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Burnt out resistor


onetenor
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have a small battery charger which has a burnt resistor Cannot read bands to identify. Voltage across the two connector holes for the res is 19 v. I need to drop this voltage low enough to fully charge a AAA 6 pack 7.2 v battery pack Sizewise it looks like a 1/2 watt . Tried a couple of on line calculators but they do not seem to recognise decimal points so got some crazy results. Can anyone help. I tried a 1 ohm and dropped voltage out to about 11 1/2 v. Another 1 ohm in series only dropped it few tenths more. I am puzzled any one any ideas?

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Depends on the capacity. None of these figures are that critical, but lets say for example that the AAA pack is 800mAh, so the c/10 rate is 80mA.

Say the charged pack will be about 8v.

If your supply is 19v that leaves 11v across the resistor (19-8)
R=Volts / Amps = 11/.08 = 137 ohms, so use a 150 ohms which is the nearest preferred value.

However, trickle-charging NiMh is a Bad Thing. They should be fast charged with peak detection for cut-off.

Cheers
Phil

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Thanks Phil Great stuff. That makes sense but what value would you reccomend for the wattage!/2 w or 1 w> I do have PROPER chargers but at present I am house bound with very bad hips so cannot get to the cave With them I have created a large no of different connectors for all my different packs These chargers cover anything from one cell to 12 or more and 12v wet cell. Good enough??

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As above but don't forget to multiply the current (in amps, not miliamps) that you wish to charge at by the volts you want the resistor to drop (minimum battery volts minus power supply volts. This will give you the Minimum wattage of the resistor. Best pick one at least double that.

e.g. current 80mA = 0.08 amps. Volts across resistor (to use Phil's example) 11v

11 x 0.08 = 0.88 Watts Say 1 watt. So your resistor needs to be 150 ohm at 1 watt. If it were me I'd go for a 2 watt resistor at 150 ohms.

Apart from the rectifier diodes is the resistor the only component in the charger?

Ian

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