RT9206 データシートの表示（PDF） - Richtek Technology

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RT9206

High Efficiency, Synchronous Buck with Dual Linear Controllers

Richtek Technology 

RT9206

The design procedure as following :

(1). Selecting the zero crossover frequency fC is 1/10 ~

1/20 switching frequency. Then according equation (24)

set the resistor RC1 to determine the zero crossover

frequency.

RC1 ≈

Vr × L × fC VOUT

×

(Ω)

VIN × gm × rC VREF

(24)

(2). Place the zero of compensator is 70% fp that is

resonance frequency of power stage. The compensator

capacitor Cc1 can be selected to set the zero. The

equation is shown in following :

L × CO

CC1 =

0.7 × RC1

(F) (25)

(3). Set a second pole to suppress the switching noise.

Assume the pole is one half of switching frequency

fs, which results in capacitor Cc2 as shows in following:

1

1

CC2 =

π × RC1× fs -

1

≈

(F) (26)

π × RC1× fs

CC1

Design example

Design example of type 2 compensator: the schematic is

shown in Figure 4, where the parameters as following:

VIN=12V, VOUT=5V, IOUT=5A, switching

frequency=200kHz, L=15µH, CO=940µF, rC=22mΩ, the

parameters of RT9206 as following: gm=1.6ms, ramp

amplitude=1.9V,and reference voltage Vref=0.8V.

Step1. Determine the power stage poles and zeros. The

pole caused by the output inductor and output capacitor is

calculated as :

1

1

fP =

=

= 1.34kHz

2π L × CO 2π 15µ × 940µ

1

1

fZ =

=

= 7.7kHz

2π × rC × CO 2π × 22mΩ × 940µF

Step2. Determine the zero crossover frequency and

compensated type.

Select desired zero-crossover frequency :

fC ≤ fS/10 ~ fS/20

Select fC = 20kHz

Step3. Determine desired location of poles and zeros for

type2 compensator.

Select:

fCZ = 0.7 × fP = 0.7 × 1.34kHz = 938Hz

Assume

fS

fCP = = 100kHz

2

Step4. Calculate the real parameters-resistor and

capacitors for type2 compensator.

From equation (21), the RC1 is calculated as following :

RC1 =

fC × L × Vr

VOUT

×

rC × VIN × gm VREF

= 20kHz × 15 µH × 1.9 × 5V = 8.4kΩ

22m Ω × 12V × 1.6m s 0.8V

Select RC1 = 8.2kΩ

Calculate CC1 from equation (25)

L × CO 15µ × 940µ

CC1 =

=

= 20.7nF

0.7 × RC1 0.7 × 8.2k

Select CC1 = 22nF

Second capacitor CC2 can be calculated using equation

(26)

CC2

=

1

π ×RC1× fS

=

1

π × 8.2kΩ× 200kHz

= 194pF

Select CC2 = 220pF

DS9206-11 March 2007

www.richtek.com

19

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