can I burn out an LED without burning out the resistor

Discussion in 'General Electronics Chat' started by opeets, Apr 7, 2015.

  1. opeets

    Thread Starter Member

    Mar 16, 2015
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    I am on a working on a science experiment with my 5th grade son and my EE knowledge is a bit rusty. It's a basic circuit which involves turning on a 3mm red 20mA LED using Ohm's law to calculate R given different voltage values (3V, 4.5V, 6V, 9V - basically different combinations of AA batteries).

    The three goals are to show that when the current is too low (high R value) the LED will not turn on. When current is just right at 20mA (using a calculated R value for the corresponding voltage) the LED lights up nice and bright. When the current is too high (low R value) the LED will (hopefully) burn out.

    If I am using basic 1/4 watt resistors, the most I can push through a 100 Ohm resistor is 50mA, right? So if I am using a 4.5V battery configuration and a 100 Ohm resistor, that should get me 45mA.

    Will that be enough to intentionally fizzle out my LED?
     
  2. upand_at_them

    Active Member

    May 15, 2010
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    Current is (Vsupply - Vf)/R, where Vf is the forward voltage drop of the LED. With 100 ohms you'll get around 25mA, depending on the LED.

    You can get high wattage low ohm resistors. You can also parallel the resistors you have to lower the R and spread the power dissipation. If you put ten 100 ohm resistors in parallel you get an effective R of 10 ohms. That makes the current around 250mA, which should kill a regular LED. Each resistor then carries only 1/10th of that current, so 25mA, which makes the power dissipated by each resistor .025^2 x 100 = 62.5mW.

    I should point out that this relies on the power supply being able to supply that much current. Some batteries can't, such as watch batteries. AA cells should work fine.
     
    Last edited: Apr 7, 2015
  3. GopherT

    AAC Fanatic!

    Nov 23, 2012
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    Maybe and maybe not instantly. Also, a 1/4 watt resistor will take a bit more time to pop than an LED if you really give it the power. For example a 1 ohm resistor will let it pop before the resistor gets hot.

    Also, it is a nice experiment but you can actually get the LED to shatter (depending on many things) so I recommend safety glasses. The MAKE Beginning electronics book recommends this very experiment and they claim safety glasses are recommended as well. Their theory is that nothing teaches failure modes like failure.
     
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  4. ScottWang

    Moderator

    Aug 23, 2012
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    For the light fades and usage life, you should calculate only 80% of the led rated current(20mA) as 16mA.
     
  5. DC_Kid

    Distinguished Member

    Feb 25, 2008
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    so, what you have is basically two resistors. the LED has rated current at specific fwd voltages. at that rated current you can size the resistor so that ohms law balances out and the voltage drop across your resistor leaves the rated voltage of LED on anode of the LED. from there if you lower R you'll increase amps, if you raise R you'll lower the amps. I^2R is watts, if they exceed the watt rating of the item then bad things will occur, which one 1st depends on many factors.

    i would Excel doc this with say 100 values of R above and below the balanced circuit. but recall, the LED on low power side will be on as long as the fwd voltage on the anode is above the LED diode threshold, whatever that spec is.

    and just to note, as long as the voltage applied is above LED diode threshold the LED is always "on", but not enough current to "see" anything, etc.
     
  6. WBahn

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    Mar 31, 2012
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    I think some of the replies are making this needlessly difficult. You have the right idea (but do take precautions against an exploding LED).

    I'm assuming that the ideal situation would be to pick a single value of resistor and apply the different voltages and see the following sequence:

    3V -- not very bright
    4.5V -- decent brightness
    6V -- really bright
    9V -- burns out

    That may not be possible with that set of voltages.

    The unknowns at play are how much power can the resistor take above its rated power before it fails and how much current can the LED take above its rated current before it fails. Both of these are unknowns and further complicating things is that the resistor failure is due to heat which is a relatively slow process compared to the normal overcurrent failure mode of LEDs which is quite a bit quicker (usually) as it is due to either thermal runaway or glass transition of the epoxy resin.

    The use of a 100Ω resistor and a 4.5V battery will not get you 45mA because you have to take into account the forward voltage drop across the diode which, depending on the diode, is probably about 2V. So that is going to reduce you to about 25mA, which the LED can probably tolerate for quite some time.

    Let's come at it from a slightly different way. What is the smallest resistor that you can use at 9V and not exceed the 1/4 W rating. You will have about 7V across the resistor and you want the power to be

    P = V²/R <= 250 mW

    R >= (7V)² / (250mW) = 196Ω (so call it 200Ω)

    At this point the current will be 7V/200Ω = 35mA which may be enough to burn out the LED, but I wouldn't be surprised if it isn't.

    Try it and see.

    If it survives, then go with your 100Ω resistor which will dissipate about 0.5W at 9V. It will heat up pretty quickly but with 70mA flowing through the diode it might fail first. Have your son place a bet on which he thinks will happen first.

    A twist on this experiment that will get the same point across and perhaps even more is to start with the 9V battery and a 1kΩ resistor and you'll have about 7mA in the LED which should be fairly dim. The resistor will only be dissipating about 50mW so it will stay pretty cool. Then put additional 1kΩ resistors in parallel with the first and you will see the LED get brighter and brighter. Each resistor will still only have about 7mA flowing in it and so they will stay pretty cool and the airflow around them should be more than enough to carry away the total heat. At some point the LED will fail and you will have a decent idea, based on how many resistors you have when that happens, of how much current caused it to fail.
     
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  7. Dr.killjoy

    Well-Known Member

    Apr 28, 2013
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    You could try using a pot to change the resistance and a ddm to show it ..It will show you the min and max for the leds which is pretty cool.
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    IF you have a pot and IF you have a DMM, I agree. I'm assuming you probably don't (though DMMs are getting awfully cheap and you can often get them for $5 or even free at Harbor Freight -- not a good one, but more than good enough for what you are doing).

    If you have two DMMs you could plot out the voltage versus current curve and show him how it looks and ask him to estimate what the current will be for voltages you haven't taken measurements at (and vice-versa, voltages at currents you haven't taken measurements at). It also becomes a way to introduce him to the concept of exponentials. He probably won't see that in school for a few years, but he is probably old enough to grasp the concept, especially if it is in the context of something he has seen and experienced.
     
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  9. WBahn

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    Mar 31, 2012
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    Oh, and please follow-up with how you actually do the experiment, what you found out, and how well your son responded to it and what concepts he seemed to grasp and what ones seem to be above his head.
     
  10. vikasbly44

    New Member

    Jan 19, 2015
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    DEAR..
    You can get high wattage low ohm resistors. You can also parallel the resistors you have to lower the R and spread the power dissipation. If you put ten 100 ohm resistors in parallel you get an effective R of 10 ohms.
    led raw material

    Thank you....
     
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  11. opeets

    Thread Starter Member

    Mar 16, 2015
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    Thanks for the info. I understood most of it but I need to simplify it so I can explain this to my son.

    The LEDs that we purchased from Radio Shack are all identical 3mm red LEDs with Vf = 1.8V and I = 20mA.

    If we use 4.5V (3 AA batteries) and a 50 Ohm resistor, we should get an approximate reading of 54mA ((4.5-1.8)/50) in the circuit (with the DMM between the resistor and the LED) and the power dissipated across the resistor would be approx 146mW (P = .054 * 2.7) so the resistor would be safe. But 54mA may not be enough to fizzle out the LED, right?

    If we then use two 50 Ohm resistors in parallel we would have an equivalent R of 25 Ohms (and thus approx 108mA in the circuit) but each resistor would still dissipate 146mW right? This configuration I would assume will cause the LED to burn out.

    Does this sound like a reasonable first approach?
     
  12. upand_at_them

    Active Member

    May 15, 2010
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    Yes, that's reasonable. But 108mA probably isn't enough to burn out the LED. You could successively show your son how hot it gets with each resistor added in parallel.

    Maybe showing him how the resistors in series will reduce the current and dim the LED. The resistors in parallel will increase the current and brighten the LED...until it burns out.

    With three in parallel it should be very hot and might burn out. Should definitely burn out with four in parallel.
     
  13. #12

    Expert

    Nov 30, 2010
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    One problem is that most of us don't study how to burn up parts. Some old rule of thumb says double the lifetime for every 10% less power. If you don't actually melt the LED you are working with a, "shortened" life span. How short? I think you are going to be the pioneer on this research. This seems a stupid way to look at it, but how many times do you have to cut the LEDs life in half to get from 100,000 hours to 10 seconds? Even if you can do the math, the answer is not going to be reliable.

    You're just going to have to burn a few of them up.
     
  14. opeets

    Thread Starter Member

    Mar 16, 2015
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    Kind of defeats the whole premise of what we were trying to demonstrate. Basically we wanted to show that using Ohm's law we could calculate the R value to achieve 20mA and see the LED light up. We also wanted to show that if R is too high, the LED would not light up and if R was too low, the LED would burn out thus proving the importance of Ohm' law. Based on your inputs I can see that LEDs have a great tolerance even when the current supplied is quadruple the normal operating current. Perhaps we need to re-state the problem to account for the tolerance range with respect to Ohm's law and it's importance (otherwise the teacher - who has no knowledge of electrical circuits - may miss the point of the experiment). I am open to suggestions.
     
  15. opeets

    Thread Starter Member

    Mar 16, 2015
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    That's why I bought several LEDs of the same type. I knew this was going to be a trial and error thing.
     
  16. #12

    Expert

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    How about little incandescent bulbs? They demonstrate incredible brightness for a few seconds before they self-destruct.
     
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  17. opeets

    Thread Starter Member

    Mar 16, 2015
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    Can you suggest any? Something that I could use in a breadboard like an LED?
     
  18. upand_at_them

    Active Member

    May 15, 2010
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    Huh? I thought that was exactly what you wanted to demonstrate. You're going to need lower R in order to burn out the LED for the final demonstration.


    20mA is only one valid operating current for the LED. Run it at 10mA and the LED will be dimmer and last longer.
     
  19. #12

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  20. upand_at_them

    Active Member

    May 15, 2010
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    Why abandon the LED? Just calculate R for different operating currents. Such as 1, 10, 20, 50, 100, 200 mA. You'll see a change in brightness and will end with the LED burned out.
     
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