FR-4, by far the most common substrate for PCBs, is a flame-retardant fiberglass-reinforced epoxy laminate. That makes it a strong, rigid, easy-to-machine material for making circuit boards, but a poor conductor of heat. To demonstrate how bad FR-4’s thermal properties are, let’s calculate its thermal resistance using the formula :
Rth sp-pcb = l/ (k x A) [1] >
where:
l = FR-4 thickness (m);
k = Thermal conductivity (W/mK);
A = The area normal to the heat source.
Taking an example where the LED’s thermal pad is mounted on a solder pad on the PCB measuring 10 x 10 mm on a PCB 1.6 mm thick with a thermal conductivity of 0.2 W/mK, the thermal resistance comes out as 80°C/W. (Note that this calculation is an approximation as it doesn’t take into account the conductivity of the interface between LED and PCB, heat spreading, convection thermal resistances, or boundary conditions.)
This is a very high thermal resistance and would likely result in rapid overheating of a high power LED.
But there is a relatively simple modification that can dramatically improve the situation. Adding so-called thermal via holes to the PCB below the solder pad mounting for the LED (Figure 3) lowers the thermal resistance of the substrate. This technique is much more cost effective because many vias are routinely drilled as circuit elements during PCB manufacturing – so adding a few more to the design incurs minimal additional cost.
Figure 3 : FR-4 cross-section with thermal vias (not to scale). (Courtesy: Cree)
For best results, LED manufacturers recommend an array of thermal vias situated directly below, and normal to the LED’s thermal pad. This minimizes the resistance by providing several thermal paths working in cooperation as well as limiting heat spreading to the rest of the PCB.
In addition to optimizing the geometry, the efficacy of thermal vias depends on their construction. For example, if the vias are left hollow, they have a higher thermal resistance to those filled with solid copper, solder, or conductive epoxy. Consider a typical 0.6-mm diameter via filled with solder (which typically happens when a copper-plated through via is passed through a soldering machine).
The area of a 0.6-mm-diameter via normal to the LED’s thermal pad is 0.28 mm². The thermal conductivity of solder is about 58 W/mK. Using the formula [1], the thermal resistance of such a via in a 1.6-mm-thick PCB is 97.5°C/W.
Each individual via may not seem to impact heat dissipation but collectively, with several vias under the LED’s thermal pad, the vias will affect the temperature. For instance, if the array comprises five thermal vias, the area normal to the heat source increases by a factor of five and hence the combined thermal resistance drops to 19.5 W/K.
Each individual via may not seem to impact heat dissipation but collectively, with several vias under the LED’s thermal pad, the vias will affect the temperature. For instance, if the array comprises five thermal vias, the area normal to the heat source increases by a factor of five and hence the combined thermal resistance drops to 19.5 W/K.
To keep the calculation relatively simple, assume the thermal resistances of the vias and the FR-4 again act in parallel. The thermal resistance can then be calculated from this formula :
Rth = 1/(1/Rth via + 1/ Rth FR-4)
The result is 15.7°C/W, or 80 percent better than FR-4 alone. (Note: This calculation ignores the area of the solder pad taken up by the vias when calculating FR-4 thermal resistance).
This is a simplified case, which assumes that the underside of the PCB is maintained at a reasonable ambient temperature. In practice, an additional heat sink may be required to dissipate heat from the underside. By adding thermal vias to the PCB and lowering the thermal resistance of the substrate, a smaller, less expensive heat sink could be specified than would otherwise be required.