>-----Original Message-----
>From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] 
>Sent: 01 July 2005 10:34
>To: piclist@mit.edu
>Subject: RE: [EE]: How to efficiently run a PIC from 12V battery?
>
>
>William "Chops" Westfield wrote :
>
>> A type 394 watch battery
>(PROBABLY 
>> available in alkaline (general purpose), silver (camera),
>> and zinc air (hearing aid) with slightly different part numbers) is
>3.6mm tall (so you couldn't fit 10 of them in the space of a
>> A23, but
>4 and a spacer would be no problem) and has
>> about twice the mAh
>capacity.
>
>Hi.
>Just thinking out loud here...
>
>Other posts have also 
>been talking about getting a lower
>voltage battery with a higher mAh 
>value (in this case "twice"
>the capacity).
>
>Now, Let's say we have a 
>12V/100mAh battery and one
>4.5/200mAh battery. Now, the 4.5V battery 
>*seems* to have
>double capacity (which most equals with "lifetime" 
>under
>similar conditions). But, if using a well designed step-down
>converter, will that not draw *less* current from the higher
>voltage 
>source, then the target is using ? Or in other words,
>if one could 
>expect a 100% efficient converter, is it not realy
>the V x mAh value 
>that is interesting, not ? Even if the
>converter runs at something like 
>80-85%, I think that
>one can not just compare the mAh values 
>directly...
>
>I'm not claiming to be an expert in this, just liked to 
>know...
>
>Jan-Erik.

You are correct, the total energy available from a 12v 100mAh battery is greater than that from a 4.5 200mAh battery.  The only way to make use of the extra eneregy is, as you suggest, a switching converter.  Obviously a 100% efficienct converter does not exist, but they can achieve very high efficiencies.  If the design can tolerate the cost, board space and potential noise of the switcher then it may be a good solution.

Regards

Mike

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