Alan B
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| posted on 11/11/11 at 02:32 PM |
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LED basics
Guys,
I'm working on a project that is a hand held device that requires two small white LEDs mounting for illumination purposes.
Being a simplistic mechanical guy I'm imagining it needs a small battery and some kind of switch.
However, knowing electrickery there will be some need for circuitry, resistors and other assorted electrical gizmos....
So basically what do I need to get the job done?
Cheers Alan
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tegwin
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| posted on 11/11/11 at 02:52 PM |
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Find out the maximum forward current and voltage of the LED emittors(from their datasheet) and then use a simple calculator like this one:
http://ledz.com/?p=zz.led.resistor.calculator
to work out what resistor you need. Worth noting... if you have high power LED you will need a capacitor rated to handle the power (W).
The above should suffice for simple LEDs, assuming you have a reasonably low power supply to begin with. If you have larger power LEDs then a more
"cunning" switched power supply can be used.
Assuming a "standard" LED. Rated at 10ma at 1.5V. Using a 9V battery you would need an 820 ohm resistor rated at .125W 
[Edited on 11/11/11 by tegwin]
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Alan B
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| posted on 11/11/11 at 04:17 PM |
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Great info. cheers
Are forward voltage and voltage drop the same thing?
Thanks Alan
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tegwin
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| posted on 11/11/11 at 04:25 PM |
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Yes, in this case they are
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MikeRJ
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| posted on 11/11/11 at 05:35 PM |
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quote:
Originally posted by tegwin
to work out what resistor you need. Worth noting... if you have high power LED you will need a capacitor rated to handle the power (W).
He'll need a suitable resistor as well 
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tegwin
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| posted on 11/11/11 at 07:23 PM |
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quote:
Originally posted by MikeRJ
quote:
Originally posted by tegwin
to work out what resistor you need. Worth noting... if you have high power LED you will need a capacitor rated to handle the power (W).
He'll need a suitable resistor as well
DOH!.... 
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Alan B
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| posted on 11/11/11 at 07:57 PM |
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quote:
Originally posted by MikeRJ
quote:
Originally posted by tegwin
to work out what resistor you need. Worth noting... if you have high power LED you will need a capacitor rated to handle the power (W).
He'll need a suitable resistor as well
Electrical humour? or am I missing something?
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MikeRJ
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| posted on 11/11/11 at 09:06 PM |
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quote:
Originally posted by Alan B
quote:
Originally posted by MikeRJ
quote:
Originally posted by tegwin
to work out what resistor you need. Worth noting... if you have high power LED you will need a capacitor rated to handle the power (W).
He'll need a suitable resistor as well
Electrical humour? or am I missing something?
You need a resistor to limit DC current flow, not a capacitor!
You can think of an LED as having a roughly constant voltage drop (it does change a bit with current but not much). You simply subtract this voltage
drop from your supply voltage, and the number left is the voltage you need to drop across a resistor. Then simply apply Ohms law, and your desired
LED current (data sheet will give max permissible value) and you get the resistor value you need.
e.g. a white LED may have a forward voltage (Vf) of 3.5v, and you want to power it from a 12v supply.
12 - 3.5 = 8.5v
You look in the LED datasheet and see this LED has a maximum forward current of 30mA, and you want the most light possible so you will run it flat
out.
Ohms law says: R = V/I where R is resiatance, V is voltage and I is current
R = 8.5 / 0.030 = 283.3 Ohms
Resistors come in a fixed range of values, and 283.3 isn't one of them! Simply choose the next higher value, in this case 300 Ohms for the most
common range of resistors, the E24 series (prefered resistor values). You can
combine resistors to get an exact value, but this is rarely necessary.
Tegwin post mentioned power rating, this is the other important resistor parameter. When you use a resistor to drop a voltage, it converts that
voltage drop into heat. The resistors power rating says how much power it can dissipate (at a fixed ambient temperature) without going black and
crispy.
P = V*V/R or P = I*I*R
In your case you are dropping 8.5v over your 300 Ohm resistor so
P = (8.5*8.5)/300 = 0.24 Watts
A 1/4 watt resistor is almost certainly not going to be man enough for the job here, so consider going for a 1/3 or 1/2 Watt resistor. The higher
power resistors are physically larger as you'd expect.
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The Shootist
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 posted on 11/11/11 at 09:26 PM |
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Depends on the LED nowdays...
Some of the higher end, high powered LEDs now come as a package with the voltage regulation and inrush limiting built into a base package.
Many of these are very high powered (upwards of 1500 lumins) and run over $35 USD each.
I have considered use in an automotive headlamp. Imagine the last bulb you'll ever change.
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tegwin
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| posted on 11/11/11 at 09:36 PM |
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Shootist... you would think that wouldnt you... but unless you do your thermal management properly, the LEDs wont last long atall!
For the record, I did mean resistor earlier... was thinking about something else and typed capacitor instead... im simple you know!
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www.verticalhorizonsmedia.tv
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bi22le
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| posted on 11/11/11 at 10:55 PM |
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Also LEDs only flow one way! they are not like a bulb. hence the name light emitting DIODE.
If its the wrong way round it will not light and will not cause any damage. if the correct way round you have light.
good luck. its not complicated. You could use an alarm key fob battery as they are low voltage. thats how we used to test them the bays and leds at my
old shop.
[Edited on 11/11/11 by bi22le]
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Alan B
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| posted on 14/11/11 at 02:20 AM |
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Cheers guys...good info, just as I expected.
Alan
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ChrisW
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| posted on 14/11/11 at 11:53 AM |
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It's worth pointing out that higher powered LEDs should really be run with a constant current generator.
If you've got a well stabilised power supply and use a current limiting resistor this should suffice, but running them in a car where the
voltage swings around you really need something a bit better. Having your lights dim when you turn the engine off just doesn't look cool. Or
worse, exceeding maximum current when the alternator is charging will shorten the life of the LEDs significantly.
The old 'back to back transistor' method is the one I like to use in this situation, but a 78xx series regulator with current limiting
resistor also works.
Chris
My gaff my rules
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