HC55143IM Intersil, HC55143IM Datasheet - Page 17

• Manufacturer Part Number: HC55143IM
• Description: IC SLIC UNIVERSAL LP 32-PLCC
• Manufacturer: Intersil
• Series: UniSLIC14r

Specifications of HC55143IM

• Function: Subscriber Line Interface Concept (SLIC)
• Number Of Circuits: 1
• Voltage - Supply: 4.75 V ~ 5.25 V
• Current - Supply: 2.25mA
• Power (watts): 1.4W
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 28-PLCC
• Includes: Battery Tracking Anti-Clipping, Loop and Ground Key Detection, Ring Control
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Interface: -

example R
next closest standard value is 41.2kΩ.
The true value of ISH-, for the selected value of R
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.
for ISH- into Equation 9 and solving for R
terms of I
Equation 10 can be used to determine the actual ISH- value
resulting from the R
should be the next standard value that is lower than that
calculated. This will insure meeting the I
requirement. ROH for the above example equals 39.1kΩ.
The current limit is set by a single resistor and is calculated
using Equation 11.
overhead voltage. If R
requirements you are done. If the loop length needs to be
longer, then consider adjusting one of the following: 1) the
R
R
R
R
ISH- =
D
OH
OH
LIM
V
=
V
SAT
V
OH(off)
BH
=
=
------------
I
V
=
500
SHD
DC FEED CURVE
DC FEED CURVE
500
--------- - (0.6)
SAT
R
-----------------------------
I
----------------------------------------------------------- -
R
V
LOOP(max)
500
--------- -
I
D
OH
BH
D
LOOP(min)
1000
I
ISH-
D
LOOP(min)
2.5V
=
is calculated to be 41.6kΩ (500/12mA). The
LOOP CURRENT
LOOP CURRENT
R
------------------------------------------- -
I
2.5V
HC55120, HC55121, HC55130, HC55140, HC55142, HC55143, HC55150
LOOP(min)
D
I
OH
500
I
I
LOOP(min)
LOOP(min)
D
- 500(.6)
and R
500
resistor selected. The value of R
LOOP(MAX)
OFF HOOK
OVER HEAD
V
OH(off)
- ISH-
D
.
17
meets the loop length
The R
is used to set the offhook
overhead voltage, is
calculated using
Equations 9 and 10.
I
difference between the
I
Substituting Equation 8
The maximum loop
resistance is calculated
using Equation 12. The
resistance of the
protection resistors
(2R
to obtain the maximum
loop length to meet the
required off hook
OH
LOOP(min)
P
LOOP(min)
is defined as the
) is subtracted out
OH
OH
defines R
resistor, which
and ISH-.
D
is given
(EQ. 10)
(EQ. 11)
(EQ. 9)
(EQ. 7)
(EQ. 8)
D
OH
in
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
I
Substitute Equation 16 into Equation 17
Substitute Equation 18 into Equation 19
Substituting -V
to solve for V
where:
V
V
current detector and the impedance matching networks.
I
between the input receive current and the feedback current.
I
R
Z
impedance.
V
R
V
V
Loop Equation
V
V
Node Equation
------------ - -
500k
Loop Equation
I
V
V
V
X
X
M
X
A
T
A
A
RX
TX
LOOP(max)
TX
TR
TR
TR
P
RX
500k - V
= Internal current in the SLIC that is the difference
=
= The AC metallic current.
= An external resistor/network for matching the line
= I
= An internal node voltage that is a function of the loop
= A protection resistor (typical 30Ω).
=
-I
′
⎛
⎜
⎝
´
=
= The input voltage at the VRX pin.
------------ - -
500k
1
V
= The tip to ring voltage at the output pins of the SLIC.
M
=
M
------- Z
I
2
RX
+
------------ - = I
500k
M
2R
I
×
V
2V
M
Z
---------- -
(
Z
A
2R
TR
P
Z
TX
L
RX
TR
TR
-----------------------------------------
+ V
=
I
S
⎞
⎟
⎠
′ + I
M
TR
–
V
------------------------------------------------------------------------------ - -2R
×
–
–
=
(
X
TR
BH
TX
Z
2R
--------- -
80k
I
2V
1000k
X
M
– V
TR
results in Equation 21
1
/Z
2
500k = 0
′ = 0
–
(
P
RX
Z
[
×
)
L
–
V
TR
RX
SAT
200
2R
into Equation 20 for I
I
LOOP(min)
–
P
2R
(
)
+
Z
TR
2V
P
)
–
+
2R
V
OH off
P
)
(
×
5
)
]
M
and rearranging
P
June 1, 2006
(EQ. 16)
FN4659.13
(EQ. 12)
(EQ. 13)
(EQ. 14)
(EQ. 15)
(EQ. 17)
(EQ. 18)
(EQ. 19)
(EQ. 20)
(EQ. 21)