# 3s in compare to 1s battery circuit

Hi,
A general question about current consumption of batteries.

1. If I am using 1 lithium battery(~3.7) to drive a dc motor(12V) or any other load, using a boost converter(to 12V of course).
Where are my power losses in comparison to using 3 batteries(~12V) other than the boost itself?
What I'm asking is what is my motivation of using 3 batteries instead of 1?
2. In contrary to that, the same question about doing the opposite thing - using 3 batteries to drive a 3V dc motor(and a 3v regulator to match the voltage).
Hope it's clear enough.
Thanks,
Omri.

omri_saporta:
What I'm asking is what is my motivation of using 3 batteries instead of 1?

The wires dont need to be so thick as they are carrying less current.

You save the cost and reliability impact of using a boost converter.

The simple answer is that 3 batteries have 3 times the stored energy.

Power is calculated as voltage x current. (Energy is power over a period of time.)

Switching regulators & boost converters are nearly 100% efficient. So, assuming 100% efficiency you need 3 times as much current going into the boost converter as you get out and your battery will last 1/3rd as long.

A switching regulator that's cutting the voltage to 1/3rd can put-out 3 times as much current as you are taking out of the battery.

With linear regulators, the current-in is the (about) same as the current-out so any voltage dropped across the regulator results in energy is wasted as heat. But, still there can be an advantage of a higher-voltage battery because if you are regulating a 12V battery down to 3V the thing will work until the battery is drained down to nearly 3V, way beyond the battery's normal life. (But, you'd get even more life with a switching regulator.)