# LED circuit Heat

Discussion in 'LEDs and Optoelectronics' started by SniffTheGlove, Dec 22, 2016.

1. ### SniffTheGlove

9
0
Dec 22, 2016
Hello,

I have built an LED Illuminator using IR940nm LEDs (100ma, 1.3Vf) but it gets so hot that it melted some other parts in the enclosure, so I am looking for a way to cool it all down or help in redesigning the circuits.

I am currently using 5 banks of 8 LEDs. Each block of 8 LED's is a self contained circuit with a 1/2w 10Ohm Resistor

This is all driven by 12vdc.

I used several online LED calculators to work out the number of LEDs per block and the resistor value.

What is tyhe best way to either cool the whole lot (fan??) or use a different resistor or change the qty of LEDs per block

Thank you

2. ### cjdelphi

1,096
104
Oct 26, 2011
10ohms 12v?

12 - 2 .. 10v/10ohms = 1amp

10 ohm resistors?

3. ### Audioguru

3,245
702
Sep 24, 2016
No. The LEDs produce the infrared, not the resistors.
The IR LEDs are 1.3V each so eight of them in series need 10.4V. The supply is 12V so the LEDs and 10 ohm current-limiting resistor have a current of (12V - 10.4V)/10 ohms= 160mA and the resistor dissipates 160mA squared x 10 ohms= 256mW which is fine for a 1/2W resistor..
The 100mA LEDs are very overloaded.

4. ### AnalogKid

2,496
718
Jun 10, 2015
8 x 1.3 = 10.4
12 - 10.4 = 1.6
1.6^2 / 10 = 0.256 W
A 0.5 W resistor will be pretty warm to the touch, but safe.

i = e / r
i = 1.6 / 10 = 160 mA
P = e x i = 1.3 x 0.16 = 0.21 W
That's fine for a small transistor like a TO-92 package, but a lot of heat to pack into something the size of an LED die.

LED part number / data sheet? Many high brightness LEDs are built for attachment for a heat sink.

ak

1,096
104
Oct 26, 2011

6. ### SniffTheGlove

9
0
Dec 22, 2016
Hi, Thank for the replies.

I am using http://ledcalc.com/ for the calculations.

The IR LED specs are this.
5mm Round Clear Lens Light Emitting Diode IR 940nm
940nm 5mm IR Infrared Launch Emission Tube Diode Emitting LED Lamp
Specifications
- Size: 5mm
- Wavelength: 940nm
- Lens color: water clear
- Forward voltage: 1.2-1.4V
- Forward current: 100mA
- Emitted color: infrared

So for the 8 IR LED Circuit
I have116mW per LED dissipation and 142mW dissipation on the Resistor. I only had 10 Ohm 1/2w resistors so hoped that would do.

So would reducing the qty to 4 LEDs per circuit.
So using 4 I would need a 1W 82 Ohm resistor I have108mW per LED dissipation and 564mW dissipation on the Resistor which is hotter I think.

Would I need to got to 2 or 3 leds per circuit or even just 1 LED and it's resistor and them multiple this by 40 separate circuits to get me illumination.

Thanks

7. ### Harald KappModeratorModerator

11,439
2,625
Nov 17, 2011
The calculator tells you to use 16 Ω. You can build 15 Ω, which is much closer to 16 Ω, from 3*10 Ω resistors:
-- 10 --|
----- 10 -- ---------
-- 10 --|
1* 10 Ω in series with 2*10 Ω in parallel. This will reduce your total power dissipation from 12V*160mA = 1.92W to 12V*107mA=1,3W (numbers rounded)
That's possible but terribly inefficient as you now dissipate more power in the resistor instead of producing IR in the LEDs.

8. ### Audioguru

3,245
702
Sep 24, 2016
What is 100mA? The recommended current or the absolute maximum current if you cool the LED somehow?
Do you realize that your 10 ohm resistors are killing the LEDs with 60% more current?

How will you cool the LEDs? Please post their datasheet and part number.

9. ### SniffTheGlove

9
0
Dec 22, 2016
Hi, Thank for the replies.

I am using http://ledcalc.com/ for the calculations.

The IR LED specs are this.
5mm Round Clear Lens Light Emitting Diode IR 940nm
940nm 5mm IR Infrared Launch Emission Tube Diode Emitting LED Lamp
Specifications
- Size: 5mm
- Wavelength: 940nm
- Lens color: water clear
- Forward voltage: 1.2-1.4V
- Forward current: 100mA
- Emitted color: infrared

So for the 8 IR LED Circuit
I have116mW per LED dissipation and 142mW dissipation on the Resistor. I only had 10 Ohm 1/2w resistors so hoped that would do.

So would reducing the qty to 4 LEDs per circuit.
So using 4 I would need a 1W 82 Ohm resistor I have108mW per LED dissipation and 564mW dissipation on the Resistor which is hotter I think.

Would I need to got to 2 or 3 leds per circuit or even just 1 LED and it's resistor and them multiple this by 40 separate circuits to get me illumination.

Thanks

I don't know how to cool, this was one of my original questions.
Also, the only data I have on the IR LEDs is what I have posted here.

Looking at all the equations is quite confusing, so if someone could explain in basic detail this would help me,

For example 12vdc - (8x1.3 = 10.7) = 1.6v / 0.100A = 16Ohms

So if I use 15Ohms what happens with the equation, and what would happen if I used 17Ohms

This is so I can understand if using a lower resistor (ie 15Ohms) it produces more heat, more current??? or does it use less??

Also how can I drive these at a non-overload as Audioguru said they were being overloaded?

This electronics is sometimes a bit hard to understand if you have never done much.

10. ### Chemelec

291
47
Jul 12, 2016
12 VDC, From a Car Battery or Power Supply?

A Car Battery is typically 12.6 Volts, But in a Car when charging it can be up to 14.6 Volts.

So What are you using?

11. ### Audioguru

3,245
702
Sep 24, 2016
I do not believe that your 5mm LEDs will survive a continuous current of 100mA without melting.

Ohms Law says that 1.6V/16 ohms is 100mA.
1.6V/15 ohms is 107mA. There is no 17 ohms but 18 ohms gives a current of 1.6V/18 ohms= 89mA.
You said you used 10 ohms so your current was 1.6V/10 ohms= 160mA if the LEDs are actually 1.3V (the current is 240mA if the LEDs are 1.2V).

The datasheet for name-brand 5mm LEDs say that the operating current is 20mA, the maximum continuous current is 30mA and the maximum momentary current is 100mA. The forward voltage range of 1.2V to 1.4V is when the current is only 20mA and is higher at 100mA.

Since the LEDs have a range of forward voltage then enough voltage must be across the resistor to safely allow 1.2V LEDs or 1.4V LEDs. First you need to know the maximum allowed continuous current and what forward voltage range is at that current. If the current is 30mA and the range is 1.2V to 1.5V then try eight 1.2V LEDs at 30mA to get a total voltage of 9.6V then the resistor value for 30mA is (12V - 9.6V)/30mA= 80 ohms, 82 ohms is the nearest value. Then calculate the current if all eight LEDs are 1.5V. It is 12 - (8 x 1.5V)/82 ohms= 0mA so you cannot use eight 1.5V LEDs. Try six 1.5V LEDs then for 30mA the resistor is 160 ohms for 1.2V LEDs and the current for 1.5V LEDs is 18.8mA which is rather low. Try five 1.2V LEDs at 30mA to get a resistor value of 200 ohms then with 1.5V LEDs the current is 22.5mA which is probably fine.

Here are the detailed specs for a name-brand IR LED:

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12. ### SniffTheGlove

9
0
Dec 22, 2016
I am using a 12Vdc 7Ah Gel Battery to power my rig. The current voltage as I type this is 12.4Vdc

13. ### Chemelec

291
47
Jul 12, 2016
Slightly on the low side, Normal is 12.6.
It will Increase when you charge it.

If your circuit is to be used Also when charging the battery, you need to Consider at least 13.6 Volts.

14. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,490
2,831
Jan 21, 2010
It sounds like you should be running your LEDs at 20mA. We have a resource describing how to use LEDs that you should look at.

For higher current LEDs (which you may want to use in your application) you are better off with a constant current driver. The same resource has some simple circuits for this.