Electronic – How to calculate battery life allowing for duty cycle

My calculation, probably missing something, but here's what I did:

$$ 1 \mathrm{\ \mu A} + (50 \mathrm{\ mA} \times 0.1\%) + 25 \mathrm{\ \mu A} = 76 \mathrm{\ \mu A} $$

$$ \frac{76 \mathrm{\ \mu A}}{ 80 \, \% \mbox{ efficiency}} = 88 \mathrm{\ \mu A} $$

Round up to \$100 \mathrm{\ \mu A} = 0.1 \mathrm{\ mA}\$

$$ \frac{2700 \mathrm{\ mAh}}{ 0.1 \mathrm{\ mA}} \approx 3 \mbox{ years} $$

If you're using rechargeable batteries, they'll discharge on their own long before that. Or if any of your other calculations are off (like maybe it's a 98% instead of 99.9% sleep), that will affect it a lot too.

Best guess is that the misleading / verging on wrong example circuit diagrams in the data sheet have lead you to reverse the FET drain & source connections.

The FET source should connect to Vin, not to battery, despite the FET-symbol on their diagrams suggesting otherwise. This is a P Channel FET (both their one and your replacement.) The symbol used incorrectly implies an N channel FET with drain to input.

For the FET to act as a FET the body diode must be as shown on their diagrams but to do this the drain must be to battery when a P channel FET is used. If d & s were reversed the FET could not be controlled as the d-s diode always conducts.

As Vin is 10V and Vbat max = 8.4V & less if depleted then charge I may lift Vbat due to IR drop & ...