HC55120, HC55121, HC55130, HC55131, HC55140, HC55141, HC55142, HC55143, HC55150, HC55151
DC FEED CURVE
VBH
VSAT
2.5V
VOH(off)
OFF HOOK
OVER HEAD
IOH
ISH-
ILOOP(min)
LOOP CURRENT
The ROH resistor, which
is used to set the offhook
overhead voltage, is
calculated using
Equations 9 and 10.
IOH is defined as the
difference between the
ILOOP(min) and ISH-.
Substituting Equation 8
for ISH- into Equation 9 and solving for ROH defines ROH in
terms of ILOOP(min) and RD.
ROH = I-5-O--0---H0-- = -I-L---O----O----P----(-5-m--0--i-n0---)-------I--S----H------
(EQ. 9)
Equation 10 can be used to determine the actual ISH- value
resulting from the RD resistor selected. The value of RD
should be the next standard value that is lower than that
calculated. This will insure meeting the ILOOP(min)
requirement. ROH for the above example equals 39.1kΩ.
ROH = R-----D----I--L---O----O---R-P----(D--m--5--i-n0---)-0------5---0----0---(--.--6---)
(EQ. 10)
The current limit is set by a single resistor and is calculated
using Equation 11.
RLIM = -I-L---O----1O---0-P--0--(--m0----a---x---)
(EQ. 11)
DC FEED CURVE
The maximum loop
VBH
VSAT
2.5V
VOH(off)
resistance is calculated
using Equation 12. The
resistance of the
protection resistors
R LOOP(MAX)
(2RP) is subtracted out
to obtain the maximum
ILOOP(min)
LOOP CURRENT
loop length to meet the
required off hook
overhead voltage. If RLOOP(MAX) meets the loop length
requirements you are done. If the loop length needs to be
longer, then consider adjusting one of the following: 1) the
SHD threshold, 2) minimum loop current requirement or 3)
the on and off hook signal levels.
RLOOP(max) = V-----B---H----–----[--V----S----IA--L--T-O----+O----P-2---(-Vm-----i+-n---)-V----O-----H----(--o---f--f--)--]- -2RP
(EQ. 12)
SLIC in the Active Mode
Figure 17 shows a simplified AC transmission model. Circuit
analysis yields the following design equations:
VA = IM × 2RS × 8----01----k- × 200(ZTR – 2RP) × 5
(EQ. 13)
VA = --I-2-M---(ZTR – 2RP)
(EQ. 14)
Node Equation
5-V---0-R--0--X--k- - 5---V-0----A0---k- = IX
Substitute Equation 14 into Equation 15
IX = 5-V---0-R--0--X--k- - --I--M-----(--Z--1--T-0---R-0----0–---k--2---R-----P----)
(EQ. 15)
(EQ. 16)
Loop Equation
IX500k - VTX′ + IX500k = 0
Substitute Equation 16 into Equation 17
VTX′ = 2VRX – IM(ZTR – 2RP)
Loop Equation
VTR-IM2RP + VTX′ = 0
Substitute Equation 18 into Equation 19
VTR = IMZTR – 2VRX
(EQ. 17)
(EQ. 18)
(EQ. 19)
(EQ. 20)
Substituting -VTR/ZL into Equation 20 for IM and rearranging
to solve for VTR results in Equation 21
VTR
1
+
-Z--Z-T---L-R--
=
– 2 VRX
(EQ. 21)
where:
VRX = The input voltage at the VRX pin.
VA = An internal node voltage that is a function of the loop
current detector and the impedance matching networks.
IX = Internal current in the SLIC that is the difference
between the input receive current and the feedback current.
IM = The AC metallic current.
RP = A protection resistor (typical 30Ω).
ZT = An external resistor/network for matching the line
impedance.
´ VTX = The tip to ring voltage at the output pins of the SLIC.
VTR = The tip to ring voltage including the voltage across the
protection resistors.
ZL = The line impedance.
ZTR = The input impedance of the SLIC including the
protection resistors.
(AC) 4-Wire to 2-Wire Gain
The 4-wire to 2-wire gain is equal to VTR/VRX.
From Equation 21 and the relationship ZT = 200(ZTR-2RP).
G4-2 = V-V----RT---R-X- = -2Z----L-----+-Z----L-Z----T----R-- = –2-Z---L-----+--------2---Z----0Z----T--0--L-----+-----2---R-----P-----
(EQ. 22)
Notice that the phase of the 4-wire to 2-wire signal is 180o
out of phase with the input signal.
4-16