Intel 386 manual Impedance Matching Example

PHYSICAL DESIGN AND DEBUGGING

as other terminations. A common solution is to place active terminations at both ends of the bus. This helps to maintain the uniform drive levels along the entire length of the bus, and it reduces crosstalk and ringing.

Table 11-2 shows the comparisons of different termination techniques.

11.4.2.1.7 Impedance Matching Example

Previous sections discuss the techniques for calculating characteristic impedances (using transmission line theory) and the termination procedures used match impedances. This section describes an impedance matching example that utilizes these techniques. Figure 11-18 shows a simple interconnection which acts like a transmission line as shown by the calculations.

In this example the different values are given as follows:

Zs = source impedance = 10 ohms

trs = source rise-time = 3 nsec (normalized to 0% to 100%)

Z1 = load impedance = 10 Kohms

trl = load rise-time = 3 nsec (normalized to 0% to 100%)

I= length of interconnection = 9 in. trace = micro-strip

er = dielectric constant = 5.0

h = .008 in.

w = .01 in.

t = .0015 in. (1 oz. Cu)

v = 6 in./nsec

The interconnection will act as a transmission line if (as was shown in Section 11.4.1).

I0:: (tr·V)/8

(tr • V)/8 = 2.25

The value of 1=9 in., thus the interconnection acts like a transmission line. The imped- ance of the transmission line is calculated as follows:

ZO = 87/ v(er + 1.41) x In (5.98h/(.8w+t))

=34.39 In 5.035 = 55.6 ohms

Table 11·2. Comparison of Various Termination Techniques

Termination

# of Extra

RL Power Consumption Prop

Components

Series1

Parallel1

Thevenin2

A.C.*2

Active1

* A.C. also uses a capacitor of 200 pf to 500 pf.

		Delay
2o-Zout	Low	Yes
20	High	No
220	High	No
20	Medium	No
220	Medium	No

11-22