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Rainbow Electronics MAX8514 User Manual

Page 26

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MAX8513/MAX8514

Wide-Input, High-Frequency, Triple-Output Supplies
with Voltage Monitor and Power-On Reset

26

______________________________________________________________________________________

This gain between f

Z1

and f

Z2

is also set by the ratio of

R3/R1, where R1 is selected in the OUT1 Voltage
Setting
section. Therefore:

And similar to Case 1, C5 can be calculated as:

Set the error-amplifier third pole, f

P3

, at approximately

1/2 the switching frequency. The gain of the error
amplifier at f

C

(between f

P2

and f

P3

) is set by the ratio

of R3/R

I

and is also equal to:

where R

I

is:

Therefore:

Similar to Case 1, R4, C11, and C12 can be calculated as:

Below is a numerical example to calculate the error-
amplifier compensation values for Case 2:

V

IN

= 12V (nomimal input voltage)

V

RAMP

= 1V

V

OUT1

= 3.3V

V

FB1

= 1.25V

L1A = 6.2µH

C4 = 560µF/ 10V OS-Con capacitor, with ESR = 0.015Ω
f

S

= 300kHz

Pick R2 = 8.06kΩ. Then:

Pick f

C

= 50kHz, which is less than f

S

/ 5.

Use 20kΩ.

Use 12nF.

Use 2.2kΩ.

Use 3.9nF.

Pick f

P

3 = f

S

/ 2 = 150kHz.

C

R

f

k

kHz

nF

ZESR

11

1

2

4

1

2

2 2

18 95

3 82

.

.

.

=

Ч

Ч

=

Ч

Ч

=

π

π

R

R

G

k

k

R

R

R

R R

k

k

k

k

k

I

MOD fc

I

I

.

.

.

.

.

.

.

( )

=

Ч

=

Ч

=

=

Ч

=

Ч

=

3

20

0 0923 1 846

4

1

13 3

1 846

13 3

1 846

2 14

1-

-

C

R

f

k

kHz

nF

PMOD

5

2

3

2

20

2 7

11 8

.

.

=

Ч

Ч

=

Ч

Ч

=

π

π

G

kHz

kHz

kHz

G

f

f

G

kHz

kHz

R

R

G

k

k

MOD DC

EA fZ

fZ

PMOD

ZESR MOD fc

EA fZ fZ

(

)

(

)

( )

(

)

.

.

.

.

.

.

.

.

.

.

=

Ч

Ч

=

=

=

Ч

=

=

Ч

=

Ч

=

12

2 7

18 95

50

0 0923

2 7

18 95

0 0923

1 543

3

1

13 3

1 543 20 48

2

1

2

1

2

-

R

k

V

V

k

G

V

V

MOD DC

IN

RAMP

1

8 06

3 3

1 25

1

13 3

12

.

.

.

.

(

)

=

×

=

=

=

-

f

L A

C

H

F

kHz

f

R

C

F

kHz

PMOD

ZESR

ESR

=

Ч

=

Ч

=

=

Ч

=

Ч

Ч

=

.

.

.

.

1

2

1

4

1

2

6 2

560

2 7

1

2

4

1

2

0 015

560

18 95

π

π

µ

µ

π

π

µ

R

R

R

R

R

C

R

f

C

C

C

R

f

I

I

ZESR

P

4

1

1

11

1

2

4

12

5

2

5

3

1

3

=

Ч

=

Ч

Ч

=

Ч

Ч

Ч

-

-

π

π

R

R

G

G

I

MOD fc

D

( )

=

Ч

Ч

3

R

R

R

R

R

I

=

×

+

1

4

1

4

G

G

EA fc

MOD fc

( )

( )

=

1

C

R

f

PMOD

5

2

3

=

Ч

Ч

π

R

R

f

f

G

PMOD

ZESR

MOD fc

3

1

( )

=

Ч
Ч