My mouse requires AA 1.5 volt batteries to work. I accidentally used a 3.6 volt Li-SOCl2 battery. Now it works fine with the 3.6 volt battery but doesn't work at all with a 1.5 volt AA battery.

I have this mouse and keyboard combi: https://www.microsoft.com/accessories/en-nz/products/keyboards/sculpt-ergonomic-desktop/l5v-00027

How is this even possible?

Perhaps some traces are burnt enough so that no current can flow through them with 1.5 volt?


closed as off-topic by brhans, Elliot Alderson, RoyC, DoxyLover, Phil G Jun 11 at 21:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Questions on the repair of consumer electronics, appliances, or other devices must involve specific troubleshooting steps and demonstrate a good understanding of the underlying design of the device being repaired. See also: Is asking on how to fix a faulty circuit on topic?" – brhans, Elliot Alderson, RoyC, DoxyLover, Phil G
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    \$\begingroup\$ If you would burn a trace, an additional 2 V over what's specified would not be enough. We can only guess, but some front-end LDO could be toast and you are now feeding it though some body diode. You need to open it and probe it to find out. \$\endgroup\$ – winny Jun 3 at 7:51

Without knowing the design, what really happened can only be guessed at.

I'm assuming that the mouse originally worked with only one 1.5 V AA cell, I mean, that it did not require two cells in series.

You probably damaged the DCDC boost converter chip inside the mouse. This boost converter is needed as 1.5 V usually isn't enough to power all circuits. This is an example of a typical boost converter circuit:

enter image description here

Note the inductor L1 and the diode D1, they are nearly always present in boost converters. It is possible that D1 is an internal (on chip) transistor with a diode in parallel.

If the IC is damaged and doesn't switch but L1 and D1 are intact then applying 3.6 V at the input will result in about 3 V at the output. That could be enough to power all circuits and make the mouse operate.


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