EMC Question of the Week: March 25, 2019
A 3.3-volt, 100 Mbps digital signal is routed across a printed circuit board on a microstrip trace. At the board's edge, it transitions to a twisted wire pair. The characteristic impedance of both the trace and the twisted wire pair are 100 ohms and the signal has a matched termination. The common-mode voltage driving the cable relative to the board is
- 0 V
- half the signal voltage
- 3 mV at 100 MHz
- 3 V at 100 MHz
Answer
The best answer is "b." From imbalance difference modeling, we know that the common-mode voltage developed at an interface is always equal to the differential-mode voltage times the change in the imbalance factor at the interface. In this case, the microstrip trace is nearly perfectly unbalanced and the twisted wire pair is nearly perfectly balanced. The change in the imbalance factor is approximately 0.5, so the common-mode voltage developed is approximately half the signal voltage.
Note that driving an unshielded cable relative to a circuit board with a 1.65-volt, 100 Mbps digital signal is almost certain to be a radiated emissions problem. A general rule for high-speed data transmission is, "If the signal is balanced, stay balanced. If the signal is unbalanced, stay unbalanced." You can learn more about electrical balance in the tutorial, Introduction to Imbalance Difference Modeling.
Finally, what about the amplitude of the harmonic at 100 MHz? The first harmonic of a 1.65-volt digital signal would have an rms amplitude of about 740 mV or 117 dB(μV).
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