MPC9449FA IDT, Integrated Device Technology Inc, MPC9449FA Datasheet - Page 6

• Manufacturer Part Number: MPC9449FA
• Manufacturer: IDT, Integrated Device Technology Inc
• Type: Clock Dividerr

Specifications of MPC9449FA

• Number Of Clock Inputs: 3
• Output Frequency: 200MHz
• Output Logic Level: LVCMOS
• Operating Supply Voltage (min): 2.375V
• Operating Supply Voltage (typ): 2.5/3.3V
• Operating Supply Voltage (max): 3.465V
• Package Type: LQFP
• Operating Temp Range: -40C to 85C
• Operating Temperature Classification: Industrial
• Signal Type: LVCMOS/LVPECL
• Mounting: Surface Mount
• Pin Count: 52
• Lead Free Status / RoHS Status: Not Compliant

IDT™ / ICS™ 1:15 PECL/LVCMOS CLOCK FANOUT BUFFER
MPC9449
3.3V/2.5V 1:15 PECL/LVCMOS CLOCK FANOUT BUFFER
Driving Transmission Lines
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 Ω 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 Ω resistance to V
thus only a single terminated line can be driven by each
output of the MPC9449 clock driver. For the series terminated
case however there is no DC current draw, thus the outputs
can drive multiple series terminated lines.
an output driving a single series terminated line versus two
series terminated lines in parallel. When taken to its extreme
the fanout of the MPC9449 clock driver is effectively doubled
due to its capability to drive multiple lines.
results of an output driving a single line versus two lines. In
both cases the drive capability of the MPC9449 output buffer
is more than sufficient to drive 50 Ω 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 MPC9449. The output waveform
in
by the impedance mismatch seen looking into the driver. The
parallel combination of the 36 Ω series resistor plus the
IN
IN
Figure 4
The MPC9449 clock driver was designed to drive high
This technique draws a fairly high level of DC current and
The waveform plots in
Figure 3. Single versus Dual Transmission Lines
MPC9449
MPC9449
14Ω
Output
Output
Buffer
Buffer
14Ω
shows a step in the waveform, this step is caused
R
R
S
S
R
= 36 Ω
= 36 Ω
S
Figure 4
CC
= 36 Ω
÷2.
show the simulation
Z
Z
Z
O
O
O
= 50 Ω
= 50 Ω
= 50 Ω
Figure 3
APPLICATIONS INFORMATION
illustrates
OutA
OutB0
OutB1
6
output impedance does not match the parallel combination of
the line impedances. The voltage wave launched down the
two lines will equal:
unity reflection coefficient, to 2.6 V. It will then increment
towards the quiescent 3.0 V in steps separated by one round
trip delay (in this case 4.0 ns).
1.
cause any false clock triggering, however designers may be
uncomfortable with unwanted reflections on the line. To better
match the impedances when driving multiple lines the
situation in
terminating resistors are reduced such that when the parallel
combination is added to the output buffer impedance the line
impedance is perfectly matched.
3.0
2.5
2.0
1.5
1.0
0.5
0
At the load end the voltage will double, due to the near
Since this step is well above the threshold region it will not
14 Ω + 22 Ω
Final skew data pending specification.
Figure 5. Optimized Dual Line Termination
Figure 4. Single versus Dual Waveforms
MPC9449
Figure 5
t
D
Output
Buffer
14Ω
25 Ω = 25 Ω
= 3.8956
V
Z
R
R
V
OutA
||
2
0
L
L
In
S
0
22 Ω = 50 Ω
= V
= 50 Ω || 50 Ω
= 36 Ω || 36 Ω
= 14 Ω
= 3.0 (25 ÷ (18 + 17 + 25)
= 1.31 V
should be used. In this case the series
S
4
(Z
0
R
||
R
÷ (R
S
S
50 Ω
= 22 Ω
= 22 Ω
6
Time (nS)
S
t
D
+ R
MPC9449 REV 5 MARCH 14, 2007
= 3.9386
OutB
8
0
Z
Z
+ Z
O
O
= 50 Ω
= 50 Ω
0
))
10
12
14