HC5503CB Intersil, HC5503CB Datasheet - Page 7

• Manufacturer Part Number: HC5503CB
• Description: IC,SLIC,2-4 CONVERSION,BIPOLAR,SOP,24PIN,PLASTIC
• Manufacturer: Intersil

Specifications of HC5503CB

• Rohs Compliant: NO

The value of Va, as a result of feedback through R
T
divider equation between resistors R
combination of resistors; R
resistor R
the sum of the voltage drops across resistors R
that is gained up by 2 to produce an output voltage at the
V
Where: V
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.
Substituting the value of 600∆IL for V
dividing both sides by ∆IL results in Equation 3.
Setting Va/∆IL equal to 150Ω and solving for R
R
R
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 R
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
R
V
Z
--------
∆I
X
FeedbackFactor
TX
V
1
2
1
Tipfeed
a
a
, R
L
= 10kΩ , R
output, is given in Equation 1. Equation 1 is a voltage
to match the input impedance of 600Ω is determined to
=
pin that is equal to -4R
=
3
------------------------------------------------ -
R
------------------------------------------------ -
R
and R
1
1
R
TX
INTERNAL
R
90kΩ R
=
1
90kΩ R
1
Z
90kΩ R
= -4R
90kΩ R
INTERNAL
Ringfeed
INTERNAL
B1
3
S
3
=
+
∆IL = -600∆IL.
. The Voltage on the transmit out (T
+
3
and R
3
R
X
R
2
=
2
=
×
= 90kΩ and R
, reference Figure 2.
×
--------
∆I
V
V
------------------------------------------------ -
R
600
a
B2
1
L
TX
1
S
, R
R
∆IL.
=
90kΩ R
1
should remain at 150Ω to
7
3
150Ω
90kΩ R
and the internal 90kΩ
2
3
TX
3
+
and the parallel
3
= 150kΩ the value of
R
in Equation 1 and
2
2
B1
, given that
2
and R
from the
(EQ. 4)
(EQ. 2)
(EQ. 1)
(EQ. 3)
X
B2
) is
HC5503
The voltage that is feed back into the RX pin is equal to the
voltage at V
Where V
So:
But, from Equation 2:
Therefore:
Equation 8 shows that 1/4 of the T
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 R
from the Tip feed + Ring feed amplifiers).
Setting V
given that R
the value of R
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 R
only requirement is that it be large enough to have little
effect on the parallel combination between R
(90kΩ) and R
The selection of the value of 10kΩ for R
The only requirement is that the value be small enough to
offset any process variations of R
enough to avoid loading of the CODEC’s output. A value of
10kΩ is a good compromise.
V
X
--------
∆I
X
V
a
a
L
=
=
=
=
------------------ -
∆I
---------- -
V
V
V
TX
150Ω
V
L
a
TX
FIGURE 2. FEEDBACK EQUIVALENT CIRCUIT
600
TX
a
a
R
T
/∆IL equal to 300Ω and solving for R
=
( )
X
X
X
1
TX
is equal to -4R
150
--------- -
600
= 10kΩ , R
1
2
T
24.9kΩ
X
times the feedback factor (Equation 5).
to match the input impedance of 900Ω is
(10kΩ). R
FEED BACK
= -4R
=
R
1
-- -
4
2
S
∆IL
INTERNAL
3
S
should be greater then 90kΩ .
∆IL (R
150kΩ
R
3
INTERNAL
= 90kΩ and R
S
X
= 150Ω)
output voltage is
10kΩ
B1
R
1
1
+ R
3
is also arbitrary.
is arbitrary. The
and large
B2
INTERNAL
2
in Equation 3,
R
and 600Ω
3
INTERNAL
90.0kΩ
= 150kΩ
(EQ. 5)
(EQ. 6)
(EQ. 7)
(EQ. 8)