Datasheet
= 2.45V
V =
(P x R)
=
(80 x 10
-3
x 75)
TRACE WIDTH (mm)
VARIATION OF H
r
0.5
1.0 1.5 2.0 2.5 3.0 3.5
4.0
0
10
20
30
40
70
80
90
100
110
120
IMPEDANCE (:)
60
50
0
10
20
30
40
70
80
90
100
110
120
60
50
IMPEDANCE
(
:
)
1.0
2.3 3.6 4.9 6.2 7.5 8.8
10.1
VARIABLE RELATIVE DIELECTRIC
CONSTANT
WIDTH = 2.85mm
h = 1.6mm
VARIABLE TRACE WIDTH
H
r
= 4.7
h = 1.6mm
87
x ln
Z =
(5.98 x h)
(0.8W)
Hr + 1.41)
LMH6559
www.ti.com
SNOSA57C –APRIL 2003–REVISED MARCH 2013
(2)
With this formula it is possible to calculate the line impedance vs. the trace width. Figure 40 shows the
impedance associated with a given line width. Using the same formula it is also possible to calculate what
happens when ε
r
varies over a certain range of values. Varying the ε
r
over a range of 1 to 10 gives a variation for
the Characteristic Impedance of about 40Ω from 80Ω to 38Ω. Most transmission lines are designed to have 50Ω
or 75Ω impedance. The reason for that is that in many cases the pcb trace has to connect to a cable whose
impedance is either 50Ω or 75Ω. As shown ε
r
and the line width influence this value.
Figure 40.
Next, there will be a discussion of some issues associated with the interaction of the transmission line at the
source and at the load.
Connecting A Load Using A Transmission Line
In most cases, it is unrealistic to think that we can place a driver or buffer so close to the load that we don't need
a transmission line to transport the signal. The pcb trace length between a driver and the load may affect
operation depending upon the operating frequency. Sometimes it is possible to do measurements by connecting
the DUT directly to the analyzer. As frequencies become higher the short lines from the DUT to the analyzer
become long lines. When this happens there is a need to use transmission lines. The next point to examine is
what happens when the load is connected to the transmission line. When driving a load, it is important to match
the line and load impedance, otherwise reflections will occur and this phenomena will distort the signal. If a
transient is applied at T = 0 (Figure 41, trace A) the resultant waveform may be observed at the start point of the
transmission line. At this point (begin) on the transmission line the voltage increases to (V) and the wave front
travels along the transmission line and arrives at the load at T = 10. At any point across along the line I = V/Z
0
,
where Z
0
is the impedance of the transmission line. For an applied transient of 2V with Z
0
= 50Ω the current from
the buffer output stage is 40mA. Many vintage opamps cannot deliver this level of current because of an output
current limitation of about 20mA or even less. At T = 10 the wave front arrives at the load. Since the load is
perfectly matched to the transmission line all of the current traveling across the line will be absorbed and there
will be no reflections. In this case source and load voltages are exactly the same. When the load and the
transmission line have unequal values of impedance a different situation results. Remember there is another
basic which says that energy cannot be lost. The power in the transmission line is P = V
2
/R. In our example the
total power is 2
2
/50 = 80mW. Assume a load of 75Ω. In that case a power of 80mW arrives at the 75Ω load and
causes a voltage of the proper amplitude to maintain the incoming power.
(3)
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