HC5526 Datasheet, PDF (10/18 Pages) Intersil Corporation – ITU CO/PABX SLIC with Low Power Standby

HC5526
I_M
TIP
R_F
Z_L
+
Z_TR
+
V_{TR -}
V_TX
-
+_-
1
+
E_G
-
I_M
RING ^RF
HC5526
A = 250
A = 250
A=4
V_TX
+
V_TX
-
Z_T
RSN
I_M
1000
Z_RX
+
V_RX
-
FIGURE 16. SIMPLIFIED AC TRANSMISSION CIRCUIT
Therefore:
Z_T = 1000 • (Z_TR – 2R_F)
(EQ. 16)
Equation 16 can now be used to match the SLIC’s
impedance to any known line impedance (Z_TR).
Example:
Calculate Z_T to make Z_TR = 600Ω in series with 2.16µF.
R_F = 20Ω.
Z_T = 1000 • _600 + -_j-_ω------_•----₂----_.-₁---¹-₆-----_•----₁---₀----_–---₆- – 2 • 20_
Z_T = 560kΩ in series with 2.16nF.
(AC) 2-Wire to 4-Wire Gain
The 2-wire to 4-wire gain is equal to V_TX/ V_TR.
From Equations 9 and 10 with V_RX = 0:
A_{2 – 4} = _V-^V----_TT---_R-^X-- = _Z----_T-----_⁄-^Z--₁-^T--₀---^⁄₀---¹-₀--⁰---+⁰----⁰²----_R-----_F-
(EQ. 17)
(AC) 4-Wire to 2-Wire Gain
The 4-wire to 2-wire gain is equal to V_TR/V_RX.
From Equations 9, 10 and 11 with E_G = 0:
For applications where the 2-wire impedance (Z_TR,
A_{4 – 2} = _V-^V----_RT----^RX-- = –_Z---^Z-_R---^T--_X-- • -_1-------Z0------₀--_T------₀------+----^Z-₂--^L-_R-----_F-----+-----_Z---_L--
(EQ. 18)
Equation 15) is chosen to equal the line impedance (Z_L), the
expression for A_4-2 simplifies to:
A_{4 – 2} = –_Z---^Z-_R---^T--_X- • ¹²--
(EQ. 19)
(AC) 4-Wire to 4-Wire Gain
The 4-wire to 4-wire gain is equal to V_TX/V_RX.
From Equations 9, 10 and 11 with E_G = 0:
A_{4 – 4} = _V-^V----_RT----^XX-- = –_Z---^Z-_R---^T--_X- • -_1-------Z0------₀--_T-----^Z_-0----^L--+---+--₂---^2R----^R-_F---^F--+-----_Z---_L--
(EQ. 20)
Transhybrid Circuit
The purpose of the transhybrid circuit is to remove the receive
signal (V_RX) from the transmit signal (V_TX), thereby preventing
an echo on the transmit side. This is accomplished by using an
external op amp (usually part of the CODEC) and by the
inversion of the signal from the 4-wire receive port (RSN) to the
4-wire transmit port (V_TX). Figure 17 shows the transhybrid
circuit. The input signal will be subtracted from the output signal
if I₁ equals I₂. Node analysis yields the following equation:
R^{-V----TT----XX-- + V---}Z^--R--_B^{--X-- = 0}
(EQ. 21)
The value of Z_B is then:
Z_B = –R_TX • ^V-_V----^RT----_XX--
(EQ. 22)
Where V_RX/V_TX equals 1/ A_4-4.
Therefore:
Z_B = R_TX • ^Z---_Z-^R---_T--^X- • -^1-------Z0------⁰--^T-----_Z^-0----_L--+---+--²---_2R----_R-^F---_F--+-----^Z---^L--
(EQ. 23)
Example:
Given: R_TX = 20kΩ, Z_RX = 280kΩ, Z_T = 562kΩ (standard
value), R_F = 20Ω and Z_L = 600Ω.
The value of Z_B = 18.7kΩ.
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