TS4994EIKJT STMICROELECTRONICS [STMicroelectronics], TS4994EIKJT Datasheet - Page 24

• Manufacturer Part Number: TS4994EIKJT
• Description: 1.2 W differential input/output audio power amplifier with selectable standby
• Manufacturer: STMICROELECTRONICS [STMicroelectronics]

Application information
4.4
24/35
the values of V
following formula:
with
The result of the calculation must be in the range:
If the result of the V
used.
Example: With V
higher than 2.5V - 0.9V = 1.6V, so input coupling capacitors are required. Alternatively, you
can change the V
Low and high frequency response
In the low frequency region, C
with a -3dB cut-off frequency. F
In the high-frequency region, you can limit the bandwidth by adding a capacitor (C
parallel with R
While these bandwidth limitations are in theory attractive, in practice, because of low
performance in terms of capacitor precision (and by consequence in terms of mismatching),
they deteriorate the values of PSRR and CMRR.
The influence of mismatching on PSRR and CMRR performance is discussed in more detail
in the following sections.
Example: A typical application with input coupling and feedback capacitor with F
and F
neglected. If we sweep the frequency from DC to 20kHz we observe the following with
respect to the PSRR value:
●
From DC to 200Hz, the C
C
CH
feed
= 8kHz. We assume that the mismatching between R
impedance is high enough to be neglected. Due to the tolerance of C
feed
CC
CC
, R
ic
. It forms a low-pass filter with a -3dB cut-off frequency. F
value.
ICM
=2.5V, R
in
and R
calculation is not in this range, an input coupling capacitor must be
V
ICM
feed
in
V
in
F
in
= R
=
ic
CL
CH
starts to have an effect. C
impedance decreases from infinite to a finite value and the
(
F
V
------------------------------------------------------------------------- -
=
A
CL
feed
is in Hz.
CC
=
V
0.6V V
Diff
------------------------------------------------------ -
). To have a good estimation of the V
2
=
×
2
×
= 20k and V
2
input+
R
π
×
≤
×
in
×
π
(
R
ICM
R
+
×
in
feed
2
+
R
2
1
1
≤
in
+
Diff
×
V
×
R
×
V
CC
C
feed
C
ic
input-
ic
in
feed
×
–
= 2V, we find V
)
0.9V
R
(
Hz
feed
(
in
)
Hz
forms, with R
(V)
)
in1,2
(V)
and C
ICM
= 1.63V. This is
in
ICM
feed1,2
, a high-pass filter
CH
value, use the
is in Hz.
can be
in1,2
TS4994FC
CL
feed
= 50Hz
, we
) in