L3234 - L3235
Figure 15: Simplified AC Circuits
Two wire impedance
To calculate the impedance presented to the two
wire line by the SLIC including the protection re-
sistors Rp and defined as ZS let:
Vrx = 0
Il/50’ = Il/50 (in first approximation)
Rp = 50Ω
ZS = ZAC/25 + 2RP
ZAC to make ZS = 600Ω
ZAC = 25 • (ZS - 2RP)
ZAC = 25 ⋅ (600 - 100)
ZAC = 12.5KΩ
Two wire to four wire gain (Tx gain)
Let Vrx = 0
Gtx
Vtx
Vl
=
=
Vtx
Vl
2⋅
ZAC
ZAC
+
+
RPC
50RP
Example: Calculate Gtx making RPC = 50 ⋅ RP
Gtx
=
2
⋅
ZAC
ZAC
+
+
50
50
⋅
⋅
RP
RP
=
2
As you can see the RPC resistor is providing the
compensation of the insertion loss introduced by
the two external protection resistors RP.
Four wire to two wire gain (Rx gain)
Let Eg = 0
Grx
=
Vl
Vrx
=
50 ⋅ Zl
25⋅ (Zl + 2RP ) + ZAC
Example:
Calculate Grx making ZAC = 25 ⋅ (ZML - 2 ⋅ RP)
Grx
=
25
⋅
(Zl
+
50 ⋅ Zl
2RP − 2RP
+
ZML)
16/26
Grx
=
2 ⋅ Zl
Zl + ZML
In particular for ZS = Zl: Grx = 1
Hybrid function
To calculated the transhybrid loss (Thl) let: Eg = 0
Thl =
=
VTx
VRx
=
4
(
ZB
ZB
+
ZA
−
50
50
⋅
⋅
(
(
2
2
⋅
⋅
RP
RP
+
+
Zl
Zl
)
)
−
−
2RPC
2RAC
)
Example:
Calculating Thl making RS = 50 ⋅ RP, ZS = 25 ⋅
(ZSlic - 2 ⋅ RP)
Thl
=
4
⋅
(
ZB
ZB + ZA
−
Zl
Zl
+ ZML
)
In particular if ZA
ZB
=
ZS
Zl
Thl = 0
From the above relation it is evident that if ZS is
equal to the Zl used in Thl test, the two ZA, ZB im-
pedances can be two resistor of the same value.
AC transmission circuit stability
To ensure stability of the feedback loop shown in
block diagram form in figure 15 two capacitors are
required. Figure 16 includes these capacitors Cc
and Ch.
AC - DC separation
The high pass filter capacitor CAC provides the
separation between DC circuits and AC circuits. A
CAC value of 100mF will position the low end fre-
quency response 3dB break point at 7Hz,
fsp =
1
2π ⋅ 220Ω ⋅ CAC