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I have mouse that take two AAA batteris, so it take 3V.

Is it OK put one 10440 of 3.7V and one spacer instead?

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    Probably not... But maybe yes. Depends on how the mouse was designed. How could we know? – dim Jul 8 '16 at 12:57
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    Try it and see whether it smokes. If it does, no. You could make a special spacer containing a series silicon diode for 0.7v drop. It still wouldn't be very clever, because I'll bet the mouse will work down to maybe 2.5, perhaps even 2v, which would damage your 10440 through undervoltage if you let the mouse tell you when it was low. – Neil_UK Jul 8 '16 at 13:08
  • I wonder why would you want to do this? Batteries are cheap. – Eugene Sh. Jul 8 '16 at 13:15
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    More than a specified tolerance is considered big. And I don't see a reason for a consumer electronic device powered from battery to have the tolerance more than +/-5%. – Eugene Sh. Jul 8 '16 at 13:20
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    @Dubon The fact is: nobody can give a definitive answer. We didn't design the mouse. – dim Jul 8 '16 at 13:23
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While a fresh alkaline cell would typically have an open circuit reading of a little more than 1.5v leading to a series pair producing more than 3v, this would never be as high as 3.7v in practice.

You'll also find that your 3.7v cell is actually closer to 4.2v when fully charged.

Although it's possible that the designers of your mouse used ICs with a wide voltage range, it's very unlikely as this would probably be more expensive.

As such you'll probably find that your mouse's max operating voltage is 3.3v and might survive a short term peak of up to 3.6v.

So your 3.7v nominal (4.2v peak) lithium cell is very likely to do some damage - don't do it.

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  • Batteries do not force current. The circuit will only draw the voltage it needs. I think the real danger is a potential lack of overdraw protection when using a Li-ion battery – Yorik Jul 8 '16 at 14:54
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    @Yorik - Your phrase "draw the voltage it needs" makes no sense. Load circuits draw current, not voltage. Voltage is determined by the source supplying it. Current is what is drawn by the load. Semiconductors have maximum voltage ratings because exceeding breaks down internal oxide insulating layers. – brhans Jul 8 '16 at 15:02

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