L6246 데이터 시트보기 (PDF) - STMicroelectronics

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L6246

12V VOICE COIL MOTOR DRIVER

STMicroelectronics 

L6246

ERROR AMPL. GAIN

(dB)

120

100

80

60

40

20

OPEN LOOP GAIN

D95IN271

1

10 100 1K 10K 100K 1M 10M

Using the compensation network of the draw of

pag.8, we have a error amplifier transfer function

of:

VO

VI

=

−

ZC

R1

=

−

R2

R1

⋅

1

+

1 + scR3

sc (R3 + R2)

so:

Gmax

(DC)

=

R2

R1

=

1000

=

60dB

with R1 = 1MΩ and R2 = 1KΩ

zero = 1

2 π R3C

pole

=

2

π

1

(R3 +

R2)

C

Note: Fpole is lower than Fzero

The best choice is to cancel the pole of the load

(at around 1KHz) with the zero of the compensa-

tion.

ACERR (dB)

120

100

80

X

60

DIFFERENTS

POLES

X EXAMPLES

40

20

CLOSED LOOP

ACERR

X

D95IN272

1

10 100 1K 10K 100K 1M 10M

As can be seen the choice of the pole influence

overall in fixing the gain at high frequency.

The gain at high frequency must be choosen in

order to not create instability problem, because

more higher is this gain and lower is the second

pole that we have at high frequency.

If this pole is taken close to the other that we

have already seen at 130KHz and 210KHz, insta-

bility problems can arise.

Adding together AX | dB and ACERR | dB we ob-

taine the Aloop:

ACERR

(dB)

COMPENSATION AT 3Hz

60

COMPENSATION AT 100Hz

40

20

AX(dB)

19

ALOOP (dB)

79

IS STABLE

60

IS NOT STABLE

40

20

D95IN273 10 100 1K 10K 100K 1M 10M

So the choice of the compensation network must

be done in order to fix at the beginning the Gmax

of the error amplifier depending on the ratio R2.

R1

To calculate the R3 and C values satisfying the

following system:

1

2

π

1

R3C

=

RL

2

+

πL

Rsense

Error amplifier zero equal to load pole

2

π

1

(R3 +

R2)C

=

Admissible Bandwith

Gloop

=

130KHz

=

10

8912

= 1.5Hz

This example is for crossing the 0dB one decade

before the first pole of the Power Amplifier

(130KHz), starting with a Gloop max of 79dB.

10/12

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