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MPC9600AE Ver la hoja de datos (PDF) - Integrated Device Technology

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MPC9600AE
IDT
Integrated Device Technology IDT
MPC9600AE Datasheet PDF : 15 Pages
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MPC9600 Data Sheet
LOW VOLTAGE, 2.5V AND 3.3V LVCMOS PLL CLOCK DRIVER
Due to the statistical nature of I/O jitter a RMS value (1 ) is
specified. I/O jitter numbers for other confidence factors (CF) can
be derived from Table 11.
Table 11. Confidence Factor CF
CF
Probability of Clock Edge Within the Distribution
1
0.68268948
2
0.95449988
3
4
0.99730007
0.99993663
5
0.99999943
6
0.99999999
The feedback trace delay is determined by the board layout
and can be used to fine-tune the effective delay through each
device. In the following example calculation a
I/O jitter confidence factor of 99.7% (3) is assumed, resulting
in a worst case timing uncertainty from input to any output of –
261 ps to 341 ps relative to CCLK (VCC = 3.3 V and fVCO = 200
MHz):
tSK(PP) = [–60 ps...140 ps] + [–150 ps...150 ps] +
[(17 ps @ –3)...(17 ps @ 3)] + tPD, LINE(FB)
tSK(PP) = [–261 ps...341 ps] + tPD, LINE(FB)
Above equation uses the maximum I/O jitter number shown in
the AC characteristic table for VCC = 3.3 V (17 ps RMS). I/O jitter
is frequency dependant with a maximum at the lowest VCO
frequency (200 MHz for the MPC9600). Applications using a
higher VCO frequency exhibit less I/O jitter than the AC
characteristic limit. The I/O jitter characteristics in Figure 7 can be
used to derive a smaller
I/O jitter number at the specific VCO frequency, resulting in tighter
timing limits in zero-delay mode and for part-to-part skew tSK(PP).
Maximum I/O Jitter versus Frequency
18
16
14
VCC = 3.3 V
12
10
VCC = 2.5 V
8
6
4
2
0
200 220 240 260 280 300 320 340 360 380 400
VCO FREQUENCY (MHz)
Figure 7. I/O Jitter versus VCO Frequency for
VCC = 2.5 V and VCC = 3.3 V
Driving Transmission Lines
The MPC9600 clock driver was designed to drive high speed
signals in a terminated transmission line environment. To provide
the optimum flexibility to the user the output drivers were designed
to exhibit the lowest impedance possible. With an output
impedance of less than 20 the drivers can drive either parallel
or series terminated transmission lines. For more information on
transmission lines the reader is referred to Freescale
Semiconductor application note AN1091. In most high
performance clock networks point-to-point distribution of signals
is the method of choice. In a point-to-point scheme either series
terminated or parallel terminated transmission lines can be used.
The parallel technique terminates the signal at the end of the line
with a 50 resistance to VCC 2.
This technique draws a fairly high level of DC current and thus
only a single terminated line can be driven by each output of the
MPC9600 clock driver. For the series terminated case however
there is no DC current draw, thus the outputs can drive multiple
series terminated lines. Figure 8 illustrates an output driving a
single series terminated line versus two series terminated lines in
parallel. When taken to its extreme the fanout of the MPC9600
clock driver is effectively doubled due to its capability to drive
multiple lines.
MPC9600
Output
Buffer
IN
14
RS = 36
ZO = 50
OutA
MPC9600
Output
Buffer
IN
14
RS = 36
ZO = 50
RS = 36
ZO = 50
OutB0
OutB1
Figure 8. Single versus Dual Transmission Lines
The waveform plots in Figure 9 shows the simulation results of
an output driving a single line versus two lines. In both cases the
drive capability of the MPC9600 output buffer is more than
sufficient to drive 50 transmission lines on the incident edge.
Note from the delay measurements in the simulations a delta of
only 43 ps exists between the two differently loaded outputs. This
suggests that the dual line driving need not be used exclusively to
maintain the tight output-to-output skew of the MPC9600. The
output waveform in Figure 9 shows a step in the waveform, this
step is caused by the impedance mismatch seen looking into the
driver. The parallel combination of the 36 series resistor plus the
output impedance does not match the parallel combination of the
line impedances. The voltage wave launched down the two lines
will equal:
VL = VS (Z0 (RS + R0 + Z0))
Z0 = 50 || 50
RS = 36 || 36
R0 = 14
VL = 3.0 (25 (18 + 17 + 25)
= 1.31 V
MPC9600 REVISION 6 JANUARY 7, 2013
10
©2013 Integrated Device Technology, Inc.

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