
Two wire impedance
To calculate the impedance presented to the two
wire line by the SLIC including the protection re-
sistorsR
p
 and definedas Z
S
 let:
V
rx
 = 0
Il/50’ = Il/50 (in first approximation)
Rp = 50
Z
S
= Z
AC
/25 + 2R
P
Z
AC
 to make Z
S
 = 600
Z
AC
= 25
 
 (Z
S
- 2R
P
)
Z
AC
= 25
 
 (600 - 100)
Z
AC
= 12.5K
Two wire to four wire gain
 (Txgain)
Let V
rx
 = 0
G
tx
 =V
tx
V
l
V
tx
V
l
Z
AC
 +
 50R
P
Example: Calculate Gtx making R
PC
 = 50
 
 R
P
G
tx
=
 2
 
Z
AC
+
 50
 R
P
Z
AC
 +
 50
 
 R
P
=
 2
As you can see the RPC resistor is providing the
compensation of the insertion loss introduced by
the two external protectionresistors R
P
.
= 2
 
Z
AC
 +
 R
PC
Four wire to two wire gain
 (Rx gain)
Let Eg = 0
G
rx
=V
l
V
rx
25
 (
Z
l
+
 2R
P
)
 +
 Z
AC
Example:
Calculate G
rx
 making Z
AC
 = 25
 
 (Z
ML
 - 2
 
 R
P
)
50
 
 Z
l
25
 
 (
Z
l
+
 2R
P
 2R
P
+
 Z
ML
)
=
50
 
 Z
l
G
rx
 =
G
rx
=
2
 
 Z
l
Z
l
+
 Z
ML
In particular for Z
S
= Z
l
: G
rx
= 1
Hybrid function
To calculatedthe transhybridloss (Thl) let: Eg = 0
Thl =
=
VT
x
50
 
 (
 2
 
 R
P
 +
 Z
l
 ) 
 2R
AC
)
Example:
Calculating Thl making R
S
 = 50
 
 R
P
, Z
S
 = 25
 
(ZSlic - 2
 
 R
P
)
Z
B
Z
B
+
 Z
A
In particularifZ
A
Z
B
Z
l
Thl = 0
From the above relation it is evident that if Z
S
 is
equal to the Z
l
 used in Thl test, the two Z
A
, Z
B
 im-
pedancescan be two resistor of the same value.
VR
x
=
 4
 (
Z
B
Z
B
+
 Z
A
50
 
 (
 2
 
 R
P
+
 Z
l
) 
 2R
PC
Thl = 4
 
 (
Z
l
Z
l
+
 Z
ML
)
=
Z
S
ACtransmissioncircuit stability
To ensure stability of the feedback loop shown in
block diagram form in figure 15 two capacitorsare
required. Figure 16 includes these capacitors Cc
and Ch.
AC - DCseparation
The high pass filter capacitor C
AC
 provides the
separationbetween DC circuitsand AC circuits. A
CAC value of 100mF will position the low end fre-
quencyresponse 3dBbreak pointat 7Hz,
fsp =
1
2
π
 
 220
 
 C
AC
Figure15:
 Simplified AC Circuits
L3234 - L3235
16/26