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HC5517_00 Datasheet, PDF (7/19 Pages) Intersil Corporation – 3 REN Ringing SLIC For ISDN Modem/TA and WLL
HC5517
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 (VRX to VTR) is set by the
feedback loop shown in Figure 3. The feedback loop senses
the loop current through resistors R13 and R14, sums their
voltage drop and multiplies it by 2 to produce an output voltage
at the VTX pin equal to +4RS∆IL. The VTX voltage is then fed
into the -IN1 input of the SLIC’s internal op amp. This signal is
multiplied by the ratio R8/R9 and fed into the tip current
summing node via the OUT1 pin. (Note: the internal VBAT/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:
IOUT1
=
–
4----R-----S----∆----I-L--
R
RR-----89- 
(EQ. 5)
Equation 6 is the node equation for the tip amplifier summing
node. The current in the tip feedback resistor (IR) is given in
Equation 7.
–
IR
–
-4---R-----RS----∆----I--L-



RR-----89- 
+ -V---R-R----X--
=
0
(EQ. 6)
IR
=
–
4----R-----RS----∆---I--L--



R-R----89-
+ -V---R-R----X--
The AC voltage at VC is then equal to:
VC = (IR)(R)
(EQ. 7)
(EQ. 8)
VC
=
–4
RS
∆IL



RR-----89- 
+
VRX
(EQ. 9)
and the AC voltage at VD is:
VD
=
4
RS
∆IL



RR-----89- 
– VRX
(EQ. 10)
The values for R8 and R9 are selected to match the
impedance requirements on tip and ring, for more
information reference AN9607 “Impedance Matching Design
Equations for the HC5509 Series of SLICs”. The following
loop current calculations will assume the proper R8 and R9
values for matching a 600Ω load.
The loop current (∆IL) with respect to the feedback network,
is calculated in Equations 11 through 14. Where R8 = 40kΩ,
R9 = 40kΩ, RL = 600Ω, R11 = R12 = R13 = R14 = 50Ω.
∆IL = -R----L-----+----R-----1---1----+-V----CR-----1–---2--V--+--D---R-----1---3----+-----R----1---4--
(EQ. 11)
IR
=
–---4-----(--R---R--S----∆----I-L-----)



RR-----89--
+ V-----RR----X---
IR
R
+ ∆IL - + ∆IL -
R11
R13
TIP
A
-
+
-
+ VC
R
VRX
1VP
OUT1
R
R/20
VRING
R/2
VC
=
–4
RS∆IL



RR-----89--
+ VRX
†
+- 2VDC
IOUT1 =
4RS∆IL
2R
R8
R9
-
VTR RL
∆IL
+
R11 = R12 = R13 = R14 = RS
+RS∆IL
-
+
∆VIN
+-
-
∆IL
+
B
R12
RING
- ∆IL +
R14
- ∆IL +
90kΩ 90kΩ
+
-
-RS∆IL
-
+
+
- VD
VD
=
4RS
∆I
L



RR-----89--
– VRX
R9
+
-
2
VTX
+
4R-S∆IL
R8
-IN1
-
+
-
4RS∆IL
+
R8
R9
†
-
+
VBAT
2
† GROUNDED FOR AC ANALYSIS
FIGURE 3. AC VOLTAGE GAIN AND IMPEDANCE MATCHING
7