Hi,
I'm doing a project which is ran by a lead acid battery. It's one with Arduino pro mini which does have a regulator too.
So question is what regulator to use? I need it to be very low power when not doing anything. And of course to save battery when running.
It will be on sleep mode when idle so i dont want the regulator to use much current.
I've dig up couple results for me.
LM2937ET-5.0 seems to take 2mA and L4941BV takes 4mA if i read the datasheets right. And i do need a regulator to fit in a veroboard so smd is not an option.
And i prefer one with reverse battery protection which is available on LM2937ET, but not necessary.
Only load will be 16x2lcd and one or two relays. 100-200mA
So should i just use the regulator IN the pro mini or cut it away to use something more efficient? Supply voltage is 12V lead acid so 11.5V - 14V.
When trying to reach best efficiency, and you should when using battery power, you should not be looking at any linear voltage regulator but rather a switching regulator. The reduction from 12v to 5v using a linear regulator means that you waste more power in the regulator then the arduino board is using. The example below is of a switching regulator that will allow you the best battery running duration.
http://www.ebay.com/itm/QZO-Buck-Converter-Step-Down-Adjustable-Converter-Power-Module-Regulator-LM2596-/131143304399?pt=LH_DefaultDomain_0&hash=item1e88c004cf
Thanks for the reply!
I was reading the datasheet and there is a graph about Iq and load. If i read it correctly i see that under 5mA load it shuts down and uses only uA:s and switches off at about 16mA load? What happens between it?
I actually looked om the datasheet of pro minis regulator. Its a micrel 5219. Just wondering why not to use it?
If im right it takes about 350uA when load is 50mA and 80uA if load is 100uA. so if arduino could sleep using only 100uA then it would use only 180uA when idling.
And would it be ok to place a diode before the regulator to act as a reverse polarity protection?