 |
Ajit K. Chowdhury, senior development manager, BioIonix Inc., McFarland, WI
|
|
The thermal laminate used in the
package shown in Figure 1 consists of a copper layer, a dielectric
layer, and a thicker metal base of aluminum or copper that serves as
a heat sink or as a highly heat conductive link to a heat sink. The
top copper layer is circuitized, typically in a print-and-etch
process, to form the leads to the LED electrodes. This circuitry is
usually not very fine. The metal base needs to be protected from the
etchant chemistry during the print and etch process. High thermal
conductivity in all layers is desirable for good heat removal. The
thermal conductivity (‘k”) of copper is very good. It is measured
in W/mK (watts per meter-Kelvin). The relationship between °C and °K
is T(°K) = T(°C) + 273.15. The value of k for copper is 390W/mK,
for aluminum it is 138W/mK, and for dielectrics it is in the range of
0.25 to 2 W/mK, depending on the chemical nature of the dielectric,
typically an organic resin, and depending on the nature and loading
level of fillers the dielectric may contain. A common filler for
improved thermal conductivity of dielectrics is alumina (Al2O3). Less
common, and more expensive, are aluminum nitride (AlN) and boron
nitride (BN). More important than good thermal conductivity “k”
of the material in the dielectric layer is the layer’s thermal
impedance, “I”. A high thermal impedance is detrimental to the
heat removal rate. It is defined as I = dielectric thickness/thermal
conductivity. “I” can also be expressed as °C-m2/W. A related
term is the thermal resistance, “R,” which takes into account the
surface area, A, of the layer: R = dielectric thickness/dielectric
thermal conductivity x A. So it becomes obvious that heat removal
through a thicker layer is poorer, and it is important to keep the
dielectric layer as thin as possible and its thermal conductivity as
high as possible. However, there are trade-offs in balancing several
desirable properties. As we reduce the dielectric thickness, it
becomes more difficult to achieve a high breakdown voltage and low
leakage current. Likewise, as we increase the filler loading to
improve thermal conductivity, it becomes more difficult to achieve
good dielectric strength, high breakdown voltage, low leakage
current, and good adhesion of the dielectric resin to the copper
layer.
The following is an example of how
combinations of different dielectric thickness values and thermal
conductivities can yield different thermal performance.
We can see that the material with the
best thermal conductivity did not yield the lowest thermal impedance
because of the thickness factor. Based on these numbers, the thermal
performance of an LED package was simulated. While maintaining the
same junction temperature as Material C, Material A and Material B
allowed 25 and 40 percent high power input, respectively. Conversely,
at the same power input, Material C gave a junction temperature of
120°C while Material A gave a junction temperature of 114°C and
Material B gave 106°C. These are significant differences with regard
to the performance and lifetime of an LED.