NX2415 Просмотр технического описания (PDF) - Microsemi Corporation

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NX2415

TWO PHASE SYNCHRONOUS PWM CONTROLLER WITH INTEGRATED FET DRIVER AND DIFFERENTIAL CURRENT SENSE

Microsemi Corporation 

NX2415

C2

Vout

R3 C1

R2

Fb

Ve

R1

Vref

Figure 4 - Type II compensator

power stage

40dB/decade

FLC = 2× π ×

1

LEFF × COUT

=

1

2×π× 0.75uH×10800uF

= 1.768kHz

FESR

=

1

2 × π × ESR × COUT

=

1

2 × π ×13mΩ ×1800uF

= 6.801kHz

2.Set

R

2

equal

to10kΩ

and

calculate

R.

1

R1=

R2 ×

VOUT

VREF

-VREF

=

10kΩ × 0.8V

1.2V-0.8V

=

20kΩ

3. Set crossover frequency FO=15kHz.

4.Calculate R3 value by the following equation.

loop gain

20dB/decade

R3

=

VOSC

Vin

×

2 × π × FO × L EFF

ESR

× R2

= 1V × 2 × π × 15kHz × 0.75uH × 10kΩ

12V

2.16m Ω

=27.3kΩ

compensator

Gain

FZ FLCFESR FO FP

Choose R3 =27.4kΩ.

5. Calculate C1 by setting compensator zero FZ

at 75% of the LC double pole.

C1=

2

×

π

1

× R3

×

Fz

=

1

2× π × 27.4kΩ ×0.75×1.768kHz

=4.4nF

Figure 5 - Bode plot of Type II compensator

For this type of compensator, FO has to satisfy

FLC<FESR<< FO and FO <=1/10~1/5Fs.

Here a type II compensator is designed for the case

which has six electrolytic capacitors(1800uF, 13mΩ) and

two 1.5uH inductors.

1.Calculate the location of LC double pole F

LC

and ESR zero F .

ESR

Choose C1=4.7nF.

6.

Calculate

C

2

by

setting

compensator

pole

Fp

at half the swithing frequency.

1

C 2 = π × R 3 × Fs

=

1

π × 27.4kΩ × 100kH z

=116pF

Choose C2=100pF.

Rev.4.8

12

05/06/08

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