TS4972 datasheet(22/28 Pages) STMICROELECTRONICS

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TS4972

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The charge time of Cb is directly proportional to

the internal generator resistance 50k

Ω.

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

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

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

τb

Disch ≈ 3xCbx100kΩ (s).

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

τb

Disch≈300ms >> tdischCs.

s 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.

tDischCs =

5Cs

Icc

-------------- = 83 ms

)

(

Vcc

max

Pdiss

G V =

VOUTPP

VINPP

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

22 RL P OUT

VINPP

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

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