I'm trying to replace the AAA batteries that came with my BT cordless phone, as they are terrible quality and have quickly degraded (they now only last about 20 minutes after full charge). They are 1.2V Ni-Mh, 550mAh, and similar to these.

I have some good Duracell AAA rechargeables, also 1.2V Ni-Mh, 750mAh. My question is, why don't these work as a replacement? Putting these into the phone does not give power to the phone, whereas swapping them out with the old batteries turns the phone on straight away.

As far as I can see there are no listed properties of the batteries that differ apart from the mAh, and I know enough about electronics to know that it shouldn't make a bit of difference to compatibility.

Is there some unlisted property of batteries that is causing this?


closed as off topic by Kortuk Oct 16 '12 at 0:13

Questions on Electrical Engineering Stack Exchange are expected to relate to electronics design within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\begingroup\$ It honestly wouldn't surprise me if BT were putting some trickery in there to force you to use "official BT batteries". \$\endgroup\$ – Polynomial May 9 '12 at 8:09
  • 1
    \$\begingroup\$ Can you elaborate - in what way don't they work? Do already charged batteries power the phone, and then not charge? Or do they not work at all. I have seen some devices need resetting after all power has been removed. \$\endgroup\$ – Cybergibbons May 9 '12 at 8:24
  • \$\begingroup\$ Sure - installing the fully charged Duracell batteries means that the phone is completely dead. Putting the phone on the base to recharge does nothing. Replacing with the old batteries turns the phone back on instantly, even when not fully charged. \$\endgroup\$ – Jon Cairns May 9 '12 at 8:31
  • \$\begingroup\$ Thanks. I find it interesting that the amazon page mentions that the mAh is designed for cordless phones. I can conceive that the charger could struggle to charge the batteries but can't think why it wouldn't work full stop. \$\endgroup\$ – Cybergibbons May 9 '12 at 9:50
  • \$\begingroup\$ Consumer electronics support is off-topic here, we are about design. On a brighter note you already fixed it. \$\endgroup\$ – Kortuk Oct 16 '12 at 0:14

You should check that your new batteries REALLY touch the terminals in the battery holder. I had to phisically modify the battery holder or add a metal washer in many cordless and LPR radios to have them working with other batteries...

  • \$\begingroup\$ Yes, and check with a voltmeter that the Duracell ones are indeed charged. There is no way for the cordless phone to know the brand of the batteries. \$\endgroup\$ – Telaclavo May 9 '12 at 10:05
  • \$\begingroup\$ Thanks @Axeman, I'll check that when I get home tonight. It would be strange though - aren't they all the same size? \$\endgroup\$ – Jon Cairns May 9 '12 at 10:57
  • \$\begingroup\$ @JonathanCairns Unfortunately they're not all alike. The positive pole can vary in diameter and thickness. If the battery holder is really tight, or poorly done, some battery will not touch the holder spring. \$\endgroup\$ – Axeman May 9 '12 at 11:03
  • 1
    \$\begingroup\$ You were absolutely right about the terminals! Basically, the batteries that came with the phone had a taller positive terminal that was pushing a metal contact into place, creating a connection. Standard AAA batteries don't have a tall enough contact on the positive terminal to work. Not only that, but BT have added little plastic guards to stop normal batteries pushing up enough to push the contact into place. How cheeky! I'm going to try and break the guards, but if I can't I'll have to buy their batteries... \$\endgroup\$ – Jon Cairns May 10 '12 at 8:05
  • \$\begingroup\$ @JonathanCairns You can also use a circular rare earth magnet of the same diameter of the positive terminal. It works, usually. \$\endgroup\$ – Axeman May 10 '12 at 9:15

Not the answer you're looking for? Browse other questions tagged or ask your own question.