Hi all - I have a row of these cheap solar garden led light alongside a path that is mostly in the shade, so the batteries very seldom charge sufficiently to give more than about 30 minutes of light at dusk, and are completely exhausted by the time it really gets dark and we could actually benefit from them!
I am thinking about attempting to wire them all to a common 1S3P 18650 li-lion power pack in my shed, which would in turn be charged from a much better placed solar panel via a ‘proper’ charge controller.
So I am guessing the easy part is simply removing the batteries from the lights, and wiring them to the li-ion pack instead. My question is whether I should be leaving the remainder of the circuit as is, but regulating the input voltage down to 1.5v at each light, or can the YX8018 operate from a 3-4.2v supply without releasing the magic smoke, and if so, should I replace the inductor (between Vcc and LX pins) with a current limiting resistor instead?
Any thoughts, or alternative solutions with the emphasis on cheap (I.e. as close to free as is practical), and minimally invasive. Given that the lights were only $2 each, spending $$ on a solution kind of defeats the object of the exercise really.
TIA
Grum
I am thinking about attempting to wire them all to a common 1S3P 18650 li-lion power pack in my shed, which would in turn be charged from a much better placed solar panel via a ‘proper’ charge controller.
So I am guessing the easy part is simply removing the batteries from the lights, and wiring them to the li-ion pack instead. My question is whether I should be leaving the remainder of the circuit as is, but regulating the input voltage down to 1.5v at each light, or can the YX8018 operate from a 3-4.2v supply without releasing the magic smoke, and if so, should I replace the inductor (between Vcc and LX pins) with a current limiting resistor instead?
Any thoughts, or alternative solutions with the emphasis on cheap (I.e. as close to free as is practical), and minimally invasive. Given that the lights were only $2 each, spending $$ on a solution kind of defeats the object of the exercise really.
TIA
Grum