EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: iandusud on January 01, 2015, 12:23:41 pm
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Hi again. Following on from my earlier post I'm trying to identify a burnt out resistor. It is in series on the output from a rectifier on a 24V cordless drill charger (see attached diagram). I get a reading of 26V and 2.5A off the rectifier. The charger is supposed to have a charge rate of 200-240 mA. The internal resistance of a discharged battery is 2.5 ohms, so negligible. I calculate that a 120 ohm resistor would drop the current to about 216mA which would good. However such a resistor would have to handle 6W which doesn't corespond to the original which appears to be a metal film resistor measuring approx 9mm x 3.5mm (see attached photo).
I'm probably doing some screwy thinking here so please help me out.
Thanks, Ian
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Now do the same math but include the remaining voltage of the discharged battery...
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Please illucidate - that means explain to school-boy physics numpty like me |O
Thanks, Ian
BTW the voltage on the discharged battery is 0. I have another partially charged battery which reads 24V but has very little charge on it.
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If the battery is fully discharged to 0V, it's most likely dead and it's unlikely you'll be able to fully recharge it and keep its charge for a significant amount of time.
It wouldn't be surprising if the original resistor was undersized and it's better to fit a higher power resistor than necessary. The power in the resistor can be calculated usig Ohm's law:
P = IV
or
P = I2R
or
P = V2/R
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Hi hero999,
That's how I calculated it for a 120 ohm resistor. 26V/120ohms=0.216A. 26Vx.216A=5.6W
I just wondered if I was missing something.
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Please illucidate - that means explain to school-boy physics numpty like me |O
Thanks, Ian
BTW the voltage on the discharged battery is 0.
:palm:
That's why you broke your charger. Many more advanced chargers will just refuse to charge such batteries.
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If the remaining voltage across the battery is zero, the battery is dead and that is probably why the resistor let the magic smoke out.
You didn't say what kind of battery it is, but for instance a lead acid battery is empty when the voltage has dropped to about 23 volt.
In that case the resistor will not have 26 volts across it but only 3 volts. Now do the math.
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Please do the math for me!!!
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For example discharged battery voltage is 20V, you have 26V. 26-20=6V. Then P=U2/R -> 62/120=0.3W
Probably battery isn't completely dead yet. However, due to it containing multiple batteries in series, imbalance inside it will be very high after such torture. Complete discharging severely damages the battery.
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Wow, that's really helpful and explains the relatively low power rating of the resistor. I will however replace it with one with a reasonable power rating to help avoid the same thing happening again. Thanks. Ian
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I did intially wonder about that until I measured the current but at 2.5 it would be trying to charge a 1.2Ah battery in 30 mins when the supposed charge time is 6 hours.
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Oops sorry, I guess you were replying to me but I had deleted my post after re-reading the thread, then your reply popped up.