Author Topic: LED Experiment Question  (Read 911 times)

Offline Denholm

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Re: LED Experiment Question
« Reply #15 on: September 20, 2009, 02:14:38 PM »
Well, probably not the exact parts, but somewhere close to it. As mentioned, it's to get me more familiar with the assembly of electronics by doing the searching, pairing, and assembling myself.
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Offline OOZ662

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Re: LED Experiment Question
« Reply #16 on: September 20, 2009, 02:24:34 PM »
You'll only need diodes if your battery voltage is greater than the rated voltage of the LED.  That is, let say you used a 6V battery, you'd need a diode to lower that voltage down to stop the LED from burning up.  Otherwise, the with the right battery voltage, the LED is only going to draw as much power as it needs.

Um...I believe you meant resistors. :) An LED is a diode as it is, hence its name...meaning putting a diode in front of it would be redundant and a tiny voltage sink. A diode (as far as I know) is simply used to restrict current flow to one direction in a circuit, thereby preventing backflow and blowing up whatever you're working with.

You do need a resistor to step down the voltage, though, as was stated before this reply.

EDIT: Denholm, you may be interested in buying a "breadboard." It's akin to the big bases used in Legos; it has many tiny holes to put leads into and "wires" running between them to make circuits. It basically removes having to solder stuff if you're just playing around.
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« Last Edit: September 20, 2009, 02:28:25 PM by OOZ662 »
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Offline Fulmar

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Re: LED Experiment Question
« Reply #17 on: September 20, 2009, 02:29:30 PM »
Yeah, resistor.  Not sure why I had diode on the brain.
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Offline Jenks

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Re: LED Experiment Question
« Reply #18 on: September 20, 2009, 05:06:09 PM »

You do need a resistor to step down the voltage, though, as was stated before this reply.

First of all, good catch on the diode/resistor slip Fulmar posted. (I have to fix my dyslexics and Freudians all the time Fulmar  :) )

Secondly, I'm nitpicking. :D :t ;)

What you meant was, you need a resistor to drop voltage thereby limiting current flow through
the LED.

I know to the neophyte it may sound like the same thing, but in reality it's not.

It's trivial thing I know, but if Denholm is going to do this he should understand there is a difference.

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Offline OOZ662

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Re: LED Experiment Question
« Reply #19 on: September 20, 2009, 05:07:47 PM »
My knowledge of electronics in this area comes from two highschool level classes. I wouldn't know the difference.
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Offline bj229r

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Re: LED Experiment Question
« Reply #20 on: September 20, 2009, 06:48:52 PM »
First of all, good catch on the diode/resistor slip Fulmar posted. (I have to fix my dyslexics and Freudians all the time Fulmar  :) )

Secondly, I'm nitpicking. :D :t ;)

What you meant was, you need a resistor to drop voltage thereby limiting current flow through
the LED.

I know to the neophyte it may sound like the same thing, but in reality it's not.

It's trivial thing I know, but if Denholm is going to do this he should understand there is a difference.

Jenksie
Ya never limit current, it is always the same throughout a series circuit, unless you make multiple paths (You CAN limit voltage getting to diode by dropping it over resistor though)--- buy a pot, start high, back it off until the proper voltage is dropping across the led, yank the pot out of the circuit, measure ohms, then go buy a fixed resistor of that size. (ORRRR....do all those silly equations;)
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Offline Jenks

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Re: LED Experiment Question
« Reply #21 on: September 21, 2009, 11:39:29 AM »
Ya never limit current, it is always the same throughout a series circuit, unless you make multiple paths (You CAN limit voltage getting to diode by dropping it over resistor though)--- buy a pot, start high, back it off until the proper voltage is dropping across the led, yank the pot out of the circuit, measure ohms, then go buy a fixed resistor of that size. (ORRRR....do all those silly equations;)

Au contraire, you are limiting current. Any good text will tell you that. You are correct that current measured anywhere in a series circuit is the same. I.E. current through the LED and current through the resistor are the same. But the resistor is there to limit current, plain and simple.

Resistors are used to do 3 things:

1. To induce a voltage drop

2. To limit current.

3. To disipate heat.



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Offline Dragon

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Re: LED Experiment Question
« Reply #22 on: September 21, 2009, 12:06:13 PM »
I'm trying to think back to 6th grade LED project I made for science class, I made a metal box, drilled out and installed 9 LED's on the top and mounted 6 push type and 3 toggle switches on the sides. Powered by 2 D batteries on a holder for 3 volts.  Using alligator clips to connect to the LED leads provided an easy war to change which lights got power with any given switch push. The whole thing was maybe 6 x 6 inches.  The idea was to figure out how to first light 1 LED, then 2, then 3...and up to 9.
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Offline Jenks

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Re: LED Experiment Question
« Reply #23 on: September 21, 2009, 12:26:19 PM »
Here is a pretty descent little tutorial on LEDs  http://www.theledlight.com/ledcircuits.html   There is also an LED calculator on that page, that tells what size resistor to use in series and parallel circuits with multiple LEDs, etc.
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Offline Shuffler

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Re: LED Experiment Question
« Reply #24 on: September 21, 2009, 12:40:11 PM »
I'm trying to think back to 6th grade LED project I made for science class, I made a metal box, drilled out and installed 9 LED's on the top and mounted 6 push type and 3 toggle switches on the sides. Powered by 2 D batteries on a holder for 3 volts.  Using alligator clips to connect to the LED leads provided an easy war to change which lights got power with any given switch push. The whole thing was maybe 6 x 6 inches.  The idea was to figure out how to first light 1 LED, then 2, then 3...and up to 9.

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Offline swoose

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Re: LED Experiment Question
« Reply #25 on: September 21, 2009, 12:48:49 PM »
Here are a few good calculators. http://ledcalculator.net/       http://led.linear1.org/1led.wiz

Good luck and have fun!
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Offline swoose

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Re: LED Experiment Question
« Reply #26 on: September 21, 2009, 01:00:16 PM »
When I was in grade school all we had was fire.

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Offline bj229r

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Re: LED Experiment Question
« Reply #27 on: September 21, 2009, 08:24:53 PM »
Au contraire, you are limiting current. Any good text will tell you that. You are correct that current measured anywhere in a series circuit is the same. I.E. current through the LED and current through the resistor are the same. But the resistor is there to limit current, plain and simple.

Resistors are used to do 3 things:

1. To induce a voltage drop

2. To limit current.

3. To disipate heat.



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5ma or 100 ma will STIIILLLLLll fry a 2.1 volt led if it has 12 vdc introduced to it :noid
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Offline Jenks

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Re: LED Experiment Question
« Reply #28 on: September 22, 2009, 11:44:28 AM »
5ma or 100 ma will STIIILLLLLll fry a 2.1 volt led if it has 12 vdc introduced to it :noid

5ma won't, but 100ma prolly will, but of course 12v across a standard red LED would force far more current than a few ma's.. Unless its one of those purpose built 12V LEDs, yes they make them. The limiting resistor is built in.

bj, the voltage in the circuit is still considered to be whatever the supply voltage is (in our discussion 12V), but since there are 2 components (loads) on the supply(12v) they share(drop) the voltage between them. You shouldn't look at it as limiting the voltage. When you have a battery, the voltage in your circuit is determined by whatever the battery voltage is. After you total up all the voltage drops in the circuit they MUST add up to the battery voltage. You can't change that.  But what you can change (limit) is the current I=E/R

I think we may be arguing semantics here.

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Offline bj229r

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Re: LED Experiment Question
« Reply #29 on: September 22, 2009, 08:11:38 PM »
5ma won't, but 100ma prolly will, but of course 12v across a standard red LED would force far more current than a few ma's.. Unless its one of those purpose built 12V LEDs, yes they make them. The limiting resistor is built in.

bj, the voltage in the circuit is still considered to be whatever the supply voltage is (in our discussion 12V), but since there are 2 components (loads) on the supply(12v) they share(drop) the voltage between them. You shouldn't look at it as limiting the voltage. When you have a battery, the voltage in your circuit is determined by whatever the battery voltage is. After you total up all the voltage drops in the circuit they MUST add up to the battery voltage. You can't change that.  But what you can change (limit) is the current I=E/R

I think we may be arguing semantics here.

Respectfully ,
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