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I hope someone can help with this very simple circuit, I have traced from the pcb, my electronics theory has really gone out the window, The circuit shown is from a car battery charger and has a 2 transistor circuit to monitor the battery turning the 'Full' LED on when charge is complete, however the resistor in series was blown due to an accidental short, I have no idea of it's value as it was fried! I have replaced it with a 0.5 ohm 1.5watt resistor as a rough guess, the resistor in the circuit was about 0.6 watt.

The charger output is marked up as 12volts 2.7Amps (4A RMS)

Hope you can help me out with the correct resistor value. Charger circuit

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[link] (i324.photobucket.com/albums/k331/midge_ebay/autocratix_cbc4.jpg) here you will find the schematic I traced from the PCB... Thankyou one again ;-) –  George May 16 '12 at 22:36
    
Are you sure you got the legs of the transistors right? It would seem to make more sense if the Zener/19/200 voltage divider was feeding the base of the transistor 9014. –  Kaz May 16 '12 at 23:31
    
Hi Kaz, thanks for the input, I have just re-checked the diagram to the board again and it is right, I must admit I thought it strange design myself, wierd design! it's only a cheap charger but 1 resistor is less than the cost of a new charger :/ Any ideas for the value of the blown resistor, I just can't seem to work it out myself for some reason. –  George May 16 '12 at 23:55
    
I mean, are you sure in identifying the transistor terminals: base, collector, emitter. The configuration varies among transistors. some are CBE, some are EBC, etc. –  Kaz May 17 '12 at 0:08
    
What is the power rating of the 19R resistor? –  madrivereric May 17 '12 at 1:34

1 Answer 1

up vote 1 down vote accepted

Does the charger list a "trickle" or "float charge" current? Also, aside from the "accidental short", has this charger properly worked?

I believe that this circuit limits the current via the transformer's impedance, Rblown and battery's internal resistance and is protected by the fuse and is uncontrolled in voltage. As you mention in your description, the circuit is there to control the battery full light.

Starting backwards from the light: The zener diode, 19R and 200R form a supply to allow the "full" diode to turn on when Q9014 is off. Q9014 is controlled by the output of the current sense amplifier formed by Rblown, 10k and Q9015. The voltage across Rblown from the charging current turns Q9015 on. When Q9015 is on, current flows through the 2.7k and 100k which turns on Q9014 (and turns off the "full" led). As the battery charges, Q9015 will start to turn off, thus reducing the current through 2k7 and 100k, which will result in Q9014 turning off (and the full led turning on). The electrolytic capacitor provides some filtering to assure that Q9014 stays on and prevent flickering by the "full" led.

The 4148's and 4k7 resistor form a supply for 9014 to pull its collector up above the voltage formed by the zener & resistors. The second 4148 that is connected to Q9014's collector prevents the power LED from turning off when Q9014 turns on. Note that I believe this supply is intentionally connected to the ac input because power diode bridge has a higher Von than the 4148's.

So, the value of Rblown is determined by the "full" current level and the required voltage threshold to cause Q9014 to turn on and is limited by the power dissipation at full charging current.

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So the circuit diagram isn't mistaken; the transistors really don't control the charging in any way, and only operate the LED. No wonder it looked wacky. This thing is basically just low-current transformer driving a battery. I might as well delete my answer. –  Kaz May 17 '12 at 16:06
    
Fantastic Madrivereric & Kaz, The Rblown just happens to be a '0' link (1 black band in the center) that looks like a resistor, I by some miricle happened to pick up a charger exactly the same!! what were the chances of that! After all that, that transformer seems to have burnt secondary windings so there is a total lack of current, and you are correct about the circuit, it's just a battery monitor really and the battery itself is just charged via direct DC from the transformer and rectifier, your circuit description is brilliant and I thank you so much for all the help you have both given me. –  George May 18 '12 at 12:23
    
Sounds good. Note that the "0" link is being used as a shunt resistor, the design is just utilizing the links small resistance. The "0" link is probably much cheaper than an actual shunt (and also less accurate). –  madrivereric May 18 '12 at 14:01

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