and it's only 3 volts!

HTH

3 or four of these in series might work

why not just use 3 AA batteries? - simple, cheap
 « Last Edit: June 08, 2011, 02:19:46 am by mmcp42 » Logged

there are only 10 types of people
them that understands binary
and them that doesn't

Offline
Newbie
Karma: 0
Posts: 8
 « Reply #2 on: June 08, 2011, 02:56:04 am » Bigger Smaller Reset

Thanks for the information. I will keep looking.

I cant use 3 x AA batteries simply because they are too thick to fit in my case. I need batteries no more then about half the height thickness of an AA battery.
 « Last Edit: June 08, 2011, 03:36:04 am by cruzah » Logged

SW Scotland
Offline
Edison Member
Karma: 17
Posts: 1377
Arduino rocks
 « Reply #3 on: June 08, 2011, 03:08:54 am » Bigger Smaller Reset

In that case use a half-sizeAA battery.  These are litium based and provide 3.6 volts.  So either 1 of these or several in parallel will meet your dimensional needs.  Google 1/2AA for information of these devices.

One query on your use of a test meter.  You say it was across +5 and ground and the system supplied 3.9mA.   Are you sure you know what you were doing or did you miss something out in your description.  You cannot measure current by applying a current measurement across a voltage supply, the meter MUST be in series with ONE of the power leads.
 Logged

Leighton Buzzard, UK
Offline
Edison Member
Karma: 21
Posts: 1339
 « Reply #4 on: June 08, 2011, 03:09:03 am » Bigger Smaller Reset

3 AAA then
crikey what size case is it

oh and what's the project??
 Logged

there are only 10 types of people
them that understands binary
and them that doesn't

Offline
Newbie
Karma: 0
Posts: 8
 « Reply #5 on: June 08, 2011, 03:37:47 am » Bigger Smaller Reset

Thanks again for the input.

I connected the multimeter in series to the +5 and ground.

Sorry, height isn't the issue with AA its the thickness. I'm using an old calculator as my case.

Im just trawling the web for batteries now.
 Logged

SW Scotland
Offline
Edison Member
Karma: 17
Posts: 1377
Arduino rocks
 « Reply #6 on: June 08, 2011, 05:19:11 am » Bigger Smaller Reset

+5 on supply (battery or whatever) to ground on supply is OK for measuring voltage  -  but NOT current.  To measure current you need to connect meter +ve lead to +5 on supply and meter -ve lead to the +ve terminal of your device (arduino or whatever).  The device -ve lead connects to the supply ground
 Logged

Austin, TX
Offline
Faraday Member
Karma: 71
Posts: 6141
Baldengineer
 « Reply #7 on: June 08, 2011, 10:07:50 am » Bigger Smaller Reset

Quote
3.9mA

A bare Arduino board typically draws on the order for 20-30mA.  Did you properly measure current:  break the current path and put the multiple in between?  (or did you try measuring current in parallel like voltage?)
 Logged

Capacitor Expert By Day, Enginerd by night.  ||  Personal Blog: www.baldengineer.com  || Electronics Tutorials for Beginners:  www.addohms.com

 Pages: [1]   Go Up
Jump to: