Does the current drop after the five second timeout?
No, it stays at about 7 mA even after the 5 seconds timeout.
Regarding the checkBatteryState(), I will get rid of the R2 resistor to definitely remove any current that could flow in R2/R3 (I made a mistake as Q1 is referenced to VBAT which is 6V, and I cannot turn it off with a 3.3V signal). However, this current is about 6V/(200K + 100K) = 20 uA and should be the cause of the 7 mA drain.
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No, it stays at about 7 mA even after the 5 seconds timeout.
Regarding the checkBatteryState(), I will get rid of the R2 resistor to definitely remove any current that could flow in R2/R3 (I made a mistake as Q1 is referenced to VBAT which is 6V, and I cannot turn it off with a 3.3V signal). However, this current is about 6V/(200K + 100K) = 20 uA and should be the cause of the 7 mA drain.
Here some parts of the schematics.
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