Black and Decker 14.4V Battery Charger Repair (revisited and improved)
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 [http://www.interfacebus.com/Resistor_Derating_Guide.html] 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!