66

AC Voltage Gain Design Equations

The HC5517 uses feedback to synthesize the impedance at

the 2-wire tip and ring terminals. This feedback network

defines the AC voltage gains for the SLIC.

feedback loop shown in Figure 3. The feedback loop senses

voltage drop and multiplies it by 2 to produce an output volt-

then fed into the -IN1 input of the SLIC’s internal op amp.

tip current summing node via the OUT1 pin. (Note: the inter-

+2V reference (tip feed amplifier) are grounded for the AC

analysis.)

The current into the OUT1 pin is equal to:

Equation 6 is the node equation for the tip amplifier summing

Equation 7.

impedance

requirements

on

tip

and

ring,

for

more

information reference AN9607 “Impedance Matching Design

Equations for the HC5509 Series of SLICs”. The following

values for matching a 600

Ω load.

The loop current (

Equation 12 simplifies to:

Solving for

Equation 14 is the loop current with respect to the feedback

network. From this, the 4-wire to 2-wire and the 2-wire to

4-wire AC voltage gains are calculated. Equation 15 shows

the 4-wire to 2-wire AC voltage gain is equal to one.

Equation 16 shows the 2-wire to 4-wire AC voltage gain is

equal to negative one-third.

Impedance Matching

The feedback network, described above, is capable of

synthesizing both resistive and complex loads. Matching the

SLIC’s 2-wire impedance to the load is important to maxi-

mize power transfer and minimize the 2-wire return loss. The

2-wire return loss is a measure of the similarity of the imped-

ance of a transmission line (tip and ring) and the impedance

at it’s termination. It is a ratio, expressed in decibels, of the

power of the outgoing signal to the power of the signal

reflected back from an impedance discontinuity.

Requirements for Impedance Matching

Impedance matching of the HC5517 application circuit to the

transmission line requires that the impedance be matched to

points “A” and “B” in Figure 3. To do this, the sense resistors

back network to make it appear as if the output of the tip and

ring amplifiers are at points “A” and “B”. The feedback network

takes a voltage that is equal to the voltage drop across the

sense resistors and feeds it into the summing node of the tip

amplifier. The effect of this is to cause the tip feed voltage to

become more negative by a value that is proportional to the

same time the ring amplifier becomes more positive by the

I

OUT1

4R

∆

R

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

–

R

8

R

9

-------







=

(EQ. 5)

I

R

–

4R

∆

R

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

R

8

R

9

-------







–

V

RX

R

-----------

+

0

=

(EQ. 6)

I

R

4R

∆

R

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

R

8

R

9

-------







–

V

RX

R

-----------

+

=

(EQ. 7)

V

C

I

R

() R

()

=

(EQ. 8)

V

C

4

– R

∆

R

8

R

9

-------







V

RX

+

=

(EQ. 9)

V

D

4R

∆

R

8

R

9

-------







V

RX

–

=

(EQ. 10)

I

L

∆

V

C

V

D

–

R

L

R

11

R

12

R

13

R

14

++++

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

=

(EQ. 11)

I

L

∆

24

– R

∆

R

8

R

9

-------







V

RX

+







×

R

L

R

11

R

12

R

13

R

14

++++

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

=

(EQ. 12)

I

L

∆

2V

RX

400 I

L

∆

–

800

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

=

(EQ. 13)

I

L

∆

V

RX

600

-----------

=

(EQ. 14)

A

4W

2W

–

V

TR

V

RX

-----------

I

∆

()

V

RX

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

V

RX

600

-----------

600

()

V

RX

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

1

==

=

=

(EQ. 15)

A

2W

4W

–

V

OUT1

V

TR

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

4

– R

∆

R

8

R

9

-------







I

∆

()

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

200

–

V

RX

600

-----------

1

()

V

RX

600

-----------

600

()

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

1

3

---

–

==

=

=

(EQ. 16)

HC5517