MP2918

—4V TO 40V SYNCHRONOUS STEP-DOWN CONTROLLER

MP2918 Rev. 1.02

www.MonolithicPower.com

24

5/31/2017

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© 2017 MPS. All Rights Reserved.

The system has two important poles: one from

the compensation capacitor (C6) and the output

resistor of the error amplifier, and the other from

the output capacitor and the load resistor (see

Figure 9). These poles can be calculated with

Equation (16) and Equation (17):





m

P1

O

G

f

2

π C6 A

(16)

LOAD

P2

R

Co

2

π

1

f







(17)

Where

is

the

error

amplifier

transconductance (500

μA/V), and Co is the

output capacitor.

The system has one important zero due to the

compensation capacitor and the compensation

resistor (R5), which can be calculated with

Equation (18):





Z1

1

f

2

π C6 R5

(18)

The system may have another significant zero if

the output capacitor has a large capacitance or

high ESR value and can be calculated with

Equation (19):

ESR

ESR

R

Co

2

π

1

f







(19)

In

this

case,

a

third

pole

set

by

the

compensation

capacitor

(C7)

and

the

compensation resistor can compensate for the

effect of the ESR zero. This pole is calculated

with Equation (20):





P3

1

f

2

π C7 R5

(20)

The goal of the compensation design is to

shape the converter transfer function for a

desired

loop

gain.

The

system

crossover

frequency where the feedback loop has unity

gain

is

important,

since

lower

crossover

frequencies result in slower line and load

transient

responses,

and

higher

crossover

frequencies lead to system instability. Set the

Follow

the

steps

below

to

design

the

compensation.

1. Choose R5 to set the desired crossover

frequency with Equation (21):







C

OUT

m

CS

FB

2

π Co f

V

R5

G

G

V

(21)

2. Choose C6 to achieve the desired phase

margin.

For

applications

with

typical

inductor values, set the compensation zero

margin. C6 is then calculated with Equation

(22):





C

4

C6

2

π R5 f

(22)

3. C7 is required if the ESR zero of the output

Equation (23) is valid:

2

f

R

Co

2

π

1

SW

ESR







(23)

at the location of the ESR zero. Determine

C7 with Equation (24):





ESR

Co R

C7

R5

(24)