electronics engineer

tonyzero · 10 Feb 2007 at 13:53

hello

could someone please tell me what resistor i need to reduce 12v to 2.5v or 3v

or if there is a online converter for doing this

Thanks

Tomsk · 10 Feb 2007 at 13:57

We need to know how many amperes are being drawn.

Freefaller · 10 Feb 2007 at 14:00

Can you tell us the power supply power i.e. how many watts it is or at least how much current you're drawing?

tonyzero · 10 Feb 2007 at 14:01

sorry its a L.E.D. with a draw of 25mA

tonyzero · 10 Feb 2007 at 14:05

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

m3csl2004 · 10 Feb 2007 at 14:16

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

tonyzero · 10 Feb 2007 at 14:24

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

tonyzero · 10 Feb 2007 at 17:25

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

megatron · 10 Feb 2007 at 18:22

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.

m3csl2004 · 10 Feb 2007 at 18:28

the resistor wouldnt be dropping 12v - it would be dropping 12 - 2.2

cyborg · 10 Feb 2007 at 19:04

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.

big_white_dog84 · 10 Feb 2007 at 20:22

tonyzero said:
hello

could someone please tell me what resistor i need to reduce 12v to 2.5v or 3v

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.

csmager · 10 Feb 2007 at 22:08

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.

Adam_151 · 10 Feb 2007 at 22:27

430R 1/2 watt is what you want, could use 390R if you want to drive the diode right at its limit!

big_white_dog84 · 10 Feb 2007 at 22:36

csmager said:
Unless you're putting an amplifier with infinite input impedance straight after it, then that's (to all intents and purposes) utterly useless.

The initial post made no reference to the purpose. I answered the question of how to reduce 12V to 3V

csmager · 10 Feb 2007 at 22:38

big_white_dog84 said:
The initial post made no reference to the purpose.

Maybe not, but the 4th and 5th did.