ST AN2583 Application note

AN2583

M41T00AUD

ST RTC STAIO

AIO

制和滤波。

ST M41T00AUD 2 1ST M41T00

8 kHz16 3 dB300 mW 8 Ω VCC 3.3 V

M41T00AUD

2007 12

Rev 1

1/18

www.st.com

	AN2583

1

1. M41T00AUD

M41T00AUD

VDD

VDD

VBACK

TRICKLE

CHARGE

VCC

AUTOMATIC

BATTERY

VINT

SECS

SWITCHOVER

MINS

uC

& DESELECT

HOURS

DATE

REFERENCE

DAY

VPFD=2.80V

MONTH

YEAR

WRITE

CENTURY BITS

PROTECT

CALIBRATION

I2C

OUT

IRQ/FT/OUT

I2C

400kHz I2C

OSCILLATOR

(SDA,

2

FAIL DETECT

INTERFACE

SCL)

256/512Hz

OSCI

32KHz

OSCILLATOR

OSCO

FBK

AUDIO

AOUT+

LPF

ADJ

Audio-in

AIN

VDD

GAIN

AOUT–

2

VBIAS

VSS

ai13927

2/18

AN2583
	AIO
	256/512 Hz
	2
2.	M41T00AUD
				M41T00AUD Audio section
				RTC Timing
				Chain
			Rf	256/512Hz
	uC	PWM / DAC	Ri1 C	LPF ADJ	AOUT+
		TONE
		GENERATOR		GAIN	AOUT–
Phone	Fax		Ri2
		MODEM
Line	Chipset
					ai13928
	Rin Rf
	C M41T00AUD
	VDD/2 M41T00AUD
	VDD/2 M41T00AUD
	Rin = 20 kΩ C = 0.1 uF 80 Hz

8 kHz 80 Hz 8 kHz

RTC 256/512 Hz4 3 dB –33 dB +12 dB

M41T00AUD / 300 mA VCC = 3.3 V300 mW 8 Ω

3/18

	AN2583

2

出功率。以下是这些特性的简略介绍。

同。

3.

VIN VDD	T
	f0 =	1
VDD / 2		T

0

VDD

VOUT

VDD / 2

IDEAL

0

VDD

VOUT

VDD / 2

WITH

DISTORTION

0

Ideal output is simply a linear scale of the input.

Asymmetric nature of waveforms due to imperfect symmetry of complementary P and N transistors

Reshaping of waveforms due to non-linear nature of transistors

Crossover distortion because switchover from P to N channel drivers does not occur perfectly

ai13929

4

3

4/18

AN2583

4.

F(VOUT )

VOUT, FUNDAMENTAL

VOUT EVEN HARMONICS

VOUT ODD HARMONICS

0

f

f0

2f0

3f0

4f0

5f0

6f0

7f0

ai13930

THD THD

5 P1 P2 P7

THD

5.

P1

F (POUT)

P3

P2

P4

P5

P6

P7

0

f

f0

2f0

3f0

4f0

5f0

6f0

7f0

ai13931

THD

THDP =	harmonic powers	=	P2 + P3 + P4 + P5	+ P6 + P7 +L
	power of fundamental frequency		P1

5/18

AN2583

6.

F (VOUT)

V1

V3

0

V2

V4

V5

V6

V7

f

f0

2f0

3f0

4f0

5f0

6f0

7f0

ai13932

THD

THDV =

∑(harmonic voltages)2

=

V2

+ V2

+ V2 + V2

+ V2

+ V2

+L

2

3

4

5

6

7

voltage of fundamental frequency

V1

P V2

V2

+ V2

+ V2 + V2 + V2 + V2 +L

1

V2

+ V2

+ V2

+ V

2

+ V

2 + V

2 +L

THDV =

=

R

2

3

4

5

6

7

2

3

4

5

6

7

V1

1

V1

R

V2

V2

V2

V2

V2

V2

2

+

3

+ 4

+

5

+

6

+

7

L

P

+ P

+ P

+ P

+ P + P +L

=

R R R R R R

=

2

3

4

5

6

7

V1

P

1

R

THDV = THDP

THD 1 THDV THDP THDP

6/18