參數(shù)資料
型號(hào): TS4871IST
廠商: STMICROELECTRONICS
元件分類: 音頻/視頻放大
英文描述: 1 W, 1 CHANNEL, AUDIO AMPLIFIER, PDSO8
封裝: MSOP-8
文件頁數(shù): 15/28頁
文件大?。?/td> 1491K
代理商: TS4871IST
TS4871
22/28
the internal generator resistance 50k
.
Then, the charge time constant for Cb is
τb = 50kxCb (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 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
τb
Disch ≈ 3xCbx100k (s).
In the majority of application, Cb=1F, then
τb
Disch≈300ms >> tdischCs.
s 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
tDischCs =
5Cs
Icc
-------------- = 83 ms
)
W
(
R
Vcc
2
max
Pdiss
L
2
π
=
G V =
VOUTPP
VINPP
--------------------- =
22 RL P OUT
VINPP
------------------------------------ = 5.65
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