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A question regarding resistors and LED's.

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Mint.:
Today I was doing a few basic tests on LED's and resistors.
I did the following list of things:
-Attempted to burn out a Green 3v led by hooking it up to a 9v battery, why didn't it burn out?
-Looked online and read that when testing a LED a resistor should always be used to control the current.
Current?
I didn't know that resistors affected current also, I thought it was only voltage. What is the formula?
-I also saw a photo:

As I understand the LED takes 2v and leaves 7v for the resistor. Is this the same effect like the silicon diode that takes 0.7v?
Does this mean if we connect 3x2v LED's to 4 AA batteries and a resistor at the end that the resistor somehow gets 0v?
While building another circuit recently I found out that the resistor was placed at the negative side of the LED connected in series, aren't we supposed to limit the voltage going in to the LED's not going back into the battery?
Here is the circuit below, just a few corrections: I used only 3 LED's,a 9v instead of the 12v and placed a 51 ohm resistor at the end, because the guys at JayCar suggested me to do it like that, again why was the resistor necessary?

Website I used was: http://www.doctronics.co.uk/resistor.htm and http://www.kripkrap.ru/home/svetyashhijsya-korob/

Psi:
Note:  Different leds require different voltages,  red/orange/yellow leds are around 2V but other colors like blue and white need more volts.

--- Quote from: MintyCondition on December 12, 2011, 09:04:56 am ---As I understand the LED takes 2v and leaves 7v for the resistor.

--- End quote ---

It's around the other way. You put the resistor in there to burn-off 7V so that there's only 2V left for the led.
Because the led will take what its given even if that blows it up.

--- Quote from: MintyCondition on December 12, 2011, 09:04:56 am ---Does this mean if we connect 3x2v LED's to 4 AA batteries and a resistor at the end that the resistor somehow gets 0v?

--- End quote ---

If you connected 3x2v leds they would consume 6V and if you had exactly 6V then you wouldn't need the resistor.
But.. If you drop the led voltage to say 1.9V you wont just get a little less light compared to 2V. Infact the brightness drops off quite a lot.
And if you put 2.1V into a 2V led the brightness will increase quite a bit meaning you aren't just pushing the led a little past its recommended rating you're pushing it a lot past it's recommended rating.

So using your example of 4 AA batteries, when they were brand new they would be 1.6V so you'd have 6.4V but very soon after you started using them they would drop to 1.5V (6v total). After a while they might be 1.4V but still got quite a bit of life in them.  If you had them connected without any resistor the leds might be damaged by the 1.6V and even if they weren't as soon as the voltage dropped below 6V they would become quite dim. The batteries would still be ok but the leds wouldn't work any more.

In short. You need a voltage really close to 2V for one led.
It's really hard to get the exact voltage you need. So instead of trying to limit the voltage you limit the current instead.
This is much easier because a little bit more or less current doesn't change the resulting led voltage by a large amount.

So you pick a resistor that will result in the current that you need.
If you have a 2V led and a 9V battery you have 7V to get rid of.
So 7V / 0.02A  = 350ohms    (0.02A = 20mA which is quite normal for a led)

--- Quote from: MintyCondition on December 12, 2011, 09:04:56 am ---building another circuit recently I found out that the resistor was placed at the negative side of the LED connected in series, aren't we supposed to limit the voltage going in to the LED's not going back into the battery?

--- End quote ---

It really makes no difference.
Think of water flowing through a pipe. If you put a flow limiter at the start of the pipe it will limit the flow just the same as if you put it at the end of the pipe. The flow will be limited just the same.

firewalker:
Have a look at Ohm's law.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmlaw.html

Mint.:
Perfect explanation thanks a heap! :D

Jon Chandler:
Here are a couple links that may help.

LED Calculations

LED Calculations - The Lab Section

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