# I need to power a device that requires 12V and 0.43A, with as small of a battery as possible

Basically, I have a device that MUST be powered by battery and it requires according to the item description an operating voltage of 12V and 0.43 Amperes. There are small batteries (A23 batteries) that provide a voltage of 12V. I feel like these batteries are too small and may not work as intended. Am I wrong? Will the device simply drain the battery(s) very quickly when used? Or will it not work at all?

I'd like it to last 2-3 days

• How long does it need to be powered for? What is the acceptable range of voltage? Take a look at a datasheet for (example) this LR6/AA battery: media.digikey.com/pdf/Data%20Sheets/… It will give you a set of discharge curves for different rates and you can plan from there.
– vir
Jun 21 at 22:57
• Please describe how long the battery should last. An hour? A week? Ten years? That significantly affects how much amp-hours you need. Jun 21 at 23:13
• As a first order-of-magnitude calculation: If you insist on using that A23 batteries, you will need to put 550 of them in there. Just dumping them in parallel won't work, as a duff cell will pull the whole contraption down. I seriously, seriously advise looking at a bigger battery. Think car battery sized. Jun 22 at 11:40
• Maybe a nuclear battery? Jun 22 at 14:33
• You need to tell us more about the application and device. It may not draw full current at all times. If it does you need an immensely large battery compared to the A123. Q1 what is the powered equipment? Q2 describe the application in detail Q3 any web links? Jun 22 at 20:58

You have a 430mA load and want it to last 2-3 days. That's 48 to 72 hours. Doing the math, you need a 21 to 31 amp-hour battery. That's approxmately one large motorcycle battery.

• Check the time stamp on my comment about 31ah and the motorcycle battery. Just saying. Jun 22 at 15:02
• Given that this is not a start-and-recharge application, a 12V ups battery would be a good choice as they are built for deep-discharge usage. In any event the OP will need a way to recharge it.
– CCTO
Jun 22 at 15:44

For a battery-powered device, consider both the peak drain current and the average drain current, as well as the desired run time.

The peak drain current is the maximum amount of current the load needs to draw; the battery has to have low enough internal resistance to supply the minimum required voltage at the peak drain current.

Most battery-powered devices are designed to enter a lower-power mode when idle, so it's also important to consider the average drain current. The battery can only supply a limited amount of energy before its internal resistance becomes too high and the battery becomes "dead". Voltage x current equals power, which is the rate at which energy is provided. Lower average current means lower average power, so the battery's energy supply will last longer.

Put another way, when selecting a battery, you need to consider the acceptable voltage range, the peak current (which determines how much internal resistance is acceptable), and the average current (which determines the rate at which the battery's energy is consumed), as well as how long the battery must sustain the device (more run time requires a battery with more internal energy capacity).

Ideally we'd like a battery with zero internal resistance (so unlimited peak drain current), and an unlimited amount of internal energy capacity (so unlimited run time), and we'd like it to cost nothing and be as small as we want. But more realistically there is a range of performance that is acceptable. If the battery can only run for 7 minutes, that's probably not useful. If the battery can run for 10 hours, that might be useful in some applications (bench lab equipment maybe), but would not be useful for hard-to-reach or remote equipment. It all depends on the specific application.

It's worth browsing the available battery data sheets, these often have performance curves showing typical runtime for various types of loads. For example here is the Energizer A12 battery datasheet

• Small aside in terms of "useful time" the entire hobby and racing quadcopter "drone" "market", lives in the 1-7 minutes of useful flight time off a battery charge. Jun 22 at 0:51

The A23 battery is rated at 55mAh. Your draw is 430mA. That's much more than this small battery can handle - it will die in about 7 minutes.

I wasn't able to quickly locate internal resistance data for A23. That said, this battery is composed of a stack of small 1.5V coin cells internally; supplying 430mA is going to make the cell heat up, further reducing its life.

MORE: You clarified your run-time requirement, which we now understand to be up to 3 days, so (at least) 31Ah. That's a much larger class of power than even the largest common primary batteries. It will not be economical, let alone feasible, to use them. You should consider a rechargeable (secondary) battery instead, be it a sealed lead-acid, AGM, LiFePO4 pack, or similar.

There exists a possible off-the-shelf solution: use a 'power bank' type of product. Some even have 12V output capability, like this one: https://www.xtpower.de/Power-Bank-MP-50000

Speaking of which, if you have a bit of flexibility in your design at this stage, you can choose to leverage USB type C power delivery, which supports 5,9,15 or 20V. This would allow you to leverage the ever-increasing ecosystem of USB-C power solutions, including power banks and other secondary battery solutions (some of which even include solar charging.)

• So if I were to put say 8 of these batteries in parallel would that help mitigate the heating issue (and also increase capacity obviously)? Jun 21 at 23:05
• Yes, but it will still power your device for only an hour total. It would be more economical to use AA batteries (about 2.5Ah for decent ones), 8 in series. If density is your concern you could consider lithium primary cells. A good choice would be 4x CR123A 3.0V cells. Jun 21 at 23:13
• More: 3 days at 0.43A is 31Ah. That's roughly the capacity of a large motorcycle or small-ish car battery. Jun 21 at 23:46