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Basic LED help please!

mowgli2084

Jun 21, 2014
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Hello,

I'm very new to electronics and aside from a brief overview in a physics class at school have limited knowledge. I'm trying to learn from the internet but I've come up against a problem.

I'm trying to power the following IR LED from a 12v power source.

http://uk.rs-online.com/web/p/ir-leds/7845787/

In looking at the spec sheet I read the power required is 2W and the forward current is 1A making the voltage 2V right? I've been looking at resistors to use for this setup and have used a calulator to find that I should be using a 10 ohm resistor, I guess to provide me with 1.2A.

When I do use a 10 ohm resistor it blows up by smoking. I've checked with a camera, the LED is powered but the resistor burns out. What am I doing wrong?

Any help would be really appreciated.

Thank you, Tom
 

Arouse1973

Adam
Dec 18, 2013
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What wattage resistor are you using sounds like it's too small it should be at least 5W? Also that is the maximum current the device can handle, do you need it to be that powerful? You can run them at a lower power level.
Adam
edit: yes sorry power rating of resistor should greater than 10W.
 
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KrisBlueNZ

Sadly passed away in 2015
Nov 28, 2011
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Right, it sounds like your resistor needs to be rated for a much higher power dissipation.

Your LED requires 1A (that's a LOT for a LED, especially continuously!) and dissipates 2W, so yes, the voltage across the LED will be 2V.

Voltages in series add together, so your resistor will drop the rest of the voltage, i.e. (12V - 2V) = 10V. And it will have 1A flowing through it. The amount of power it will dissipate can be calculated as P = V I = 10 x 1 = 10W. That's why it's going up in smoke.

Can you give us a link to the part number and/or data sheet for the LED? I think you're running too much current through it.
 

BobK

Jan 5, 2010
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You would be much better off powering it from a lower voltage. You are wasting over 80% of the power when driving it from 12V.

The forward voltage of IR LEDs is typically lower than the 2V you calculated. I am using one that take 200mA and its forward voltage is 1.35V.

Bob
 

mowgli2084

Jun 21, 2014
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Hi all,

Thank you for your replies, really appreciate your help. Sorry for being slow in replying, I've just come back from a trip.

I think that a lower voltage power supply would be really helpful. The only thing is that the LEDs are going to provide the light source for an IR camera on a RC quadcopter which is running from 12V and the extra weight of another battery would be a big issue.

The data sheet for the IR LED is here:

http://docs-europe.electrocomponents.com/webdocs/121b/0900766b8121b287.pdf

Honestly, I didn't realise that resistors could have power ratings as well. It sounds like I'm putting way too much through the resistors and that's why they're burning out. So is what I need a 10 ohm, 10W resistor? Is that something that exists?

I thought they would require a high current as they're pretty bright LEDs. I basically chose the ones that had the highest light output I could afford.

Thanks again, Tom
 

Anon_LG

Jun 24, 2014
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10 watt resistor help.jpg If you put 4 resistors with a resistance of 47 ohms each and a power rating of 3 watts (these will be called wirewound resistors) in parallel this will give 11 3/4 ohms, you could use a 10 ohm, 10 watt resistor however this resistor will be operating at the very top of its power capability.Or if you wanted the safe option event closer to 10 ohms you could put 3 47 ohm and 1 27 ohm 3 watt resistors in parallel to give 9 117/128 ohms (pretty much 10)

Resistors in parallel follow the rule 1/(1/R1)+(1/R2)+(1/R3)+(1/R4)... (carry on depending on how many resistors there are) to give the total resistance

Resistors in series follow a much simpler rule R1+R2+R3+R4... (carry on depending on how many resistors there are) to give the total resistance
 
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Arouse1973

Adam
Dec 18, 2013
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Hi all,

Thank you for your replies, really appreciate your help. Sorry for being slow in replying, I've just come back from a trip.

I think that a lower voltage power supply would be really helpful. The only thing is that the LEDs are going to provide the light source for an IR camera on a RC quadcopter which is running from 12V and the extra weight of another battery would be a big issue.

The data sheet for the IR LED is here:

http://docs-europe.electrocomponents.com/webdocs/121b/0900766b8121b287.pdf

Honestly, I didn't realise that resistors could have power ratings as well. It sounds like I'm putting way too much through the resistors and that's why they're burning out. So is what I need a 10 ohm, 10W resistor? Is that something that exists?

I thought they would require a high current as they're pretty bright LEDs. I basically chose the ones that had the highest light output I could afford.

Thanks again, Tom

Yep they do exists. Here are a few.
http://uk.rs-online.com/web/p/through-hole-fixed-resistors/6661724/
http://uk.rs-online.com/web/p/panel-mount-fixed-resistors/6835997/
http://uk.rs-online.com/web/p/through-hole-fixed-resistors/6253402/

Adam
 

Harald Kapp

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You are wasting a lot of (precious) power in the series resistors. The ideal thing to use would be a switch mode current source which wil reduce losses greatly. In the light of your limited electronics knowledge, I suggest another way, less difficult than a swizch mode regulator but more efficient than a series resistor:
Instead of using a single LED at high current, use many LEDs in series at a lower current. If you use, for example. 10 LEDs rated at 2V and 100mA, you have 0.2W per LED and a total of 2W for all ten LEDs. Your series resistor needs to drop 2V (12Vbattery-10Vled) at 100mA which makes for 20 Ohm at 0.2W. Instead of "burning" the power from the excess voltage in a series resistor, you are now putting it to work in the LEDs.

Apart from saving money (10 smal IR LEDs are proably less expensive than 1 high power LED) you may also save weight:
  • A wirewound power resistor is comparatively heavy.
  • A 2W LED cannot be operated without a heatsink which will add dead weight to your 'copter. At 0.2W the smaller LEDs probybly will not need a heatsink.
Plus you can distribute the smaller LEDs in a way that illuminates a larger area than a single high power LED does (at least without proper optics).
 

Arouse1973

Adam
Dec 18, 2013
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You are wasting a lot of (precious) power in the series resistors. The ideal thing to use would be a switch mode current source which wil reduce losses greatly. In the light of your limited electronics knowledge, I suggest another way, less difficult than a swizch mode regulator but more efficient than a series resistor:
Instead of using a single LED at high current, use many LEDs in series at a lower current. If you use, for example. 10 LEDs rated at 2V and 100mA, you have 0.2W per LED and a total of 2W for all ten LEDs. Your series resistor needs to drop 2V (12Vbattery-10Vled) at 100mA which makes for 20 Ohm at 0.2W. Instead of "burning" the power from the excess voltage in a series resistor, you are now putting it to work in the LEDs.

Apart from saving money (10 smal IR LEDs are proably less expensive than 1 high power LED) you may also save weight:
  • A wirewound power resistor is comparatively heavy.
  • A 2W LED cannot be operated without a heatsink which will add dead weight to your 'copter. At 0.2W the smaller LEDs probybly will not need a heatsink.
Plus you can distribute the smaller LEDs in a way that illuminates a larger area than a single high power LED does (at least without proper optics).

Yes good idea as long as he has the space for this. He might also want just one LED for a reason. Maybe he can let us know.
Adam
 

mowgli2084

Jun 21, 2014
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Firstly, thank you for all the ideas and help with this.

I've made a quick trip to Maplin just to test the 10 ohm 10 watt resistor option and they are rather big and heavy but the whole system does work (the resistors are quite warm but I presume this is normal).

I'm using a lower number of high power LEDs (I have two) because I thought it would be quite weight and space efficient but I see with the heatsinks and heavy resistors that this isn't the case.

I think I'll try the option with a higher number of lower power LEDs and see how that works out.

Cheers all, Tom
 

gorgon

Jun 6, 2011
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I hope you are aware that 1A is absolute max current for continuous use. There should be a safety margin when using a resistor for current control.
 
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