Is there a component that requires a higher voltage? Wouldn't they last longer on AA/AAA batteries?

  • \$\begingroup\$ The current drain of a smoke detector is so low that the battery life of a 9V battery is near its shelf life. \$\endgroup\$
    – Hot Licks
    Commented Feb 24, 2015 at 20:03
  • \$\begingroup\$ how do you figure @HotLicks? is that supposed to be an answer to the question? please post it as such so people can vote on it. \$\endgroup\$
    – Malachi
    Commented Feb 24, 2015 at 20:13
  • \$\begingroup\$ @Malachi: sensor derates after a couple of years, maybe holds up to ten years or so. Many models have fixed installed batteries that hold the whole time. Replacing them after 10 years is not viable since the sensor isnt working anymore anyways. \$\endgroup\$
    – PlasmaHH
    Commented Feb 24, 2015 at 21:12
  • \$\begingroup\$ that makes sense, I never thought of the the actual sensor having a shelf life, but Americium probably doesn't have a very long half-life.. \$\endgroup\$
    – Malachi
    Commented Feb 24, 2015 at 21:32
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    \$\begingroup\$ Nobody has suggested afaics: (I would expect) the battery clip on a 9v battery to be somewhat more reliable statistically over many years. While batteries (cells) should not "leak" some do, and a small trace of electrolyte can wick between the battery contacting surface and the contact and cause unreliable operation. As current is small "wetting" is unlikely to keep this path clean. 9V batteries have cells within an outer shell so are less likely to contaminate the outer contact, and the click action battery connector is liable to offer a higher reliability of connection. [[B, IMBW]] \$\endgroup\$
    – Russell McMahon
    Commented Mar 3, 2015 at 23:19

1 Answer 1


The sensor (typically an ionization chamber with some radioactive Americium 241) is normally specified at 9V. I think it would work at a lower voltage, but with less sensitivity.

More information can be found here.

Since the battery drain is very low, it's not necessarily true that cells with higher ampere-hour capacity would last significantly longer- much of the battery drain is self-discharge, and if the voltage is stepped up for the detector, then the efficiency of that must be taken into account.

Edit: Also, the piezo horn will provide more sound volume at a higher voltage. It's still possible to use lower voltages but it would require something like an inductor or a transformer to get an acceptable volume level for smoke alarm purposes.

Early (1970s) single-station ionization smoke detectors used expensive batteries adding up to 12V (with standard 4000-series CMOS used internally). See, for example, patent US4004288. Modern products use CMOS ASICs.

Photoelectric smoke detectors which operate on the basis of scattering of light are also used (because they detect certain types of fires better, and because of concerns about the radioactive sources ending up in landfills). There is no reason to use higher than a few volts for the photoelectric sensor and source, however the piezo beeper issue remains. 9V batteries are also used in this type of smoke detector.

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    \$\begingroup\$ The fewer parts and active components, the more reliable: probably a consideration for a safety device. \$\endgroup\$
    – Nick T
    Commented Feb 24, 2015 at 22:05
  • \$\begingroup\$ In addition -- commercial smoke detector ICs run directly off 9V using an internal regulator. \$\endgroup\$ Commented Feb 25, 2015 at 2:42
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    \$\begingroup\$ @ThreePhaseEel, Because commercial smoke detectors use 9V, therefore commercial smoke detector ICs are designed to run directly off of 9V. \$\endgroup\$
    – caveman
    Commented Feb 25, 2015 at 4:28
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    \$\begingroup\$ This answer is incomplete. In Europe ionization detectors are not common anymore and have been replaced by optical detectors. These also run off 9V batteries. \$\endgroup\$
    – realtime
    Commented Feb 25, 2015 at 15:13
  • \$\begingroup\$ @realtime Paragraph added. If you have direct experience with these devices any corrections would be useful and appreciated. \$\endgroup\$ Commented Feb 25, 2015 at 15:28

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