On Thu, Jun 16, 2011 at 04:06:58PM -0400, Paul Alfille wrote: > Marcus has very practical suggestions. > > If you really want to measure voltage, just use a voltage divider -- > two resistors in series and measure the midpoint voltage. > > For instance, 10K Ohm resistors will allow 12V/2*10^4 = 0.6 mAmp > current with a midpoint voltage of 6V (or so, the bttery voltage might > start a little higher).
However, if the supply voltage is 5V, then this would result in a continuous drain of 5/20000 = 250uA which is probably a lot more than all the semiconductors in the circuit. For monitoring batteries you really want to use the largest resistances you can get away with, or you'll just flatten your battery! njh ------------------------------------------------------------------------------ EditLive Enterprise is the world's most technically advanced content authoring tool. Experience the power of Track Changes, Inline Image Editing and ensure content is compliant with Accessibility Checking. http://p.sf.net/sfu/ephox-dev2dev _______________________________________________ Owfs-developers mailing list Owfs-developers@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/owfs-developers
