Power system design engineers always want to achieve higher power density on smaller PCB areas, and need to support data center servers and LTE base stations with high current loads such as FPGAs, ASICs and microprocessors, which are increasingly expensive Especially so. To achieve higher output current, the use of multiphase systems is increasing. To achieve a higher current level on a smaller PCB area, the system design engineer begins to abandon the discrete power solution and select the power supply module. This is because the power supply module provides a welcome choice for reducing the complexity of power supply design and addressing printed circuit board (PCB) layout issues associated with DC / DC converters.

This paper discusses a multi-layer PCB layout method that uses a through-hole arrangement to maximize the heat dissipation performance of a two-phase power supply module. One of the power modules can be configured as two 20A single-phase output or single 40A dual-phase output. An example circuit board design with a through-hole is used to heat the power supply module to achieve higher power density without the need for a heat sink or fan to work.
Two 20A outputs of the ISL8240M circuit
Figure 1: ISL8240M circuit with two 20A outputs

So how can the power module achieve such a high power density? The power module shown in the circuit diagram of Figure 1 provides an extremely low thermal resistance of only 8.5 ° C / W, since the substrate uses copper material. To heat the power supply module, the power supply module is mounted on an efficient thermal circuit board with direct mounting characteristics. The multilayer circuit board has a top wire layer (the power supply module is mounted thereon) and two buried copper planes connected to the top layer using vias. The structure has a very high thermal conductivity (low thermal resistance), so that the power module cooling is easy.

To understand this phenomenon, let's analyze the ISL8240MEVAL4Z evaluation board implementation (Figure 2). This is a power module evaluation board that supports dual 20A output on a four-layer board
ISL8240MEVAL4Z power supply module evaluation board
Figure 2: ISL8240MEVAL4Z power module evaluation board

The circuit board has four PCB layers, the nominal thickness of 0.062 inches (± 10%), and the use of stacked, as shown in Figure 3.
The ISL8240M power supply module uses a four-layer stack of 0.062 "circuit boards
Figure 3: ISL8240M power supply module used in the four layers of 0.062 "circuit board stacking

The PCB is mainly composed of FR4 circuit board material and copper, and a small amount of solder, nickel and gold. Table 1 lists the thermal conductivity of the main material.
 Thermal conductivity of PCB materials
PCB Design for Solving Heat Supply Problem of Power Module

SAC305 * is the most popular lead-free solder consisting of 96.5% tin, 3.0% silver and 0.5% copper. W = Watts, in = inches, C = degrees Celsius, m = m, K = Kelvin

We use Equation 1 to determine the thermal resistance of the material.
Calculate the thermal resistance of the material
Equation 1: Calculate the thermal resistance of the material

To determine the thermal resistance of the copper layer at the top of the circuit board in Figure 3, we take the thickness (t) of the copper layer and divide it by the product of the thermal conductivity and the cross-sectional area. For calculation convenience, we use 1 square inch as the cross-sectional area, then A = B = 1 inch. The copper layer has a thickness of 2.8 mils (0.0028 inches). It is 2 ounces of copper deposited in a 1 square inch thickness of the circuit board area. The coefficient k is the W / (in- ° C) coefficient of copper and its value is equal to 9. Thus, for this 1 square inch 2.8 mil copper heat flow, the thermal resistance is 0.0028 / 9 = 0.0003 ° C / W. We can use the size of each layer shown in Figure 3 and the corresponding k-factor in Table 1 to calculate the thermal resistance of each 1-square-inch circuit board area. The results are shown in Fig.
1 square inch circuit board layer thermal resistance
Figure 4: 1 square inch circuit board layer thermal resistance

From these figures, we know that the thermal resistance of the 33.4 mil (t5) layer is the highest. All the figures in Figure 4 show the total thermal resistance of the four-layer 1-square-inch circuit board from the top to the bottom. What if we add a through-hole connection from the top of the board to the bottom? Let's analyze the case where the hole connection is added.

The hole size of the through hole used for the circuit board is about 12 mils (0.012 inches). When making this through hole, drill a hole with a diameter of 0.014 in. And then copper it, which adds about 1 mil (0.001 inch) thick copper wall inside the hole. The circuit board also uses the ENIG plating process. This adds about 200 microinches of nickel to the outer surface of the copper and about 5 microinches. We ignore these materials in the calculation, using only copper to determine the thermal resistance of the through-hole.

Equation 2 is the formula for calculating the thermal resistance of a cylindrical tube.
Calculate the thermal resistance of the cylindrical tube
Equation 2: Calculate the thermal resistance of the cylindrical tube

The variable l is the length of the cylindrical tube, k is the thermal conductivity, r1 is the larger radius, r0 is the smaller radius.

We use r0 = 6 mil (0.006 in.), R1 = 7 mil (0.007 in.) And K = 9 (copper) for 12 mil (diameter) holes.
12 cm through hole surface size
Figure 5:12 Surface diameter of the stud hole

The variable l is the length of the through hole (from the top copper layer to the bottom copper layer). There is no solder mask on the circuit board where the power supply module is sold, but for other areas, the PCB design engineer may require that the solder resist layer be placed at the top of each via, otherwise the area above the vias will be empty. As the through-hole only connected to the outer copper layer, so its length is 63.4 mil (0.0634 inches). The total thermal resistance of the total through-hole length is 167 ° C / W, as shown in Equation 3.
Calculate the thermal resistance of a through hole (12 mil)
Equation 3: Calculate the thermal resistance of a through hole (12 mil)

Figure 6 shows the thermal resistance of each of the vias connecting the layers of the circuit board.
Connect the thermal resistance of the through-hole of each layer of the circuit board
Figure 6: Thermal resistance of the through-hole segments connecting the layers of the circuit board

Note that these values ​​are only a thermal resistance of the through-hole itself, and it is not considered that each segment passing through the circuit board is laterally connected to the material surrounding it.

If we analyze the thermal resistance values ​​of the individual board layers in Figure 4 and compare them to the thermal resistance of a through-hole, it appears that the thermal resistance of the vias is much higher than the thermal resistance of each layer, but note that one Through hole only 1 square inch circuit board area of ​​less than 1/5000. If we decide to compare smaller board areas, such as 0.25 inches x 0.25 inches (which is 1/16 of the front board area), each thermal resistance in Figure 4 will increase to 16 times the original. For example, the thermal resistance of t4 and 33.4 mil thick FR4 layers increases from 5.21875 ° C / W to 83.5 ° C / W. Adding only one vias to the 0.25 inch x 0.25 inch area will reduce the thermal resistance through the 33.4 mil FR4 layer by nearly half (83.5 ° C / W and 90.91 ° ​​C / W). The area of ​​the 0.25 inch x 0.25 inch box is about 400 times the area of ​​a through hole. So what if there are 16 holes in the area? The effective thermal resistance of all parallel vias will be reduced by 16 times compared to a through hole. Figure 7 compares the thermal resistance of each 0.25 inch x 0.25 inch circuit board layer with 16 through holes. The thermal resistance of the 33.4 mil thick FR4 layer of the 0.25 inch x 0.25 inch circuit board is 83.5 ° C / W. 16 parallel vias have an equivalent thermal resistance of 5.6821 ° C / W.

The 16 vias only account for less than 1/25 of the 0.25 inch x 0.25 inch circuit board area, but can significantly reduce the thermal resistance from the top to the lower.
Comparison of thermal resistance values
Figure 7: Comparison of thermal resistance values

Note that when the heat flows down through the through-hole and reaches another layer, especially the other copper layer, it will laterally diffuse into the material layer. Adding more and more vias will eventually reduce the effect, since the heat from a through hole laterally diffuses to the nearby material will eventually meet with heat from another direction (from another through hole). The size of the ISL8240MEVAL4Z evaluation board is 3 inches x 4 inches. There are 2 ounces of copper on the top and bottom of the circuit board, and two inner layers each containing 2 ounces of copper. In order for these copper layers to function, the circuit board has 917 12-mil diameter through-holes, all of which help to diffuse heat from the power module to the underlying copper layer.

Concluding remarks

Advanced power management solutions such as the ISL8240M power supply module for increasing the number of viable rails and higher performance microprocessors and FPGAs, helping to improve efficiency by providing greater power density and lower power consumption. Through-hole in the power module circuit board design of the optimal implementation, has become a higher power density to achieve an increasingly important factor.