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| 型號： | HC55150CMZ |
| 廠商： | Intersil |
| 文件頁數(shù)： | 10/36頁 |
| 文件大?。?/td> | 0K |
| 描述： | IC SLIC UNIVERSAL LP 28-PLCC |
| 標(biāo)準(zhǔn)包裝： | 925 |
| 系列： | UniSLIC14 |
| 功能： | 用戶線路接口概念（SLIC） |
| 電路數(shù)： | 1 |
| 電源電壓： | 4.75 V ~ 5.25 V |
| 電流 - 電源： | 2.25mA |
| 功率（瓦特）： | 2W |
| 工作溫度： | 0°C ~ 70°C |
| 安裝類型： | 表面貼裝 |
| 封裝/外殼： | 28-LCC（J 形引線） |
| 供應(yīng)商設(shè)備封裝： | 28-PLCC（11.51x11.51） |
| 包裝： | 管件 |
| 包括： | 電池跟蹤抗削頂失真，回路和接地鍵檢測，振鈴控制 |
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18
FN4659.13
June 1, 2006
VTR = The tip to ring voltage including the voltage across the
protection resistors.
ZL = The line impedance.
ZTR = The input impedance of the SLIC including the
protection resistors.
(AC) 4-Wire to 2-Wire Gain
The 4-wire to 2-wire gain is equal to VTR/VRX.
From Equation 21 and the relationship ZT = 200(ZTR-2RP).
Notice that the phase of the 4-wire to 2-wire signal is 180°out
of phase with the input signal.
(AC) 2-Wire to 4-Wire Gain
The 2-wire to 4-wire gain is equal to VTX/EG with VRX = 0
From Equation 18 with VRX = 0
Substituting Equation 24 into Equation 23 and simplifying.
By design, VTX = -VTX, therefore
A more useful form of the equation is rewritten in terms of
VTX/VTR. A voltage divider equation is written to convert
from EG to VTR as shown in Equation 27.
Rearranging Equation 27 in terms of EG, and substituting
into Equation 26 results in an equation for 2-wire to 4-wire
gain, that’s a function of the synthesized input impedance of
the SLIC (ZTR) and the protection resistors (RP).
Notice that the phase of the 2-wire to 4-wire signal is in
phase with the input signal.
(AC) 4-Wire to 4-Wire Gain
The 4-wire to 4-wire gain is equal to VTX/VRX, EG = 0.
From Equation 18.
Substituting -VTR/ZL into Equation 29 for IM results in
Equation 30.
G
4-2 =
V
TR
V
RX
----------- = -2
Z
L
Z
L + ZTR
-------------------------
2
Z
L
Z
L
Z
T
200
----------
2R
P
+
+
----------------------------------------------
–
=
(EQ. 22)
VTX
VRX
TIP
RING
ZTR
VTR
EG
VTX
IM
VTX
UniSLIC14
RP
+
-
+
-
+
-
+
VRX
+
-
IM
ZL
FIGURE 17. SIMPLIFIED AC TRANSMISSION CIRCUIT
+
-
500K
RS
+
-
500K
RS
ZT
500K
PTG
+
-
IX
VA = IM(ZTR-2RP)
IX
+
-
IX
+
-
5
+
-
+
-
IM
+
-
20
20
1/80K
= 200 (ZTR - 2RP)
A = 1
IX
2
RINT
20
RINT
20
E
–
G
Z
LIM
2R
PIM
V
TX′
–
+
0
=
Loop Equation
(EQ. 23)
V
TX′
I
M ZTR
2R
P
–
()
–
=
(EQ. 24)
E
G
I
M ZL
Z
TR
+
()
=
(EQ. 25)
G
2-4 =
V
TX
E
G
---------- =
I
M ZTR
2R
P
–
()
I
M ZL
Z
TR
+
()
----------------------------------------
Z
TR
2R
P
–
()
Z
L
Z
TR
+
()
---------------------------------
=
(EQ. 26)
V
TR =
Z
TR
Z
TR
Z
L
+
------------------------
E
G
(EQ. 27)
G
2-4 =
V
TX
V
TR
----------- =
Z
TR - 2RP
Z
TR
-----------------------------
(EQ. 28)
V
TX′
V
–
TX
2
– V
RX
I
M ZTR
2R
P
–
()
+
==
(EQ. 29)
V
TX
2
– V
RX
V
TR ZTR
2R
P
–
()
Z
L
---------------------------------------------
–
=
(EQ. 30)
HC55120, HC55121, HC55130, HC55140, HC55142, HC55143, HC55150
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