Hello everybody,
Before all, thank you for the Arduino project that (I hope) helps me to like electronics
I have a question concerning the first project (yes ...).
To remember, we have a 5V input and a 220 ohm resistor.
At the end of the project, they said : "That value (e.g 23 mA) is just about the maximum you can safely use with these LEDs, which is why you used a 220-ohm resistor."
I don't understand why they said that 23mA is the maximum value (safely use) when datasheet said 20mA as If .... Can someone explains me that ?
I see this : "The next thing we need to know is the I, which is current we want to drive the LED at. LEDs have a maximum continuous current rating (often listed as If, or Imax on datasheets). This is often around 25 or 30 mA. What this really means is that a typical current value to aim for with a standard LED is 20 mA to 25 mA—slightly under the maximum current." on this website : Basics: Picking Resistors for LEDs | Evil Mad Scientist Laboratories
I'm okay with that but how can I check it in the datasheet ?...
"The first project" is as specific as "my first car". There are many starter guides out there
And yes, 20mA is the max of most simple leds, 23mA is over driving the led a bit. It will most likely don't die immediately but it's better to stay under it. And maybe they rely on the fact that the voltage is most likly a bit lower then 5V when connected to USB.
And are you sure that's the datasheet for it? Don't see the 20mA
But for normal leds 20mA is around the maximum current. But you don't have to drive them @ 20mA. Most modern leds are very bright at 20mA. 1mA is enough to have a very visible led.
And when you don't have a datasheet with the leds (happens like 99%..) good rule of thumb for normal leds is
20mA max, less is okay, just play with it
+-2V drop across for red, IR, orange and (diffuse) green leds
+-3V drop across other colors
When in doubt, just measure the voltage. Just put a 500Ohm (or 1k, as long as it's not very low to be safe) in series with the led, connect it to 5V and measure the voltage across the led.
The LED has a voltage drop associated with it as well, typically ~2v for a red led and ~3 for a blue or green one. So in fact, it's more like 2~3v / 220 ohms = 9~14mA
Hi everyone and thank you for your quickly answers :
AWOL:
Which first project?
I follow the beginner guide (Arduino Projects Book) and the first project name is "Get to know your tools"
septillion:
"The first project" is as specific as "my first car". There are many starter guides out there
And yes, 20mA is the max of most simple leds, 23mA is over driving the led a bit. It will most likely don't die immediately but it's better to stay under it. And maybe they rely on the fact that the voltage is most likly a bit lower then 5V when connected to USB.
And are you sure that's the datasheet for it? Don't see the 20mA
But for normal leds 20mA is around the maximum current. But you don't have to drive them @ 20mA. Most modern leds are very bright at 20mA. 1mA is enough to have a very visible led.
And when you don't have a datasheet with the leds (happens like 99%..) good rule of thumb for normal leds is
20mA max, less is okay, just play with it
+-2V drop across for red, IR, orange and (diffuse) green leds
+-3V drop across other colors
When in doubt, just measure the voltage. Just put a 500Ohm (or 1k, as long as it's not very low to be safe) in series with the led, connect it to 5V and measure the voltage across the led.
Thank you for the advice of 20mA as a reference. I was not thinking that the voltage could be lower than 5V ! I don't see the 20mA value in this datasheet but in another one on Google. Here, it's most about 11,5mA if I remember ?
DrAzzy:
5v / 220 ohms = 23 mA
But there is NOT 5v across that resistor!
The LED has a voltage drop associated with it as well, typically ~2v for a red led and ~3 for a blue or green one. So in fact, it's more like 2~3v / 220 ohms = 9~14mA
I'm okay with what you are saying (I have 2.8V~3.2V for the resistor and 1.8~2.2V for the LED). So it's not a problem to apply the Ohm law with 5V as input voltage ?....
We have a circuit with 5V as input, a resistor and a LED. We want to know the value of the resistance in order to avoid sending too much current to the LED.
When I go on the datasheet of the LED red, I see that this LED consumes ~2V and has a maximum current of 20mA.
So the resistor will consumes 5V(input)-2V(LED) = 3V. If we apply the Ohm law, we found using the maximum current of 20mA for the LED : R=V/I=3/(20*10^(-3))=150 ohm. This is for the minimum value for resistor (because of maximum value for current) so we take a higher resistor. That why we are using 220 ohm.
It's okay or am I wrong ? Did I have to think in another way ?
That's correct! But keep in mind it's the max current, no one says you must drive them close to that. Like I said, modern leds are bright. 1k in series with a red led is still very (too) bright as an indicator led.
I see that this LED consumes ~2V and has a maximum current of 20mA.
Well, not exactly. We wouldn't say that, no. We would say the led has a forward voltage of 2V@20 mA.
(it doesn't "consume" 2V , it CONDUCTS (TURNS ON) at 2V. There is a big difference.
5v / 220 ohms = 23 mA
Led current is based on a parameter called "forward voltage" This leads you to a parameter called "forward current" which is the current for which the forward voltage is specified. Forward voltage is specified as "X" voltage @ "Y" current:
ie; Vf = 2.2V @ 20 mA
You don't start with a 220 ohm resistor and then calculate the current .
You start with the forward voltage (from the datasheet) and then calculate the resistor.
This is just basic led electronics (LEDS 101).
Just to be clear, you do not HAVE to use 140 ohms, because you do not HAVE to draw 20 mA.
The 20 mA is like a maximum continuous current rating so if you want a bright led you CAN use 20 mA but if you prefer the led not to be so bright or you want to reduce current you can decrease the current by increasing the resistance to 220 ohms.
septillion:
That's correct! But keep in mind it's the max current, no one says you must drive them close to that. Like I said, modern leds are bright. 1k in series with a red led is still very (too) bright as an indicator led.
Ok, I know that I don't have to be close to the max value, it's only to know what is the minimal value for my resistor to use.
raschemmel:
Well, not exactly. We wouldn't say that, no. We would say the led has a forward voltage of 2V@20 mA.
(it doesn't "consume" 2V , it CONDUCTS (TURNS ON) at 2V. There is a big difference.
Led current is based on a parameter called "forward voltage" This leads you to a parameter called "forward current" which is the current for which the forward voltage is specified. Forward voltage is specified as "X" voltage @ "Y" current:
ie; Vf = 2.2V @ 20 mA
You don't start with a 220 ohm resistor and then calculate the current .
You start with the forward voltage (from the datasheet) and then calculate the resistor.
This is just basic led electronics (LEDS 101).
Just to be clear, you do not HAVE to use 140 ohms, because you do not HAVE to draw 20 mA.
The 20 mA is like a maximum continuous current rating so if you want a bright led you CAN use 20 mA but if you prefer the led not to be so bright or you want to reduce current you can decrease the current by increasing the resistance to 220 ohms.
Did you want to say that a lesser voltage than 2V doesn't conduct and so can cause bad effects on the circuit ?
Ok so if I understand, it's the same above response ? It's only an indicator of what is the minimal value for the Resistor to use to avoid problems ? And we have to use a higher resistor to not be close to 20mA (with 2.2V@20mA).
Thank you for your explaination about forward current.
If you want less current, say 5mA, you can calculate the resistor for that.
(5V - Vf)/current = resistor
(5V - 2.2V)/.005 = 560 ohm
You can then measure the voltage across the resistor and see how close you are:
Vr/560 = current
Ideally, Vr would be (5 -2.2V) = 2.8V and if R = 560 then 2.8V/560 ohm = 5mA.
5V might not really be 5V, Vf may vary, the resistor may vary, but they'll be close.
CrossRoads:
If you want less current, say 5mA, you can calculate the resistor for that.
(5V - Vf)/current = resistor
(5V - 2.2V)/.005 = 560 ohm
You can then measure the voltage across the resistor and see how close you are:
Vr/560 = current
Ideally, Vr would be (5 -2.2V) = 2.8V and if R = 560 then 2.8V/560 ohm = 5mA.
5V might not really be 5V, Vf may vary, the resistor may vary, but they'll be close.
