I know it is an easy question, but I have forgotten all of this stuff and am trying to play around with something in my garage.

I need to reduce a AA battery's voltage, probably by half (1.5 - .75) What is the best way to do so, and what is the formula or theory behind it ?

Thanks

 

For what purpose? How you do it will depend on what you'll do with it, though your options will be somewhat limited considering the very low voltage. The easiest way would be a voltage divider (two equal resistors in series with the voltage, tap the 1/2 Voltage between them) but that will only go so far...

j.

 

A diode placed in series with load will do if it doesn't have to be exactly .75v..

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gerty

 

hi,

both suggestions would do in reducing the voltage supply by about half. the diode would give a voltage drop of about 0.65v. for a voltage divider if you use a 1% tolerance you would probably get half of whatever voltage your battery have and you have to consider the heat to be dissipated by the resistors depending on you load.

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As John said - WHY! - any possible solution will depend greatly on what you are wanting it for, and what the current demands are - I suspect you may well be asking the completely wrong question?.

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