A microstrip line with alumina substrate ε_r = 9 has a strip width w = 3 mm. Substrate thickness h = 0.5 mm.

What is the approximate characteristic impedance of the line, assuming TEM wave propagation and negligible fringing field?

Concept:

A lossless microstrip transmission line consists of a trace of width ‘w’ as shown:

The characteristic impedance of the transmission line is given as:

\({Z_0} = \sqrt {\frac{L}{C}} \) , where \(C = \frac{{\varepsilon A}}{d}\)

But here d = t, A ≅ w, and ε = εeff

\(\therefore C = \frac{{{\varepsilon _{eff}}w}}{t}\)

The above is the actual capacitance, with no fringing taken into account.

\(\therefore {Z_0} = \sqrt {\frac{L}{{\frac{{{\varepsilon _{eff}} \cdot w}}{t}}}} = \sqrt {\frac{{L.t}}{{w\;{\varepsilon _{eff}}}}} \)

Let the characteristic impedance of a practical transmission line be Z0’.

Since in practice, the capacitance C will be smaller than the actual capacitance due to the fringing effects, i.e. since εeff < ε0 εr, we can write:

Z0’ > Z0, i.e.

\({Z_0} < \sqrt {\frac{{Lt}}{{{\varepsilon _0}{\varepsilon _r}w}}} \)

This is because the capacitance is inversely related to the characteristic impedance, i.e. smaller the C, the larger is the impedance. 

Characteristics impedance of micro strip line 

When, w/h >>1,

\(Z_{microstrip}=\frac{Z_0}{\sqrt{\epsilon_r }(\frac{w}{h})}\)

Z₀ → free space impedance = 377 Ω 

Calculation:

Given:

ε_r = 9, w = 3 mm, h = 0.5 mm

As, w/h = 6 > 1

\(Z_{microstrip}=\frac{377}{\sqrt{9 }(\frac{3}{0.5})}\)

≈ 21 Ω