Hello, I want to know how I can calculate the voltage of the LEDs, I have not the datasheet,
and calculate the resistance-
I know the LED is 5mm and 20mA
as knowing the voltage, the LED and how to calculate the resistance?
as knowing the voltage, the LED and how to calculate the resistance?
The diameter of the LED is useless information. Ohm's law (V = I * R) tells you how to calculate R, knowing V and I.
Another value that you need to calculate the resistance is the forward voltage.
Assuming that voltage is 3V, the resistance will be:
R = V / I = (5 - 3) / 20mA = 2 / 20 mA = 0,1 kOhm = 100 Ohm
I believe that 20mA is to much current (or to much bright).
EDIT: I don't see that you are asking for the voltage. The voltage is a value that is given in the data sheet of your LED.
luisilva:
Another value that you need to calculate the resistance is the forward voltage.
Assuming that voltage is 3V, the resistance will be:R = V / I = (5 - 3) / 20mA = 2 / 20 mA = 0,1 kOhm = 100 Ohm
I believe that 20mA is to much current (or to much bright).
EDIT: I don't see that you are asking for the voltage. The voltage is a value that is given in the data sheet of your LED.
but how to calculate the voltage of the led?
For the first datasheet that I found in google:
http://www.msc-ge.com/download/opt/datasheets/3294-15-t2c9-1hmb.pdf
You can find in page 2 3 that the forward current is between 3 and 4V for a current of 20mA.
EDIT: For different values of current you have the graph "Forward Current vs. Forward Voltage" in page 6.
luisilva:
For the first datasheet that I found in google:
http://www.msc-ge.com/download/opt/datasheets/3294-15-t2c9-1hmb.pdfYou can find in page
23 that the forward current is between 3 and 4V for a current of 20mA.EDIT: For different values of current you have the graph "Forward Current vs. Forward Voltage" in page 6.
but the information is not accurate for my leds.
there is a practical way of knowing which is the, precise voltage led up correctly?
Yes. Using the value inaccurate (100 Ohm). And after that take measures with a multimeter.
luisilva:
Yes. Using the value inaccurate (100 Ohm). And after that take measures with a multimeter.
so?
R = V / I = (5 - 3) / 20mA = 2 / 20 mA = 0,1 kOhm = 100 Ohm
by chance the resistance is 100 
Is that a question?
as resistance is connected properly?. do I need one or two resistance?
Or a 200ohm??
You need two resistances.
You have to measure what the voltage drop is for your LED, there is absolutely no way to calculate it from ohms law.
Your example:-
R = V / I = (5 - 3) / 20mA = 2 / 20 mA = 0,1 kOhm = 100 Ohm
by chance the resistance is 100 smiley
Is incorrect.
Where did that 3 come from?
That should be the forward voltage that you have to measure if you don't have the data sheets. You measure the forward voltage drop at various currents ( that means using different resistors and measuring the current ) and plot a graph and join up the dots. Then with the graph look up the current you want to use and see what the forward voltage is.
Grumpy_Mike:
You need two resistances.You have to measure what the voltage drop is for your LED, there is absolutely no way to calculate it from ohms law.
Your example:-R = V / I = (5 - 3) / 20mA = 2 / 20 mA = 0,1 kOhm = 100 Ohm
by chance the resistance is 100 smileyIs incorrect.
Where did that 3 come from?
That should be the forward voltage that you have to measure if you don't have the data sheets. You measure the forward voltage drop at various currents ( that means using different resistors and measuring the current ) and plot a graph and join up the dots. Then with the graph look up the current you want to use and see what the forward voltage is.
Where did that 3 come from?
First of all, I don't know why you can have 2 LED's connected at the same pin (it just don't make sense).
Second, I don't know where do you learn electronics and how you can say that the 2 assemblies (with one and two resistors) are equivalents.
Third take a look to this. Your drawings are very, very crappy.
luisilva:
First of all, I don't know why you can have 2 LED's connected at the same pin (it just don't make sense).
Second, I don't know where do you learn electronics and how you can say that the 2 assemblies (with one and two resistors) are equivalents.
Third take a look to this. Your drawings are very, very crappy.
:D.
I have to connect 13 leds arduino. Two. By pin
Why 2 by 2? If you are using 20mA per LED and 2 per pin and 13 LED's, maybe with a little lucky you can burn your Arduino.
each pin is 40mA. two LEDs per pin 20mA* 2 = 40mA
you say, that does not use pin 2 and 13?
I think that is too much current for Arduino, and I think that the "40mA" is too close to the limit of the Arduino. Are you 100% sure that the "40mA" are not 42mA, for example?
luisilva:
I think that is too much current for Arduino, and I think that the "40mA" is too close to the limit of the Arduino. Are you 100% sure that the "40mA" are not 42mA, for example?
no, I'm not sure. neither as check
Bi-pin LEDs
see attachment
