10-23
PHYSICAL DESIGN AND SYSTEM DEBUGGING
Impedance Matching Example
We have already discussed the techniques for calculating characteristic impedances (using trans-
mission line theory) and the termination procedures used to avoid impedance mismatching. This
section describes an impedance matching example that utilizes these techniques. Figure 10-18
shows a simple interconnection which acts like a transmission line, as shown by the calculations.
Figure 10-18. Impedance Mismatch Example
In this example the different values are given as follows:
Z
S
= source impedance = 10 ohms
t
rs
= source rise-time = 3 ns (normalized to 0% to 100%)
Z
L
= load impedance = 10 Kohms
t
rl
= load rise-time = 3 ns (normalized to 0% to 100%)
L = length of interconnection = 9"
trace = micro-strip
e = dielectric constant = 5.0
H = .008"
W = .01"
T = .0015" Cu (1 oz. Cu) thickness
v = 6"/ns
The interconnection acts as a transmission line if (as was shown in Section 10.3.1, “Transmission
Line Effects”).
l ≥ (tr x v) / 8 ≥ (3 x6)/8 ≥ 3".
The value of l = 9", thus the interconnection acts like a transmission line.
The impedance of the transmission line is calculated as follows:
Z
0
= 87 / x ln (5.98H/(.8W +T))
= 34.39 ln 5.05 = 55.6 ohms
Because Z
S
= 10 ohms, the termination techniques described previously are needed to match the
difference of 45.6 ohms. One method is to use a series terminating resistor of 45.6 ohms or use
AC termination where r = 55.6 ohms and c = 300 pF. The terminated circuit of Figure 10-18 is
shown in Figure 10-19.
Source Load
Z
S
= 10 Kohms
L = g"
Z
L
= 10 Kohms
t
rt
= 3 ns
Trace is Micro-strip
t
rs
= 3 ns
e
r
1.41+
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