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FAN6520B Rev. 1.0.3

Figure 5. PC Board Small Signal Layout Guidelines

Figure 5 shows the circuit traces that require additional

layout consideration. Use single point and ground plane

construction for the circuits shown. Minimize any leakage

current paths on the COMP pin and locate the resistor,

decoupling between VCC and GND pins. Locate the

capacitor, CBOOT as close as practical to the BOOT and

PHASE pins. All components used for feedback compen-

sation should be located as close to the IC as practical.

Feedback Compensation

Figure 6 highlights the voltage-mode control loop for a

synchronous-rectified buck converter. The output voltage

with the oscillator (OSC) triangular wave to provide a

pulse-width modulated (PWM) wave with an amplitude of

The modulator transfer function is the small-signal trans-

The DC Gain of the modulator is simply the input voltage

∆V

The following equations define the modulator break fre-

quencies as a function of the output LC filter:

Figure 6. Voltage Mode Buck

Converter Compensation Design

1.

The compensation network consists of the error

amplifier (internal to the FAN6520B) and the imped-

pensation network is to provide a closed loop

transfer function with the highest 0dB crossing fre-

margin is the difference between the closed loop

below relate the compensation network’s poles,

zeros and gain to the components (R1, R2, R3, C1,

C2, and C3) in Figure 6.

Q2

VCC

SW

BOOT

GND

+5V

FAN6520B

Q1

Vin

F

LC

1

2

π LC

×

-------------------------

=

(15)

F

ESR

1

2

π ESR

×

C

×

------------------------------------

=

(16)

Z

FB

COMP

FB

Q2

L

OUT

+5V

VIN

SW

0.8V

ERROR

AMP

PWM

OSC

DETAILED COMPENSATION

COMPONENTS

COMP

FB

0.8V

ERROR

AMP

C1

R2

C3

R3

C2

R1

Z

IN

V

OUT

F

Z1

1

2

πR

----------------------

=

(17)

F

P1

1

2

πR

2

C

C

1

C

2

+

--------------------





-----------------------------------------

=

(18)

F

Z2

1

2

πC

R

3

+

()

----------------------------------------

=

(19)

F

P2

1

2

πR

----------------------

=

(20)