Using 33uH inductor in the circuit, LED can draw 110mA current. This is the not full power required to get full brightness of 1W LED but 110mA gave good light.
Thus, 1W LED works with QX5252F in the solar garden light circuit.
This thread can be closed now.
Here is what I have gathered here from the answers, comments, and some additional research:
1) The cable that shipped with the device should work for my purposes. However, I will need to be able to install the correct driver(s). Upon further research, I was able to figure out that the device manufacturer downloads include a Windows installer for this ...
Is it necessary to do the ethernet cable splicing as indicated in the
above link, or should I simply be able to use the provided USB cable?
If you can find drivers for the R Pi then you can use the converter. I'd start by finding out what chip is in the propriety RS485 to USB, they might have a linux driver.
Is it likely that the shipped cable is ...
A solar cell has a point at which it supplies maximum power. While you can get away with a switching regulator, it is not optimal for drawing power from the cell.
One problem is you are using a switching regulator connected directly to the solar cell. To use sufficient power from the cell, an switching regulator with Maximum Peak Power Tracking (MPPT) ...
You've made a mess of the charger circuit: -
Your "switch" node doesn't have a schottky diode to ground and it appears that your boost capacitor is inappropriately connected.
It looks like your reverse battery protection will work but why over-complicate a reliable standard circuit such as this one: -
Is it because of the charger circuit might keep the ...
A 5mm White LED draws 15 ~20mA @3V or 45mW ~60mW.
Your Solar panel is 2V 40mA = 80 mW max
Thus you 1W/50mW =20x bigger power LED all components, Solar Cell, & battery Ah capacity need to be increased by 20x
If your 1W LED is a single white LED with a voltage around 3V, it will work in place of any other white LED. If it's made up of several LEDs in series, then it probably won't. Check the voltage rating specified by the manufacturer/seller.
However, don't assume that you will get the full 1W from your circuit. 1W would require nearly 1A from the battery, ...