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TS4871
22/28
the internal generator resistance 50kΩ.
Then, the charge time constant for Cb is
τb = 50kΩxCb (s)
As Cb is directly connected to the non-inverting
input (pin 2 & 3) and if we want to minimize, in
amplitude and duration, the output spike on Vout1
(pin 5), Cin must be charged faster than Cb. The
cha rge time constant of Cin is
τin = (Rin+Rf eed)xCin (s)
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 kΩ,
Cin=72nF (in fact 82nF or 100nF).
With Cb=1µF, 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 ve ry smal l
50 kΩx1µF >> 44kΩx1 00nF (50m s >> 4.4ms).
Increasing Cin value increases the pop and click
phenomena to an unpleasant sound at power
supply ON and standby function ON/OFF.
W hy Cs is n ot important in pop and click
consideration ?
Hypo thesis :
• Cs = 100µF
• Supply voltage = 5V
• Supply voltage internal resistor = 0.1Ω
• Supply current of the amplifier Icc = 6mA
At power ON o f the su pply, 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 50µ s to 100µs << τb in the
majo rity of application.
At power OFF of the supply, Cs is discharged by a
constant current Icc. The di scharge time from 5V
to 0V of Cs is:
Now, we must consider the disc harge time of Cb.
At power OFF or standby ON, Cb is discharged by
a 100kΩ resistor. So the discharge time is about
τbDisch ≈ 3xCbx100kΩ (s).
In the majority of application, Cb=1µF, then
τbDisch≈300ms >> tdischCs.
■Power amplifier design examples
Given :
• Load impedanc e : 8Ω
• Output pow er @ 1% THD+N : 0.5W
• Input impedance : 10kΩ min.
• Input voltage peak to peak : 1Vpp
• Bandwidth frequ ency : 20Hz to 20 kHz (0, -3dB)
• Ambient temperature max = 50°C
• SO8 package
First of all, we m ust cal culate t he m inimum p ower
supply voltage to obtain 0.5W into 8Ω. 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
equat ion
with 3.5V we have Pdissmax=0.31W .
Refer to power derating curves (fig. 20), with
0.31W the maxim um ambien t temperature wi ll 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 > 10kΩ. Let's take Rin = 10kΩ, th en
Rfeed = 28.25kΩ. We could use for Rfeed = 30kΩ
in nor mal ized va lue and th e gain will be Gv = 6.
In lower frequency we want 20 Hz (-3dB cut off
frequency). Then:
So, we could use for Cin a 1µF capacitor value
tDischCs = 5Cs
Icc
-------------- = 83 ms
)W(
R
Vcc2
maxPdiss L
2
2
π
=
GV = VOUTPP
VINPP
--------------------- = 22R
L P OUT
VINPP
------------------------------------ = 5.65