参数资料
| 型号： | HC55143IM |
| 厂商： | Intersil |
| 文件页数： | 9/36页 |
| 文件大小： | 0K |
| 描述： | IC SLIC UNIVERSAL LP 32-PLCC |
| 标准包装： | 30 |
| 系列： | UniSLIC14 |
| 功能： | 用户线路接口概念（SLIC） |
| 电路数： | 1 |
| 电源电压： | 4.75 V ~ 5.25 V |
| 电流 - 电源： | 2.25mA |
| 功率（瓦特）： | 1.4W |
| 工作温度： | -40°C ~ 85°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 32-LCC（J 形引线） |
| 供应商设备封装： | 32-PLCC |
| 包装： | 管件 |
| 包括： | 电池跟踪抗削顶失真，回路和接地键检测，振铃控制 |
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17
FN4659.13
June 1, 2006
example RD is calculated to be 41.6k (500/12mA). The
next closest standard value is 41.2k
.
The true value of ISH-, for the selected value of RD is given
by Equation 8:
For the example above, ISH- equals 7.28mA (500 x 0.6/
41.2K). Verify that the value of ISH- is above the suspected
line leakage of the application. The UniSLIC family will
provide a constant on hook voltage level for leakage currents
up to this value of line leakage.
The ROH resistor, which
is used to set the offhook
overhead voltage, is
calculated using
Equations 9 and 10.
IOH is defined as the
difference between the
ILOOP(min) and ISH-.
Substituting Equation 8
for ISH- into Equation 9 and solving for ROH defines ROH in
terms of ILOOP(min) and RD.
Equation 10 can be used to determine the actual ISH- value
resulting from the RD resistor selected. The value of RD
should be the next standard value that is lower than that
calculated. This will insure meeting the ILOOP(min)
requirement. ROH for the above example equals 39.1k
.
The current limit is set by a single resistor and is calculated
using Equation 11.
The maximum loop
resistance is calculated
using Equation 12. The
resistance of the
protection resistors
(2RP) is subtracted out
to obtain the maximum
loop length to meet the
required off hook
overhead voltage. If RLOOP(MAX) meets the loop length
requirements you are done. If the loop length needs to be
longer, then consider adjusting one of the following: 1) the
SHD threshold, 2) minimum loop current requirement or 3)
the on and off hook signal levels.
SLIC in the Active Mode
Figure 17 shows a simplified AC transmission model. Circuit
analysis yields the following design equations:
Substitute Equation 14 into Equation 15
Substitute Equation 16 into Equation 17
Substitute Equation 18 into Equation 19
Substituting -VTR/ZL into Equation 20 for IM and rearranging
to solve for VTR results in Equation 21
where:
VRX = The input voltage at the VRX pin.
VA = An internal node voltage that is a function of the loop
current detector and the impedance matching networks.
IX = Internal current in the SLIC that is the difference
between the input receive current and the feedback current.
IM = The AC metallic current.
RP = A protection resistor (typical 30).
ZT = An external resistor/network for matching the line
impedance.
VTX= The tip to ring voltage at the output pins of the SLIC.
R
D =
500
I
SHD
------------
(EQ. 7)
ISH- =
500
R
D
---------- (0.6)
(EQ. 8)
VBH
VSAT
VOH(off)
2.5V
OFF HOOK
TIP
T
O
RING
VOL
T
AGE
LOOP CURRENT
ILOOP(min)
DC FEED CURVE
ISH-
IOH
OVER HEAD
R
OH
500
I
OH
----------
500
I
LOOP(min)- ISH-
--------------------------------------------
==
(EQ. 9)
R
OH =
R
D 500
R
DILOOP(min)
- 500(.6)
------------------------------------------------------------
(EQ. 10)
R
LIM =
1000
I
LOOP(max)
-----------------------------
(EQ. 11)
VBH
VSAT
VOH(off)
2.5V
TIP
T
O
RING
VOL
T
AGE
LOOP CURRENT
ILOOP(min)
DC FEED CURVE
RLOO
P(M
AX)
R
LOOP(max) =
V
BH
V
SAT
2V
V
OH off
()
++
[]
–
I
LOOP(min)
------------------------------------------------------------------------------- -2R
P
(EQ. 12)
V
A = IM
2R
S
1
80k
----------
200
×
Z
TR
2R
P
–
()
×
5
×
(EQ. 13)
V
A
I
M
2
------- Z
TR
2R
P
–
()
=
(EQ. 14)
V
RX
500k
------------- -
V
A
500k
------------- = I
X
Node Equation
(EQ. 15)
I
X
V
RX
500k
------------- -
I
M ZTR
2R
P
–
()
1000k
-----------------------------------------
=
(EQ. 16)
I
X500k - VTX′ + IX500k = 0
Loop Equation
(EQ. 17)
V
TX′
2V
RX
I
M ZTR
2R
P
–
()
–
=
(EQ. 18)
V
TR -IM2RP + VTX′ = 0
Loop Equation
(EQ. 19)
V
TR
I
MZTR
2V
RX
–
=
(EQ. 20)
V
TR 1
Z
TR
Z
L
-----------
+
2
– V
RX
=
(EQ. 21)
HC55120, HC55121, HC55130, HC55140, HC55142, HC55143, HC55150
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