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參數(shù)資料
| 型號(hào)： | HC5526IP |
| 廠商： | INTERSIL CORP |
| 元件分類： | 模擬傳輸電路 |
| 英文描述： | ITU CO/PABX SLIC with Low Power Standby |
| 中文描述： | TELECOM-SLIC, PDIP22 |
| 封裝： | PLASTIC, DIP-22 |
| 文件頁數(shù)： | 10/18頁 |
| 文件大小： | 176K |
| 代理商： | HC5526IP |
66
Therefore:
Equation 16 can now be used to match the SLIC’s
impedance to any known line impedance (Z
TR
).
Example:
Calculate Z
T
to make Z
TR
= 600
in series with 2.16
μ
F.
R
F
= 20
.
-----------------------------------------
+
Z
T
= 560k
in series with 2.16nF.
(AC) 2-Wire to 4-Wire Gain
The 2-wire to 4-wire gain is equal to V
TX
/ V
TR
.
From Equations 9 and 10 with V
RX
= 0:
V
TR
(AC) 4-Wire to 2-Wire Gain
The 4-wire to 2-wire gain is equal to V
TR
/V
RX
.
From Equations 9, 10 and 11 with E
G
= 0:
For applications where the 2-wire impedance (Z
TR
,
V
RX
Equation 15) is chosen to equal the line impedance (Z
L
), the
expression for A
4-2
simplifies to:
(AC) 4-Wire to 4-Wire Gain
The 4-wire to 4-wire gain is equal to V
TX
/V
RX
.
From Equations 9, 10 and 11 with E
G
= 0:
Transhybrid Circuit
The purpose of the transhybrid circuit is to remove the receive
signal (V
RX
) from the transmit signal (V
TX
), thereby preventing
an echo on the transmit side. This is accomplished by using an
external op amp (usually part of the CODEC) and by the
inversion of the signal from the 4-wire receive port (RSN) to the
4-wire transmit port (V
TX
). Figure 17 shows the transhybrid
circuit. The input signal will be subtracted from the output signal
if I
1
equals I
2
. Node analysis yields the following equation:
The value of Z
B
is then:
Where V
RX
/V
TX
equals 1/ A
4-4
.
Therefore:
Example:
Given: R
TX
= 20k
, Z
RX
= 280k
, Z
T
= 562k
(standard
value), R
F
= 20
and Z
L
= 600
.
The value of Z
B
= 18.7k
.
V
TX
RSN
TIP
RING
I
M
1000
Z
TR
V
TR
E
G
-
V
TX
I
M
V
TX
Z
RX
1
HC5526
R
F
R
F
A = 4
+
-
+
-
+
+
-
Z
T
+
-
V
RX
+
-
A = 250
A = 250
I
M
Z
L
FIGURE 16. SIMPLIFIED AC TRANSMISSION CIRCUIT
Z
T
1000
Z
TR
2R
F
–
(
)
=
(EQ. 16)
Z
T
1000
600
j
ω
2.16
10
6
–
2
20
–
=
A
2
4
–
-----------
Z
1000
T
F
--------------------------+
=
=
(EQ. 17)
A
4
2
–
-----------
Z
RX
-----------
–
Z
------------
2R
F
Z
L
+
+
--------------------------------------------
=
=
(EQ. 18)
A
4
2
–
Z
RX
-----------
–
1
2
--
=
(EQ. 19)
A
4
4
–
V
RX
-----------
Z
RX
-----------
–
Z
2R
+
------------
2R
F
Z
L
+
+
--------------------------------------------
=
=
(EQ. 20)
V
TX
-----------
V
B
-----------
+
0
=
(EQ. 21)
Z
B
R
–
TX
V
TX
-----------
=
(EQ. 22)
Z
B
R
TX
Z
T
-----------
Z
------------------+
------------
2R
F
F
Z
L
+
+
L
=
(EQ. 23)
HC5526
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