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Fig.6 thd as a function of frequency, Fig.8 svrr as a function of frequency, Fig.9 p – Philips TDA8542 User Manual

Page 10: As a function of v

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1998 Apr 01

10

Philips Semiconductors

Product specification

2

×

1 W BTL audio amplifier

TDA8542

Fig.6 THD as a function of frequency.

P

o

= 0.5 W, G

v

= 20 dB.

(1) V

CC

= 5 V, R

L

= 8

.

(2) V

CC

= 9 V, R

L

= 16

.

handbook, halfpage

10

1

10

1

10

2

MGD892

10

10

2

10

3

10

4

THD

(%)

f (Hz)

10

5

(1)

(2)

Fig.7

Channel separation as a function of
frequency.

V

CC

= 5 V, V

o

= 2 V, R

L

= 8

.

(1) G

v

= 30 dB.

(2) G

v

= 20 dB.

(3) G

v

= 6 dB.

handbook, halfpage

100

90

80

70

60

MGD893

10

10

2

10

3

10

4

10

5

f (Hz)

α

cs

(dB)

(1)

(2)

(3)

Fig.8 SVRR as a function of frequency.

V

CC

= 5 V, R

s

= 0

, V

r

100 mV.

(1) G

v

= 30 dB.

(2) G

v

= 20 dB.

(3) G

v

= 6 dB.

handbook, halfpage

80

60

40

20

MGD894

10

2

10

10

3

SVRR

(dB)

f (Hz)

10

4

10

5

(1)

(2)

(3)

Fig.9 P

o

as a function of V

CC

.

(1) THD = 10%, R

L

= 8

.

(2) THD = 10%, R

L

= 16

.

handbook, halfpage

0

4

8

Po

(W)

VCC (V)

12

2.5

0

2

1.5

1

(1)

(2)

0.5

MGD895