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參數資料
| 型號： | ADM5104 |
| 廠商： | Analog Devices, Inc. |
| 英文描述： | Four-channel FastFast Ethernet 110Base-TX Transceiver(四通道110Base-TX快速以太網收發器) |
| 中文描述： | 四通道FastFast以太網110Base -得克薩斯州收發器（四通道110Base -得克薩斯州快速以太網收發器） |
| 文件頁數： | 7/8頁 |
| 文件大小： | 75K |
| 代理商： | ADM5104 |
ADM5104
–7–
REV. 0
DATA
TEMPERATURE –
°
C
TECHNICAL
V
OL
MAX
differentially (typically the interface IC “self-biased” PECL
Inputs require ac coupling), the differential voltage swing be-
tween V
OH
and V
OL
determine compatibility. Yet, simply combin-
ing the V
OH
min specification with the V
OL
max specification
confuses a compatibility analysis since both V
OH
and V
OL
track
with temperature. T his means that V
OH
min and V
OL
max do
not occur simultaneously, but at opposite temperature extremes
(refer to Figure 2). Note that the different voltage swing (V
OH
–
V
OL
) remains
≥
0.93 V over temperature.
T est results of identical PECL outputs over temperature reveal
that minimum differential voltage swings at –40
°
C and at +85
°
C
equal 0.72 V and 0.81 V, respectively (with 6 sigma confidence).
T ypical interface IC PECL inputs have V
IH
min and V
IL
max dc
specifications that require differential drive (V
IL
max)
≥
0.4 V to
≥
0.6 V. T he ADM5104 PECL outpu min – V
ferential swing (0.72 V minimum) to drive these inputs with
sufficient margin.
0
–1.2
–2.0
–60
140
–40
V
–20
0
20
40
60
80
100
120
–0.2
–1.0
–1.4
–1.8
–0.6
–0.8
–1.6
–0.4
0.93V
0.94V
V
OH
V
OL
V
OH
MIN
5V SYSTEM, PECL TERMINATION,
330
TO V
EE
, VOLTAGE REFERENCED TO V
CC
Figure 2. PECL Output Levels Chart
5 V PE CL to 3.3 V PE CL Interface Analysis
T he following three equations need to be satisfied for this inter-
face (in the following example Rh = Resistor connected to
PECL 5, Rm = Resistor connected between termination line
and destination pin, and Rl = Resistor connected to ground):
1. T ermination Impedance must match trace impedance:
Termination Impedance
= (
Rh
×
[
Rm
+
Rl])
/(
Rh
+
Rm
+
Rl
) = 50
2. Resistors need to provide the correct voltage levels:
(Vsource – Vterm_destination)/Rm = Vterm_destination/Rl
where:
Vsource
= 3.67 V (PECL 5 V midpoint) and
Vterm_destination
= 2.0 V (PECL 3.3 V midpoint).
3. Desired driver current of 25 mA:
Idrive =
([4
– Vdest_high
])
/Rm
)
–
([5
–
4]
/Rh
)
,
where:
Vdest_high
= (4
×
Rl)/(Rm + Rl) and Idrive = 0.025
T he midpoints are used to ensure that the waveforms are cen-
tered at the critical levels. T he waveform is attenuated at the
destination because of the voltage divider. Rounding the resistor
values to the nearest standard 5% resistors results in circuit of
Figure x. A 3.2 V to 4.0 V input swing into this circuit creates
an output swing between 1.8 V and 2.2 V.
Figure x.
3.3 V PE CL to 5 V PE CL Interface Analysis
T he common-mode rejection area of the ADM5104 line re-
ceiver input requires the input signal voltage swing to be above
2.6 V. T his is lower that standard PECL and helps simplify the
termination resistor network (less drive current is required). In
the following example: Rh = Resistor connected to PECL 5 V,
Rm = Resistor connected between termination line and destina-
tion pin, and Rl = Resistor connected to ground. T he following
three equations need to be satisfied for this interface:
1. T he voltage swings need to be centered at the correct voltage
levels:
(
5 – Vmid_destination
)/
Rh =
(
Vmid_destination – Vmid_source
)
/Rm
where:
Vmid_destination
= 3.0 V and
Vmid_source
= 2.0 V.
2. T ermination Impedance must match trace impedance:
Termination Impedance
=
(
Rl
×
[
Rm + Rh
])
/
(
Rh + Rm + Rl
) = 50
3. T he voltage for the driver should be within 5% of 1.7 V for
the proper swing:
Vsource =
(
5
×
Rl
)/(
Rh + Rm + Rl
)
where:
= 1.7 V
Using the equations and rounding the resistor values to the
nearest standard 5% resistor results in the circuit of Figure x.
T his circuit will result in a source voltage swing between 1.66 V
and 2.4 V, and a destination voltage swing between 2.79 V and
3.28 V. T his exceeds the minimum required voltage swing suffi-
cient margin.
4. Also, the current required by the driver must be less than 17mA:
Idrive
= (2.4/
Rl
) – ([5 –
Vterm_destination_high
]/
Rh
)
Figure x.
In this case, the current of the driver is 15 mA.
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