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參數(shù)資料
| 型號(hào)： | HC5517CM |
| 廠商： | HARRIS SEMICONDUCTOR |
| 元件分類： | 模擬傳輸電路 |
| 英文描述： | 3 REN Ringing SLIC For ISDN Modem/TA and WLL |
| 中文描述： | TELECOM-SLIC, PQCC28 |
| 文件頁(yè)數(shù)： | 7/18頁(yè) |
| 文件大小： | 175K |
| 代理商： | HC5517CM |
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66
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
RX
to V
TR
) is set by the
feedback loop shown in Figure 3. The feedback loop senses
the loop current through resistors R
13
and R
14
, sums their
voltage drop and multiplies it by 2 to produce an output volt-
age at the V
TX
pin equal to +4R
S
I
L
. The V
TX
voltage is
then fed into the -IN1 input of the SLIC’s internal op amp.
This signal is multiplied by the ratio R
8
/R
9
and fed into the
tip current summing node via the OUT1 pin. (Note: the inter-
nal V
BAT
/2 reference (ring feed amplifier) and the internal
+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
R
) is given in
Equation 7.
The AC voltage at V
C
is then equal to:
and the AC voltage at V
D
is:
The values for R
8
and R
9
are selected to match the
impedance
requirements
on
information reference AN9607 “Impedance Matching Design
Equations for the HC5509 Series of SLICs”. The following
loop current calculations will assume the proper R
8
and R
9
values for matching a 600
load.
tip
and
ring,
for
more
The loop current (
I
L
) with respect to the feedback network,
is calculated in Equations 11 through 14. Where R
8
= 40k
,
R
9
= 40k
, R
L
= 600
,
R
11
= R
12
= R
13
= R
14
= 50
.
Substituting the expressions for V
C
and V
D
:
R
8
R
L
Equation 12 simplifies to:
Solving for
I
L
results in:
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
11
, R
12
, R
13
and R
14
must be accounted for by the feed-
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
11
and R
13
. At the
same time the ring amplifier becomes more positive by the
I
OUT1
4R
I
L
--------------------
–
R
8
R
9
------
=
(EQ. 5)
I
R
–
4R
I
L
--------------------
R
8
R
9
------
–
V
-----------
+
0
=
(EQ. 6)
I
R
4R
I
L
--------------------
R
8
R
9
------
–
V
-----------
+
=
(EQ. 7)
V
C
I
R
(
)
R
( )
=
(EQ. 8)
V
C
4
–
R
S
I
L
R
8
R
9
------
V
RX
+
=
(EQ. 9)
V
D
4R
S
I
L
R
8
R
9
------
V
RX
–
=
(EQ. 10)
I
L
V
V
D
+
–
L
11
R
12
R
13
R
14
+
---------+
=
(EQ. 11)
I
L
2
4
–
R
S
I
L
------
V
RX
+
×
11
12
13
14
+
+
+
---------+
=
(EQ. 12)
I
L
2V
----------------------------------------
400 I
L
–
=
(EQ. 13)
I
L
V
-----------
=
(EQ. 14)
A
4W
2W
–
V
RX
-----------
I
R
L
RX
(
)
--------------------
V
--------------------------
-----------
600
(
)
RX
1
=
=
=
=
(EQ. 15)
A
2W
4W
–
V
TR
------------------
4
–
R
S
I
L
-------------------------------------
R
8
R
------
L
L
200
---------------------------------
–
V
-----------
1
( )
-----------
600
(
)
1
3
--
–
=
=
=
=
(EQ. 16)
HC5517
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