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June 18, 2011
about 4 years ago
I find it kind of sad that a 6 volt battery is required to supply 3.3 volts, when with just a couple of changes, the voltage dropout can be lowered to under 0.3v, and the circuit can run off of a single 3.6v Li-Ion cell. The changes are:
1) Replace the voltage regulator VR1 (MCP1702 with 0.525v dropout), to the MCP1700 with 0.18v dropout. This saves 0.35v of drop.
2) Short-out the diode array D1, so as to connect the battery directly to VR1. This saves 0.8v of drop.
3) Connect the +5v USB supply to a battery charger IC such as the low-cost MCP73811T. This will float both the battery and the regulator-input to 4.2v at up to 500mA, within the USB 2 spec.
The parts cost to do this is very low. Interestingly, even though Li-Ion is typically a more expensive battery chemistry, with this more efficient circuit and lower cell count, the battery cost is about the same! A single 14500 AA, 3.6v 800mAh Li-Ion cell is about $4. A pack of 4 AAA, 1.2v 800mAh NiMH cells is also about $4.
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