MP1567 – 1.2A SYNCHRONOUS RECTIFIED STEP-DOWN CONVERTER

MP1567 Rev. 2.3

www.MonolithicPower.com

8

1/3/2006

MPS Proprietary Information. Unauthorized Photocopy and Duplication Prohibited.

© 2006 MPS. All Rights Reserved.

TM

75KHz or lower to insure stable operation.

Lower crossover frequencies result in slower

response and worse transient load recovery.

Higher crossover frequencies degrade the

phase and/or gain margins and can result in

instability.

Choosing the Compensation Components

The values of the compensation components

given in Table 1 yield a stable control loop for

the output voltage and capacitor given.

Table 1—Compensation Values for Typical

Output Voltage/Capacitor Combinations

C2

R3

C3

C4

1.8V

4.7µF Ceramic

3.3kΩ

2.2nF

None

2.5V

4.7µF Ceramic

5.1kΩ

1.5nF

None

3.3V

4.7µF Ceramic

6.8kΩ

1.2nF

None

1.8V

10µF Ceramic

7.5kΩ

1nF

None

2.5V

10µF Ceramic

10kΩ

820pF

None

3.3V

10µF Ceramic

10kΩ

820pF

None

1.8V

47µF Tantalum

(300mΩ)

10kΩ

2.2nF

1.5nF

2.5V

47µF Tantalum

(300mΩ)

10kΩ

3.3nF

1.5nF

3.3V

47µF Tantalum

(300mΩ)

10kΩ

4.7nF

1.5nF

To optimize the compensation components for

conditions not listed in Table 1, use the

following procedure.

Choose the compensation resistor to set the

desired crossover frequency. Determine the

value by the following equation:

FB

CS

EA

C

OUT

V

G

G

f

V

2

C

2

3

R

×

×

×

×

×

π

=

Putting in the known constants and setting the

crossover frequency to the desired 75KHz:

OUT

8

V

2

C

10

36

.

4

3

R

×

×

×

≈

In this case, the actual crossover frequency is

less than the desired 75KHz, and it is

calculated by:

OUT

FB

CS

EA

C

V

2

C

2

V

G

G

3

R

f

×

×

π

×

×

×

=

Choose the compensation capacitor to set the

zero to one fourth of the crossover frequency.

Determine the value by the following equation:

FB

CS

EA

2

OUT

V

G

G

3

R

V

2

C

4

3

C

×

×

×

×

×

=

Determine

if

the

second

compensation

capacitor, C4, is required. It is required if the

ESR zero of the output capacitor occurs at less

than four times the crossover frequency, or:

1

f

R

2

C

8

C

ESR

≥

×

×

×

π

of the output capacitor.

If this is the case, then add the second

compensation capacitor. Determine the value

by the equation:

3

R

R

2

C

4

C

)

MAX

(

ESR

×

=

output capacitor.

For Example:

Given:

C2 = 10µF Ceramic (ESR = 10mΩ max.)

Calculate:

Ω

=

×

µ

×

≈

k

85

.

7

)

V

8

.

1

(

)

F

10

(

10

36

.

4

3

R

8

(Use the nearest standard value of 7.5kΩ.)

nF

05

.

1

V

8

.

1

F

10

10

9

.

1

3

C

14

=

×

µ

×

=

−

(Use 1nF since it is a standard value.)

19

.

0

f

R

2

C

8

C

ESR

=

×

×

×

π

which is less than 1, therefore the second

compensation capacitor (C4) is not required.