In previous articles, I have touched upon the use of adjacent plane layers rather than discrete capacitors. In essence, when it comes to the successful PDN design, the ground plane and power plane have some parasitic inductance, which contributes to the overall PDN inductance. This value is then related to the PDN capacitance and is very low. For this reason, the capacitance provided by adjacent plane layers is able to support the very high switching events associated with driving transmission lines. These are the frequencies at which the discrete capacitors cease to function well.
Also, in previous articles, I described the role of the inductance of the capacitor, its footprint, the vias going from the capacitor footprint to the power planes, and how all of this comes into play to affect the overall PDN inductance. To best round out the topic, it’s important to understand why there is inductance in plane pair and some solutions to reduce its effects on signal integrity.
As described previously, there are four major characteristics that contribute to PDN inductance.
They include:
Table 1 provides the mounted ESL (inductance) for a variety of capacitors. For all of the capacitors, the mounted ESL of the capacitor is a function of via length to the power plane pair.
When it comes to determining power plane inductance, there are several factors that are taken into consideration. For any given conductive material, a sheet of the inductance of that material is measured in Henrys per square. Note: A Henry is the primary unit of inductance for a component that stores energy in the magnetic field that surrounds it. A Henry is defined as the inductance for which the induced voltage (in V) is numerically equal to the rate of change in current (in A/s). These same formulations are used in determining resistance, based on the plane's sheet resistance (Ω per square). Four different equations are used to determine the inductance of a plane pair. The first thing that is done is to determine the parallel plane capacitance, as shown in Equation 1. The reference example in Equation 1 is based on the power plane between two ground planes, as depicted in Figure 1.
Here, there is a pair of conductors with a uniform cross-section from a transmission line with characteristic impedance Z0 (characteristic impedance). Assuming the wire is a signal trace with a width of about 3.75 mils and a dielectric thickness of 3 mils, then Z0 = 50 Ω. If the width of the signal wire is increased, the Z0 of the line will decrease. For example, a signal wire with a width of 1.0 cm (0.4 in) and the same dielectric thickness of 3 mils will have a Z0 = 1.07 Ω.
Next, the impedance of the transmission line impedance versus the capacitance can be determined using a simple LC model, as shown in Equation 2. Here, the assumption is that the transmission line has a width of 1.0 cm with losses and dispersion ignored.
Because a transmission line consists of uniform segments of capacitance and inductance, the impedance of the line can be calculated as depicted in Equation 3.
Based on the information obtained from the foregoing, it is now possible to convert Equation 3 to solve for inductance. Since the values of Z0 and C are known, Equation 4 can be used to determine the inductance of the transmission line.
The inductance for a 1 cm long and 1 cm wide transmission line can be thought of as inductance per square since the equations will produce the same value so long as the width and length remain the same. 71.6 pH is a very small inductance compared to the ESL values of mounted capacitors.
Figure 2 shows an integrated circuit mounted on a PCB with a ring of 0603 decoupling capacitors surrounding the IC's BGA package. In this conceptual example, the IC has 2000 pins and is mounted on a board with 4 power planes. There are 48 capacitors, but they can be divided equally among the four power planes in the IC package. Thus, there are only 12 capacitors for each plane.
In Figure 2, the BGA package is a 1.5” square; the IC chip is a 0.8” square, and the capacitors are placed ¼” from the edge of the package. The inductance of the power plane can be approximated using Equation 5 for the inductance of two parallel planes from an inner radius, R1, to the outer radius R2.
At 580 pH per capacitor, the 12 capacitors in parallel would have an effective inductance of 48 pH if the capacitors are on the same side of the PCB as the BGA, and are connected to the V1/Ground plane pair. The addition of 10.4 pH for the inductance of the power planes is small compared to the capacitors’ inductance. If the capacitors are on the backside of the PCB, then vias would be needed to connect to the capacitors' SMD pads. This would bring each capacitor’s ESL value up to 1.15 nH. The 12 capacitors in parallel would have an inductance of 95 pH—nearly double its effective inductance when the capacitors are on the same side as the BGA.
Overall, the ground/power plane inductance adds to the overall PDN inductance and impedance. When the inductance is larger, any slope associated with inductive impedance in the PDN impedance spectrum is steeper, and resonances associated with the planes occur at lower frequencies. In high speed PCBs, this is the opposite of what we want to see in PDN impedance. In addition to interplane capacitance, this should illustrate why the use of adjacent plane layers in a PCB stackup is ideal for controlling PDN impedance to low values.
The extremely high frequency of switching events associated with today’s technologies requires a more innovative approach for dealing with PDN inductance than can be handled with discrete capacitors. The low capacitance that is inherent in power plane pairs is a much better solution for handling this inductance.
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1. Ritchey, Lee W. and Zasio, John J., “Right The First Time, A Practical Handbook on High-Speed PCB and System Design, Volumes 1 and 2.”