i attached a usb charger to my solar panel(it has an output of about .2 volts) but i read on my multimeter its only using about 8 volts however on the conventional wall charger that i use to charge it has an output of about 5. volts.... i believe that my mp3 player is not charging it does not display the charging icon when it is on or off ....is this becuase the output is higher ? do i need to install a volt dropping diode to prevent it from putting to much energy at once? (my mp3 charges fine on the wall charger but does not seem to respond to the solar panel so it the circuits are fine....)
5.3V is the open voltage or? So the full wattage of the photograph voltaic panel is 45W or so? i assume you are able to no longer make all your small A/C home equipment into DC ones. so which you would be able to think of bearing directly to the 2V DC enter 00W inverter. this type of inverter might have a some bit extensive enter voltage variety from like V-20V some situations.
You description is not clear as to what is supplying and what is consuming the .2, 8, and 5. volts you mention. I get that your solar panel output is .2 VDC (volts DC), but is that open circuit or under load? Also, is the charger output AC or DC? What's using 8 volts? Circuits don't selectively use a portion of the voltage supplied. They use all or nothing. Is the polarity correct from your solar panel to your MP3 player? If your MP3 player requires DC for charging and you're feeding it the correct polarity, but too high a voltage, an overvoltage protection circuit in your MP3 charging circuit might be blocking the charging current from your solar panel. Go through everything again and if you need to post another question on YA, be specific as to what you're measuring, where, and whether it's the source or load you're measuring.
Have okorder . This could undoubtedly assist person!
