Before I break anything

I've got the Adafruit ARDX - v1.3 Experimentation Kit for Arduino (Uno R3). The mounting board for the breadboard and microcontroller was kind of weird which I hope isn't a bad omen but I've got that set up. Before I break anything, there are a few practicalities I want to get out of the way. My knowledge of electronics is largely theoretical (which is why I got this thing), so a laundry list of questions around the Arduino system, mainly the breadboard. For reference, here is the very first project described in my kit:

http://www.oomlout.com/oom.php/products/ardx/circ-01

  • Do the pin headers serve any purpose circuit-wise, or are they just there to hold down the paper?
  • How do I know that I need a 560 ? resistor rather than any other sort of resistor? More generally, how can I determine likely tolerable parameters of operation for any of my components?
  • Out of curiosity, do I actually need to connect the + bus strip to the 5V output on the microcontroller or could I attach it to the - bus strip instead? In other words, is this merely a matter of convention?
  • Why is the 5V connection there at all? Why is it necessary? As far as I can see, it does nothing in that configuration.

TIA!

The headers pper to hold that other sheet in place, Ohms law allows you to fugure out which value resistor you need to power your led here's what's happening 5v - vdrop of the led ~2.7 volts= 2.3 volts / 560 ohms = 4ma led forward current which is within the 20ma spec so its safe for the led if the current was too high due to a lower( or non existent resistor) the led would burnout quickly, and in the case of no resistor probably so fast you don't even see it blink The 5v line. Doesn't do anything at the moment and the only difference would be when the led was on if the led was hooked up properly from pin to ground, when its set high it will be on, if it was hhooked up from pin to 5v it will be on when the pin is set low

winner10920: Ohms law allows you to fugure out which value resistor you need to power your led here's what's happening 5v - vdrop of the led ~2.7 volts= 2.3 volts / 560 ohms = 4ma led forward current which is within the 20ma spec so its safe for the led

Of course I know Ohm's law but where did you find out that the LED only tolerates 20ma or less?

winner10920: The 5v line. Doesn't do anything at the moment and the only difference would be when the led was on if the led was hooked up properly from pin to ground, when its set high it will be on, if it was hhooked up from pin to 5v it will be on when the pin is set low

Sorry?

  1. You dont need that paper. I think you have the circuit printed in it. Its only for your reference.
  2. For this LED, 560 is not mandatory 220 ? even 10k ? can be used. It depends on the color of the LED. You have Ohms law. Use it for your circuits, it works for almost anything.
  3. That is your supply voltage. you dont need it in this circuit. (As far as I can see.)

-SaRaH-: 2. For this LED, 560 is not mandatory 220 ? even 10k ? can be used. It depends on the color of the LED. You have Ohms law. Use it for your circuits, it works for almost anything.

Again, I am quite familiar with Ohm's law but some pieces of the puzzle are missing.

I take it the digital pins are all 5V, yes?

How do I know that the large LED tolerates only 20mA and no more?

20ma is the common forward current for a regular led, yull find the real number in the datasheet also sorry my phone messes up a lot typing and I thought you were talking about using the +5v rail instead of 0v, so long as current is flowing in the proper direction across the led it will light, no matter if you switch the positive or negative side but for the extent of this tutorial that doesn't matter as for using the + or - rail they are both the same, its more a matter of clear labeling so you don't forget and make a mistake for small projects it doesn't help much but for large complicated projects proper labeling and sticking to it is very important, might as well get used to it the digital pins will output the vcc of the mcu(max vcc is 5.5v) and usually runs at 5v if you were to run it on 3.3v the output would be 3.3v, not that the arduino uno can easily run on other voltages as the entire board is designed for 5v

Ok... you dont need to know the tolerance of a Led. Your purpose here is to reduce the voltage provided to the LED. the resistor here, merely acts as a voltage divider circuit. If I am correct, the max current from the arduino is only around 20mA. Trust me on this, connect ur led directly to the pin without the resistor, it works fine

Arduino UNO pins are 5V.

You need the datasheet for the LED to determine it's characteristics. There are lots of different LED type/sizes/colours but as you supplied a link to oomlout in your first post I went there looking for the datasheet and found this http://oomlout.com/parts/LED-10B-F-datasheet.pdf. Page 2 top entry in first table Forward Current 20mA max.

In referencd to the post by sarah NO! Do not do that that can permanently damge the pin and the led, more importantly the pin the pin can supply in excess of 40ma for a period of time, but that will severly shorten the life don't expect it to last very long if you do that do the math, the led is a diode and offers very little resistance That 2.3 volts left will only be limited by the resistance of the pin and wires, you will draw alot more than the mcu can handle, definetly out of spec the resistor limits the current to protect the pin as well as the led if you need to draw more than 20ma, you should get a transistor or logic level mosfet as well as get a highper power led that can handle more

It is always necessary to look at the data sheets before u start anything. But for LEDs you are wasting your time!! Is wat I think,,

Leds are cheap, the mcu is less than cheap And its not worth the pain of trying to figure out why the led only lasted n hour or less, then replace it to realize you burn the pin

Trust me on this, connect ur led directly to the pin without the resistor, it works fine

Don't do that.

When it's on, the forward voltage across a diode or an LED (light emitting diode) is relatively constant. An Arduino output is a nominal 5V. A red LED has a forward voltage of 2V or so. So, you have a series circuit, a resistor and an LED, fed by 5V. With 2V then across the LED, the rest of the 5V (3V) is where? It's across the resistor. What's the current then? It's the voltage across the resistor, 3V, divided by that resistance.

Walking that back, take the same LED and you want to design for a desired current. What resistor to use for 10mA? x ? = (5V source voltage - 2V LED) / 0.01A x ? = 3 / 0.01A x ? = 300

-SaRaH-: Ok... you dont need to know the tolerance of a Led. Your purpose here is to reduce the voltage provided to the LED. the resistor here, merely acts as a voltage divider circuit. If I am correct, the max current from the arduino is only around 20mA. Trust me on this, connect ur led directly to the pin without the resistor, it works fine

No, your statements are not correct. An Arduino output pin has no means to limit it's output current. It does have a safety rating that you the user must insure that you do not attempt to draw more then the pin can safely provide. If your load circuit (led in this case) does indeed draw more current then the safety rating of the arduino output pin then you risk damage to the pin and chip.

An led is a current operated device. Once it is turned on by applying a voltage equal to or in excess of it's forward voltage rating, it acts like a 'short circuit' and has no way to limit it's current draw to it's recommended continuous forward current specification. Therefore the external circuit must have a way to keep the current fixed at a specific amount, say 20ma for a common red led. This is usually done by either having a proper calculated series resistor or driving the led via a constant current source. When working with leds one must deal with current control and limits, not just voltage values.

Lefty

-SaRaH-: If I am correct, the max current from the arduino is only around 20mA. Trust me on this, connect ur led directly to the pin without the resistor, it works fine

Connect this LED directly to a pin without a resistor will most likely damage the LED & your arduino. The pin can take 40mA max but driving it this hard will probably shorten it's life.

Hmm… I am familiar with transistors(I use opamps to increase the amps though). But, isnt current load dependent? If Led is the load, mcu will only supply the the current which led requires… Rite?

-SaRaH-: Hmm.. I am familiar with transistors(I use opamps to increase the amps though). But, isnt current load dependent? If Led is the load, mcu will only supply the the current which led requires.. Rite?

No, that is the point. Once a led 'sees' enough voltage equal to or higher then it's forward voltage drop specification, it turns on fully and has no control or means to limit the current that will try and flow through it to it's recommended ratings. An led is not like a lamp bulb where if you just supply it with it's rated voltage the proper current amount will flow. An led is a diode, a much different animal, once it is turned on it acts like a short circuit to current flow and something external to the led must limit the current to a specific value. A diode is a form of a solid state switch with only two modes, fully off or fully on, the circuit must provide proper driving current as the led cannot.

Does that help?

Lefty

Oh ok.. I get it.. If i got this rite, the LED doesnt really exist if 5v is supplied, which makes it a closed loop providing 40mA to the arduino pins.. Correct?

But, I have another doubt now. If this is the case, y doesnt it work when it is reversed? I mean the LED

-SaRaH-: Oh ok.. I get it.. If i got this rite, the LED doesnt really exist if 5v is supplied, which makes it a closed loop providing 40mA to the arduino pins.. Correct?

That statement makes no sense, maybe you can reword it?

But, I have another doubt now. If this is the case, y doesnt it work when it is reversed? I mean the LED

A led is a diode, if wired with proper polarity and at a voltage equal to or in excess of it's forward voltage rating it will turn on and conduct current. If a diode has a voltage applied at less then that or if the voltage is of reverse polarity, then the led will turn off and not conduct any current. Again a diode is a switch, a led is a diode that happens to emit light.

You also have a misunderstanding about the arduino 40 ma rating. It's like a maximum temperature rating, you must take steps to keep the temperature of the chip less then it's maximum rated operating temperature, the chip has such a rating but can't do anything to keep you from subjecting it to temperatures higher then it's limit. So also for a arduino output pin it has a similar safety ratings, but cannot prevent you from actually exceeding that safety rating, that's up to you with what ever circuit load you wire up to the output pin.

Led is a mere wire at 5v. All we are doing is short circuiting arduino pins.. rite?

At reverse breakdown voltage of LED, it becomes short circuit rite? Does it still damage the pins when it is reverse biased and voltage exceeds the Breakdown voltage???

..if you may want to learn more practical electronics, just do experiment (within a safe area). You are asking why to use 560ohm resistor - you may use any resistor bigger than the recommended minimal value. The max LED current is typicaly 20mA, but the today's LED will lit fine with 0.3mA as well.

Just take few resistors - for example 10k, 6k8, 4k7, 3k3, 2k2, 1k, 560 ohm, and do try how the LED will lit. It has been discussed many times - the people use "20mA rule" for LED's resistor calculation because it was so in seventhies/eighties. Today's modern LEDs lit nice with 10k resistor at 3V3 (they have much bigger efficiency and most of them include internal optical reflector, etc). Most people learned electronics through experimentation (ie Tesla, Marconi, and others..) :)