Using the compensation network of the draw of

pag.8, we have a error amplifier transfer function

of:

=−

ZC

R1

=−

R2

R1

⋅

1

+ scR3

1

+ sc (R3 + R2)

so:

Gmax (DC) =

R2

R1

= 1000 = 60dB

with R1 = 1M

Ω and R2 = 1KΩ

zero =

1

2

π R3C

pole =

1

2

π(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.

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.

taine the Aloop:

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 R3C

=

1

2

π L

Error amplifier zero equal to load pole

1

2

π(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.

ERROR AMPL. GAIN

(dB)

40

60

20

1K

10K

100K

D95IN271

100

120

80

1

10

100

1M

10M

OPEN LOOP GAIN

ACERR (dB)

40

60

20

1K

10K

100K

D95IN272

100

120

80

1

10

100

1M

10M

CLOSED LOOP

ACERR

X

X

DIFFERENTS

POLES

EXAMPLES

X

ACERR

(dB)

20

COMPENSATION AT 3Hz

AX(dB)

19

D95IN273

40

60

20

40

60

79

10

100

1K

10K

100K

1M

10M

COMPENSATION AT 100Hz

IS STABLE

IS NOT STABLE

L6246

10/12