TS4871
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The charge time of Cb is directly proportional to
the internal generator resistance 50kΩ.
Then, the charge time constant for Cbis
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
charge time constant of Cin is
τin = (Rin+Rfeed)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 valuesof Cin, the pop and click phenomenaat
power supply ON or standby function ON/OFF will
be very small
50 kΩx1µF >> 44kΩx100nF (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 = 100µF
•Supply voltage = 5V
•Supply voltage internal resistor = 0.1Ω
•Supply current of the amplifier Icc = 6mA
At powerON 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 50µs to 100µs<<τb in the
majority of application.
At power OFF of the supply, Csis dischargedby a
constant current Icc. The discharge time from 5V
to 0V of Cs is:
Now, we must consider the discharge time of Cb.
At powerOFF or standby ON, Cb is discharged by
a 100kΩresistor. 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 impedance : 8Ω
•Output power @ 1% THD+N : 0.5W
•Input impedance : 10kΩmin.
•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 8Ω. Withcurves
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:
tDischCs =5Cs
Icc
-------------- = 83 ms
)W(
R
Vcc2
maxPdiss L
2
2
π
=
GV=VOUTPP
VINPP
---------------------=
22RLPOUT
VINPP
----------------------------------- - = 5.65