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ICS843023 bảng dữ liệu(PDF) 5 Page - Integrated Circuit Systems

tên linh kiện ICS843023
Giải thích chi tiết về linh kiện  3.3V, 250MHZ LVPECL CLOCK GENERATOR
PDF  10 Pages
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nhà sản xuất  ICST [Integrated Circuit Systems]
Trang chủ  http://www.icst.com
Logo ICST - Integrated Circuit Systems

ICS843023 bảng dữ liệu(HTML) 5 Page - Integrated Circuit Systems

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843023AG
www.icst.com/products/hiperclocks.html
REV. A NOVEMBER 30, 2004
5
Integrated
Circuit
Systems, Inc.
ICS843023
FEMTOCLOCKS™CRYSTAL-TO-
3.3V, 250MHZ LVPECL CLOCK GENERATOR
PRELIMINARY
APPLICATION INFORMATION
Figure 1. CRYSTAL INPUt INTERFACE
CRYSTAL INPUT INTERFACE
The ICS843023 has been characterized with 18pF parallel
resonant crystals. The capacitor values, C1 and C2, shown in
Figure 1 below were determined using a 25MHz, 18pF parallel
resonant crystal and were chosen to minimize the ppm error.
The optimum C1 and C2 values can be slightly adjusted for
different board layouts.
C1
33p
X1
18pF Parallel Crystal
C2
27p
XTAL_OUT
XTAL_IN
TERMINATION FOR 3.3V LVPECL OUTPUT
The clock layout topology shown below is a typical termina-
tion for LVPECL outputs. The two different layouts mentioned
are recommended only as guidelines.
FOUT and nFOUT are low impedance follower outputs that
generate ECL/LVPECL compatible outputs.Therefore, terminat-
ing resistors (DC current path to ground) or current sources
must be used for functionality. These outputs are designed to
FIGURE 2B. LVPECL OUTPUT TERMINATION
FIGURE 2A. LVPECL OUTPUT TERMINATION
drive 50
Ω transmission lines. Matched impedance techniques
should be used to maximize operating frequency and minimize
signal distortion.
Figures 2A and 2B show two different layouts
which are recommended only as guidelines. Other suitable clock
layouts may exist and it would be recommended that the board
designers simulate to guarantee compatibility across all printed
circuit and clock component process variations.
V
CC - 2V
50
Ω
50
Ω
RTT
Z
o = 50Ω
Z
o = 50Ω
FOUT
FIN
RTT =
Z
o
1
(V
OH + VOL / VCC – 2) – 2
3.3V
125
Ω
125
Ω
84
Ω
84
Ω
Z
o = 50Ω
Z
o = 50Ω
FOUT
FIN



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