Datasheet
5.98 x h
e
A
w =
where A =
Z
O
x
(H
r
+ 1.41)
87
87
x ln
Z =
(5.98 x h)
(0.8W)
Hr + 1.41)
C
L
= 100pF, R
S
= 50:
C
L
= 47pF, R
S
= 50:
C
L
= 10pF, R
S
= 50:
C
L
= 0pF
V
S
= 10V
100k 1M
10M 100M 1G
FREQUENCY (Hz)
-48
-42
-36
-30
-24
-18
GAIN (dB)
-12
-6
0
6
LMH6559
SNOSA57C –APRIL 2003–REVISED MARCH 2013
www.ti.com
Figure 46.
USING GROUND PLANES
The use of ground planes is recommended both for providing a low impedance path to ground (or to one of the
other supply voltages) and also for forming effective controlled impedance transmission lines for the high
frequency signal flow on the board. Multilayer boards often make use of inner conductive layers for routing
supply voltages. These supply voltage layers form a complete plane rather than using discrete traces to connect
the different points together for the specified supply. Signal traces on the other hand are routed on outside layers
both top and bottom. This allows for easy access for measurement purposes. Fortunately, only very high density
boards have signal layers in the middle of the board. In an earlier section, the formula for Z
0
was derived as:
(7)
The width of a trace is determined by the thickness of the board. In the case of a multilayer board the thickness
is the space between the trace and the first supply plane under this trace layer. By common practice, layers do
not have to be evenly divided in the construction of a pcb. Refer to Figure 47. The design of a transmission line
design over a pcb is based upon the thickness of the different internal layers and the ε
r
of the board material.
The pcb manufacturer can supply information about important specifications. For example, a nominal 1.6mm
thick pcb produces a 50Ω trace for a calculated width of 2.9mm. If this layer has a thickness of 0.35mm and for
the same ε
r
, the trace width for 50Ω should be of 0.63mm, as calculated from Equation 8, a derivation from
Equation 7.
(8)
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