Fixing my Black and Decker 14.4V Battery Charger

UPDATE!!! – I revisited this repair. Read the history below then go here:
Black and Decker 14.4V Battery Charger Repair (revisited and improved)
History: I own a 14.4V Black and Decker cordless drill with a laser. My wife bought it for me for some reason; probably because I liked it, and she wanted me to build something.
Measuring the voltage drop across the 10Ohm resistor

Measuring the voltage drop across the 10Ohm resistor

A little over a year ago, I was given a 14.4V B&D Firestorm drill that quit working. The Firestorm charger quit charging, as verified by the transformer windings giving me an open reading on the “Oh-meter”. To clarify, “‘Oh’, that’s why it isn’t charging.”

I assumed the batteries were good so I hooked them to my regular 14.4V B&D charger. The LED charging indicator glowed for a few minutes, then it smoked somewhere and died. I never really discovered where the smoke came from until more recently.

How I Tested It: I tested the voltage at the charger’s terminals where the battery mates up to it. It measured 16 volts under no load. Next, I hooked a 100 Ohm resistor in series with my DMM at the terminals to measure the current. Calculating the current [I=V/R] it should have been 160mA. The charger should have been able to deliver this amount of current without a problem. Its rating is 17.4v @ 210mA. The current I measured was next to 0mA. Something was wrong.

Disassembling the Wall Wart: I took a razor blade to the wall wart. I usually use a rotary tool, but I couldn’t find it. Rotary tools are a bit safer than using a box cutter. Once inside, I immediately noticed a charred resistor. It was burned open. I’m not really sure what the resistor does, but I am assuming it is either a feedback or simply a current limiting resistor. Honestly, I really didn’t take much time to trace the parts to get an idea. I just wanted my battery charger to charge.

The Fix: I removed the burnt resistor with solderwick and a soldering iron. It was burned so bad that the color bands were unreadable. Not that I could read them with any accuracy anyways due to being colorblind, but it could have at least given me somewhere to start.

I replaced the unkown resistor with a 1kohm resistor. I began charging a dead battery and measured the voltage drop across the resistor and concluded that it wouldn’t catch fire [I didn't exceed the power rating of .25W]. I’m writing this long after the fact, and I don’t remember the voltage. I’ll explain my final calculation in a minute. I decided that since the battery didn’t hold a charge for more than a few screws were driven that I would replace that resistor with something smaller.

The second time I used a 120Ohm resistor. I followed the same procedure as before to make sure that the power rating wasn’t being exceeded for the quarter-watt resistor. After a week of charging, I used the drill for a longer period of time. It seemed to hold the charge for a reasonable amount of time. Then I tried a battery that charged for 24 hours. It died fairly quickly.

Tonight, March 16, 2010, after a bunch of Calculus homework, I decided that I wanted to change the resistor value again, and then I wanted to write about it. I changed the 120Ohm resistor for a 10Ohm .25W resistor. This is where I actually started to take notes.

I soldered in the new resistor leaving enough room to clip leads to it to measure the values. I docked the charger to the dead battery, and plugged in the charger. The battery is now charging. I hooked my DMM up in parallel to the 10Ohm resistor to measure the voltage drop. The voltage drop was about 1.6 Volts DC.

Current Through the Resistor:Voltage = Current x Resistance [V=IR]. V=1.6V, I=?, R=10Ohm. Rearranging the formula to find the current gives us I=V/R. I=1.6V/10Ohm. I=160mA.

Power Dissipation by the Resistor: Power = Current x Voltage [P=IE or P=IV]. V=1.6V, I=160mA, P=?. Simply plugging into this formula P=160mA * 1.6V= 256mW. This is the same as .256 Watts. This is pretty close to the rating of the resistor.

Another Way to Figure Power Dissipation: I could have skipped the first step in finding the current through the resistor. P=V^2 / R. P=1.6^2 / 10. P=256mW.

Final Thoughts:I did this test with a battery that was even more drained, and the voltage drop across the 10Ohm resistor started at 3.8 volts, and quickly dropped to 1.8 volts. This exceeds the power rating of the resistor, but I’ll stick with the 10Ohm until it burns out. At 3.8 volts, the power dissipated by the resistor is 1.444 watts. This is way too much current for a .25W resistor. The leads acted as a heat sink, which probably kept the resistor from turning black.

I believe that I will try a 15Ohm resistor if this one burns out. I expect that it will eventually burn out based on the starting power levels. The resistors can take it for a little while, but it will eventually degrade them. This site [] has a graph at the top that shows what happens to the power rating as the ambient temperature rises. As the power dissipated by the resistor rises, so does the temperature. I assume that the temperature of the resistor was over 150 degrees F. That’s not such a big deal, but it just nudges it one step closer to a premature death. This will be grounds for another experiment later on!

About robbie

I am an electronics enthusiest and a ham radio operator (W1RCP). I like to play with electronics. It's fun and educational. I looked forward to working in the engineering field in the future. I have a BS in Electronics Engineering Technology from DeVry University. I also have an Associate's degree in Marketing Management from Moultrie Tech, and a diploma in Electronics from MTC.

26 thoughts on “Fixing my Black and Decker 14.4V Battery Charger

  1. nice catch! I’m having a similar problem, except that my adapter is different. I have the version that has only a transformer in the wall-wart. the tiny circuit board is contained in the part that slides onto the battery. my problem is the transformer. it is no longer supplying any DC voltage to the battery. it is rated 17.4V DC at 210mA. i assume there is no practical way to repair the transformer. is there a way to use a different wall adapter and splice it on to the battery cap? can i use a slightly different voltage or current? thanks!

  2. Yes, you can use a transformer that is a little higher than your battery’s voltage. Jameco part number 121216 – 18volt AC adapter It’s about $11, and is probably built much better than the original equipment.

    You should be able to cut the end off this wall-wart and splice it into the existing circuit.

    I am assuming that your transformer box only had a transformer, and the diodes were connected on the battery end’s circuit board.

    Check and make sure there are no shorts in the circuit (make sure the diodes are good, resistors are good, and that the caps have a high ohmage rating). Install this, and you should be ready to go.

    Let me know if there’s anything else I can help with.


  3. thanks for the quick response! you are correct about the diode placement. I’m assuming that the circuit in the battery end is ok…it’s just the transformer that’s bad.

    the adapter you listed has an amperage of 1000mA. is that not a fire hazard or detrimental to the battery? the original had an output of 210mA.

    also, it says that the output is AC, while the original was DC. is this a problem?

    would an adapter with an output of 18V and 400mA, or something similar, suffice?

    thanks so much for your help.

  4. I might be a little confused, and I don’t want to send you in the wrong direction. I think a photo shoot for your parts might be in order. If you could, email me, instead. My email is:

    Email address edited out to avoid spam :)

    The amperage noted for the Jameco part number is the maximum allowance for the transformer. The actual battery circuitry limits this current using a resistor. If you do find an 18 volt transformer that is 400mA, it will work and probably be a few dollars cheaper. I was working on a detailed Instructable, and didn’t look very hard. I apologize for that.

    Shoot me an email with pictures. Either cell phone, digital camera, webcam, and let’s see if I can help solve your problem. I’ll need a picture of the inside of the wall-wart itself, and the picture of the part that connects to the battery to be charged. Several views if you have time.


  5. I am having problems with my 14.4V B&D battery charger. My B&D origional charger died and would not charge. Even the LED light would not come on with or without the battery. I purchased another charger on Ebay. Unfortunately I did not open it for a month and that too did not work. The LED indicator light did not come on with or without the battery. I would appreciate any suggestion. By the way how do you cut open the wall socket?

  6. To open it, you’ll need to cut carefully along the perforation along the side of the black wall box. I usually use a cutoff rotary tool and basically melt mine apart. I used a box cutter on this box. If you have a multimeter, simply check the parts for a normal tolerance. I found that there was a severely burned resistor inside that I replaced. I recommend a 1/2 to 1 watt resistor since it may get hot. You may also need to replace the LED, but only testing will tell.

  7. Thanks for the instructions. I have a firestorm 18v charger and almost threw it away when it stopped charging. On a whim, i opened up the wallwart and sure enough the resistor was burnt beyond recognition.

    Hopefully, a visit to Radioshack will have this puppy up and running soon.

  8. Hey, I just ran across this site and I took apart my wall wart and found a burned resistor. Which is the best to replace it with? is it still the .25w 10ohm?

  9. I put in a 1/2w 10 ohm, resistor and put my pack on charge. About 2 minutes I started smelling something and looked and my resistor was smoking, so I unplugged the charger removed the batt, and then plugged it back into the wall. I am getting about 15v on the charger end without the batt plugged in. Any ideas? And should I replace the smoked resistor even though it is still measuring 15v?

  10. I did a little research last night and I think I just got lucky with my resistor not burning (also didn’t solder my resistor close to the PCB, so I have some free air around it). Depending on how low the battery voltage gets, the current through the resistor could be higher. If you have a 15V output from the transformer and your battery is at 10v, there would be a 5v difference across that resistor. 5v^2/10ohms = 2.5watts. I am going to add these calculations to this article soon.

    Also, you might be able to help me out on something else. Is there a 14-pin integrated circuit in your wall wart? I tried to reverse engineer the charger, but mine has the part numbers sanded off. My transistors are also unlabeled. If you have the time, could you email me some close up pictures of your charger and possibly list any numbers that are written on the black parts to robbie[at @]

    You can also email me there and I can work you through this charger issue. If you have NiCd batteries, there is a way to charge them with just a power supply and a hi-power (5w) resistor, which is what I feel that we are doing in this case.

  11. If you have a multimeter, start at the plug and check for continuity between the spades. If you get a continuity beep, open up the charger and look for burns. You may want to plug the wall-wart in and check for voltage at the charger end. It’s a hide and seek game, but I believe that most issues are fixable.

    If you have any questions or need help I can be reached here —> robbie [et,at,@]

    If you email me, send pics of your circuits. Also, let me know what kind of testing tools that you have around and what your level of electronics expertise is. I look forward to hearing from you.

  12. Excellent write-up. Was hoping for some insight on my problem, with a B & D 12v charger. When plugged in, I get 0.2 volts at the slots where the battery would plug in to the charger. Should have 14.5v there, per the back of the charger. I can’t see anything burnt on the inside of the charger. Any ideas where to look? The good news is I can use a motorcycle trickle charger and some alligator clips to charge the batteries, as they are 12V, although I’d prefer to get the dedicated charger working. Thanks.

  13. Hi and thank you for your insightful comments. I am not that tech savy and I have a question.

    I have lost my wall charger which gets connected directly to the back of the drill and charges the 14.4 V B&D battery. Now I cannot recharge the battery!

    What is the least expensive way to get around this. Does any of you suggest an inexpensive charger which can replace the B&D charger?

    Thank you

  14. The transformer of my charger is not working, I want to replace it but I don’t know what the output voltage is. does anybody know? -It is a firestorm 14.4, 1A charger DC output 18V-

  15. I have a similar problem, though my charger seems a bit different from the pictures in your newer article.

    My BW 14.4v battery has a charger that consist of a transformer that plugs into the wall ( 17.4v AC) , joing by a wire to a plastic cover that snaps onto the battery. This cover contains a sm circuit board which is just a standard bridge rectifier circuit with an LED ( and its resistor) and a current limiting resistor of probably 1/4 or 1/2 watt. This resistor is charred beyond recognition.

    My initial problem was that the wall transformer died and i replaced this with a locally purchased 18v transformer. this worked fine for a while , untill the resistor decided to die.

    I’m in two minds – replaced the resistor ( witn a 10Ohm , 2W) or build a better charger , using a 3-pin regulator ( 15v or 18v) IC.

  16. There’s enough room charger to solder up to four 1/2W resistors. When resistors are in parallel the wattage is added. Try using nominal rated resistances in parallel such as four 100 Ohm rated at 1/2W. Assuming there’s a voltage potential difference of 7V between the output of the charger and the battery, this will be borderline with the 2W resistor bank used above.

    If you battery is drained to 10V, one might expect severe heating and or the initiation of funky aromas from the bank. If you need to steer clear of damaging the bank from excessive current also try a slower charge time using three 150W 1/2W resistors with a estimated final trickle current of 100mA and assuming 18V charger stays at a 18V with the charger as a load.

  17. Good points. You could use four 39 ohm resistors, too. The original resistor that was in the charger was something near a 10-ohm resistor. I revisited this repair here ->

    I later found out that a transistor was burned out in my charger, and a few others have reported the same thing. Changing both the transistor and the resistor fixed a handful of chargers. I used a 1-watt resistor and it has worked fine for quite some time.

    I like your idea of the trickle charging. It would be a neat mod to have a switch to select a resistor. Even so, the switching mechanism also charges through the transistor Q1. The excessive heat is what caused it to let the smoke out. The only limit to the amount of power sent though Q1 is the PWM circuit.

    Hope all this helps those with B&D drills and chargers.

  18. I have a charger Automatic power B&D 550 watts and I think that there is a problem is no longer the charger as it was before (the index reading has become close to the end, green at 90) with all batteries of 9am/h
    Gives a short time and quickly stop the charging process and gives the message with the completion of charging ( the batt is full charged ) the battery it does not continue to operate for a long time and soon emptied

  19. I have the 100 year anniversary B&D cordless drill.
    I think I paid $100.00 about 8 years ago.
    It has very little usage so I just went for a new charger for $29.99
    B&W wanted $69.00

  20. I hav same problem in black & decker charger 14.4 volt
    i replace transformer a 15volt ac 1000mA
    the drill also work but its too weak.
    anybody can help. thanks

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