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I 'm trying to build a portable device running on battery. It needs power at 2 different voltage: 3v and 4.5v Is doing something like this a good idea ?

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if so, what type of diode should be used here ?

Thank you

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  • \$\begingroup\$ Alkaline: 1.6V new. 1V orless exhausted. |NimH. NiCd: 1.3V maybe new. 1.2V average.1.0V or less exhausted. \$\endgroup\$
    – Russell McMahon
    Commented Jul 2, 2014 at 5:42

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You don't show where the grounds are, but if it's a common ground, then no diode would be required, just tap off two batteries for about 3V and three batteries for 4.5V.

It may be not be a desirable thing to do because the 3V batteries will be exhausted before the "extra" battery and this might cause strange behavior in your circuit unless you separately detect voltage brownout conditions on both supply lines. If you don't care about the user having to replace the 3rd battery unnecessarily, it may be okay, but you should also parallel (not series) the 3V pair with a reverse-biased diode so that the 3V bus can't go negative if the 3V batteries are exhausted and a heavy load is applied to the 4.5V bus.

The diode type would depend on the maximum current (including fault current). If you have a polyfuse in there, a 1N5819 may be fine. If the batteries can be inserted backwards, the diode will conduct large current and may get very hot and fail unless you have something like a polyfuse in series.

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  • \$\begingroup\$ Thank you, alternatively, is it a good idea to connect the 3 batteries in series to provide 4.5v to the 1st module and connect a voltage regulator like this one (jameco.com/1/1/…) to provide 3.3V to the 2nd module ? The 2 modules are sleeping 90% of the time, they only draw power when they are active. Will the voltage regulator draw constant power or only when it 's required ? \$\endgroup\$
    – user46624
    Commented Jul 3, 2014 at 5:12
  • \$\begingroup\$ Yes, that is the conventional way. You can pick a CMOS regulator that draws only a couple uA - less than the internal self- discharge of the batteries. The ( otherwise good) LT1084 draws 1000 x more than that just sitting there, so it's a poor choice for continuous battery power. \$\endgroup\$ Commented Jul 3, 2014 at 10:25

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