ISL5585FCRZ Intersil, ISL5585FCRZ Datasheet - Page 13

• 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

Substituting Equation 19 into Equation 17 (V
defining ∆I
Combining Equations 36 and 37 results in Equation 38.
A more useful form of the equation is rewritten in terms of
V
from E
Substituting Z
terms of E
Substituting Equation 40 into Equation 38 results in an
equation for 2-wire to 4-wire gain that’s a function of the
synthesized input impedance of the SLIC and the protection
resistors.
If Z
498.76Ω (66.5kΩ/133.33), and R
results in a 2-wire to 4-wire gain of 0.416 or -7.6dB.
When the protection resistors are set to zero, the transmit
gain is -6dB.
Transhybrid Gain
The transhybrid gain is defined as the 4-wire to 4-wire gain
(G
E
V
G
V
G
G
TX
G
TX
2W
2-4
44
2-4
44
L
= 2V
/V
).
=
is set to 600Ω, Z
=
=
=
=
2W
G
V
---------- =



V
----------- -
G
E
----------- - =
V
V
--------------------------------------- -
Z
2W
TX
2W
2
G
to V
2W
42
TX
L
. A voltage divider equation is written to convert
Z
G
M
+
×
O
Z
Z
----------------------- -
results in Equation 40.
= -V
2W
G
–
+ 2
O
L
L
–
----------------------------------------------- -
2 Z
24



–
Z
= Z
(
+ 2
------------------------------------------- -
(
as shown in Equation 39.
RP
2W
2R
L
Z
=
L
Z
L
RP
O
+
P
–
L
+
/Z



2R
+ 2
–



O
2R
--------- -
R
L
R
Z
E
2R
IN
O
is programmed with R
S
results in Equation 37 for VTX.
P
G
P
RP



+
P
+



Z
-------------------------------------- -
Z
Z
and rearranging Equation 39 in
O
L
13
O
)
+
)



2R
=
Z
P
=
–
O
P
is equal to 49.9Ω. This
----------------------------------------------- -
2 Z
0.416
+
(
Z
L
O
+



2R
Z
O
P
IN
S
+
=0) and
Z
to be
O
)
(EQ. 37)
(EQ. 38)
(EQ. 40)
(EQ. 39)
(EQ. 41)
(EQ. 42)
ISL5585
Understanding Phase Across the ISL5585
4-Wire to 2-Wire Phase
The phase of a signal through the ISL5585 is dependent
upon whether the source is driving the signal 4-wire to 2-wire
or 2-wire to 4-wire.
Figure 6 illustrates the phase of the input signal across the
ISL5585 when the signal is applied at the -IN pin of the
ISL5585 through the R
(TA) inverts the signal 180 degrees at the VTX pin. The
feedback around the tip amplifier inverts the signal again on
the tip lead. The input signal will cause AC loop current to
flow through the 20 Ω sense resistors in the direction from
V 1 to V2 and V3 to V4. This results in an inverted signal
(referenced from tip) on the VSA and thus the VFB pin. This
out of phase signal is the signal used by the feedback path to
match the line impedance of the 2-wire side.
2-Wire to 4-Wire Phase
Figure 7 Illustrates the phase of the input signal across the
ISL5585 when the signal is applied across tip and ring.
When you’re driving the 2-wire side with a source the
ISL5585 looks like a predetermined impedance
(programmed with resistor RS). The current flows through
the 20Ω sense resistors in the direction V2 to V1 and V4 to
V3. This results in a non-inverted signal (referenced from tip)
on the VSA and thus the VFB pin. This signal is then
inverted by the TA amplifier and the signal appearing on the
VTX putput is out of phase with the signal on tip.
Summary of the Phase Through the ISL5585
4-Wire to 2-Wire (V
2-Wire to 4-Wire (V
4-Wire to 4-Wire (V
IN
2w
IN
to V
to V
to V
IN
resistor. The Transmit Amplifier
2W
TX
TX
) is 180
) is 180
) is 180
°
°
°
out of phase
out of phase
out of phase