# Is it true that Pin 13 will not be able to power a LED as bright as other pins ?

I understand that Pin 13 already have a resistor attached to it for Easy LED Blink test to make sure that the board works.

I don't know what is the value of the resistor attached to Pin 13 [what is it anyway ?].

But assuming that I am using a LED that require a particular resistor value that is lower then the resistor build into pin 13 and all I/O pins are to be set as outputs, I would not have the ability to put in my own custom resistor the same way I can on pin 13.

If the Resistor value required is less the resistor already connected to Pin 13, then Pin 13 will never have the same brightness as the other LED no ?

This situation is made worst by nano because it actually have an LED build into the PCB connected pin 13, so the whole "thing" is already setup, connecting another LED circuit to this pin will create some sort of parallel split current thingy that is way over my head and this whole thing means Pin 13 is un-useable because of this no ?

this whole mean Pin 13 is un-useable because of this no ?

no.

Pin 13 has a 1K resistor and an LED ready connected to it. This will take less than 5mA. It will not affect any other resistor / LED combination you attach to the pin as you are probably driving those at 20mA and the pin limit is 40mA.

But do I have to use a different resistor then the other pins to achieve the same brightness because of the 1K resistor/LED already attached to pin 13 because of some ohm's theory or some crap like that ?

BlueBlondeTard: But do I have to use a different resistor then the other pins to achieve the same brightness because of the 1K resistor/LED already attached to pin 13 because of some ohm's theory or some crap like that ?

No, you use the same resistor size you calculated for any of the other pins, and the brightness will be the same. The only difference is that the pin 13 will be supplying about 5ma more current then the other pins, which as long as that is less then the maximum current rating for a output pin all will be fine.

Lefty

Wow, this is weird, by having another load attached to Pin 13 [because Pin 13 will now have my external LED like all the other pin TOGETHER with the built in LED on the PCB] actually increases the LED's received Ma. This....this is weird.

actually increases the LED's received Ma.

No it increases the current out of the pin, it does not change the current through the LED.

So the LED will just take whatever current it needs & leave the rest of the currents alone even if there are more ?

If that is the case why the need for a resistor at all ? The LED will just take what it needs no ?

I know I am getting silly here, we all know resistors are required for LED, but the electronic theories sometimes seems to fight each other.

ok ok all these theoretical nonsense aside [I got into Arduino because it promise I don't have to get into these MATH that even Bender hates].

What you all are saying is that the extra 1k resistor and LED connected to Pin 13 will not affect the brightness of the LED of my external LED that is connected to it compared to the other output pin LEDs and I don't have to change the resistor value for Pin 13 and I could use the same resistor value like what I use for all the other pins and the brightness of ALL the LEDs across all the pins will be the same.

Right ?

The LED will just take what it needs no ?

The combination of LED and resistor will take a fixed current if you put a fixed voltage across it. You have two such combinations in parallel. They both act as if they were the only thing in the circuit. However the pin is supplying current for both of them, so it supplies the arithmetic sum of the two currents.

but the electronic theories sometimes seems to fight each other.

No, it’s your understanding that is not quite right yet.

ok ok all these theoretical nonsense aside [I got into Arduino because it promise I don't have to get into these MATH that even Bender hates].

What you all are saying is that the extra 1k resistor and LED connected to Pin 13 will not affect the brightness of the LED of my external LED that is connected to it compared to the other output pin LEDs and I don't have to change the resistor value for Pin 13 and I could use the same resistor value like what I use for all the other pins and the brightness of ALL the LEDs across all the pins will be the same.

Right ?

Yes right.

This might help:- http://www.furryelephant.com/content/electricity/parallel-circuits/

I got into Arduino because it promise I don't have to get into these MATH that even Bender hates

Don't know who told you that, it's not true.

What you all are saying is that the extra 1k resistor and LED connected to Pin 13 will not affect the brightness of the LED of my external LED that is connected to it ..

. If the light in your kitchen is on and then you turn on the light in the dining room how does that affect the kitchen light? (a) the kitchen light gets brighter because there is more current being supplied by the power company. (b) the kitchen light gets dimmer because the current being supplied by the power company is being shared by the two lights. (c) the kitchen light stays the same because the power company now supplies enough extra current to power the added light.

Don

Thank you :D I understand now :)

What about it. The voltage in a parallel circuit is the same across all elements of the circuit.

I think it’s really awesome that nature makes things as they are you know, like water freezes and get lighter so that it become a warm blanket for the fish below instead of sinking and that devices connected in parallel get the same voltage but they don’t have to worry about the current because all burden is on the source to provide it.

That being said, it’s kind of silly that LED have a Ma rating huh.
I mean, you would expect LED to have just a volt rating and that’s all.

Supply the LED with the voltage and it will just take the current correct for it and no more.

The fact that LED have a Ma rating and a forward voltage seems to “suggest” that even with the correct voltage given to the LED, it may be a naughty boy and get more current then it needs, hence the need for resistors, this is in complete contradiction to the “device will just get whatever current they need and no more theory.”

I mean, I could connect my 5V plug with a higher watt to a device that requires 5V but a lower watt and it’s ok because the device will just take the current it requires and no more but for LED we cannot do that ?

I mean if a Blue LED have a 3v forward voltage and we supply a 3V 1BILLION Ma to it it will be fried ? How weird. Should it just take what it need and no more leaving more current for other devices to connect to the source in parallel ?

Giving an LED a mA rating is actually extremely important, it may not seem so when working with arduino and these small scale projects, but its still important to get the closest resistor possible to what the need is to keep the life of the board as long as possible.

With that said, it also matters because large projects with thousands and thousands of leds, the mA rating really adds up! and in the same manner, resistance and everything is best to be as specific as possible.

We are dealing with computers and math here and nothing is more important in both than having the most specific information possible to get the best answers

Supply the LED with the voltage and it will just take the current correct for it and no more.

No, the forward volts drop changes with time and from device to device and with temperature. You need to read this:- http://www.thebox.myzen.co.uk/Tutorial/LEDs.html

Supply the LED with the voltage and it will just take the current correct for it and no more.

We have been trying to tell you that it doesn't work that way. LEDs do not have a 'rated voltage' as light bulbs do, they have a 'rated current'. Due to the highly non-linear relationship between the current and voltage of an LED it just cannot be operated the way you want. If you supply it with too little voltage it will draw very little current and will not produce any light. If you supply it with too much voltage it will draw a lot of current and produce a lot of light - for a very short time. Due to the non-linearity that occurs between too little and too much voltage the operation is unstable in that region. You cannot predict how much voltage will be just right for a specific LED and even if you could it wouldn't remain 'just right' for very long. Give up.

Don

Hey now, don't be harsh. Everybody has come here to learn at one point. Plus these questions help me out since I can wonder the same things