Electronic – Does boosting voltage using MT3608 from 3.7V to 10V module reduces the Lipo battery capacity

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Suppose say, Iam trying to boost the voltage of my 3.7 Lipo battery to 10V using MT3608 STEUP UP CONVERTER and I need to run 3 LEDs(connected in series) which takes 9V (3v for each LED)

If my LIPO battery specifications are 3.7v and 750mAh, how long does my LEDs last with that battery if my voltage had been boosted to 10v?

After boosting my voltage to 10v from 3.7v, Do the capacity of battery still be 750mAh or does it reduce? If it reduces how much will be the output capacity at 10V. Explain in brief

P.S : I know that number of hours a LED can light up from my LIPO battery can be calculated by 750mAh/30mA = 25hrs (assuming my LED takes 30mA forward current and 3V forward voltage).

I would like to know the boosting circuit effect on my battery capacity

Best Answer

With a 100% efficient power conversion from 3.7 volts to 10 volts, any current taken from the boosted 10 volts will be seen as 10/3.7 (2.702) times as much on the battery so, it's probably better to just consider the battery capacity as unchanged but, in these circumstances it will be dispensing 2.702 times as much current.

Given that the efficiency of a boost converter is probably around 90%, you should account for this also. This would mean that every amp taken at 10 volts will be more likely equivalent to 3 amps at 3.7 volts.

assuming my LED takes 30mA forward current and 3V forward voltage

Three series LEDs will take the same current and this will be equivalent to 90 mA from a 3.7 volt battery. I'm assuming here that the 1 volt difference between 9 volts and 10 volts is used for a circuit for maintaining a constant current into the LEDs.

how long does my LEDs last with that battery if my voltage had been boosted to 10v

The added complexity here is that as time passes, the battery voltage diminishes and to produce the same 10 volt boosted output requires a larger step-up ratio and, in terms of current drawn from the battery this means more current. So, it's quite a complex relationship if you factor in the diminishing terminal voltage across the battery.