EMC Question of the Week: June 25, 2018
Two circuit board traces are routed 0.25 mm over a half-ounce copper current-return plane. One trace carries a 5-volt digital signal and is terminated with a 10-pF load capacitance. The minimum transition time of the digital signal is 1 nanosecond. If the shared return path resistance of the plane is 1 mΩ, what is the maximum voltage coupled to the second trace due to common-impedance coupling?
- about 50 μV
- about 500 μV
- about 50 mV
- about 500 mV
Answer
The correct answer is "a." The maximum coupled voltage due to common-impedance coupling is the shared impedance times the maximum current in the source circuit. The maximum current in the source circuit is approximately C * dV/dt. C is 10 pF. dV is about 5 volts. dt is about 1 nanosecond. Therefore, the maximum current is about 50 mA. Multiplying by 1 mΩ yields 50 μV.
Note that this is not the maximum possible coupling. It is very likely in this case, that the electric-field coupling or magnetic-field coupling will dominate.
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