參數(shù)資料
型號： HC5503CBZ
廠商： INTERSIL CORP
元件分類： 模擬傳輸電路
英文描述： Low Cost 24V SLIC For PABX / Key Systems
中文描述： TELECOM-SLIC, PDSO24
封裝： GREEN, PLASTIC, MS-013AD, SOIC-24
文件頁數(shù)： 8/17頁
文件大小： 386K
代理商： HC5503CBZ
8
2-Wire to 4-Wire Gain
The 2-wire to 4-wire gain is defined as the output voltage
V
TX
divided by the tip to ring voltage (V
TR
). Where:
V
TX
= -4R
S
IL = -600
IL and V
TR
= (RL)
IL = 600
IL. The
2-wire to 4-wire gain is therefore equal to -1.0, as shown in
Equation 9.
4-Wire to 2-Wire Gain
The 4-wire to 2-wire gain is defined as the output voltage
V
TR
divided by the input voltage, V
IN
. To determine the
4-wire to 2-wire gain we need to define V
TR
in terms of V
IN
The voltage at V
TR
is the loop current times the load
impedance Z
L
For optimum 2-wire return loss, the input impedance of the
SLIC (Z
O
) must equal the load impedance (Z
L
) of the line. All
Equations going further assume Z
L
= Z
O
.
The loop current
IL is the total voltage across the loop
divided by the total resistance of the loop. The total voltage
across the loop is the sum of the tip feed voltage (V
TF
) and
the ring feed voltage (V
RF
) where V
TF
= -V
RF
. The total
resistance is the sum of the sense resistors RB
1
and RB
2
and the load Z
L
(Z
L
+2R
S
). The total loop current is defined
in Equation 11.
From Equation 10:
Substituting Equation 12 into Equation 11 and solving for
V
TR
:
Using Superposition, the voltage at the receive input R
X
is
given as:
Where R
′
1
is the effective impedance that is formed by the
parallel combination of R
INTERNAL
(90k
), R
3
(150k
), R
1
(10k
) and is equal to 8.49k
. R
′
2
is the effective
impedance that’s formed by the parallel combination of
R
INTERNAL
(90k
), R
3
(150k
), R
2
(24.9k
) and is equal
to 17.25k
.
V
RX
for the recommended values of R
1
and R
2
is given in
Equations 15 and 16. For impedance matching to a load
other than 600
, recalculate the parallel impedances R
′
1
,
R
′
2
and substitute into Equation 15. The 4-wire to 2-wire
gain is recalculated by using the Equations below.
Substituting Equation 16 into Equation 13:
From Equation 10:
From Equation 1:
Substituting Equation 18 into Equation 19:
Substituting Equation 20 into Equation 17:
Assuming R
S
= 150
and rearranging terms:
The 4-wire to 2-wire gain (Given that: R
1
= 10k
R
2
= 24.9k
and R
3
= 150k
)
for a 600
load is:
A
2
4
–
V
TR
-----------
600
L
–
----------------------
I
L
1.0
–
=
=
=
(EQ. 9)
V
TR
I
L
Z
L
×
I
L
Z
O
×
=
=
(EQ. 10)
I
L
V
O
V
S
–
-----------+
2 V
O
(
)
2R
S
--------+
=
=
(EQ. 11)
I
L
V
O
-----------
=
(EQ. 12)
V
TR
2 V
O
)
2R
S
--------+
Z
O
=
(EQ. 13)
V
RX
V
TF
R
′
1
R
2
-----------+
V
TX
R
′
2
R
1
-----------+
V
IN
+
=
=
(EQ. 14)
V
RX
V
TF
24.9k
--------------------+
V
TX
------------------------+
V
IN
+
=
=
(EQ. 15)
V
RX
V
TF
0.25
(
)
V
TX
0.633
(
)
V
IN
+
=
=
(EQ. 16)
V
TR
2 0.25
-------------------------------------------------------------------------
)
V
O
0.633
S
(
)
V
+
)
+
Z
O
=
(EQ. 17)
I
L
V
O
-----------
=
(EQ. 18)
V
TX
4RS
I
L
–
=
(EQ. 19)
V
TX
4RS
V
O
-----------
–
=
(EQ. 20)
V
TR
2RS
V
O
-----------
–
1.266V
IN
+
Z
O
2R
S
--------+
=
(EQ. 21)
1
+
Z
O
300
-----------------------
+
V
TR
1.266Z
O
O
300
--------+
V
IN
=
(EQ. 22)
A
4
2
–
V
IN
-----------
1.266Z
O
O
600
--------+
0.633
3.96dB
–
=
=
=
=
(EQ. 23)
HC5503
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