i got alot of leds and i just need to know how many volts is needed to turn on an led.
and my second question ishow many volts is there in one pin (example pin13) when it connected to the ground
It depends on the LED but between 1.8 and 3.5V. There is 5V from a pin so you need a resistor as well as an LED to limit the current.
let say a normal 5mm red led.
how many volts does it need to turn it on.
without any resistors?
Generally ~1.7V.
It will conduct all the current you can supply it until something smokes; either the source, the LED, or the sink.
The resistor keeps the current flow in check.
Calculate it thusly:
(Vsource - VLED)/20mA = Resistor value.
Example:
(5V - 1.7V)/.02 = 165 ohm
If you have a multimeter you can put some current thru your LED and just measure the voltage drop across it.
without any resistors?
This kind of question will always raise concerns amoung many of us, because it makes many of us assume you may not understand the function the resistor serves, and might try to turn on a led without using a resistor. You would then join a large group of begineers what have collected burned out LEDs and Arduino boards with burned out digital output pins.
So can you explain why you need to know the forward voltage drop specification of a given LED? It it's for other then trying to properly calculate the value of a required series current limiting resistor, then we should talk some more on the topic.
So where are coming from on this?
Lefty
This is definately worth a read, if you are not sure about the current and voltage characteristics of an LED: http://www.thebox.myzen.co.uk/Tutorial/LEDs.html.
It is on Grumpy_Mike's webpage, and explains all about the current and voltage of an LED, as well as what happens if you ignore the advice.
Quoted from that page, when talking about connecting an LED straight to the arduino without a resistor,
One justification says “well I tried it for months and nothing has burn’t out”! This is just like saying I have crossed the road blindfold and I am not dead.
Sums it all up nicely!
Onions.
Generally It needs 1.8-2.5V. but it depends on LED. Now Resistor is necessary. LED is just a Light Emitting Diode.
So if you dont connect resistor with it then it ll burn out then, Circuit would be same like Connecting General Purpose diode with 12/5V Battery. And so maximum current ll pass and circuit ll blow!! (Due to short circuit).
So External Pullup or PullDown Resistor is necessary.
Calculate your resistor as shown above by moderators.
So External Pullup or PullDown Resistor is necessary.
I think you meant to write a series resistor is required, not a pull-up or pull-down resistor.
Lefty
Of course its required series resistor. but when using with micro controller we have to specify Pull-up resistor and pull-down resistor.
aalex_arduino:
Of course its required series resistor. but when using with micro controller we have to specify Pull-up resistor and pull-down resistor.
Please explain. Pull-ups and pull-down are sometimes required, but almost always in the context of using input pins properly. Leds are almost always used only with output pins. So as the posting question was about using LEDs I don't get your comment?
Lefty
Typically the 5mm Red LED uses about 2 volts with 20 mA max current, with the Arduino output at 5 volts then you need to use about a 150 ohm resistor (brown/green/brown) inline with the LED to keep it from burning out.
If your LED specs are different then use the calculator at: http://led.linear1.org/1led.wiz
Also, nice article previously posted*
If you are using Negative Logic then we have to connect pull-up resistor. But when you are working with positive logic then we have to connect pulldown resistor. (To turn it ON in both the case!!) it requires sometimes...not always Lefty. I agree with you.
So here as shown in below image to turn the LED on its depend on Positive/negative Logic.
If you are using Negative Logic then we have to connect pull-up resistor. But when you are working with positive logic then we have to connect pulldown resistor.
@aalex_arduino: Both of the resistors in your drawing are 'current limiting' resistors. There is no 'pull-up resistor' on your diagram nor is there a 'pull-down' resistor.
Don
The resistors as far as I know are used only for limiting the current. Pulling up or pulling down a network has to do with inputs rather than lighting LEDs.
Like if you have a switch, and when you press it you want to connect it to +5 volts, you will use a pull down resistor connected to ground so that due to noise the output does not get disturbed.
We use pull up network to get a clean Positive voltage and pull down to get clean Ground or negative.
Also about the need of a resistor I would like to add one more thing.
People often give an excuse of not using a resistor if they are just supplying enough voltage to turn the LED on.
FIrst, thing is that almost no LEDs will have exact same forward voltage as shown in the data sheet.
Secondly, the voltage source will never be as accurate as people think they are.
For example if you have an LED with fwd voltage to 2 volts and rated forward current of 20mA and you connect it directly without any resistor to a voltage source of 2.1 volts which you think is just 2 volts,which is very much possible because even your voltmeter won't give you exactly accurate answers.
Current , I = ( Vsource - Vfwd )/(Total series resistance in the circuit)
The resistance of good quality wires is very very less, still if you consider total resistance of the wires in your circuit to be 1 ohm( which I think is very much greater than usual) the current comes out to be:
I = (2.1-2.0)V/1 ohm = 0.1 A = 100mA
Now you are giving your LED which has max rated current of 20mA, a 100mA current which will burn the LED very quickly
I = (2.1-2.0)V/1 ohm = 0.1 A = 100mA
Now you are giving your LED which has max rated current of 20mA, a 100mA current which will burn the LED very quickly
Yep...
My standard for limiting resistor for LED is : 330 ohms to 470 ohms. Raison : Safety and I don't want to overload my Arduino Board / ATMega chip. Here the calculation : (5 - 2) / 330 = 9 mA or (5 - 2) / 470 = 6 mA And it work just fine. I just want something to light up. Beside, I dont need a LED flashlight here !!.