EMC Question of the Week: June 2, 2025

A certain microstrip trace pair has a differential-mode impedance equal to 100 Ω and a common-mode impedance equal to 40 Ω. Determine termination impedances Z₁, Z₂ and Z₁₂ required to match both the common-mode and differential-mode propagation on these traces.  

1. Z₁ = Z₂ = 50 Ω, Z₁₂ = 1000 Ω
2. Z₁ = Z₂ = 50 Ω, Z₁₂ = 100 Ω
3. Z₁ = Z₂ = 80 Ω, Z₁₂ = 100 Ω
4. Z₁ = Z₂ = 80 Ω, Z₁₂ = 267 Ω

Answer

The best answer is “d.” The wire pair is balanced, so the values required to match both propagation modes are,

$Z_1=Z_2=2Z_{CM}=80\text{Ω}$ ${}_{}$

$Z_{12}=\frac{4Z_{CM}{Z}_{DM}}{4Z_{CM}-Z_{DM}}=\frac{4\left(40\text{Ω}\right)\left(100\text{Ω}\right)}{4\left(40\text{Ω}\right)-\left(100\text{Ω}\right)}=267\text{Ω}\text{.}$

The common-mode impedance, Z_CM, is the parallel combination of the two 80-Ω resistances, or 40 Ω. The differential-mode impedance, Z_DM, is equal to Z₁₂ in parallel with the series combination of the two 80-Ω resistances. This is 267 Ω in parallel with 160 Ω, which is 100 Ω.

Note: These values correspond to an even-mode impedance of 80 Ω and an odd-mode impedance of 50 Ω. Terminating each trace to ground with its odd-mode impedance matches the differential mode, but not the common-mode. 

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