Craftsman car battery charger8/8/2023 This is not a nonsense like ΧωρίςΌ want us to think. My simple Ohms Law calculations, lead me to a Main current –limiting Resistor of 100Ω, with a battery internal resistance of 50Ω (which sounds too high to me) and a charging current of only 95mA.ĭoes anyone have knowledge of this type of charger circuit and the resistance (and wattage) to use to replace the burned out main resistor? I could send an email with photos of the circuit board and my derived circuit diagram, to anyone who might be able to help but will try to attach these here. But without any realistic knowledge of the Battery internal resistance, the LED resistance, and the Diode resistance, I am rather in the dark. I have been trying (by means of an Excel spreadsheet) to investigate and estimate the value of the Main Resistor, by assuming the battery EMF of 18V opposes the Input EMF of 30V, leaving a net EMF of 12V, to achieve the 23V output (to battery) and a charging current of less than 400mA. According to wording on the Base Unit its output voltage (to battery) should be 23volts, rated at 17watts. The Battery unit appears almost completely ‘flat’, with only a few volts EMF. All the colour bands look black).In parallel with the Main Resistor is a small resistor (measured at 157Ω, in series with a red LED, for which my assumed resistance is 13 Ω, although that intuitively sounds high, to me). The resistor is a now dead short and the colour coding cannot be distinguished, presumably due to its having been “fried”. The Base Unit has a small printed circuit board containing a Rectifying Diode (my assumed resistance for this was 13Ω) and a Main Resistor (BURNED OUT) in series. The Transformer output mini jack plug connects to the Battery through a 17watt Base Unit, which fastens onto the top of the battery during charging. (But my open- circuit EMF measurement was 29.5V). The Transformer pack (working) says it delivers 17watts, up to 400mA DC at 23Volts. I have tried to get information from Draper’s technical team but they were totally useless. Hello, I am surprised and delighted to find a forum where cordless drill chargers (which have stopped working ), are discussed and possibly resurrected! I have such a problem with a DRAPER REDLINE 18-volt cordless drill driver. This limits the current into the LED to keep it from dying. The charge indication LED should get about 0.02A max. BE SURE TO DO YOUR OWN CALCULATIONS - MINE ARE ONLY AN EXAMPLE and might burn your house down. Six 1/2W resistors would probably be okay for this too, but the values would be different (120 ohms instead of 60). For a DC 20V charger, the desired charge rate would have worked out to 1/16 = 0.1A and 120 Ohms. They are really only exposed to 1.6 watts. This gave me about 3.0Watts of power dissipation using 0.5W resistors-Because I used an AC charger, the 3.0W capacity is more than enough, because the resistors get to rest during the negative AC half cycle. I used a set of 6 resistors to get near my 60 ohms. 2.4Watts / 2 (because of AC tranformer ) = 1.6Watts. 12V * 0.2A = 2.4Watt of heat that will be generated. 20V charger - 8V battery = 12V difference. The fully discharged voltage of a NiCd cell is 0.8V. The nominal voltage of a fully charged NiCd cell is 1.2V. 1.6Amp hour capacity / 8 = 0.2A charge rate. Use 1/16 in the math for DC chargers, or it will kill your packs or maybe start your house on fire. Because mine is an AC output transformer, I had to double that to 1/8th. I aimed for a nice, slow 1/16th C charge rate (capacity divided by 16). I used an AC output pack, so the charger outputs effectively half that of a DC wall transformer. BATTERY PACKS WILL ACCEPT EXCESS CHARGE CURRENTS WHILE REMAINING COOL UNTIL THEY ARE FULL-AT THAT POINT THEY WILL START TO OVERHEAT AND MAY CATCH FIRE OR EXPLODE. THIS COULD BURN DOWN YOUR HOUSE, PERHAPS SETTING OFF A CHAIN REACTION OF FIRES THAT BURNS DOWN YOUR WHOLE TOWN. MISTAKES IN SELECTING COMPONENTS CAN START A FIRE. If you have any doubts on how to do this, consult an expert on the topic or quit. A diode, an LED, and a few resistors are required to make sure the battery pack is charged at a safe rate.
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