A daily life battery charging question..

[sappy2000]

Hi all,

I have a black and decker drill powered by a 18v ni cd battery. It comes with a charge which output 21.4v 210mA 9w. However the charge blown.

So we a did was charge the battery with a pc power supply with 17v terminal. It starts off with 300mA and after a day the current dropped to 60mA. The battery was only 60% full after the charge.

I tried charging it with the 24v terminal of the atx ps, the current was quite high 860mA.

So I was wondering how should I charge the battery? I am thinking using a buck converter to turn the 24v to 21v. However will that work as buck converter is a switching power source? Or what is the proper way to charge it?

Does each type of battery requires a different way of charging? Ie Li On, Ni Cd, Etc?

 

[chuckey]

At this power level, I would have thought a straight linear regulator to drop 24 -> 21.4 = 3.6V at a current of 300 mA = ~ 1.1W, Try 4 or 5 Si diodes in series? Often the proper battery chargers sense the temperature of a cell and reduce the current to keep it below some figure, so as to preserve the life of the battery.
Frank

 

[sappy2000]

Thanks, using diodes sounds like the simplest way to go.

I tried using a 5k pot to bring down the voltage to 21v but it was smoking and melting after 10 second. Why is that?

 

[chuckey]

The problem is that a 5K pot is only rated for less then 1mA of current through its slider contact, not the 300mA you are using.
Frank

 

[sappy2000]

Ah.. Thanks for the lesson, this is the second pot I blown. But the small blue one with the little gold dial on it didn't blow, I used it on a breadboard. It gets a little warm only. Does that one has a higher watt rating?

They looks like this

**broken link removed**

:razz:

 

[sappy2000]

Hi there,

I removed the resistor and just connect the 6 diode (rectifier) in series between the +24v terminal to the positive of the battery. However it's drawing 4.0A into the battery. Should I be using a resistor? What value should I use? Do I have to look at the power rating of the resistor?

thanks:shock:

 

[sappy2000]

Hope someone can help..

 

[chuckey]

Have you taken apart the charger yet?. If its a wallwart then if you carefully run a fine hacksaw around the groove you can open up the moulding. One common fault if its of this sort is the thermal fuse on the mains transformer breaks. To check this see if the primary of the transformer appears to be open circuit. If on inspection, the transformer looks alright, i.e. no signs of burning or overheating etc. then very carefully pull the tape of the transformer and you will find a little thermal fuse which is open circuit. Solder a short across it. Re-assemble the case and tightly wrap insulating tape around it to hold the two halves together. Do not operate the "repaired" unit unattended, because it has no thermal protection on it and it might catch fire.
To get back to your question, have have 24V, you need 21V at .3A, so you need to drop 3 V at .3A, so you need 3/,3 = 10 ohms, at a power of 3 X .3 = .9W, i.e.1W.
Frank

 

[sappy2000]

Hi Frank,

Thanks for the reply. However, I have a problem, I am using a Pc power supply. When i use 10 ohm resistance, the current is drawing more than 2A!! To set it to 300mA, I used a POT and found that I have to use 18.6ohm, but the voltage drop across it is 5.6volt. The psu only supplying 22volt for odd reasons, so my battery only get 15v:sad:


Thanks

 

[chuckey]

if you use a 10 ohm resistor and are running 2A through it, it means that the resistor is dropping 2 X 10 = 20V, so your battery must be less then 24 - 20 = 4V. In this case you need to drop 24 -4 = 20 V at .3A = 20/ .3 = 66 ohms. If your battery is getting 15V, as it charges up the current will drop and the available voltage will increase. Just leave it on for a couple of hours and see what happens.
Frank

 

[sappy2000]

I am using six ~100 to 150 ohms 0.25W resistors to make the current at 300mA. Does putting six 0.25W in parallel makes it into a 1.5W resistor?