The MCP1700 has a dropout voltage of approximately 0.2 volts. which means it will power an 8266 from full charge (4.2 volts) down to dropout which would be 3.3+0.2 or 3.5 volts. Using a typical lipo discharge curve it shows that about 45% of the lipo’s full charge will be available. quiescent current is 1,6 uA.
Comparing with a simple diode with a forward drop of 0.7 volts: The lipo should now power the 8266 from full charge down to 3.2 volts (3.2V – .7V = 2.5 V (which is the 8266 low voltage cutoff) This uses about 95% of the lipo’s charge and no quiescent current.
Obviously I’m missing something here as the simplicity of the diode, the non existent quiescent current and about 50% greater use of the lipo seem to make it a better choice. Where did I go wrong? Or did I?
Is there a better solution yet?
Thanks for any comments.