Divide voltage by 2

stumpfield · Dec 13, 2006

Anyone know a simple and inexpensive way to divide the output voltage by 2?
I want to charge a 6v battery. My battery charger is 12v only. I don't want to connect 2 6v batteries in series each time I want to charge them because each have different state-of-charge. Any ideas? Thanks.

RavensRoost · Dec 13, 2006

I think the only real answer is to use a 6 volt charger. Any other scheme will likely only over or undercharge the 6 volt battery. I don't think there is an easy way out....

RavensRoost

randy41 · Dec 13, 2006

seems to me that you could put a resistor in the + line to cause a 6v drop. but i dont know what size.
hmmmm....there is a relationship between volts, amps, and ohms. but i think the amperage output of the charger is not constant.

RavensRoost · Dec 13, 2006

randy41 said:
seems to me that you could put a resistor in the + line to cause a 6v drop. but i dont know what size.
hmmmm....there is a relationship between volts, amps, and ohms. but i think the amperage output of the charger is not constant.

Randy, you nailed it. You could choose a charging voltage and current and select a resistor to give the 6v battery what it wanted.....for a very brief time until it chatrged up more and now wants less current/higher voltage. I am sure there are "intelligent" devices that will do this. I bet they also cost more than a "dumb" 6 volt charger ;-)

czechsonofagun · Dec 13, 2006

randy41 said:
seems to me that you could put a resistor in the + line to cause a 6v drop. but i dont know what size.
hmmmm....there is a relationship between volts, amps, and ohms. but i think the amperage output of the charger is not constant.

yes, this is correct. If you have an electrical heating element I would connect it in series with the battery and check for the voltage on. 6V battery could be charged by something like 8V (12V is charged by 14.8V in a car).

my $0.02 only

Highbeam · Dec 13, 2006

V=IR or P=IE

Voltage (potential) = current * resistance

You've got to get the units right but that's the relationship.

Ken_CT · Dec 13, 2006

Whenever I try to do something like that, the most expensive part in the arrangement gets ruined. Kinda like a $300 picture tube blowing to protect a 10 cent fuse

.

heehaw · Dec 13, 2006

i've occasionally needed to do something like your talking about, and wired a light socket in series with whatever i'm tryin to burn up, oops, i mean charge or operate: anyway, the more current you want, the bigger the bulb you screw into the socket. actually this does work "fairly" well for a cheap motor speed control on small AC motors. what the heck, light bulbs are cheap, and "maybe" it won't cook the battery or charger.
heehaw

AndyM · Dec 13, 2006

By the time you rig up whatever it will take to drop the voltage in your charger, it would probably be easier to buy a cheap six volt charger.

ericinmich · Dec 14, 2006

Highbeam said:
V=IR or P=IE

Voltage (potential) = current * resistance

You've got to get the units right but that's the relationship.

It was stated in another post, but I'll try to clarify.

The V=IR is true, but the problem is that the I (current) would be changing during the charging. You've all seen that when you hook up the charger to a battery it reads high (10, 15, ?) amps right away, then starts coming down over the next minutes and hours. You would need to have a variable resistor and a volt meter across the battery (or resitor if you like to subtract) to keep the correct voltage for charging.

Much easier to get a charger that has a 6V setting like my 20 year old one does...