Fazer Owners Club - Unofficial
General => General => Topic started by: Panthor on 29 October 2012, 10:40:01 am
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I was wondering if someone here could help me out. I need to wire 4 leds to run off a normal household battery (eg 9v rectangle one, or something like that) with an on/off switch and as i have never really done something like this before i thought id get some advice so i do it right. Ive done a quick diagram to show what i was thinking but i have a few questions:
What size/voltage battery?
What gauge wire?
What resistors?
Also can i use any switch? The one i like is an on/off 12v toggle switch
Just wanted to add that the LEDs spec is: Forward voltage 3.2 - 3.6 Forward current 20 - 30
Thanks
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The circuit will work as you've drawn it, provided you get the diodes the right way round, but it won't be very efficient... about two thirds of the energy taken from the battery will be wasted in warming the resistors (220 ohms each is required to limit the current to about 25mA).
Little PP3 style 9v batteries don't have a huge capacity, so you could arrange things to be more efficient by having two parallel chains, each consisting of two diodes and a resistor in series. That way you'll take half as much current from the battery and have the same amount of light. Resistors in that case would each be 82 ohms.
Wire and switch requirements aren't a problem, because the currents and voltages are tiny.
(http://s10.postimage.org/jlrbm1xc9/LEDs.jpg)
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Thanks fazerider that looks simpler. just wondering if you had any idea how long one of those 9v batteries would be able to power this for (done the way you suggested) would it be many many hours or if not for a very long time would it be worth connecting up 2 batteries? Thanks
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Typical capacity for a PP3 is about 600 mAh or so. The series/parallel circuit takes 50 mA so that will give about 12 hours use. :)
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Fazerider the electrical genius strikes again.
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Apologies for being a pedant, but...
a) 4 x 220Ω resistors in parallel give an equivalent resistance of 55Ω,
and
b) 2 x 82Ω resistors in parallel give an equivalent resistance of 41Ω.
So ignoring forward voltage drop across diodes,
a) gives a current of 165mA and
b) gives a current of 219mA
I'm obviously missing summat here....
H
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Panthor,
Don't forget to double check the polarity of the diodes and battery before you switch on. Those white LEDs have a peak reverse voltage rating of 5 or 6 volts... 9v applied the wrong way will kill them.
Fazerider the electrical genius strikes again.
Aww, shucks... but "smartarse" is what people usually say. :D
Apologies for being a pedant, but...
a) 4 x 220Ω resistors in parallel give an equivalent resistance of 55Ω,
and
b) 2 x 82Ω resistors in parallel give an equivalent resistance of 41Ω.
So ignoring forward voltage drop across diodes,
a) gives a current of 165mA and
b) gives a current of 219mA
I'm obviously missing summat here....
H
Then don't ignore the diodes. :lol
Forward voltage drop with two diodes in series is roughly 7v, leaving 2v across the resistor.
The purpose of the resistor is to limit the current to 25mA, so the resistor needs to be 2/0.025=80Ω (the nearest preferred value being 82Ω).