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32 The PCB Design Magazine • February 2014 ohms, and since these are in parallel the load seen at the branch is 50 ohms. This would work fairly well if we could just halve the trace width. But, the width of the trace is not directly proportional to the imped- ance and a field solver is required to calculate the correct width required. For instance, a 24 mil trace of 52 ohms has to drop to 4 mil to double in impedance given the same substrate. This is solved by the ICD Stackup Planner Field Solver in Figure 3. The ICD Stackup Planner can be downloaded from www.icd.com.au. This configuration also requires parallel pull-down resistors of 2 x Zo at each load. These end terminators ensure that the transmitted pulse progresses once down the line and then beyond design Figure 2: The resultant waveform of the unmatched line. Figure 3: The iCD Stackup Planner solves the impedance of 24 mil trace of 52 ohms compared to a 4 mil trace of 104 ohms. EFFECTIVE RoUTING oF MULTIPLE LoADS continues