HC5503

FN4344 Rev 5.00

Page 7 of 17

June 2004

The value of Va, as a result of feedback through R2 from the

TX output, is given in Equation 1. Equation 1 is a voltage

divider equation between resistors R2 and the parallel

combination of resistors; R1, R3 and the internal 90k resistor

RINTERNAL. The Voltage on the transmit out (TX) is the sum of

the voltage drops across resistors RB1 and RB2 that is gained

up by 2 to produce an output voltage at the VTX pin that is

equal to -4RSIL.

Where: VTX = -4RSIL = -600IL.

To match a 600

 line, the synthesized tip and ring impedances

must be equal to 150

. The impedance looking into either the tip

or ring terminal is once again the voltage at the terminal (Va)

divided by the AC current

IL as shown in Equation 2.

Substituting the value of 600

IL for VTX in Equation 1 and

dividing both sides by

IL results in Equation 3.

Setting Va/

IL equal to 150 and solving for R2, given that R1

= 10k

, RINTERNAL = 90k and R3 = 150k the value of R2

to match the input impedance of 600

 is determined to be

25.47k

. (Note: nearest standard value is 24.9k

The amount of negative feedback is dependent upon the

additional synthesized resistance required for matching. The

sense resistors RB1 and RB2 should remain at 150 to

maintain the SHD threshold listed in the electrical

specifications. The additional synthesized resistance is

determined by the feed back factor X (Equation 4) which needs

to be applied to the transmit output and fed into the RX pin of

the HC5503. The feed back factor is equal to the voltage

divider between R2 and the parallel combination of R1, R3 and

RINTERNAL, reference Figure 2.

The voltage that is feed back into the RX pin is equal to the

voltage at VTX times the feedback factor (Equation 5).

Where VTX is equal to -4RSIL (RS = 150)

So:

But, from Equation 2:

Therefore:

Equation 8 shows that 1/4 of the TX output voltage is required

to synthesize 150

 at both the Tip feed and Ring feed

amplifiers.

To match a 900

 load would require 300 worth of

synthesized impedance (300

 from RB1 + RB2 and 600 from

the Tip feed + Ring feed amplifiers).

Setting Va/IL equal to 300 and solving for R2 in Equation 3,

given that R1 = 10k, RINTERNAL = 90k and R3 = 150k the

value of R2 to match the input impedance of 900 is determined

to be 8.49k

 (Note: nearest standard value is 8.45k). The feed

back factor to match a 900

 load is 1/2 (300/600).

The selection of the value of 150k

 for R3 is arbitrary. The only

requirement is that it be large enough to have little effect on the

parallel combination between RINTERNAL (90k) and R1

(10k

). R3 should be greater then 90k.

The selection of the value of 10k

 for R1 is also arbitrary. The

only requirement is that the value be small enough to offset

any process variations of RINTERNAL and large enough to

avoid loading of the CODEC’s output. A value of 10k

 is a

good compromise.

2-Wire to 4-Wire Gain

Va

R1 90k R3

R1 90k R3 R2

+

------------------------------------------------- VTX



=

(EQ. 1)

ZTipfeed

ZRingfeed

Va

IL

--------

150



==

=

(EQ. 2)

Va

IL

--------

R1 90k R3

R1 90k R3 R2

+

------------------------------------------------- 600



=

(EQ. 3)

FeedbackFactor

X

R1 90k R3

R1 90k R3 R2

+

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

==

(EQ. 4)

FIGURE 2. FEEDBACK EQUIVALENT CIRCUIT

RX

TX

R1

R3

150k



R2

24.9k



RINTERNAL

90.0k



10k



FEED BACK

TX = -4RSIL

Va

VTX X



=

(EQ. 5)

X

Va

IL600

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

=

(EQ. 6)

Va

IL

--------

150



=

(EQ. 7)

X

Va

VTX

-----------

150

600

----------

1

4

---

==

=

(EQ. 8)