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參數資料
| 型號： | HC5503CBZ96 |
| 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： | 7/17頁 |
| 文件大小： | 386K |
| 代理商： | HC5503CBZ96 |
7
The value of Va, as a result of feedback through R
2
from the
T
X
output, is given in Equation 1. Equation 1 is a voltage
divider equation between resistors R
2
and the parallel
combination of resistors; R
1
, R
3
and the internal 90k
resistor R
INTERNAL
. The Voltage on the transmit out (T
X
) is
the sum of the voltage drops across resistors R
B1
and R
B2
that is gained up by 2 to produce an output voltage at the
V
TX
pin that is equal to -4R
S
IL.
Where: V
TX
= -4R
S
IL = -600
IL.
To match a 600
line, the synthesized tip and ring impedances
must be equal to 150
. The impedance looking into either the
tip or ring terminal is once again the voltage at the terminal (Va)
divided by the AC current
IL as shown in Equation 2.
Substituting the value of 600
IL for V
TX
in Equation 1 and
dividing both sides by
IL results in Equation 3.
V
a
I
L
Setting Va/
IL equal to 150
and solving for R
2
given that
R
1
= 10k
, R
INTERNAL
= 90k
and R
3
= 150k
the value of
R
2
to match the input impedance of 600
is determined to
be 25.47k
(Note: nearest standard value is 24.9k
).
The amount of negative feedback is dependent upon the
additional synthesized resistance required for matching. The
sense resistors R
B1
and R
B2
should remain at 150
to
maintain the SHD threshold listed in the electrical
specifications. The additional synthesized resistance is
determined by the feed back factor X (Equation 4) which
needs to be applied to the transmit output and fed into the
RX pin of the HC5503. The feed back factor is equal to the
voltage divider between R2 and the parallel combination of
R
1
R
3
and R
INTERNAL
reference Figure 2.
The voltage that is feed back into the RX pin is equal to the
voltage at V
TX
times the feedback factor (Equation 5).
Where V
TX
is equal to -4R
S
IL (R
S
= 150
)
So:
But, from Equation 2:
Therefore:
Equation 8 shows that 1/4 of the T
X
output voltage is
required to synthesize 150
at both the Tip feed and Ring
feed amplifiers.
To match a 900
load would require 300
worth of
synthesized impedance (300
from R
B1
+ R
B2
and 600
from the Tip feed + Ring feed amplifiers).
Setting V
a
/
IL equal to 300
and solving for R
2
in Equation 3,
given that R
1
= 10k
R
INTERNAL
= 90k
and R
3
= 150k
the value of R
2
to match the input impedance of 900
is
determined to be 8.49k
(Note: nearest standard value is
8.45k
). The feed back factor to match a 900
load is 1/2
(300/600).
The selection of the value of 150k
for R
3
is arbitrary. The
only requirement is that it be large enough to have little
effect on the parallel combination between R
INTERNAL
(90k
) and R
1
(10k
). R
3
should be greater then 90k
The selection of the value of 10k
for R
1
is also arbitrary.
The only requirement is that the value be small enough to
offset any process variations of R
INTERNAL
and large
enough to avoid loading of the CODEC’s output. A value of
10k
is a good compromise.
V
a
R
90k
R
1
3
2
------------------------------------+
V
TX
×
=
(EQ. 1)
Z
Tipfeed
Z
Ringfeed
V
a
I
L
--------
150
=
=
=
(EQ. 2)
--------
R
90k
R
1
3
2
------------------------------------+
600
×
=
(EQ. 3)
FeedbackFactor
X
R
90k
R
1
3
2
-------------------------------------+
=
=
(EQ. 4)
FIGURE 2. FEEDBACK EQUIVALENT CIRCUIT
H
R
X
T
X
R
1
10k
R
3
150k
R
2
24.9k
R
INTERNAL
90.0k
FEED BACK
T
X
= -4R
S
IL
V
a
V
TX
X
( )
=
(EQ. 5)
X
V
L
------------------
=
(EQ. 6)
V
a
I
L
--------
150
=
(EQ. 7)
X
V
a
TX
----------
---------
1
4
--
=
=
=
(EQ. 8)
HC5503
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