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.
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