HC5503 Intersil Corporation, HC5503 Datasheet - Page 6

• Manufacturer Part Number: HC5503
• Description: Low Cost 24V SLIC For PABX/Key Systems
• Manufacturer: Intersil Corporation

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 600 line, the synthesized tip and ring impedance
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 V
dividing both sides by IL results in Equation 3.
Setting Va/ IL equal to 150 and solving for R
R
R
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 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
--------
V
FeedbackFactor
X
TX
Tipfeed
1
2
1
a
I
a
L
, R
= 10k , R
output, is given in Equation 1. Equation 1 is a voltage
to match the input impedance of 600 is determined to
=
=
pin that is equal to -4R
------------------------------------------------ -
R
3
------------------------------------------------ -
R
1
and R
1
FIGURE 2. FEEDBACK EQUIVALENT CIRCUIT
R
R
90k
=
TX
R
T
INTERNAL
1
90k
1
X
X
Z
90k
90k
= -4R
Ringfeed
INTERNAL
INTERNAL
T
24.9k
R
X
R
FEED BACK
= -4R
3
3
B1
R
R
+
S
R
=
+
3
2
. The Voltage on the transmit out (T
R
3
X
R
and R
IL = -600 IL.
S
2
=
2
=
IL
--------
= 90k and R
, reference Figure 2.
V
V
------------------------------------------------ -
R
I
600
a
L
TX
1
B2
1
S
R
, R
=
150k
90k
R
1
should remain at 150 to
3
6
150
IL.
3
90k
and the internal 90k
R
3
2
R
TX
+
3
10k
and the parallel
3
R
R
= 150k the value of
1
in Equation 1 and
2
2
B1
R
, given that
INTERNAL
90.0k
2
and R
from the
(EQ. 1)
(EQ. 2)
(EQ. 3)
(EQ. 4)
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 150 at both the Tip feed and Ring
feed amplifiers.
To match a 900 load would require 300 worth of
synthesized impedance (300 from R
from the Tip feed + Ring feed amplifiers).
Setting Va/ IL equal to 300 and solving for R
given that R
the value of R
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 R
only requirement is that it be large enough to have little effect
on the parallel combination between R
R
The selection of the value of 10k 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
10k is a good compromise.
2-Wire to 4-Wire Gain
The 2-wire to 4-wire gain is defined as the output voltage
V
V
The 2-wire to 4-wire gain is therefore equal to -1.0, as
shown in Equation 9.
V
X
--------
X
A
V
TX
TX
a
I
1
2 4
a
L
=
=
–
(10k ). R
=
=
------------------ -
---------- -
V
divided by the tip to ring voltage (V
= -4R
V
V
I
TX
=
150
V
L
a
TX
600
a
V
---------- -
V
TX
=
TR
TX
X
S
150
--------- -
600
1
TX
is equal to -4R
3
= 10k , R
IL = -600 IL and V
2
=
should be greater then 90k .
times the feedback factor (Equation 5).
to match the input impedance of 900 is
–
--------------------- -
=
600 I
600
1
-- -
4
I
L
L
INTERNAL
=
–
S
1.0
IL (R
TR
INTERNAL
= 90k and R
S
X
= (RL) IL = 600 IL.
= 150 )
output voltage is
B1
INTERNAL
1
TR
3
+ R
is also arbitrary.
is arbitrary. The
). Where:
and large
B2
2
in Equation 3,
3
and 600
(90k ) and
= 150k
(EQ. 5)
(EQ. 6)
(EQ. 7)
(EQ. 8)
(EQ. 9)