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3W white led
In designing my first PCB ever to mount a 3W white led (shown above) thermal dissipation is a concern.

It is a simple PCB and I don't have the ability to add any heat sink beyond the ground plane of the PCB to which the led will be soldered. Here's what the PCB will look like :

pcb

The lighter green you see all over the place is the ground plane and in the center of the PCB is the foot print of the led with the round center piece being the heatsinking pad. Overall the PCB is 6x6 cm, the copper is 1 oz / ft2 or 305 g/m2 and the board is 1.6mm thick. The led is the only sensitive thing on there.

How much power do you think this board can take before the led starts cooking itself? Do you think it will survive the full 3W ? Or even 0.5W ?

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    \$\begingroup\$ A big photo of your LED doesn't help much, we need to see the datasheet. \$\endgroup\$ Aug 27, 2018 at 16:07
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    \$\begingroup\$ Here's a good resource to look for your answer: we-online.com/web/en/index.php/show/media/04_leiterplatte/… \$\endgroup\$
    – D Duck
    Aug 27, 2018 at 16:10
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    \$\begingroup\$ If you are concerned about heat dissipation, you shouldn't use thermal relief on pads, and trade the convenience of soldering for better heat conductance. Also you can use as many vias as you can to bridge the pads to the opposite side of your PCB, which will serve as additional heat transfer plate. Use 2-2.5 oz copper if you can. \$\endgroup\$ Aug 27, 2018 at 16:40
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    \$\begingroup\$ @NickAlexeev of course the thickness of the copper is important, the fiberglass substrate is NOT conducting significant amounts of heat (glass is a thermal insulator) and it is shielding the whole of one side of the copper. So thickness is definitely NOT a secondary factor. Using an Al substrate would bring it perhaps in line with the same unit on an AL substrate from the datasheet. \$\endgroup\$ Aug 27, 2018 at 22:32
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    \$\begingroup\$ Oh -- I get it!! he put thermal relief on the HEAT SINKING PAD. That's a bad idea! \$\endgroup\$ Aug 30, 2018 at 21:14

3 Answers 3

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Your biggest mistake was using thermal reliefs on the thermal pad. You nearly halved the lateral thermal conductivity.

enter image description here

The most important criteria is to move the heat away from the LED. Lateral heat spreading is done mostly using thermal conduction of the component side copper plane.

2 oz copper rather than 1 oz rarely costs more and will double lateral thermal conductivity.

Orientation of the PCB will affect the effectiveness of many thermal management techniques. Vertical is best and horizontal with the LED facing down is worst. Horizontal with the LED facing up is okay.

enter image description here
Source: Constructing Your Power Supply — Layout Considerations
Section V. Thermal Considerations


It's simple. You design the PCB with good thermal management techniques then adjust the max current by measuring the LED's temperature. A temperature of 45-50°C is a good target temperature range.

The app note I have found to be the "simplest" most concise thermal reference, from Texas Instruments, is:

Thermal Design By Insight, Not Hindsight


First off you should not use an LED manufactured with archaic technology. Your LED was originally patented by Lumiled which is now expired.

An LED dissipates the electrical energy (Vf x current) in two ways, heat and light.

If you use an LED with high efficacy (200+ lumens/watt), less electrical energy is dissipated as heat.

A decent LED will dissipate 70% of the electrical energy as radiant energy and 30% as heat. A Cree XP-3G or Samsung LM301B is almost 80%/20%.

The LED you are using is likely closer to 30% radiant and 70% heat or twice the heat of a quality LED. So the best way to reduce the heat transfer requirements is to use a better LED.


You should use both sides of the PCB to dissipate the heat.
You should enlarge the thermal pad to as close to the edge of the PCB as possible. Use copper pours on both sides. Do not interrupt the lateral thermal path with signal traces.

A thinner PCB generally does not cost more than a 1.6 mm thick PCB. A 0.8 mm thick PCB will have half the thermal resistance of 1.6 mm. Thinner PCB will reduce the thermal resistance of thermal vias as well.

Cree's paper, Optimizing PCB Thermal Performance, says:

Cree recommends creating areas of 10-mil (0.254-mm) vias arranged on a 25-mil (0.635-mm) rectilinear grid.

See also: AB32: LUXEON Rebel platform Assembly and Handling Information Application Brief, Section 3.3 Thermal Via Design

For your design I would recommend 20-30 0.3 mm thermal vias, spaced on 0.65 mm centers, in a square pattern, centered on the LED's thermal pad. See fig. 20 the Luxeon application brief.



You should consider using multiple mid power LEDs rather than a single high power LED. This way you can reduce the temperature and spread the heat. For example four LM301B LEDs equal the luminous intensity of one XP-3G. The LM301Bs are more efficient and cost effective.


How much power do you think this board can take before the led starts cooking itself?

Likely will not dissipate 2W of heat.

Below are two burnt 3W Lumiled Rebel ES royal blue LEDs run at 1 amp with about 50% efficacy with about 35 thermal vias. The strip was 18 mm wide and the LEDs were spaced at 40 mm"

enter image description here



Do you think it will survive the full 3W? Or even 0.5W?

Somewhere in between, my guess would be 1 watt of heat. Depends on a variety of factors. If the PCB is mounted vertically, and a thermal pad were added to the opposite side, it may run at max current.

You cannot rely on thermal calculations. They are only estimates. The online calculators are a waste of time. PCB designs are not the same. The temperature and power dissipation gradients are non linear and the calculators often use linear calculations for very limited design criteria that is unlikely to work with your design. Just because it is on the Internet does not mean it is true.

A good source for calculations is Texas Instruments paper:
Thermal Considerations for Surface Mount Layouts

and

Constructing Your Power Supply — Layout Considerations

A simplified paper with step by step calculations from OSRAM:
Thermal Management of SMT LED

for calculations using package thermal reistance from OSRAM:
Package-Related Thermal Resistance of LEDs


Generally what I do is, do my best with the PCB design and then adjust the maximum current by temperature of the LED.

The 9 mm strips (560 mm length with 48 LM301B) I made can be run at 300 mA with no heatsink.
Measurements taken at room temperature of 23°C.

No heatsink

 mA Temp
160 36°C
300 47°C
400 70°C

Mounted to a 25 mm wide,1.6 mm thick strip of aluminum they can be run at 600 mA.

 mA Temp
400 44°C
600 60°C
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The datasheet posted specifies a max operating temperature of 80C and a Thermal resistance of the Junction to Case at 10C/W. You also can't go over 3W as that is the maximum rated power for the device.

So 3W*10C/W= 30C temperature rise above the PCB temperature, if you are operating at 3W and you have a thermal pathway good enough to dissipate 3W (or an infinite thermal sink rather).

Thermal Pathway to PCB

The heat not only needs to leave the pad, but needs to go to out to the air through a PCB or heatsink. The thermal pathway through the PCB with vias is this for 3W, the numbers below assume a 3W source on top of the vias. I've calculated the numbers for a 10C rise above the PCB like this: 3W*(Via thermal resistance)C/Watt*1/n= thermal rise above PCB.

The 1/n is the more vias you have the more the thermal pathway goes down.

With normal 1mil plating would have using the Saturn PCB design calculator:
A 5mil via has 328C/Watt,you'd need 98 vias.
A 10mil via has 179C/Watt,you'd need 53 vias.
A 15mil via has 123C/Watt,you'd need 36 vias.

Thermal Pathway from PCB to air

Use this thermal calculator to get an estimate for board area, you'd need 54cm^2, sunk to a layer that covered the whole board.

enter image description here

Total Temp rise

With the appropriate amount of vias an 54cm^2 of board area, for an estimate, there would still probably be a 30C+10C temp rise above ambient. As a worst case if ambient was 40C then that would be 80C at 3W Thermal. I would either derate the power on the LED by 33% to 50% (so 1.5W or 1W Thermal) , put a heatsink on the PCB, or increase the thermal pathway. The best thing would be to build and test after adding vias.

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  • \$\begingroup\$ Overall an okay answer with sound recommendations. You neglected to subtract the radiant watts from the electrical watts. Dissipation is only needed for thermal watts. Your thermal via resistances seem very high. 98, 5 mil vias is a bit exorbitant if not foolish. Thermal modeling is very complex with non-linear gradients. This calculator is a linear over simplification with many dependencies and considerations which you did not disclose. \$\endgroup\$ Aug 30, 2018 at 20:12
  • \$\begingroup\$ @Misunderstood Ideally you'd use a therm FEM, if you don't have one, then bound the problem. These are estimates and extreme values, the real values will fall somewhere in between. The package can take 3W of thermal, you that means you could run the LED with more than 3W of electric power. 98 vias is ridiculous but it is the most you'd ever ever need. I will change my answer to reflect that the 3W is thermal not electrical power. \$\endgroup\$
    – Voltage Spike
    Aug 30, 2018 at 20:17
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One square of standard-thickness (1.4mils, 35 microns) copper foil has Rthermal of 70 degree Centigrade per watt. For any size square.

Consider this

schematic

simulate this circuit – Schematic created using CircuitLab

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