HC5503CB Intersil, HC5503CB Datasheet - Page 8

• Manufacturer Part Number: HC5503CB
• Description: IC,SLIC,2-4 CONVERSION,BIPOLAR,SOP,24PIN,PLASTIC
• Manufacturer: Intersil

Specifications of HC5503CB

• Rohs Compliant: NO

2-Wire to 4-Wire Gain
The 2-wire to 4-wire gain is defined as the output voltage
V
V
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
4-wire to 2-wire gain we need to define V
The voltage at V
impedance Z
For optimum 2-wire return loss, the input impedance of the
SLIC (Z
Equations going further assume Z
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
the ring feed voltage (V
resistance is the sum of the sense resistors RB
and the load Z
in Equation 11.
From Equation 10:
Substituting Equation 12 into Equation 11 and solving for
V
Using Superposition, the voltage at the receive input R
given as:
Where R′
parallel combination of R
(10kΩ) and is equal to 8.49kΩ . R′
impedance that’s formed by the parallel combination of
R
to 17.25kΩ .
A
V
∆I
∆I
V
V
TX
TX
TR
TR
INTERNAL
2 4
TR
TR
RX
L
L
–
:
divided by the tip to ring voltage (V
= -4R
divided by the input voltage, V
=
=
=
=
=
=
V
--------------------------- -
V
---------- -
Z
O
Z
∆I



V
TF
TR
O
O
------------------------- -
Z
V
---------- -
V
) must equal the load impedance (Z
1
TF
L
2 V
O
S
TX
TR
+
–
is the effective impedance that is formed by the
×
(
∆IL = -600∆IL and V
2R
+
(90kΩ), R
V
L
Z
=
TF
2R
=
RF
.
L
L
S



)
(Z
---------------------- -
R′
–
--------------------- -
=
TR
S
600∆I
600
=



L
1
∆I
R′
Z
is the loop current times the load
+
------------------------- -
Z
+2R
O
L
1
∆I
2 V
O
R
×
3
(
L
L
2
+
RF
Z
(150kΩ ), R



S
TF
2R
O
INTERNAL
=
V
). The total loop current is defined
) where V
TX
)
–
S
1.0
+
8



---------------------- -
R′
TR
2
L
2
R′
IN
= Z
2
= (RL)∆IL = 600∆IL. The
+
(90kΩ), R
is the effective
TF
2
(24.9kΩ) and is equal
. To determine the
R
O
= -V
1



.
TR
TR
V
IN
RF
). Where:
in terms of V
L
) of the line. All
3
. The total
(150kΩ), R
1
and RB
TF
(EQ. 13)
(EQ. 14)
(EQ. 10)
(EQ. 11)
(EQ. 12)
(EQ. 9)
) and
X
2
is
IN
1
.
HC5503
V
Equations 15 and 16. For impedance matching to a load
other than 600Ω , recalculate the parallel impedances R′
R′
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
The 4-wire to 2-wire gain (Given that: R
and R
V
V
V
V
A
∆I
V
V


RX
1
2
RX
RX
TR
TR
4 2
TX
TX
L
+
–
and substitute into Equation 15. The 4-wire to 2-wire
for the recommended values of R
=
----------------------- -
Z
=
=
3
=
=
=
=
O
=
V
---------- -
= 150kΩ) for a 600Ω load is:






300
V
V
Z
–
–
+
TR
–
2 0.25
------------------------------------------------------------------------ -
4RS∆I
4RS
V
---------- -
TF
TF
O
V
(
300
2RS
TR
IN
(
S
=
=
 V

V
---------- -
= 150Ω and rearranging terms:
=
Z
V
---------- -


(
TR
Z
0.25
L
O
----------------------------------------------
8.49kΩ
)V
TR
TR



O
Z
----------------------- -
Z
1.266Z
TX
O
O
+
=
)V
+
8.49kΩ
+
1.266V
+
2R



TX
600
(
----------------------- -
Z
1.266Z
+
0.633
O
O
S
24.9kΩ
+



+
(
0.633
IN
=
300
)V



O
0.633
------------------------- -
Z
IN



 V

O
V
)V
)
IN



Z
+
TX
IN
Z
O
2R
=
O
+
1
–
S
1


3.96dB
= 10kΩ , R
-------------------------------------------- -
17.25kΩ
and R
17.25kΩ
2
is given in
+
2
10kΩ
= 24.9kΩ
(EQ. 16)
(EQ. 17)
(EQ. 18)
(EQ. 19)
(EQ. 20)
(EQ. 21)
(EQ. 22)
(EQ. 23)
(EQ. 15)
 V

1
IN
,