Hey guys i need help selecting a battery that would work most efficiently. The battery is powering a 200mW laser. Currently i'm using a 12v battery with a lm1117 voltage regulator. the laser runs on 3v so the voltage regulator gets very hot because there is a huge voltage difference.

So would it work better if i buy a 6v battery? my only concern is then the dropout voltage of the voltage regulator. the data sheet says its 1.15 v so (6v - 1.15v = 4.85v). so i have very little room for the battery to get flat. for example when the battery voltage goes down to 4.15v then the voltage to the laser will start dropping below 3v

My other option is to keep the 12v battery and get a buck voltage regulator like this one https://www.robotics.org.za/index.php?route=product/product&path=59_146&product_id=256 Although im not sure which is my best option. My goal is for the battery to last as long as possible

I also found this cheap switching voltage regulator? would this work? http://za.rs-online.com/web/p/switching-regulators/7773291/

im actually confused between linear, switching and buck voltage regulators?

If you are concerned about 4.85 volts from a 6 volt battery not being enough then you need to reconsider your understanding about battery abuse. At that voltage the battery is well on its way to that great battery store in the sky.

What current does the laser draw and what capacity of battery are you using. In other words battery duration is based upon AH capacity and load current.

I take it you appreciate that a 12V 5AH battery will last no longer than a 6V 5AH battery in your application.

jackrae: If you are concerned about 4.85 volts from a 6 volt battery not being enough then you need to reconsider your understanding about battery abuse. At that voltage the battery is well on its way to that great battery store in the sky.

What current does the laser draw and what capacity of battery are you using. In other words battery duration is based upon AH capacity and load current.

I take it you appreciate that a 12V 5AH battery will last no longer than a 6V 5AH battery in your application.

Sorry I do not understand what u mean by the battery is well on its way to the store in the sky?

And the laser is a 200mW laser so it draws 0.066 amps. Or 66milliamps

calvingloster: Sorry I do not understand what u mean by the battery is well on its way to the store in the sky?

If you're describing a 6V lead acid battery, drawing it down to 4.85V would be killing it. ~5.25V would be the lower limit. After 6.0V there's only about 10% capacity left so it's not worth it to take it much lower than that, and the lower you take it the less life it will have.

Chagrin:

calvingloster: Sorry I do not understand what u mean by the battery is well on its way to the store in the sky?

If you're describing a 6V lead acid battery, drawing it down to 4.85V would be killing it. ~5.25V would be the lower limit. After 6.0V there's only about 10% capacity left so it's not worth it to take it much lower than that, and the lower you take it the less life it will have.

Sorry again I'm not too sure what u mean, do you mean that if I use the battery till the voltage drops so low I am damaging the battery?

calvingloster:
do you mean that if I use the battery till the voltage drops so low I am damaging the battery?

yes.

So what am I to do? Must I just make sure I never use the battery until it gets that flat?

Basically that's correct. If you use the battery until it's dead then you have indeed shortened its life. the exception to this is NiCad which will tolerate running completely flat.

calvingloster: So what am I to do? Must I just make sure I never use the battery until it gets that flat?

Yes.

At the very least you need to measure the battery voltage on the Arduino and go into a low-power mode when it reaches danger level (ie. shut down the laser, etc).

Even better is to power the entire system via a MOSFET and drive the MOSFET gate from a comparator so that the MOSFET shuts down when the comparator input reaches a certain voltage level.

Ok that makes sense. One question though. If I'm not up for that right now. Will I be better off using a switching voltage regulator? Will my battery last longer? Compared to when I use a linear regulator

Basically yes (or maybe)

A switched mode regulator will be around 80 to 90% efficient, but that depends to some extent on the drive voltage. Input current depends upon input voltage whilst output current depends upon the specific load demand. Best to look at output wattage to determine input wattage.

Linear regulators are much less efficient. For example if you only want 3 volts and your battery is 12 volts the maximum efficiency will be 25% since 9 volts are being "lost" across the regulator and the input current is slightly greater than the output current.

If possible you'd be better to select a 3volt battery to drive the laser device directly

Ok well I'm slightly confused. To me it seams like my better will last longer if I use a 12v battery with a switching voltage regulator to convert the voltage to 3v as compared to if I use a linear regulator? Am I missing something?

calvingloster: Ok well I'm slightly confused. To me it seams like my better will last longer if I use a 12v battery with a switching voltage regulator to convert the voltage to 3v as compared to if I use a linear regulator?

Yes.

calvingloster: Am I missing something?

Yes.

Amps into the regulator vs. amps out of the regulator.

Ok last question. I am going to be using the switching voltage regulator. Now what would give me better results if I buy the a 6v 4ah battery or a 12v 4ah battery? If I understand correctly they both gana give me the same amount of ah regardless of the voltage. So is it just more efficient to use the 6v battery then?

On the basis that you use a switching regulator, the 12v 4AH battery will provide a longer duration than the 6v 4AH battery.

This is because the switching regulator "transfers" watts rather than a linear regulator which simply "transfers" volts. Note that the word "transfers" is a simple metaphor to illustrate the point.

The 6v battery stores 24WH of energy and the 12v battery stores 48WH (volts x AH)

Battery capacity is usually related to a discharge period of 10 hours, so your 4AH battery discharged at C/10 should deliver 0.4 amps for a period of 10 hours. If you reduce the discharge rate to, say C/20 you could expect to get around an extra 20% of capacity. This 20% gain factor tends to repeat for each doubling of the C factor.

Your quoted discharge of 200mW out of an 80% efficient regulator implies an input wattage of 200 / 0.8 = 250mW. For a 12 volt battery, this is a load current of around 0.02 amps. Compared to the rated battery C/10 current of 0.4 amps, this implies the battery load is C/(10 x .4 / .02) = C/200 which is very conservative. At such a low C loading you might expect the battery capacity to rise substantially, based on 4AH x 1.2 (C/10 to C20) x 1.2 (C/20 to C/40) x 1.2 (C40 to C80) x 1.2 (C80 to C160). That all works out at around 8AH.

That's the good news, now the bad :

If you are using lead-acid type batteries, these shouldn't be discharged lower than 50% capacity and ideally no more than 20% of capacity rating. Depth of discharge defines battery life. Deep discharging reduces battery life drastically. So, assuming you limit consumption to no more than 20% of capacity (1.6AH based on the increase to 8AH above) and assuming a load current of .02 amps, your battery duration will be 1.6 / 0.02 = 80 hours.

Edit : Last paragraph edited to clarify discharge limit