TS4972 datasheet(23/30 Pages) STMICROELECTRONICS

23 / 30 page

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

Ω,

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 very small

50 k

Ωx1µF >> 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 = 0.1

Ω

• 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 50µs to 100µs <<

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

Ω resistor. So the discharge time is about

τbDisch ≈ 3xCbx100kΩ (s).

In the majority of application, Cb=1µF, then

τb

Disch≈300ms >> tdischCs.

Power amplifier design examples

Given :

• Load impedance : 8

Ω

• Output power @ 1% THD+N : 0.5W

• Input impedance : 10k

Ω 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

Ω. 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

Ω. Let's take Rin = 10kΩ, then

Rfeed = 28.25k

Ω. We could use for Rfeed = 30kΩ

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 1µF 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

)

(

Vcc

max

Pdiss

GV =

VOUTPP

VINPP

--------------------- =

22R L POUT

VINPP

------------------------------------ = 5.65

CIN =

π RinFCL

------------------------------ = 795nF

Rfeed

Rin

----------------- = 3

TS4972 数据表(PDF) 23 Page - STMicroelectronics

TS4972 数据表(HTML) 23 Page - STMicroelectronics

Html Pages

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

数据表下载