
Lucent Technologies Inc.
31
Advance Data Sheet
March 1997
Relay, and Protector (SRP) for Short Loop and TA-909 Applications
L7597 Resistive Subscriber Line Interface Circuit (SLIC), Ring
ac Design
(continued)
Design Equations
(continued)
As a practical design example, design the interface for
the following set of requirements per TA-909.
RX = –4 dB
TX = –2 dB
ZT = 600
—> (R
1
= 600; R
2
= 0; C =
∞
)
Hy = 600
Thus:
R
1
′
= 600
R
2
′
= 0
C
′
=
∞
Calculate the gain constants:
K
O
=
= 1.
K
RCV
= K
O
10
RX/20
= 1 x 10
–4/20
= 0.631
K
TX
=
10
TX/20
= 1 x 10
–2/20
= 0.794
Calculate individual components:
R
RV1
=
= 95.1 k
Using a standard value component:
R
RV1
= 95.3 k
R
RV2
=
= 0
C
R
=
=
∞
Choose R
GX1
= 82.5 k
R
T1
= 21.4 k
Using a standard value component:
R
T1
= 21.5 k
C
T
=
C
T
=
∞
R
T2
=
= 0
R
GX
= 2 x K
TX
(R
GX1
+ R
T1
)
R
GX
= 2 x 0.794 (82.5 + 21.5) = 165.2 k
Using a standard value component:
R
GX
= 165 k
R
HB
=
R
HB
=
Using a standard value R
HB
= 332 k
.
Therefore, for this design example, use the following
values in the circuit shown in Figure 11.
R
T1
= 21.5 k
R
T2
= 0
R
GX
= 165 k
R
GX1
= 82.5 k
R
RV1
= 95.3 k
R
RV2
= 0
R
HB1
= 332 k
C
T
=
∞
C
R
=
∞
Figure 14 is the application circuit with the above
values.
---------------------------
----------
=
-------
1
′
--------------------
---------------------------
=
2
′
----------------------
′
---------------------
+
--------------------------------
--------------
----------
1
′
---------
–
=
+
---------------------------------------
-----------------------
----------
----------
–
=
′
----------
1
--------------
+
1
1
′
----------------------
+
′
----------------
GX
×
K
TX
K
RCV
---------R
k
0.794
0.631
×
--------165
329 k
=