Setup
Let's say I have a 12V / 1Ah rechargeable battery pack. A device that consumes 0.1A will run for 10 hours.
When the battery drops to 8V, the voltage will be insufficient and the device won't work anymore.
Most of the current flows through a transistor that sends pulses to a coil (of a metal detector). The pulse width gets shorter as the voltage increases so I guess, the consumption remains the same not matter which is the battery voltage.
Question:
Can I increase the number of operating hours for that device WITHOUT increasing the capacity (Ah) of the battery? More exactly by adding extra batteries (series) to the pack and thus increasing the voltage from 12V to 14V? (We assume that the device can operate also to 14V).
My explanation/thoughts:
The capacity of a 1Ah of a battery means that it can provide 1A current for one hour without dropping its output voltage under 75%.
If the device stops working at 8V, it means that the battery can be used from 12V downto 8V. This is a 4V difference.
By adding another battery, the total capacity is still 1Ah. But the total voltage is now 14V. So, the battery pack can be used from 14V down to 8V. This is a 6V difference! Therefore, by adding another battery, it takes longer until the voltage drops from max to min. HOWEVER, the load will "burn" more current because now it operates at 14V instead of 12V (I = U/R).
Best Answer
12V,1Ah means you've got 12Wh. 14V,1Ah will get you 14Wh.
So yes - you've got more "power" (and you did increase the capacity of the batteries by that...)
Now it depends on how your circuit will use that energy. When it needs less than 0.1Ah @ 14V, it will work longer. If it "burns" the excess voltage, and still need 0.1Ah, regardless of whether it is running at 8,12 or 14V, it will run for the same duration