Car Battery Charger Repaired
Yesterday my friend brought his friend’s Car Battery Charger. A model PRO USER- EBC4. For 12V LEAD BATTERY 4A Charging.
After opening I saw immediately why this Charger did not work. A 5W 0.05 Ohm resistor was defect. The simple circuit had a signaling circuit for the front leds with a single 8 pins LM358, 4 Power Diodes that deliver the Charging current for the Battery. With the now bad series resistor to protect the Transformer from Over Charging with too large currents. Probably because the owner replaced his Car type Fuse on the Front of this Charger with the wrong value, this resistor was now an open circuit.
I phoned my Electronics Shop in my Hometown’s Centre if they had this bigger resistor in their shop. To be sure I didn’t went there all the way for nothing , I always phone up first because most of the time I had to pre-order a component to get it weeks later. And because it was a device from someone else, I didn’t want to wait three weeks or more on a simple component.
But this time I made this resistor myself by using four 0.22 Ohm 3 Watt resistors connected parallel. This makes a replacement resistor of 1/ (1/0.22+1/0.22+1/0.22+1/0.22) = 1/18,18 <=> R = 0.055 Ohm. Or ((0.22 Ohm x 0.22 Ohm)/ 0.22 Ohm + 0.22 Ohm) = 0.11 Ohm <=> ((0.11 Ohm x 0.11 Ohm) / 0.11 Ohm + 0.11 Ohm) = 0.055 Ohm. Is almost the value of 0.05 Ohm we want. (The 0.005 Ohm difference don’t really matter. The Charger will work Fine! 0.005 is just 1/1000 times 5. That’s totally Neglectable.)
We wanted a 5 Watt 0.05 Ohm Resistor. We get from 4 parallel switched 0.22 Ohm 3 Watt Resistors a Bigger 0.055 Ohm 12 Watt Replacement Resistor. Or Calculated: Power = Voltage x Current = (Voltage^2)/Resistance= (Current^2)x Resistance. One 0.22 Ohm 3 Watt Resistor gives a Umax of 0.81240 Volt at 3.6927 Amps, because 3 Watt equals 0.22 Ohm x I^2 <=> 3 Watt / 0.22 = 13.63 Amps^2 <=> I = square root of 13.63 Amp^2 = 3.6927 A.
We know that if we switch 4 identical Resistors in Parallel that the Voltage over all of them is the same. In other words P with four resistors Calculates again 0.81240 Volts but now at 4 times the current = 3.6927 x 4 = 14.77 A. And we know that P = UxI = I^2 x 0.055 Ohm =
U^2/ 0.055 Ohm <=> gives again a Calculated replacement Resistor of 0.055 Ohm 12 Watt ! In other words: Four times the Power Dissipation of 1 Resistor of 0.22 Ohm.
Above photo shows the replacement Resistor 0.055 Ohm 12 Watt. That replaces the old defect 0.05 Ohm 5 Watt Resistor.
Above photo displays the bad resistor. An original 5W 0Ohm05J (=0.05 Ohm) resistor. Without this the Battery Charger gives no Output. Probably because of short circuiting the Battery Charging Wires.
Next photos show the Charger pcb with the Bad 0.05 Ohm 5 Watt Resistor before replacing it by the previous mentioned substitute 0.22//0.22//0.22//0.22 Ohm is 0.055 Ohm 12 Watt Resistor.
After the bad 0.05 Ohm Resistor was replaced by our self-made replacement as previous photo shows, the Charger worked as was expected. If this self constructed resistor with its 0.055 value now is a bit higher. 5 times 1/1000 of an Ohm to be exact more than the originally replaced resistor it could mean this less Current.
Also 5/1000 also means 1/200 of an Ohm less current. And furthermore that calculates to be only about 20 mA (which is about the current through one single Led). Because this Charger’s Maximum Charging Current of 4000 mA means 1/200th of it is 20 mA = 5 x (4000/1000)mA less Current.
By just calculating we can be sure to rule out any unwanted effects that we would not have anticipated. (Worst Case scenarios). Hope this was nevertheless an entertaining and not to complicated article.
Until next time.
Albert van Bemmelen, The Netherlands.
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