10

100

1000 10000 100000

0

10

-10

-20

-30

-40

-50

-60

Rl=8ohm

Frequency(Hz)

Attenuation (dB)

Figure 14

10

10

1000

10000 100000

0

10

-10

-20

-30

-40

-50

-60

Rl=4ohm

Frequency (Hz)

Attenuation (dB)

Figure 15

if a 4

Ω loudspeaker was used without changing the filter components values, a high frequency loss

would be caused (fig.15), otherwise if a 16

Ω loudspeaker was used, an high frequency peaking would

be caused (fig.16).

10

100

1000

10000 100000

0

10

-10

-20

-30

-40

-50

-60

Rl=16ohm

Frequency (Hz)

Attenuation (dB)

Figure 16

RL

C1

L1

Figure 17

Class-D Amplifier TDA7480/81/82

Appendix to the Application Note

As explained, it is quite important to correctly dimension the filter components value to obtain the best

performances from the application.

In the following section suggestions will be given for the right dimensioning. information

2 Pole Butterworth filter: (fig.17)

To avoid Underdamped or overdamped behaviour of the filter (fig.18) the amount of the ratio of induc-

tance to capacitance Q (quality factor) Q= Rload

√C/L should be chosen between 0.6 ≤ Q ≤ 0.8, in our

application we used Q = 0.707 as god compromise that takes in account the impedance variation in

real loudspeakers.

The equations to be used to calculate the filter components values are the following:

L

=

1.414

⋅ Rload

2

[1]

C

=

0.707

2

π

[2]

where: Fc=Cutoff frequenc y(30KHz;Rload =Loudspeakerimpedance)

Example: Fc = 30KHz,Rload = 8ohm

L

=

1.414

⋅ 8

2

π ⋅ 30

= 60µHC =

0.707

2

= 0.47µF

AN1013 APPLICATION NOTE

10/13