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Author Topic: Battery charger circuit - Burned resistor  (Read 1351 times)
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Today a friend gave me a broken flashlight that uses a 6V 4AH battery. and it says the cycle use is 7.25 - 7.45 V
He told me to keep it because is not working....

I took it and i open it and i found that the circuit board is burned
and now i cannot recognise the resistance value of the resistor my mulitmeter doesnt show anything
there is also a diode that connects the charger plug to the resistor. 

He also gave me the charger that he use to charge the battery
the charger is a 230V  - 50 Hz 6W max  with output DC 12 500mA

now my question is how to rebuild this small circuit in order to charge the battery?
how to calculate the resistance? what about the diode they are both burned so does the circuit board smiley-sad

Thank you very much

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Edison Member
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The 1st problem is...  Why did the resistor burn-up?   Maybe the battery is bad/shorted?

How big is the resistor, physically?  You can get a good idea of the resistor's power rating from it's size.  (Just look-up some specs/dimensions online.)

A lower value resistor will allow more current and generate more heat.   A higher value resistor will generate less heat (and is less-likely to burn-up smiley-grin ) but the battery will charge more slowely.

I'm not sure why the diode is there, unless it's to prevent you from charging the battery "backwards" (discharging it).  Or, I wonder if it's a Zener diode, which drops a fixed voltage?    A zener diode would have to be big enough to dissipate a few watts of power (maybe 3W).    A regular diode can be small and only needs something more than the 500mA current rating.   Any diode can handle 12 Volts.

6W max  with output DC 12 500mA
OK "worst case'', with minimum resistor value   and a shorted battery, you'd need a 6 Watt resistor.    If we take one of the power formulas (Voltage squared/R or Current squared x R), we can manipulate the formula and calculate the minimum resistance =   24 Ohms.

A 144 Ohm resistor with 12V applied across it (again assuming a shorted or completely dead battery with no internal resistance) would dissipate 1W.  You won't find a 144 Ohm resistor, but you could probably find 150 Ohms.

Please double-check my math before you solder anything!   smiley-razz

A Zener diode sort-of makes sense...  With, say 5V, dropped across the Zener, the resistor & battery only have to share about 7V and you can get-away with a lower-value, lower-power resistor and the battery chould charge faster.



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Shannon Member
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A photo of the board (both sides) would be useful...

[ I won't respond to messages, use the forum please ]

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