Poll
Question: Why my LEDs without series resistors do not burn out?
What is the maximum LED current? - 3 (75%)
Why LEDs do not burn in 5V? - 1 (25%)
Total Voters: 4

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Author Topic: LED - resistor & current relations  (Read 1447 times)
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Hello,

I am often using LEDs as many of you. I noticed that LEDs are operating very well without burning out even if i do not add any resistor. But when I read any information about them, it is written almost everywhere that series resistor is a must to use, otherwise LED will burn out instantly.

The point is LEDs are not burning out at all in 5V in the absence of a resistor.

Do they have extra resistance inside? If so why everybody still uses resistors in series with them?

It may sound weird but, i really couldn't find any information about this fact.

Best,
« Last Edit: April 02, 2012, 10:08:49 am by cmd1024 » Logged

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I voted for the first one. Hope this helps!
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Gosport, UK
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The resistor is not there to protect the LED, it's there to protect the Arduino.
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dx.... I am talking about the information is not matching the experiments. Look what is written about LEDs:

"So it’s vital to stay within the limits of the LED. If you would attach an LED to a 5 Volt power supply directly, you would burn it instantly. The high current would destroy the pn-gate. That’s the point where the current limiting resistor comes in."

This is not true or not complete. LEDs are ok with 5V without any resistor.
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Depends where the 5V is coming from, if its from a digital pin on the arduino, the gate on the processor is acting as the current limiter.  You're too tight to buy resistors ?
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dx.... I am talking about the information is not matching the experiments. Look what is written about LEDs:

"So it’s vital to stay within the limits of the LED. If you would attach an LED to a 5 Volt power supply directly, you would burn it instantly. The high current would destroy the pn-gate. That’s the point where the current limiting resistor comes in."

This is not true or not complete. LEDs are ok with 5V without any resistor.

I'll state that your statement is not correct. Could you give examples or sources showing where a standard led (lets say a red one) can be wired directly to a 5vdc voltage source and not suffer damage either immediatlly or after a given amount of time?

 An led, once biased above it's forward voltage drop rating, effectively becomes a short circuit and therefore there is no means to keep the forward current below it's maximum rated value, damage is sure to follow. But please to be sure to explain the technical reason you don't think this is true.

Lefty
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ok - let me try to explain this a bit more clearly...

There are 2 kinds of limits you'll deal with -
1. the current limit beyond which the LED eventually goes Kaput - for most LED's going beyond 50mA 100 mA will eventually burn them out
2. the amount of current which Arduino's pins are able to support - 40 to 50mA is about the safe limit.

So what's probably happening in your case is, you're running Arduino's pins at or beyond it's current limit - and the gate in Arduino is probably acting as the current limiter - your LED's life will be reduced... but more importantly, it's possible that your Arduino will get damaged.
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Depends where the 5V is coming from, if its from a digital pin on the arduino, the gate on the processor is acting as the current limiter.

What 'gate' on what 'processor' are you talking about? Arduino output pins comprise of simple N and P channel transistors wired to ground and +5vdc, Their junction becomes the output pin. there is no current limit to how much current a load can try and draw from the pin, just a limit on how much current will flow for how long before one or the other transistor burns open.



  You're too tight to buy resistors ?

Agreed, one must use series current limiting resistors to drive leds at their designed current level.

Lefty
« Last Edit: April 01, 2012, 12:31:23 pm by retrolefty » Logged

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I think it is because Arduino supplies voltage and current with PWM, which switches on and off the LED for very small durations. That might protect to LED to burn out. Since it is not a continuous current.

And i think the best explanation is mine ;P
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and the gate in Arduino is probably acting as the current limiter -

Again what gate are you speaking of, there is no such 'gate' involved. Arduino output pins are fully saturated on or off when in their output mode.

Lefty
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A regular led should not be powered more than 30ma or you willl get a shorter life,
An arduino pin can only provide 40ma max so the internal pin resistance  is actually acting as a series limiting resistor, except the heat will destroy the internals of your arduino instead of being dissapated in a normal resistor
if you leave 10 leds like that, most likely they will all be out in a week
some leds actually do have internal resistors and are rated for 5v, most do not
I've tried the same to se how far I can push a 1w lled before it burns out
I got it to 1.81 amps(normally rated for 350ma) without it burning out, I had a cpu heatsink and fan on it to dissapate heat, and I thought wow it survived untouched and gave off the light of a 10w led
however when I attempted to use it at the rated power again you can tell it was damaged from the high current because it was half as bright as a fresh led
im sure a regular led can't withstand 1.8 amps but say you drive 80ma thru it, you will permanently damage it even tho you see it working
spec sheets are there with max operating values for a reason
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A regular led should not be powered more than 30ma or you willl get a shorter life,
An arduino pin can only provide 40ma max

Not correct, an arduino output pin cannot limit it's output current to 40ma if the load resistance is such that more then that will flow. The 40ma specification is just a warning that pin damage is certain to follow if you attempt to cause the output current to exceed that amount by having a load resistance that is too small.

 so the internal pin resistance  is actually acting as a series limiting resistor, except the heat will destroy the internals of your arduino instead of being dissapated in a normal resistor
if you leave 10 leds like that, most likely they will all be out in a week
some leds actually do have internal resistors and are rated for 5v, most do not
I've tried the same to se how far I can push a 1w lled before it burns out
I got it to 1.81 amps(normally rated for 350ma) without it burning out, I had a cpu heatsink and fan on it to dissapate heat, and I thought wow it survived untouched and gave off the light of a 10w led
however when I attempted to use it at the rated power again you can tell it was damaged from the high current because it was half as bright as a fresh led
im sure a regular led can't withstand 1.8 amps but say you drive 80ma thru it, you will permanently damage it even tho you see it working
spec sheets are there with max operating values for a reason
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I think it is because Arduino supplies voltage and current with PWM, which switches on and off the LED for very small durations. That might protect to LED to burn out. Since it is not a continuous current.
@cmd1024
Total and utter rubbish. You are fooling yourself and damaging your arduino.
I wrote a page for deluded folk like you:-
http://www.thebox.myzen.co.uk/Tutorial/LEDs.html
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I don't think you connected the grounds, Dave.
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  I read any information about them, it is written almost everywhere that series resistor is a must to use, otherwise LED will burn out instantly.
Please show us where you read this.
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Here’s a commercial design that powers not a single LED, but up to 8+ LED’s from a single ATMega328P pin without any current limiting resistors whatsoever:

http://www.sparkfun.com/datasheets/Components/LED/7-Segment/SPI_to_4x7SD-v22.pdf

Buyers/prospects seem to be concerned with lack of standoffs/mounting holes and one guy is “shocked” with regards to the SPI SW interface implementation. Not a single comment about the basic electrical design or lack of current limiting resistors.

Assuming the SparkFun designer is ignorant to the popular opinion of this forum, how would you qualify this design in terms of ee? (Facts rather than name-calling and opinions would be refreshing).
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