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Rainbow Electronics MAX5099 User Manual

Page 22

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MAX5099

Dual, 2.2MHz, Automotive Synchronous Buck
Converter with 80V Load-Dump Protection

22

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Procedure 2 (See Figure 4)

If the output capacitor used is a low-ESR ceramic type,
the ESR frequency is usually far away from the targeted
unity crossover frequency (f

C

). In this case, Type III

compensation is recommended. Type III compensation
provides two-pole zero pairs. The locations of the zero
and poles should be such that the phase margin peaks
around f

C

. It is also important to place the two zeros at

or below the double pole to avoid the conditional stabil-
ity issue.

1) Select a crossover frequency:

2) Calculate the LC double-pole frequency, f

LC

:

3) Place a zero

where

and R

F

≥ 10kΩ.

4) Calculate C

I

for a target unity crossover frequency, f

C

.

5) Place a pole

6) Place a second zero, f

Z2

, at 0.2 x f

C

or at f

LC

,

whichever is lower.

7) Place a second pole at 1/2 the switching frequency.

Boost Converter Compensation

The boost converter compensation gets complicated
due to the presence of a right-half-plane zero
f

ZERO,RHP

. The right-half-plane zero causes a drop in

phase while adding positive (+1) slope to the gain
curve. It is important to drop the gain significantly below
unity before the RHP frequency. Use the following pro-
cedure to calculate the compensation components:

1) Calculate the LC double-pole frequency, f

LC

, and

the right-half-plane-zero frequency.

where

Target the unity-gain crossover frequency for:

f

f

C

ZERO RHP

,

5

D

V

V

R

V

I

IN

OUT

MIN

OUT

OUT MAX

=

=

( )

1

(

)

f

D R

L

ZERO RHP

MIN

OUT

,

=

( )

×

( )

1

2

2

π

f

D

L

C

LC

OUT

OUT

=

Ч

Ч

1

2

π

C

C

f

R

C

CF

F

SW

F

F

=

Ч

Ч

Ч

Ч

(

)

2

0 5

1

π

.

R

f

C

R

Z

I

I

1

1

2

2

=

Ч

Ч

π

R

f

C

I

ZERO ESR

I

=

Ч

Ч

1

2

π

,

f

R

C

at f

P

I

I

ZERO ESR

1

1

2

=

Ч Ч

π

.

,

C

f

L

C

V

V

R

I

C

OUT

OUT

OSC

IN

F

=

Ч Ч

Ч

Ч

Ч

2

π

C

f

R

F

LC

F

=

Ч

Ч

Ч

1

2

0 75

π

.

f

R

C

at

f

Z

F

F

LC

1

1

2

0 75

=

Ч

Ч

Ч

π

.

.

f

L

C

LC

OUT

OUT

=

Ч

Ч

1

2

π

f

f

SW

SW

20

R1

R

F

COMP_

V

OUT

V

REF

R2

R

I

C

I

C

F

C

CF

-

+

g

M

FB_

Figure 4. Type III Compensation Network