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ISL5586 Datasheet, PDF (15/20 Pages) Intersil Corporation – Low Power Ringing SLIC for Home Gateways
ISL5586
RP TIP
+
V2W
-
ZL IL Vtr
RP RING
200K
20
-
+
20
+
-
200K
4R
4R
4R
4R
Irx4w
200K
VRX
1:1
TA
200K
Iz0
VZO
RS
1.4R
R
VRXP
1.4R
R
VRXM
3R
-IN
8K
-
+ VSA
3R
CFB
VFB
VRX4W
FIGURE 12. SCHEMATIC FOR 4-WIRE TO 2-WIRE GAIN DERIVATION
Note that in Figure 13, VTR is referenced on the line side of the
protection resistors.
On the 2-wire side, solving for IL in terms of VIN gives
Equation 49. Equations 50 and 51 show the relationship of
VIN to the outputs of the Sense Amplifier (VSA) and the
Feedback Amplifier (VZ0) respectively.
IL=



-Z---L-----+-----Z-V--0--I--N-+-----2----R----P--
(EQ. 49)
VSA=

–

-Z---L-----+-----Z-V---0-I--N-+-----2----R----P--
×
2
×
20
×


34--
(EQ. 50)
VZ0=

–

-Z---L-----+-----Z-V---0-I--N-+-----2----R----P--
×
2
×
20
×


34--
× 8---R-K----S-Ω---
(EQ. 51)
Simplifying Equation 51 in terms of Z0 gives the following
equation.
VZ0=

–

-Z---L-----+-----Z-V---0-I--N-+-----2----R----P--
× Z--2--0--
(EQ. 52)
The resulting differential output voltage VTX4W, is shown in
Equation 53.
VTX4W = VTXP – VTXM= VZ0 – (–VZ0 )= 2VZ0
(EQ. 53)
Note that the gain from VZ0 to the differential output is outside
the impedance synthesis loop and will have no effect on the
SLIC’s programmed impedance.
Substituting Equation 53 into Equation 52 and rearranging
terms gives the gain from the 2-wire source (VIN) to the
differential output of the Transmit Amplifier.
V-----T-V---X-I--N-4---W--- =

–

-Z---L-----+-----Z--Z-0---0--+-----2----R----P--
(EQ. 54)
If the combination of the protection resistors and the
programmed impedance of the SLIC are equal to ZL the
voltage VTR will be 1/2 VIN. The 2-wire to 4-wire gain is
defined by Equation 55.
-V----VT----XT----4R---W--- =

–

-Z---L-----+-----Z2----0Z----0+-----2----R----P--
(EQ. 55)
4-WIRE TO 4-WIRE GAIN
The 4-Wire to 4-Wire gain is defined in Equation 56 and is
illustrated in Figure 14.The first term is identical to
Equation 48.
V-V----RT----XX----44---WW--- = V-----VR----X2---W-4---W---- × -V---V-T---X-2---W4----W---
(EQ. 56)
The second term is derived in a similar manner as the 2-wire
to 4-wire gain starting with Equation 57.
V2W = IL × ZL
(EQ. 57)
Moving around the loop from the 2-wire side to the 4-wire
output we solve for VSA and VZO.
VSA = –IL × 2 × 20 × 34-- = V---Z--2--L-W--- × 40 × 34--
(EQ. 58)
VZ0
=
V----Z-2--L-W---
×


8---R-K----S-Ω---
×
40
×
34--
=
-V--Z--2--L-W--- × Z--2---0-
(EQ. 59)
The relationship between VZ0 and the 4-wire output is
shown in Equation 53. Substituting Equation 59 into
Equation 53 yields Equation 60, the second term in Equation
56.
V----V-T----2X---W-4---W--- = Z-Z---0L--
(EQ. 60)
4-15