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I plan to use a rechargeable lithium 3.7V battery in my wireless mouse, which takes 2 AAAs. As I understand it, I could theoretically drop 0.7V using a diode in series with the mouse, but would it be safe for me to do so consistently? I will charge the lithium battery safely using a charging circuit (TP4056) (without the voltage dropping diode obviously) but can I safely use a 3.7V lithium with my mouse using just a diode as voltage protection? The voltage would often be between 3.7 and 4.2 volts, which leads to 3 to 3.5 volts into the mouse - and by design it would draw little current too, such that the diode should not overheat due to current draw.

Thank you.

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  • \$\begingroup\$ The little current draw could also be the problem for your mouse electronics - for really low currents, the voltage drop over a diode is far less than 0.7V. Therefore the electronics must be able to deal with at least 4V. \$\endgroup\$
    – jusaca
    Commented Mar 24, 2019 at 12:53
  • \$\begingroup\$ Ah, thanks for that. I will look into the datasheet of the diode I hope to use (1n4004) but since this looks likely, I might have to look for another solution (like a 3v regulator perhaps) \$\endgroup\$
    – QuickishFM
    Commented Mar 24, 2019 at 12:54

2 Answers 2

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I could theoretically drop 0.7V using a diode

This voltage drop depends on the current through the diode. A wireless mouse has very close to zero current flow when it is in "sleep" mode - resulting in a significant voltage rise at the output.

I recommend just using 2 sets of NiMh rechargables - the type with low self-discharge rate.

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  • \$\begingroup\$ I already use 2 Ni-MH rechargeable AAAs, though this would allow me to get a bit more capacity, since the batteries I have aren't very good. I might end up buying higher capacity/low drain ones though. Thank you for your input \$\endgroup\$
    – QuickishFM
    Commented Mar 24, 2019 at 16:37
  • \$\begingroup\$ Don't use "high capacity" cells - these have high self discharge. The longer lasting cells will have a somewhat lower capacity of around 800mAh for AAA size and are advertised as such. \$\endgroup\$
    – Turbo J
    Commented Mar 25, 2019 at 17:42
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It would be much much better to use some sort of active regulation technology - for example a Low Dropout Linear Regulator with circuitry to switch out the regulator when the voltage falls down significantly or a buck/boost regulator.

However, if you must use some sort of a voltage drop mechanism - it would be much better to just use a resistive divider. This is because the low current draw would cause the diode to reduce the voltage dropped across it, this wouldn't happen in a resistive divider and the low current would mean there wouldn't be too much heat dissipation.

One issue you would face would be that the resistive divider would drop a proportion of the voltage ( = V * (R1 / R2) ) rather than a flat amount. This would reduce the usable portion of the battery's capacity.

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