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Old 01-26-2004, 11:17 PM   #11
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So, the forward voltage is actually lost as the semiconductor gap is crossed? That's why you subtract it out of the equation for the correct resistor?

Also, I plan on running probably five LEDs in parallel. If I want to have just one resistor (since all the LEDs will be the same type), then I would use the following math...

R = V / I
R = (Applied Voltage - Forward Voltage) / (5 x Amperage)
R = (5v - 3.9v) / (5 x 20mA)
R = 1.1v / 0.1A
R = 11 ohms

Is this correct? I seem to remember reading that voltage is constant in a parallel circuit.
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Old 01-27-2004, 12:17 AM   #12
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That looks right to me, but I'm no math guru. Voltage does remain constant in a parallel circuit as long as the wiring and power supply are capable of supplying the needed power. Voltage drops in a series circuit. For instance, you can hook 10 12v bulbs up in series and connect them to 120v...they will light just fine.
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Old 01-27-2004, 12:39 AM   #13
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A friend of mine has taken interest... He, unlike me, wants to do a combination of white and blue LEDs. Is there a way to have them on the same parallel circuit? I've not heard of a way to drop the voltage yet... only the current.
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Old 01-27-2004, 12:40 AM   #14
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gauge: Just check out this link, it should give you everything you need to know. Also, remember that resistors have a wattage rating. .25 W is very common. http://metku.net/index.html?sect=vie...calc/index_eng

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Old 01-27-2004, 12:43 AM   #15
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okay, since people insist on posting while i am, i'll answer another one. Different color LED's have different voltages. In order to modify the voltage you must use a resistor, just one rated for the color that you want to use. Check this page out, it tells you about multiple colors:

http://www.theledlight.com/ledcircuits.html

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Old 01-27-2004, 01:22 AM   #16
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haha yea metku is a good place for info too. japala is a good guy. he knows what he is doing.
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Old 01-27-2004, 01:43 AM   #17
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Actually I had seen both those pages already. :p

I think that the misunderstanding I'm having involves LEDs themselves. My understanding is that the resistances of the "appliance" sets the amperage for itself. Thus, the resistance of a lightbulb naturally lets the correct amperage through. And a lower wattage bulb running on the same voltage will have a higher resistance to it.

This understanding, however, is contradicted with LEDs, though. Do they not have a resistance of their own? If there is no resistance, then there is no voltage across the appliance. Yet, they require a voltage?

How are they different?
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Old 01-27-2004, 09:32 AM   #18
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Yes there is still a voltage across the LED. Someone can correct me if i'm wrong, but you can think of the LED as almost a wire that requires a certain voltage. In order to get the specified voltage, you must have an inline resistor to reduce the voltage. A light bulb is a resistor. The tungsten filament just happens to be a resistor that gives off light, where as a regular resistor tends to just disperse the energy as heat.

The LED translates nearly 100% (?) of the voltage to light without heat. Well, there si some heat, but very minimal. Where's TheMadNucleus? He's got this stuff down. I just do mechanical and computer systems

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Old 01-27-2004, 11:17 AM   #19
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Sorry Folks,

Work got in the way.

Hey so far everything here looks ok. A few comments on your thoughts on parallel with one resistor and the resistance of LED's. They do have some level of resistance (though this is not fixed like most appliances - that is a character of diodes) but since they are current sensitive the circuits are sort of designed to be little current limitors. But you can find the resistance at a given operating voltage using ohms law as well:

for a blue LED:

Max current = 20ma at 3.9 volts so resistance = 3.9 / .020 Amps = 195 ohms

This sort makes sense as the formula I gave you earlier (Vs - Vf)/Iled is a combination of these, note 5-3.9/.020 = 55 ohms

so add the ohms 195 + 55 = 250 ohms

using vanilla version of Ohm's V=250 * .02 and we're back at our 5 volts.

So there are two objects in an LED circuit: 1.) supply the correct volatage across the LED and 2.) Limit the current in the whole circuit to 20 ma. The LED current Limiting resistor does this job.

On running with one resistor for parallel LED's this is not a good idea. This is why:
If you use a resitor in parallel you can specifically mix and match colored LED as each has it's own requirements and also you are not fixed in the number of LED's based on the resistor you chose. Other than that - you can in fact add up the current requirements for as many LED's you have, and use the same formula to arrive at the single Load/current limiting resistor in your circuit.

For example:

5v - 3.9v/ (.02 + .02 + .02...nLED)

See chart

BTW - the law that shows that the 60ma going into node one gets split to 40ma and 20 ma is Kirchoff's current law: The sum of the currents entring a node is equal to the sum of the currents leaving the node


HTH

Tom
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Old 01-27-2004, 06:29 PM   #20
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Thanks Tom.

Electricity is confusing... and painful. I think I understand, though. The reason that you subtract the voltage of the LED is so that the voltage that you're removing with the resistor is equal to the difference between the applied voltage and the desired voltage... thus yielding the desired voltage. Correct?

And, just to clarify, the resistance of an LED is proportional to the voltage going through it?
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