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?
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 LEDA 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.
Led is a mere wire at 5v. All we are doing is short circuiting arduino pins.. rite?Yes, if the led has a forward voltage rating of less then or equal to 5vdc, it will turn on and act like a short circuit, just as if you wired an output pin to ground and turned the pin to HIGH. Such a 'short circuit' has no way to limit the output pin current to 40 ma or less and pin damage will soon follow.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 the applied reverse voltage is less then it's max reverse voltage limit value then the led will just be turned off and no current will flow and no damage will happen. If the applied reverse voltage exceeds the leds maximum reverse voltage rating then the diode can be damaged and possibly high current can flow.
The only way you can bypass the resistor would be to provide it the exact voltage it needs so say 2.7 volt supply, the problem is the exact forward voltage varies with temperture, that's were constant current supplies come in, they automatically change the voltage to the exact value needed to stay at a select current
Alright, this blew up quickly. How does this sound?I now know that the pins give 5V; upon closer inspection, I was able to find specifications for the 10mm LEDs (like this: http://www.adafruit.com/products/301) and as far as I can tell, I want to stay at or below 20 mA and, let's say, 3.6 V (middle of the range given on Adafruit).So then by Ohm's law, I want a resistor of at least 250 ?. My impression is that the resistance an LED puts up under forward current is negligible compared to that of the resistor, so this is not really not something I need to worry about in terms of voltage division.Is that right?
Didn't see your last post, just wanted to tell you about another optionand yes, if you wanted to put an actual resistance to the led it would be like an open circuit if the voltage is below its forward voltage, and a near short circuit(so less than a few ohms) after it has passed its forward voltageand yes that would work5v-3.6=1.4/.02=70ohms minimum resistor(same as above just said)anything higher will work just fine so long as you have a few ma still flowing, and like someone said leds these days do great wit even less than 1mapersonally I usually use a 1k resistor, it is plenty for a wide range of voltages and at 5v its still quite bright for drawing so little
at gooby:the MINIMAL recommended resistor for a typical LED connected to the Arduino (5V) is calculated as follows:- supply Voltage 5V (from the PIN when log 1, or to the pin when log0)- maximal LED current 20mA (must be less than the mcu's pin max sink/source current)- typical forward voltage with RED diode @20mA is 1.6VoltSo:R = (5V-1.6V)/0.02A = 170ohm.That is a MINIMAL (the smallest value) resistor I would ever use with ANY LED (an LED of ANY colour) connected to Arduino.
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