Application Information
TS4972
23/30
Thus we have the relation
τin << τb (s)
The respect of this relation permits to minimize
the pop and click noise.
Remark : Minimize Cin and Cb has a benefit on
pop and click phenomena but also on cost and
size of the application.
Example : your target for the -3dB cut off
frequency is 100 Hz. With Rin=Rfeed=22 k
,
Cin=72nF (in fact 82nF or 100nF).
With Cb=1F, if you choose the one of the latest
two values of Cin, the pop and click phenomena at
power supply ON or standby function ON/OFF will
be very small
50 k
x1F >> 44kx100nF (50ms >> 4.4ms).
Increasing Cin value increases the pop and click
phenomena to an unpleasant sound at power
supply ON and standby function ON/OFF.
Why Cs is not important in pop and click
consideration ?
Hypothesis :
Cs = 100F
Supply voltage = 5V
Supply voltage internal resistor = 0.1
Supply current of the amplifier Icc = 6mA
At power ON of the supply, the supply capacitor is
charged
through
the internal
power supply
resistor. So, to reach 5V you need about five to
ten times the charging time constant of Cs (
τs =
0.1xCs (s)).
Then, this time equal 50s to 100s <<
τb in the
majority of application.
At power OFF of the supply, Cs is discharged by a
constant current Icc. The discharge time from 5V
to 0V of Cs is:
Now, we must consider the discharge time of Cb.
At power OFF or standby ON, Cb is discharged by
a 100k
resistor. So the discharge time is about
τbDisch ≈ 3xCbx100k (s).
In the majority of application, Cb=1F, then
τb
Disch≈300ms >> tdischCs.
Power amplifier design examples
Given :
Load impedance : 8
Output power @ 1% THD+N : 0.5W
Input impedance : 10k
min.
Input voltage peak to peak : 1Vpp
Bandwidth frequency : 20Hz to 20kHz (0, -
3dB)
Ambient temperature max = 50°C
SO8 package
First of all, we must calculate the minimum power
supply voltage to obtain 0.5W into 8
. With
curves in fig. 15, we can read 3.5V. Thus, the
power supply voltage value min. will be 3.5V.
Following
the
maximum
power
dissipation
equation
with 3.5V we have Pdissmax=0.31W.
Refer to power derating curves (fig. 20), with
0.31W the maximum ambient temperature will be
100°C. This last value could be higher if you follow
the example layout shown on the demoboard
(better dissipation).
The gain of the amplifier in flat region will be:
We have Rin > 10k
. Let's take Rin = 10k, then
Rfeed = 28.25k
. We could use for Rfeed = 30k
in normalized value and the gain will be Gv = 6.
In lower frequency we want 20 Hz (-3dB cut off
frequency). Then:
So, we could use for Cin a 1F capacitor value
which gives 16Hz.
In Higher frequency we want 20kHz (-3dB cut off
frequency). The Gain Bandwidth Product of the
TS4972 is 2MHz typical and doesn't change when
the amplifier delivers power into the load.
The first amplifier has a gain of:
tDi schCs =
5Cs
Icc
-------------- = 83 ms
)
W
(
R
Vcc
2
max
Pdiss
L
2
π
=
GV =
VOUTPP
VINPP
--------------------- =
22R L POUT
VINPP
------------------------------------ = 5.65
CIN =
1
2
π RinFCL
------------------------------ = 795nF
Rfeed
Rin
----------------- = 3