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should have given more info

i am using a 12v 7ah battery to power some glow string to light up my mini keyboard that sits under my TFT so i got a switch today which has an led in it that runs at 2.2v and draws 25mA i don't know what the glow string draws but it can't be much
 
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something around 400 ohms should be fine from a 12v supply


edit: wait a sec if your running a switch with an led in then i assume the switch will connect the glow strings to 12v? dont connect the glow strings to the resistor for the led. Try and find the specs for the strings
 
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on the switch is two prongs for the power so 12v in one end 12v out the other negative from the glow sting to the battery, also on the switch is the option of using the l.e.d. so all i am going to do is take a feed from the 12v at the switch which needs reducing to 2.2v and then the other end to the negative.

the switch will be lit at all times
 
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m3csl2004 said:
something around 400 ohms should be fine from a 12v supply


edit: wait a sec if your running a switch with an led in then i assume the switch will connect the glow strings to 12v? dont connect the glow strings to the resistor for the led. Try and find the specs for the strings

so 400 ohms it is then
 
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From V=IR 12/400= 30ma (milliamps). The calculation needed is volt drop, so that the resistor is used to volt drop down to 2.2v. Just atm I cant remember exactly how to do it. I might have an idea if I knew the resistance of the diode (led), this would have to be read off a spec sheet.

Im thinking Vdrop= Vbat/Rvariable + Rdiode

U also might be able to do 30ma*Rdiode(from spec sheet)= Voltage accross LED/diode.

I think the 7A of the power supply doesnt factor into the calculations since its just a limit of which the power supply can put out; which corresponds to how we usually see a computer PSU.
 
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It would be the voltage drop divided by the current draw so:
voltage drop would be 12 - 2.5 = 9.5 divided by 25ma which is 0.025
9.5 / 0.025 = 380. So the closet to 380r you can get.

Oh and just in case, if this is a car its best if you use 13.8 or so instead of 12 in that equation so it would be about 450 ohm.
 
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big_white_dog84 said:
Voltage divider rule.

You need one resistor value x and one 3x. Put them in series with the battery and the voltage across the x resistor will be 3V.
Unless you're putting an amplifier with infinite input impedance straight after it, then that's (to all intents and purposes) utterly useless.
 
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