A
Agrophobic
Guest
Pete-I Hope i can help you with your figures, the efficiency is 85% power out compared to power in. your power out calculation is correct but as the input voltage is higher the input current is lower.
At 100 % efficiency 3.8w/4.8v=0.79A(from battery)
At 85% efficiency (P in=3.8w/0.85=4.47W) 4.47w/4.8v=0.93A
so 930mA in at 4.8v gives 1A out at 3.8v
using 2650mah batteries- 2650/930=2.8 hours. (in theory)
however- efficiency improves as battery voltage drops so you'll get a bit more than that.
and........the figure of 2650mah is quoted from discharging the batterys at 500ma until they drop to 4v.
so in reality you'll get less capacity by discharging at 1A but still get usefull light when the batterys drop below 4v. the only way to really tell is to time the dishcharge to a usefull light level, as you have done.
At 100 % efficiency 3.8w/4.8v=0.79A(from battery)
At 85% efficiency (P in=3.8w/0.85=4.47W) 4.47w/4.8v=0.93A
so 930mA in at 4.8v gives 1A out at 3.8v
using 2650mah batteries- 2650/930=2.8 hours. (in theory)
however- efficiency improves as battery voltage drops so you'll get a bit more than that.
and........the figure of 2650mah is quoted from discharging the batterys at 500ma until they drop to 4v.
so in reality you'll get less capacity by discharging at 1A but still get usefull light when the batterys drop below 4v. the only way to really tell is to time the dishcharge to a usefull light level, as you have done.