Just out of curiosity what value of the resistor did you calculate?
Target voltage is 2.7v, current is
125ma. With a 3.7v battery, voltage drop across resistor is 1v.
Thus, 1=.125/R, so R=8.
I would use a 7ohm resistor instead, in order to account for the various random resistances along the way.
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If you want to prioritize runtime, why not aim for a higher resistance value and lower current?
Correct, assuming the battery is 3.7v all the time. however battery is not 3.7v at all times, it starts at 4.2, by 3v it is time to recharge. it is pretty significant fluctuation for such small load, on one end you need around 3ohm resistor on another 11, to have that current. Not saying it will not work, it will work with fluctuating output, and significant loses, honestly Iām not sure it worth the troubles. to get somewhat better performance, and no risk of leaking, just use lithium AA.
But it is your project, either way let us know how it works out.
Edit. what this graph does not say it what kind of alkaline batteries they used, cuz they are not all the same. I have seen duracell optimum, AA push out 4A after about 30 sec of "warm up"time, while regular duracells and energizers top out at 2A in that particular test.
Iām already running the emmitter close to its minimum current. If the resistance is any larger Iām worried I wonāt get any light at all.
See my comment in your other threadāthere is no such thing as a āminimum currentā, and the amount of light you receive is essentially proportional to the reciprocal of resistance. So (at least in theory) as long as the resistance is finite, you will get light.
According to this graph, there is a nearly insignificant time where the battery can actually output 4.2v - it quickly drops to 4v.
Fluctuating output is fine - it gives me a warning that the light is about to die. I donāt like how a lot of modern LED flashlights just cut out with no warning. Brightness acts as a proxy for SOC.