HC5503 Ver la hoja de datos (PDF) - Intersil

Número de pieza

componentes Descripción

Fabricante

HC5503

Low Cost 24V SLIC For PABX/Key Systems

Intersil 

HC5503

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 90kΩ

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.

Va = R-----1--R----9-1--0----9k---0Ω----k---Ω-R----3---R--+--3---R----2-- × VTX

(EQ. 1)

Where: VTX = -4RS∆IL = -600∆IL.

To match a 600Ω line, the synthesized tip and ring impedances

must be equal to 150Ω. 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.

ZTipfeed

=

ZRingfeed

=

-V-----a--

∆IL

=

150 Ω

(EQ. 2)

Substituting the value of 600∆IL for VTX in Equation 1 and

dividing both sides by ∆IL results in Equation 3.

-V-----a--

∆IL

=

-R----1--R----9-1--0----9k---0Ω----k---Ω-R----3---R--+--3---R----2-- × 600

(EQ. 3)

Setting Va/∆IL equal to 150Ω and solving for R2 , given that

R1 = 10kΩ , RINTERNAL = 90kΩ and R3 = 150kΩ the value of

R2 to match the input impedance of 600Ω is determined to

be 25.47kΩ . (Note: nearest standard value is 24.9kΩ).

The amount of negative feedback is dependent upon the

additional synthesized resistance required for matching. The

sense resistors RB1 and RB2 should remain at 150Ω 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.

FeedbackFactor = X = -R----1--R----9-1--0----9k---0Ω----k---Ω-R----3---R--+--3---R----2--

(EQ. 4)

FEED BACK

RX

R2

24.9kΩ

TX

TX = -4RS∆IL

R3

150kΩ

R1

10kΩ

RINTERNAL

90.0kΩ

FIGURE 2. FEEDBACK EQUIVALENT CIRCUIT

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

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

Va = VTX(X)

(EQ. 5)

Where VTX is equal to -4RS∆IL (RS = 150Ω)

So:

X = -------V----a-------

∆ IL 600

(EQ. 6)

But, from Equation 2:

-V-----a-- = 150Ω

∆IL

Therefore:

X = ---V----a--- = 1----5---0-- = 1--

VTX 600 4

(EQ. 7)

(EQ. 8)

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

required to synthesize 150Ω at both the Tip feed and Ring

feed amplifiers.

To match a 900Ω load would require 300Ω worth of

synthesized impedance (300Ω from RB1 + RB2 and 600Ω

from the Tip feed + Ring feed amplifiers).

Setting Va/∆IL equal to 300Ω and solving for R2 in Equation 3,

given that R1 = 10kΩ , RINTERNAL = 90kΩ and R3 = 150kΩ

the value of R2 to match the input impedance of 900Ω is

determined to be 8.49kΩ (Note: nearest standard value is

8.45kΩ). The feed back factor to match a 900Ω load is 1/2

(300/600).

The selection of the value of 150kΩ for R3 is arbitrary. The

only requirement is that it be large enough to have little

effect on the parallel combination between RINTERNAL

(90kΩ) and R1 (10kΩ). R3 should be greater then 90kΩ .

The selection of the value of 10kΩ 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

10kΩ is a good compromise.

7

Share Link: