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L3234 データシート(PDF) 16 Page - STMicroelectronics

部品番号 L3234
部品情報  HIGHLY INTEGRATED SLIC KIT TARGETED TO PABX AND KEY SYSTEM APPLICATIONS
PDF  26 Pages
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メーカー  STMICROELECTRONICS [STMicroelectronics]
ホームページ  http://www.st.com
Logo STMICROELECTRONICS - STMicroelectronics

L3234 データシート(HTML) 16 Page - STMicroelectronics

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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 = 50
Ω
ZS =ZAC/25 + 2RP
ZAC to make ZS = 600
Ω
ZAC =25
• (ZS -2RP)
ZAC =25
⋅ (600 - 100)
ZAC = 12.5K
Ω
Two wire to four wire gain (Tx gain)
Let Vrx =0
Gtx =
Vtx
Vl
Vtx
Vl
=2
⋅
ZAC
+ RPC
ZAC
+ 50RP
Example: Calculate Gtx making RPC =50
⋅ RP
Gtx
= 2 ⋅
ZAC + 50
⋅ RP
ZAC
+ 50 ⋅ 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 =
50
⋅ Zl
25
⋅
(Zl + 2RP − 2RP + ZML)
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
⋅ ( 2 ⋅ RP + Zl )− 2RPC
50
⋅ ( 2 ⋅ RP + Zl )− 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
2
π ⋅ 220Ω ⋅ CAC
Figure 15: Simplified AC Circuits
L3234 - L3235
16/26



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