Maximum Voltage when charging 12V NiCd Battery pack (from cordless drill)

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umetnic

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Hi!

I have this old 12V Black & Decker cordless drill with two Ni-Cd battery packs that are rated 12V and should probably hold 1200mAh. I also have corresponding charging 'docking station' with this circuit inside:




I am only missing 'the power supply part' that goes into outlet and transforms 230V AC to ? DC.

So I will use another DC power supply that I have lying around, but I need to be sure what output voltage do I need. Is it 12V like the battery pack or is it higher?

When testing for example with 14,5V, I get 100mA current to the battery docking station.

My question is, Can I go even higher with voltage? Or will I damage anything....:-?:grin:
 

You are missing the transformer that reduces 230VAC to ?VAC. I cannot see the color bars of the burnt resistor on your docking station. You can feed 15.8VDC into the terminals marked "AC". The rectifiers will reduce the voltage to 14.0V and the current will be zero when the battery is fully charged. The value of the burnt resistor determines the maximum charging current and how long it takes to charge.

A Ni-Cad cell is about 1.2V when it is being used. So your "12V" battery pack has 10 cells.
A Ni-Cad cell is about 1.4V when fully charged and is still charging (about 14.0V for your battery). The trickle-charge current should not exceed 1/10th its capacity of 1200mAh which is 120mA DC.
You can charge with a higher voltage but the current must be limited to 120mA or less with a resistor. Do not leave it trickle-charging all the time and never discharge the battery to below 10V.

But the original charger circuit does not have a huge capacitor for a filter so it charges with full-wave rectified pulses. Your multimeter might not be able to measure the average current accurately.
 
Thank You for fast RE, Mr. Audioguru!

I was playing with resistor calculator and trying to figure out exact colors, but it's really hard to tell exactly...and if I understand correctly i can not measure resistance of resistor directly on the circuit board...(If I Try I get only 8.2 Ohms) so that's that.

I counted the cells inside =10 it is.


If I feed 15,0V DC to 'AC terminals' the current on MM says 120mAh

If I feed 15,8V DC to 'AC terminals' the current on MM goes up to 200mAh.

I checked Wikipedia for trickle-charging and full-wave rectified pulses and I hope I understood everything in Your post. But I am still a little confused...so please, can I charge with 15,8V and 200mAh and will the current automatically go down when approaching full capacity, or do I need to go with 15V, just to be sure that when the battery will be almost full it is not charged with more than 120mAh?

p.s.
When charging with 15V...the battery gets 12,5V from the board.
When charging with 15,8V...the battery gets 12,8V from the board.
 
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Then ohm's Lawe says that the voltage across it for a current of 120mA is 8.2 ohms x .12A= 0.984v which is very low.

When charging with 15V...the battery gets 12,5V from the board.
When charging with 15,8V...the battery gets 12,8V from the board.
The voltage changes with the amount of charge. If the battery has been charging for a long time then the rectifiers reduce the 15V to 13.2V and the battery is only 12.5V so the resistor has 13.2V - 12.5V= 0.7V and the current in the 8.2 ohm resistor is 0.7V/8.2 ohms= 85mA.
The battery voltage is so low that I think the battery has one shorted cell because 9 cells at 1.4V each make 12.6V.

ALL my old Ni-Cad battery cells became shorted after a few years. It was a common problem with Ni-Cad. Ni-MH cells do not do that.
 
Dear Audioguru,
just finished test-charging for both packs. They both charge up to 13,8 and they look that they hold the charge for now, so probably all cells are ok. Still, one pack charges with 40mA more then the other...but that's normal I guess.

And it definitely has to be 15.8V, or the current goes to zero after 1h or so into charging.

Currently one battery has 13,8 and other 13,5...will see how they hold up after a few drills...and than look into possibility about bad cell that You mentioned.

Thank You again!!
 

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