The power into the regulator will depend on the power into the load and the efficiency of the device. e.g. if it was only 50% efficient, you'd have to put twice as much power in as you got out.
So, an example:
I use 2xAA batterries with 2500 mAh and the buck converter with 50% efficiency and 100 mA quiescent current. If the load takes 2400 mA, the battery life should be 0.5 hour, correct?
So, an example:
I use 2xAA batterries with 2500 mAh and the buck converter with 50% efficiency and 100 mA quiescent current. If the load takes 2400 mA, the battery life should be 0.5 hour, correct?
A switching regulator is a constant power device i.e. output power = input power * efficiency. You are talking in terms of only current, so without knowing the voltages of the supply and the load there isn't enough information to answer your question.
The quiescent current of a switching regulator is taken into account in the efficiency calculations, which is why efficiency falls as the load current falls to low values.
e.g. if your batteries are wired in series and give 3v, and your load voltage is 2v and the load draws 100mA
Load power = 2*0.1 = 0.2 Watts
Source power = 0.2/50% (i.e. 50% efficiency at the given load current) = 0.4 Watts
Source current = 0.4/3 = 0.133 Amps
Battery life = 2500/133 = 18.8 hours.
In practice battery life isn't as easy as that to calculate, since the capacity of the battery varies according to how much current to draw from it, and also the voltage at which your system stop working.
Thanks!
Yes, and also the voltage fo the batteries goes down over time...It seems a good problem to solve.