electrical question

[archer_ISIS]

i have a light that uses batteries that i would like to wire into an outlet
it runs off of 4 AA batteries hooked up in series. so i know i need a 4 volt DC power supply but as far as amperage goes i have no idea, its not listed on the batteries oh and im using standard alkaline as recommended
in the instructions

[Knightslugger]

1.2x4 = 4.8

You need a 5vdc power supply.

amperage doesn't matter so long as it meets the minimum current draw that the light demands.

1 Ampere x 5 Volts = 5000 milliWatts

[Groo]

the 5v power on a USB power plug-in thing would probably work.

[dhardy57]

Here are the specs for current draw. Since the batteries are in series not parrallel you can assume the current to be 500 mA. The chart shows you how long the different type of batteries will last at that current draw rate. The actual Voltage for 4 AA in series will be eather 4 x 1.5 (6 volts) or 4 x 1.2 (4.8 volts) depending on the battery type you use.

AA battery - Wikipedia, the free encyclopedia

[Fixxer]

Originally Posted by Knightslugger

1.2x4 = 4.8

You need a 5vdc power supply.

amperage doesn't matter so long as it meets the minimum current draw that the light demands.

1 Ampere x 5 Volts = 5000 milliWatts

NiCd = 1.2v/cell * 4 = 4.8v
Alkaline = 1.5v/cell * 4 = 6.0v

[AwPhuch]

Originally Posted by Groo

the 5v power on a USB power plug-in thing would probably work.

Yes

[Vhyrus]

Watts equals volts times amps.
W = v*i
Figure out the wattage of the bulb and divide by the voltage of the batteries (which should be 6 volts) to get the required amps. The wattage of the bulb should be listed on the bulb itself or on the flashlight.

Also remember that the wall is AC and the flashlight is DC so you need some kind of converter. An old power cord from a laptop might work.

[Knightslugger]

Originally Posted by Fixxer

NiCd = 1.2v/cell * 4 = 4.8v
Alkaline = 1.5v/cell * 4 = 6.0v

so you see why a 5v PSU will do just fine...

[AwPhuch]

Originally Posted by Vhyrus

Watts equals volts times amps.
W = v*i
Figure out the wattage of the bulb and divide by the voltage of the batteries (which should be 6 volts) to get the required amps. The wattage of the bulb should be listed on the bulb itself or on the flashlight.

Also remember that the wall is AC and the flashlight is DC so you need some kind of converter. An old power cord from a laptop might work.

converter = transformer and rectifier...without it...it will not survive long if you try to use line power from the wall...heck YOU might not survive long if you try it.

If you are talking about charging said flashlight (with Nicd or NiMH batteries) the 5vd should work, as long as the bulb isn't over 5 watts (usually a VERY large bulb) then even the output from a USB cord 5vdc at 1000mA (1A) will be enough to light the bulb.

My recommendations....find a 4x NiMH battery charger and battery pack and have at it
Power2000 4AAA NiMH 2900mAh Batteries/Charger, 110/220v XP350

[Fixxer]

Originally Posted by Knightslugger

so you see why a 5v PSU will do just fine...

Actually, I can easily envision 3 red LEDs in series, each with a 1.8 Vf along with a 30 Ohm resistor to limit the current to 20mA @ 6Vin, in which case a 5-Volt PSU will NOT work.