Electronic – arduino – Battery powered device hardware design tips

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I'm working on a prototype product(Wifi motion sensor) that will be powered from a battery source, it has a mega328(arduino bootloader, with sleep mode on and wake on interrupt,and WDT in case it hangs up), a PIR motion sensor , and a CC3000 break out board.

Power saving is done in software, but what type of battery should I use? I initially used 2AA with LT1302-5 boost converter, but I think it was not able supply enough current for the CC3000 module, it could not start the SSID scan also the LED on the break out board dimmed out a bit.

Now I'm thinking of using 4AA and using a 5v LDO regulator.

How do I go about it?, what other options I have, I need 5V to operate the circuit. I've never designed a battery powered project before. Should I use different type of battery,LiPo perhaps?

Peak current of 150mA is drawn when communication is going on, draws and 10mA when its asleep.
Let me know if you need more details.
Circuit Diagram
enter image description here
The LT1302 IC is soldered exactly as in its data-sheet.
Source code
https://github.com/hemalchevli/Sketchbook/blob/master/PIR_LP/PIR_LP.ino

This is how to looks

enter image description here

Best Answer

You need to be certain on your voltage and current specifications.
You say 150 ma peak load.

If properly configured The LT1302-5 will easily meet your stated need. If it is failing to do so there is either something wrong with your circuit or your load is far greater than you state. The Digikey parametric selector suggests that there are 150 mA and 600 mA Iout versions but this does not appear to be suppoted by the datasheet.

The LT13o2-5 will provide up to 600 mA at 5V out. The graph below (from page 1 of the LT1302 data sheet shows that peak efficiency with vIN = 2v is achieved at Iload ~= 200 mA - above your stated requirement. and they show results up to Iout = 600 mA.

The battery current will be ~+ Iout x Vout/Vin x 1/Z
where Z = converter efficiency (0 <= Z <= 1).
For say 80% efficiency, 2V in and 5V out and 150 mA load, I battery
~= 150 mA x 5V/2V x 1/0.8 =~ 500 mA.
That is well within the capabilities of Alkaline AA cells or of rechargeable NimH cells.
Battery life at 150 mA ~= Battery_Capacity / Battery_Current x K
where K is a derating factor 0 ...1 which drops as load increases and as the battery is used (due to rising internal resistance).

enter image description here


4 x AA cells will provide 5V for only a small part of an Alkaline battery's life and not at all with NimH.
5V/4 = 1.25V
You need at least 1..25V/cell to regulate it down to 5V. Alkaline and Nimh can be considered to be 'dead' at 1V.
So 5 cells and an LDO (low dropout regulator) would be a minimum.
Valkaline_new ~= 1.65V, so initially 5 cells would give you 1.65 x 5 = 8V +.