Intel Embedded Intel486 Hardware Reference Manual page 307

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.
shows a simple interconnection which acts like a transmission line, as shown by the calculations.
Source
Z = 10 Kohms
S
t = 3 ns
rs
In this example the different values are given as follows:
Z = source impedance = 10 ohms
S
t = source rise-time = 3 ns (normalized to 0% to 100%)
rs
Z = load impedance = 10 Kohms
L
t = load rise-time = 3 ns (normalized to 0% to 100%)
rl
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
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 = 87 / e 1.41 x ln (5.98H/(.8W +T))
+
0
r
= 34.39 ln 5.05 = 55.6 ohms
Because Z = 10 ohms, the termination techniques described previously are needed to match the
S
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
shown in Figure 10-19.
PHYSICAL DESIGN AND SYSTEM DEBUGGING
L = g"
Trace is Micro-strip
Figure 10-18. Impedance Mismatch Example
Figure 10-18
Load
Z = 10 Kohms
L
t = 3 ns
rt
Section 10.3.1, "Transmission
Figure 10-18 is
10-23