HC5503 データシートの表示（PDF） - Intersil
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HC5503 Datasheet PDF : 17 Pages
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HC5503
2-Wire to 4-Wire Gain
The 2-wire to 4-wire gain is defined as the output voltage
VTX divided by the tip to ring voltage (VTR). Where:
VTX = -4RS∆IL = -600∆IL and VTR = (RL)∆IL = 600∆IL. The
2-wire to 4-wire gain is therefore equal to -1.0, as shown in
Equation 9.
A2 – 4
=
-V----T---X---
VTR
=
–----6---0---0----∆----I--L--
600 ∆ IL
=
–1.0
(EQ. 9)
4-Wire to 2-Wire Gain
The 4-wire to 2-wire gain is defined as the output voltage
VTR divided by the input voltage, VIN . To determine the
4-wire to 2-wire gain we need to define VTR in terms of VIN.
The voltage at VTR is the loop current times the load
impedance ZL .
VTR = ∆IL × ZL = ∆IL × ZO
(EQ. 10)
For optimum 2-wire return loss, the input impedance of the
SLIC (ZO) must equal the load impedance (ZL) of the line. All
Equations going further assume ZL= ZO .
The loop current ∆IL is the total voltage across the loop
divided by the total resistance of the loop. The total voltage
across the loop is the sum of the tip feed voltage (VTF) and
the ring feed voltage (VRF) where VTF = -VRF . The total
resistance is the sum of the sense resistors RB1 and RB2
and the load ZL (ZL +2RS). The total loop current is defined
in Equation 11.
∆IL = -V-Z---T-O--F---+--–---2--V--R--R---S-F-- = Z----2O----(--V+----T-2---F-R---)-S--
(EQ. 11)
From Equation 10:
∆IL
=
V-----T---R---
ZO
(EQ. 12)
Substituting Equation 12 into Equation 11 and solving for
VTR :
VTR
=
Z----2O----(--V+----T-2---F-R---)-S--
ZO
(EQ. 13)
Using Superposition, the voltage at the receive input RX is
given as:
VRX
=
VTF
=
-R----′--1R----+-′--1--R-----2-
VT
X
+
R-----′--2R----+-′--2--R-----1-
VIN
(EQ. 14)
VRX for the recommended values of R1 and R2 is given in
Equations 15 and 16. For impedance matching to a load
other than 600Ω , recalculate the parallel impedances R′1 ,
R′2 and substitute into Equation 15. The 4-wire to 2-wire
gain is recalculated by using the Equations below.
VRX
=
VTF
=
-8---.--4---9----k-8--Ω-.--4---9+----k-2--Ω-4----.-9----k----Ω---
VTX
+
1----7---.--2--1-5---7-k--.-Ω-2---5---+-k---Ω1----0----k---Ω---
VI
N
(EQ. 15)
VRX = VTF = (0.25)VTX + (0.633)VIN
Substituting Equation 16 into Equation 13:
VTR
=
2----(---(--0---.--2---5----)---VZ----TO----X--+---+--2--(-R--0---S.--6---3----3---)---V----I--N-----)
ZO
From Equation 10:
∆IL
=
V-----T---R---
ZO
From Equation 1:
VTX = –4RS∆IL
Substituting Equation 18 into Equation 19:
VTX
=
–4R
S
V-----T---R---
ZO
Substituting Equation 20 into Equation 17:
VTR
=
–
2R
S
-V----T---R---
ZO
+
1.266
V I N
-Z---O------Z+----O-2----R----S--
Assuming RS = 150Ω and rearranging terms:
1
+
-Z---O----3--+-0---0-3---0----0-
VTR
=
Z-1---.O--2---6-+---6--3-Z--0--O--0-
VI
N
(EQ. 16)
(EQ. 17)
(EQ. 18)
(EQ. 19)
(EQ. 20)
(EQ. 21)
(EQ. 22)
The 4-wire to 2-wire gain (Given that: R1 = 10kΩ, R2 = 24.9kΩ
and R3 = 150kΩ) for a 600Ω load is:
A4 – 2
=
V-----T---R---
VIN
=
Z-1---.O--2---6-+---6--6-Z--0--O--0-
=
0.633
=
–3.96 d B
(EQ. 23)
Where R′1 is the effective impedance that is formed by the
parallel combination of RINTERNAL (90kΩ), R3 (150kΩ), R1
(10kΩ) and is equal to 8.49kΩ . R′2 is the effective
impedance that’s formed by the parallel combination of
RINTERNAL (90kΩ), R3 (150kΩ ), R2 (24.9kΩ) and is equal
to 17.25kΩ .
8
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