I'm planning to make a battery application with a PIR sensor. The sensor consumes 70 μA and will be supplied with 4-5 V. I have two Li-ion batteries in series which gives around 8-9 V in total.
I have 2 options:
1. Supply directly from the batteries (there is a battery protection module in between) and let the sensor's linear regulator obtain 3.3 V;
2. Put a buck converter in between and take the voltage down to 4-5 V before sending it to the sensor.
I was planning to use an LM2576 as a buck converter but I realized it draws 80 μA even in off mode so I'm sure it draws even more than that for its operation.
Let's assume 80 μA input for the IC itself and 100% efficiency from the converter to be optimistic. Input voltage will be 9 V and input current will be 70 μA * 3.3/9 = 26 μA so wasted power is {input power} - {necessary power} = 9*(80+26) μA - 3.3*70 μA = 723 μW. I don't know if the power calculation is relevant or not, but the necessary current is more than doubled with a buck converter anyway. On the other hand, a linear regulator will only waste (9-3.3) * 70 μA = 400 μW.
So is it true that linear voltage regulators can be more efficient than switch mode ones?