HC5517 Intersil Corporation, HC5517 Datasheet - Page 7

• Manufacturer Part Number: HC5517
• Description: 3 REN Ringing SLIC For ISDN Modem/TA and WLL
• Manufacturer: Intersil Corporation

AC Voltage Gain Design Equations
The HC5517 uses feedback to synthesize the impedance at
the 2-wire tip and ring terminals. This feedback network
defines the AC voltage gains for the SLIC.
The 4-wire to 2-wire voltage gain (V
feedback loop shown in Figure 3. The feedback loop senses
the loop current through resistors R
voltage drop and multiplies it by 2 to produce an output volt-
age at the V
then fed into the -IN1 input of the SLIC’s internal op amp.
This signal is multiplied by the ratio R
tip current summing node via the OUT1 pin. (Note: the inter-
nal V
+2V reference (tip feed amplifier) are grounded for the AC
analysis.)
The current into the OUT1 pin is equal to:
Equation 6 is the node equation for the tip amplifier summing
node. The current in the tip feedback resistor (I
Equation 7.
The AC voltage at V
and the AC voltage at V
The values for R
impedance requirements on tip and ring, for more
information reference AN9607 “Impedance Matching Design
Equations for the HC5509 Series of SLICs”. The following
loop current calculations will assume the proper R
values for matching a 600 load.
The loop current (
is calculated in Equations 11 through 14. Where R
R
I
–
I
V
V
V
OUT1
R
9
I
C
C
D
I
L
R
=
= 40k , R
=
=
=
–
=
BAT
–
4R
--------------------
– R
4R
------------------------------------------------------------------------- -
R
4R
--------------------
I
=
4
R
L
S
R
S
/2 reference (ring feed amplifier) and the internal
–
+
S
R
S
4R
--------------------
R
R
I
I
L
L
I
I
11
L
TX
S
R
L
L
R
------ -
R
R
------ -
R
+
R
------ -
R
= 600
V
8
9
R
------ -
R
I
8
9
L
pin equal to +4R
8
9
C
R
8
9
+
–
12
–
R
------ -
R
+
8
+
I
V
-----------
V
L
V
8
9
C
+
R
V
-----------
) with respect to the feedback network,
RX
V
RX
D
and R
R
RX
is then equal to:
R
RX
D
R
13
=
11
is:
+
0
R
= R
9
66
14
are selected to match the
12
S
= R
RX
13
I
L
8
13
. The V
and R
to V
/R
= R
9
TR
and fed into the
14
14
) is set by the
TX
R
= 50
, sums their
) is given in
voltage is
8
8
= 40k ,
(EQ. 10)
(EQ. 11)
and R
(EQ. 5)
(EQ. 6)
(EQ. 7)
(EQ. 8)
(EQ. 9)
9
HC5517
Substituting the expressions for V
Equation 12 simplifies to:
Solving for I
Equation 14 is the loop current with respect to the feedback
network. From this, the 4-wire to 2-wire and the 2-wire to
4-wire AC voltage gains are calculated. Equation 15 shows
the 4-wire to 2-wire AC voltage gain is equal to one.
Equation 16 shows the 2-wire to 4-wire AC voltage gain is
equal to negative one-third.
Impedance Matching
The feedback network, described above, is capable of
synthesizing both resistive and complex loads. Matching the
SLIC’s 2-wire impedance to the load is important to maxi-
mize power transfer and minimize the 2-wire return loss. The
2-wire return loss is a measure of the similarity of the imped-
ance of a transmission line (tip and ring) and the impedance
at it’s termination. It is a ratio, expressed in decibels, of the
power of the outgoing signal to the power of the signal
reflected back from an impedance discontinuity.
Requirements for Impedance Matching
Impedance matching of the HC5517 application circuit to the
transmission line requires that the impedance be matched to
points “A” and “B” in Figure 3. To do this, the sense resistors
R
back network to make it appear as if the output of the tip and
ring amplifiers are at points “A” and “B”. The feedback network
takes a voltage that is equal to the voltage drop across the
sense resistors and feeds it into the summing node of the tip
amplifier. The effect of this is to cause the tip feed voltage to
become more negative by a value that is proportional to the
voltage drop across the sense resistors R
same time the ring amplifier becomes more positive by the
A
A
I
I
I
4W 2W
11
2W 4W
L
L
L
, R
=
=
=
–
–
------------------------------------------------------------------------- -
R
2V
--------------------------------------- -
V
-----------
600
2
12
RX
L
RX
+
, R
=
=
– R
R
800
4
–
V
-----------
V
V
------------------ -
11
13
L
400 I
RX
V
TR
OUT1
S
+
results in:
TR
and R
R
=
I
L
12
L
-------------------- -
=
R
------ -
R
+
I
V
L
8
9
14
– R
------------------------------------ -
R
RX
4
R
13
+
L
must be accounted for by the feed-
V
S
I
+
L
RX
=
R
I
R
L
14
V
----------- 600
-------------------------- -
600
L
RX
R
------ -
R
V
8
9
C
RX
and V
=
–
--------------------------------- -
200
V
----------- 600
600
=
RX
11
D
1
V
----------- 1
600
:
RX
and R
=
13
–
(EQ. 12)
(EQ. 13)
(EQ. 14)
(EQ. 15)
(EQ. 16)
. At the
1
-- -
3