Basically I'd like to deploy a sensor temporarily for a period of about a month. The sensor communicates via a wireless protocol and sends a data packet about once every 5 minutes as far as I understand it. It is normally connected to a device which powers it via the following specifications:

50-60mA and 12-20vdc

Is it practical to do this via a battery? I was initially thinking that maybe I could use a rechargeable battery pack such as the ones that you can use to recharge your mobile phone via a USB port. A Jackery Bar for example. I have no real idea of how you match a battery to your purpose so any thoughts gratefully received. Robin

  • \$\begingroup\$ Multiply the number of hours by the amount of current to get a figure in amp-hours. That's how much battery you will need. \$\endgroup\$
    – pjc50
    Feb 24, 2016 at 15:14
  • 1
    \$\begingroup\$ Are those specifications of the power supply or of the device? And does the sensor run continuously? i.e. is it always drawing current, or can it be put to sleep or powered down most of the time? \$\endgroup\$
    – uint128_t
    Feb 24, 2016 at 15:39
  • \$\begingroup\$ Measuring the actual current drawn by the device over several cycles (ie, non-transmitting and transmitting) would be a very advisable step in evaluating actual power requirements. If you have a meter that can log data directly, great. Otherwise, these days, point a cell phone camera or webcam at the meter readout and let it record for an hour or so, then review the recording. \$\endgroup\$
    – Ecnerwal
    Feb 24, 2016 at 15:53

2 Answers 2


Batteries are rated in amp-hours, which is the amount of current (in amps or milliamps) that they can supply for one hour. This value is ratio-metric, however. For example, if you have a 1000mAH battery, it can supply:

  • 1000mA for 1 hour
  • 500mA for 2 hours
  • etc.

So, to figure out how many amp hours your battery needs to be, simply take the number of hours you need to run for and multiply by the current, or in this case:

30 days = 720 hours, 720 hours * 60mA = 43.2AH

A 43AH battery would be gigantic, so clearly that wont work. My guess is that your sensor does not draw 60mA continuously, it probably only draws that much current during transmit mode. So, to figure out what size battery you need, you need to compare the active current (assuming 60mA) and the inactive or sleep current (for example, lets assume 100uA) and then figure out what the active time duty cycle is. In this case, you say it transmits every five minutes, and I'll assume that the transmit time takes 100ms. That would give us a duty cycle of 100ms/5000ms = 2% duty cycle.

What this means is that it only needs 60mA for two percent of that month, and 100uA for the other 98%. So to determine the battery size, we just modify our calculations a bit:

30 days = 720hrs, 0.02 * 720h * 60mA + 0.98 * 720hr * 100uA = 0.86Ah + 0.07Ah = 0.93Ah

Lets call that a 1Ah battery, which is a much more reasonable number. Now, your mileage may vary, and I had to make some assumptions about your system. However, even if your sleep current was much higher, like 1mA, the battery still wouldn't need to be massive. Hope that helps!

  • \$\begingroup\$ MOST batteries (at least large storage batteries, which realm this job is in) are rated at the "20 hour" rate, so 1000 mAh would be 50 mA for 20 hours, and you'd be disappointed if you expected to get 1000 mA for an hour. In any case, check the data sheet to see what discharge rate the stated capacity is for. You generally get more aH at a rate slower than rated, and fewer aH at a rate faster than rated. \$\endgroup\$
    – Ecnerwal
    Feb 24, 2016 at 15:45
  • \$\begingroup\$ Sure but the charge and discharge rate depends a lot on the chemistry, and I figured at 60mA worst case current draw it wouldn't really matter, and I didn't want to over complicate my answer with a description of all the charge/discharge possibilities. As always though, RTFM. \$\endgroup\$ Feb 24, 2016 at 15:52

The way to approach this is to figure out how much energy you need, then see if there's a battery big enough to support it.

You need 60mA for 720 hours (24hours*30 days.)

That is 720hours*0.06 Amperes= 43.2 Ampere Hours.

Since your sensor can operate on 12Volts, you need a 12Volt battery rated for 43 AmpereHours (Ah).

A quick look around google says that batteries of that size are used in cars and trucks.

A battery pack like you would use to recharge a phone via USB provides 5Volts, so it won't work without extra parts (a boost regulator to transform 5V to 12V.) It also won't have the needed capacity.

Powerbanks are rated in milliAmpereHours. To compare to your needed capacity, you have to multiply by 1000. So, you need a powerbank rated for 43000mAh.

Well, no. You need more than that because you have to trade current for voltage to get the 5Volts from the powerbank up to the 12V needed for your sensor. So, at least (12/5)*43000=103200mAh. You won't find one that size for a price you will like, and maybe not at all as a separate unit.

So, your best bet is to use a fully charged 12V auto battery of at least 43Ah capacity.

In response to the comment from @Ecnerwal:

A battery made for deep discharge (such as a trolling motor battery) would work better. Trolling motor batteries are available at 12V. The difference is that trolling motor batteries are designed to be discharged to the limit without being damaged. A regular car battery used this way might not be able to hold a charge properly after being run down - you would have to replace it. If this is a one off operation and you can get a car battery cheap it might not matter. If you will need to use the battery again you might want to look into getting a battery rated for deep discharge.

You might get away with less if the sensor shuts down most of the time and only draws 60mA when transmitting. You didn't say, and maybe you don't know. The data sheet for the sensor would probably say, so it would be nice of you to post a link to the datasheet for the sensor.

  • \$\begingroup\$ A "deep cycle" battery (trolling motors being the most common mass application, or golf carts, depending on location) would be a better choice than an automotive starting battery. 3x 6V golf cart batteries would be an excellent fit, given the voltage range. \$\endgroup\$
    – Ecnerwal
    Feb 24, 2016 at 15:48

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