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This might sound silly but I am new to electronics and trying to understand the difference between watt-hour vs ampere-hour when it comes to battery life per charge.

So let's consider this scenario.

I have a circuit that runs for 7 days when connected to a 3.7 V (1200 mAh) battery. Here the watt-hour of the battery is (3.7 * 1.2) 4.44 Wh.

Now if I connect a 12 V (1200 mAh) battery which provides (12 * 1.2) 14.4 Wh and use a voltage regulator, will I be able to run the circuit for more than 7 days on a single charge?

EDIT:

The circuit voltage requirement is 3.7 to 4.4 V and consumes 500 mA when active for 10 seconds and then it goes into a deep sleep for 5 minutes. During deep sleep, it only consumes around ~50 uA. My goal here is to get the maximum battery life (more than 60 days) from a single charge.

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    \$\begingroup\$ SI units named after a person have their symbols capitalised but are lowercase when spelled out. 'W' for watt. 'V' for volt. 'A' for ampere. Capitals matter. SI standard also recommends a space between the number and the units (same as "5 cups" rather than "5cups"). \$\endgroup\$
    – Transistor
    Dec 21, 2020 at 12:50
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    \$\begingroup\$ Depends on the regulator how much it consumes power. And what is the usable voltage range on battery. Do you have a specific battery types, specific regulator, and specific load in mind? If no, we are just guessing here and the question is not answerable. \$\endgroup\$
    – Justme
    Dec 21, 2020 at 12:54
  • \$\begingroup\$ @Transistor Specifically a non-breaking space, though it's not really worth the trouble most of the time. \$\endgroup\$
    – Hearth
    Dec 21, 2020 at 13:03
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    \$\begingroup\$ If the regulator is a linear regulator then that regulator will simply "burn" (turn into heat) the extra energy from the 12 V battery. With a linear regulator the current consumed stays the same and since the battery capacity remains 1200 mAh, the amount of current * time does not change and the 12 V battery will give roughly the same runtime on a single charge (assuming you're not depleting the 12 V battery down to 3.7 V as that damages the battery!). \$\endgroup\$ Dec 21, 2020 at 13:04
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    \$\begingroup\$ However if the regulator is a switching regulator and it is an efficient enough design then it is possible that the 12 V battery will give you a longer runtime. A switching regulator will not just "burn" the extra voltage but it will convert the voltage down more efficiently resulting in a lower power consumption compared to the linear regulator situation. So with a switched regulator and a battery that has a higher amount of energy, a longer run time is possible. I'm not saying: "You will always get a longer runtime" as the design needs to be correct. \$\endgroup\$ Dec 21, 2020 at 13:09

1 Answer 1

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Watt hours is the amount of energy the battery can deliver. Ampere hours is the amount of charge it can deliver. Which of those maps better to your load run time depends on the type of load.

If you have a load capable of using all of the energy efficiently, that is, your load starts with a good switchmode power supply, then the time it can run will be governed by the Watt hours. A switchmode regulator can deliver a higher current than it receives if it is regulating the voltage down, so the total charge in the battery does not relate to the run time.

If on the other hand you have an inefficient linear regulator that wastes energy, while it outputs essentially the same current as it receives, then the battery Ampere hours is more useful. As the battery voltage varies, the regulator will waste variable power, but the run time will still be governed by the charge.

If you have a low voltage load, and a high voltage battery, then running the load from a switch mode power supply will get you significantly higher load run time than running it from a linear regulator.

If your load draws large peaks, and then sleeps for a long time, you may need to choose your switch mode power supply carefully, so that its quiescent current doesn't dominate your load's sleep current. Designing for and verifying load life in such circumstances is non-trivial.

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  • \$\begingroup\$ This was helpful. Thanks. \$\endgroup\$
    – Jarvis
    Dec 21, 2020 at 15:26

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