HC5503

FN4344 Rev 5.00

Page 8 of 17

June 2004

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.

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.

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.

From Equation 10:

Substituting Equation 12 into Equation 11 and solving for VTR:

Using Superposition, the voltage at the receive input RX is

given as:

Where R

1 is the effective impedance that is formed by the

parallel combination of RINTERNAL (90k), R3 (150k), R1

(10k

) and is equal to 8.49k. R2 is the effective impedance

that’s formed by the parallel combination of RINTERNAL

(90k

), R3 (150k), R2 (24.9k) and is equal to 17.25k.

VRX for the recommended values of R1 and R2 is given in

Equations 15 and 16. For impedance matching to a load other

than 600

, 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.

Substituting Equation 16 into Equation 13:

From Equation 10:

From Equation 1:

Substituting Equation 18 into Equation 19:

Substituting Equation 20 into Equation 17:

Assuming RS = 150 and rearranging terms:

The 4-wire to 2-wire gain (Given that: R1 =10k, R2 = 24.9k

and R3 = 150kfor a 600 load is:

A24

–

VTX

VTR

-----------

600

–

IL

600

IL

----------------------

1.0

–

==

=

(EQ. 9)

VTR

IL ZL

IL ZO



==

(EQ. 10)

IL

VTF VRF

–

ZO 2RS

+

----------------------------

2VTF



ZO 2RS

+

--------------------------

==

(EQ. 11)

IL

VTR

ZO

-----------

=

(EQ. 12)

VTR

2VTF



ZO 2RS

+

--------------------------







ZO

=

(EQ. 13)

VRX

VTF

R

1

R

1 R2

+

-----------------------







VTX

R

2

R

2 R1

+

-----------------------







VIN

+

==

(EQ. 14)

VRX

VTF

8.49k



8.49k

 24.9k

+

----------------------------------------------



 V

TX

17.25k



17.25k

 10k

+

---------------------------------------------



 V

IN

+

==

(EQ. 15)

VRX

VTF

0.25

VTX 0.633

VIN

+

==

(EQ. 16)

VTR

20.25

VTX 0.633

VIN

+



ZO 2RS

+

-------------------------------------------------------------------------







ZO

=

(EQ. 17)

IL

VTR

ZO

-----------

=

(EQ. 18)

VTX

4RS

IL

–

=

(EQ. 19)

VTX

4RS

VTR

ZO

-----------

–

=

(EQ. 20)

VTR

2RS

VTR

ZO

-----------

–

1.266VIN

+







ZO

ZO 2RS

+

--------------------------

=

(EQ. 21)

1

300

ZO 300

+

------------------------

+



 V

TR

1.266ZO

ZO 300

+

------------------------







VIN

=

(EQ. 22)

A42

–

VTR

VIN

-----------

1.266ZO

ZO 600

+

------------------------







0.633

3.96dB

–

==

=

=

(EQ. 23)