R. Ortiz

I have a (wall wart) battery charger that puts out 4.8v at 50mah. I need to reduce the current output of this charger to a maximum of 40mah (preferably 35mah) in order to safely charge 400 mah batteries at 10%C.
Can I accomplish this by placing a resistor between the charger and the battery? If so, what size (value) of resister would I need and should I place it on the positive side or the negative. Would placing a resistor affect the voltage output.
My knowledge of electronics is very limited so I will be grateful for any suggestions that you are willing to share. Thanks

guver

My Feedback: (325)

My electronic knowledge is limited too , but anything you do to add resistance in the complete circuit will raise the voltage and thus lower the current on that type of charger. A full batery will be going with less current than a discharged one (maybe you can measure this AND you may also find that the actual current is less than expected anyways)

Wouldn't it be much easier to reduce the time instead of the current though? Inotherwords charge for 12 or 13 hrs instead of 15 hrs?

Rodney

Here is an excellent site to get info on how to use the wall warts for other than there intended purpose as well as much other good advice on battery care.
http://www.hangtimes.com/redsbatteryclinic.html
Basically the answer to your question is yes, a resistor in series with the output will lower the current.

exeter_acres

My Feedback: (2)

while I do not have an answer to the original question other than a digital charger or something...

50mAh is absolutely fine for a 400mAh pack and will not harm it at all

porterdog

FWIW, I just now charged a Gens Ace 5300 mAh 2s lipo at 20A.  It took about 18 minutes and the battery didn't heat at all.

iCharger 208 FTW.

bgosselin

Short answer: yes, a resistor will limit the current.


Long answer: Yes, a resistor will limit the current.

But you did not state what chemistry you are using. If lipo, I would not use it. Lipo needs a charge scheme that uses a constant-current charge up to exactly 4.2v per cell, and then switches over to a constant-voltage charge until the current dwindles to a pre-set value (usually around 10% of the initial charge current). So, even if the current value of the wall-wart is safe, it still doesn't have the 4.2v voltage limiting circuit lipos NEED.

If charging NiMHs, then it depends. How many cells are you charging? If any more than 3, they may never reach full charge. 4 NiMH cells are 4.8v nominal, but peak higher (~6v) before charge end. As the battery voltage climbs towards full charge, the charge rate will be less and less (current would be defined by the deltaV divided by battery and transformer resistances). Since NiMH's can take 0.1C charge rate just about indefinitely, this might work for you. But it might not.

Those wall warts use a simple transformer (to reduce voltage) and rectifier (to turn AC into DC). The combination of the load and the transformer's output impedance creates a voltage divider network whose voltage at the load varies with current being drawn. So, even though it might be rated at 4.8v @ 50mA, it will most likely measure quite a bit higher voltage unloaded, and lower voltage when loaded to draw more than 50mA. Adding a resistor to reduce current will lessen the load and you might be getting 5v output instead. You could use various load resistances and determine the transformer's output impedance and then do the calculation to see if it will be sufficient for your needs, but that's probably more than you want to do. They're simply not very precise and are used for devices where they draw a known current and/or aren't sensitive to voltage fluctuations.

R. Ortiz

Thank you very much for the information. I just wanted to use the wall charger for the initial charge to form the new NIMH batteries. After that, was planning charge them with peak detection charger. I was told never to use a peak detection charger for the initial charge or go over C/10 when charging with the wall charger. I had been thinking of putting a 150 OHM resistor between the charger and the battery because I assumed the output numbers on the wall charger were stable through out the charging cycle and the resistor would bring down the current below 40 mah.
I guess there is more to it than that. Like I said before, my knowledge of electricity is very limited so I very grateful for all of your imput.

KB9STD

My Feedback: (1)

YEs,No and "HELL YES" You're an electronic Genius! I hope I got those in the right order,But it doesn't matter.A resistor in series with the "wall-wart" will indeed lower the charging current to the battery.It will also raise the voltage supplied a little bit,But that doesn't matter,The battery doesn't care about voltage/You are using current to charge the battery,and the battery is determining it's voltage based on it's chemistry and it's present state of charge.
How many of You have seen one of the older RC radios from when Ni-Cad batteries first became common? They placed the transmitter and receiver batteries in series(9.6 + 4.8 volts=14.4 volts) and charged them straight from the 120 volt wall outlet.No transformer or voltage control of ANY kind.Just a diode(to ""rectify" the AC to DC) and a resistor(a "current-Limiting" resistor).A very dangerous practice,by the way.That resistor did exactly what the Poster is suggesting with the resistor between the "Wall-Wart" and the battery.
I don't know if the 150 OHMs you suggest will be the right size or not(I would guess You don't need that much resistance,for the small change you are after). The size of the resistor depends of the specific charger and even the battery you are using.You will have to try it, and measure the current it puts into the battery.Just get it to around 1 tenth the battery's rating.Then charge the battery for at least 18 hours(I'd recommend a FULL 24 hours).You want to put in 1 and a half times the battery's capacity or more ,SLOWLY and Gently.That overcharges some cells in the pack,But slowly enough that they don't overheat,while the rest of the cells(the "slow-Learners") come up to full charge.