I have an electromagnetic device designed to hold a fire door open until either the circuit is cut or the building's electrical supply goes out. At which point the door is released and will close.

It is rated as 12 V/80 mA. What does that mean in terms of the length of time that the device will draw the 80 mA?

Should I read such a rating as 12 V/80 mAh or 12 V/.0080 h?

I'm doing some rough calculations as to how long a given battery combination would keep this device energized.

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    \$\begingroup\$ "Should I read such a rating as 12v/80mah or 12V/.0080h" - neither. It will draw 80mA from a 12V supply indefinitely. \$\endgroup\$ – brhans Feb 18 at 1:06
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    \$\begingroup\$ The slash "/" is throwing you off. It is not a divide. 80 mA @ 12V would be a better way to spec it. \$\endgroup\$ – Mattman944 Feb 18 at 1:11
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    \$\begingroup\$ No, the slash isn't representing division, it is simply the way the rating was shown and I copied and pasted it. Okay, how about 12v 80ma? \$\endgroup\$ – BobNll Feb 18 at 1:17
  • \$\begingroup\$ I was not speaking of a power supply but of batteries. \$\endgroup\$ – BobNll Feb 18 at 1:23
  • \$\begingroup\$ By 'supply' I wasn't referring to any particular kind of power source. It's whatever you choose to use to supply power to your device - batteries, wall-wart, solar panel, thermopile, whatever. \$\endgroup\$ – brhans Feb 18 at 2:06

It will draw the 80mA current for the period it is on, ie, has power applied.

If you are powering it from a battery, that's when the time comes into play.

For example, an 8 Ah battery would power the device for 8 Ah / 0.080 A = 100 hours.


  • A battery's storage is often quoted under various conditions which might not be continuous discharge (and are often given as multiple values at different currents).
  • You might well also find that the holding voltage is lower than 12V, so that as your nominal 12V battery's voltage drops below 12V the device still holds the door open, it will depend on the strength of the spring or other door closer.
  • Battery charge capacity varies considerably with age, charging history and battery technology
  • If the device is a simple electromagnet, its current will vary with voltage (basically linearly)
  • Be aware that batteries and devices in general might give nominal, maximum, minimum, typical, average or "specification" values on the label without saying which. And caveat utilitor, nobody really expects to see the manufacturer's Highway Mileage or even Urban Mileage on their car. For a fire door, make sure your system is fail-to-safe (and tested!) and be clear about which responsibility is whose.
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  • \$\begingroup\$ Thanks, that is exactly what I thought but I was not sure. You are no doubt correct that the holding power is lower at lower voltages. That means it would be a matter of the force being exerted against the coil's pull that would determine when the device would release the item being held by the coil, given the voltage applied as the battery drains. \$\endgroup\$ – BobNll Feb 18 at 1:22
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    \$\begingroup\$ @BobNll, it would be a good idea to measure the actual current just to make sure. Either that, or generously pad the size of the battery. Also, keep in mind that batteries go bad over time. If you just leave a battery there for 5 years, it may not work correctly at the end of that time. Somehow you need to test the battery or just automatically replace it periodically. \$\endgroup\$ – mkeith Feb 18 at 2:24
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    \$\begingroup\$ @mkeith - I'm actually going to set up a test and see just how long various batteries last with device. I'll try to post my results here but don't hold our breath as I may not get around to the test for a few days (weeks?) \$\endgroup\$ – BobNll Feb 18 at 3:34
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    \$\begingroup\$ If that's the spec for the power supply the device needs, it doesn't need to mean it draws the full 80 mA all the time, constantly, just that it may need that much available at some point of its operation (in general). \$\endgroup\$ – ilkkachu Feb 18 at 10:42
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    \$\begingroup\$ @BobNll Be aware that most (rechargeable) batteries (especially lead) don’t like deep discharge. \$\endgroup\$ – Michael Feb 18 at 12:04

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