ISL5585FCRZ Intersil, ISL5585FCRZ Datasheet - Page 11

• Manufacturer Part Number: ISL5585FCRZ
• Description: SLIC 1-CH 53dB 45mA 3.3V/-18V/-24V/-28V 32-Pin QFN EP
• Manufacturer: Intersil

Specifications of ISL5585FCRZ

• Package: 32QFN EP
• Number Of Channels Per Chip: 1
• Polarity Reversal: Yes
• Longitudinal Balanced: 53(Min) dB
• Loop Current: 45 mA
• Minimum Operating Supply Voltage: 2.97|-16 V
• Typical Operating Supply Voltage: 3.3|-18|-24|-28 V
• Typical Supply Current: 9.3 mA

The gain of the Transmit Amplifier, set by R
the programmed impedance of the device. The capacitor
C
ground symbols in the model represent AC grounds, not
actual DC potentials.
The Sense Amplifier is configured as a 4 input differential
amplifier with a gain of 3/4. The voltage at the output of the
sense amplifier (V
V
resulting from V2 and so on (reference Figure 4).
Where ∆V is equal to I
V
V
V
V
V
V
FB
SA
SA
SA
SA
SA
SA
SA
1 is the voltage resulting from V1, V
1
2
3
4
blocks the DC component of the loop current. The
=
=
=
=
=
=
[
2
(
–
3
-- - V
4
–
3
-- - V
4
(
V
∆
3
-- - V
4
3
-- - V
4
(
(
2
I
(
(
M
–
2
4
V
1
3
×
)
)
V
+
-
2W
)
)
1
20
)
+
)
SA
3
-- -
4
I
+
(
-
M
E
-
V
G
=
RING
) is calculated using superposition.
4
TIP
M
Z
–
∆
L
I
V
R
M
3
SENSE
30
-
+
)
]
V
R
R
3
-- -
4
I
I
M
TR
M
11
P
P
=
+
-
+
-
[
(R
∆
Z
V
0
SENSE
V
+
2
V
3
∆
-
V
R
R
+
SA
SENSE
SENSE
20Ω
]
I
20Ω
FIGURE 4. AC SIGNAL TRANSMISSION MODEL
I
M
3
-- -
4
M
2 is the voltage
S
= 20Ω)
+
, determines
-
V
4
-
V
1
AMPLIFIER
AMPLIFIER
FEED
FEED
I
R
+
X
I
+
R
(EQ. 11)
(EQ. 12)
(EQ. 14)
(EQ. 15)
(EQ. 16)
X
(EQ. 13)
4R
4R
4R
4R
+
+
-
-
-
INTERSIL
ISL5585
ISL5585
I
FEEDBACK
AMPLIFIER
X
3R
+
-
3R
RECEIVE BLOCK
The voltage at VTX is equal to:
V
Setting V
and substituting it into EQ18 will enable the user to
determine the required feedback to match the line
impedance at V
2-Wire Impedance Matching
Z
Z
resistor. R
V
V
1:1
V
Z
Z
R
O
L
TR
SA
TX
O
O
TR
S
is the line impedance and R
is the source impedance of the device and is defined as.
= ∆I
SENSE
AMPLIFIER
=
=
=
is defined in Figure 4, note polarity assigned to V
=
=
----------------- - R
133.33
Z
133.33 Z
8k
TA
M
–
2 V
L
V
30
IN
(
1
–
I
IN
S
X
2R
TX
equal to zero in EQ 18, defining Z



is defined as:
--------- -
R
(
R
R
P
)
IN
S
R
S
L
=
2W



–
–
2 V
2R
V



as shown in Equation 19.
AUX
V
V
SA
P
TX
TX
IN
+
-IN
)
-
VFB


R
C



R
------- -
8k
S
FB
--------- -
R
R
S
IN
S





=
+
–
∆ (
P
V
I
is the external protection
IN
M
R



30
IN
--------- -
R
R
IN
)
S
FROM CODEC


R
------- -



8k
–
S
∆ (





V
I
IN
M
O
30
= -V
)


R
------- -
8k
S
(EQ. 18)
TR
(EQ. 19)
(EQ. 20)
(EQ. 21)
(EQ. 17)


TR
/∆I
:
M