charge controller

Question for you all. I have a solar powered spot light I really like, when it's charged up. It has a 6vdc 4ah battery in it and I wish to re-charge it from my 12vdc system. Source power will be my 12vdc 160 ah battery and I'm thinking of using my extra 12vdc charge controller with a 3.5 ohm / 1 watt (or better) resistor attached to reduce the power to 6vdc. I believe that will work, but I don't know if the controller will know when the battery has reached capacity? The other thought is to just by-pass the 6vdc battery and run the light from the 12vdc system (using the resistor to reduce the power). Any thoughts?

olddawg (author) 8 years ago
Okay..?? Work with me here for I'm still new at this. I follow what you're saying, but not following the math. If I'm re-charging a 6vdc battery, don't I need 6vdc to pass? If I restrict 8vdc, don't I now only have a 4vdc charge? If indeed I need 6vdc to pass and I use the new panel I just bought (11watt @ 16vdc) . I figure I need a 8.82ohm / 4 watt resistor. Am I even close here?
the resistor does not limit the voltage. it limits current. the current it allows depends on the voltage difference between charge source and battery

if you leave the battery connected with resistor too long its possible that it'd reach 12 V. but it severely damages it

an important thing i forgot in the previous reply is a diode. you connect a standard diode in series with the resistor. it prevents the battery from discharging when the charge source is off or down

lets calculate again what we need

the panel is 16 V 11 A. the max current we can take from it is 11 W / 16 V = about 0.7 A

discharged battery is about 4 V. the resistor gets 16 V - 4 V = 12 V. it drops to 16 V - 6 V = 10 V as the battery charges to 6 V

we want to limit the max charge current (battery at 4 V) to 0.7 A

12 V / 0.7 A = about 17 ohm. the closest resistor is 18 ohm

12 V * 0.7 A = about 8 W. we need a resistor of 5/4 X = 10 W (it'll get hot) or 2 X = 16 W (it'll get warm)

the efficiency of it is less than 1/4 in the beginning of the charge and rises to about 1/3 when its charged. most of the energy is wasted in the resistor

it'd be more efficient to convert it to ac and use a step down transformer - all this just to be able to use smaller resistor
olddawg (author)  110100101108 years ago
Thank you very much for hanging in there with me. This has helped greatly!
110100101108 years ago
the controller won't stop the charge. it makes no sense in using a controler then

just connect it to 12 V with the resistor. and you need way higher ohm and watt resistor

you want to charge with current X (for sealed lead acid i guess X = 0.5 A is ok)

the battery is 6 V (about 4 V when discharged). we need the resistor to take 8 - 9 V

8 V / 0.5 A = 16 ohm

8 V * 0.5 A = 4 W

4 W * 5/4 = 5 W (hot resistor)
4 W * 2 = 8 W (warm resistor)

and disconnect it when its charged enough

btw this is extremely inefficient. more power is wasted on the resistor than going into the battery. try to find nother way to charge it with 6 V