參數(shù)資料
型號(hào)： L8560
廠商： Lineage Power
英文描述： Low-Power SLIC（electronic subscriber loop interface circuits ） with Ringing(帶振鈴的低功耗用戶線接口（SLIC）)
中文描述： 低功耗用戶接口（電子用戶環(huán)路接口電路）與振鈴（帶振鈴的低功耗用戶線接口電路（SLIC））
文件頁(yè)數(shù)： 39/46頁(yè)
文件大?。?/td> 1060K
代理商： L8560
Lucent Technologies Inc.
39
Data Sheet
April 2000
L8560 Low-Power SLIC with Ringing
Applications
(continued)
Design Examples
In the preceding examples, use of a first-generation
codec is shown. The equations for second- and third-
generation codecs are simply subsets of these. There
are two examples below. The first shows the simplest
circuit, which uses a minimum number of discrete com-
ponents to synthesize a real termination impedance.
The second example shows the use of the uncommit-
ted op amp to synthesize a complex termination. The
design has been automated in a DOS-based program,
available on request.
Example 1, Real Termination
The following design equations refer to the circuit in
Figure 37. Use these to synthesize real termination
impedance.
Termination impedance:
z
T
=
Receive gain:
Transmit gain:
Hybrid balance:
h
bal
= 20log
h
bal
= 20log
To optimize the hybrid balance, the sum of the currents
at the VFX input of the codec op amp should be set to
0. The expression for ZHB becomes:
Example 2, Complex Termination
For complex termination, the spare op amp may be
used (see Figure 38).
= 2R
P
+ k(Z
T5
)
g
tx
=
The hybrid balance equation is the same as in Exam-
ple 1.
Example 3, Complex Termination Without Spare Op
Amp
The gain shaping necessary for a complex termination
impedance may be done without using the spare op
amp by shaping across the Ax amplifier at nodes TG
and VTX. This is a recommended approach.
I
T/R
–
------------
z
T
2R
P
R
GP
1
--------
R
RCV
-----------
+
+
----------------------------------
+
=
g
rcv
V
FR
-----------
=
g
rcv
1
R
T3
-----------
R
GP
-----------
+
+
1
Z
T/R
--------
+
------------------------------------------------------------------
=
g
tx
V
T/R
----------
=
g
tx
X
R
T6
--------
Z
T/R
--------
×
=
R
HB1
--------------
g
tx
–
g
rcv
×
V
FR
--------------
R
HB1
k
(
)
×
g
tx
g
rcv
-------------------
=
z
T
2R
P
R
GN
1
--------
R
RCV
-----------
+
+
-----------------------------------
T5
R
T4
Z
(
)
+
=
g
rcv
1
RCV
R
T3
R
RCV
R
GN
R
+
+
1
T
Z
T/R
-z
+
-------------------------------------8
=
R
T6
---------
Z
T/R
----------
×
R
T4
---------
×
1
Z
T5
-------------
R
G
N
||
R
RCV
R
T3
+
---------------------------------------------------
+
+
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