R2J20751NP

Preliminary

R07DS0240EJ0100 Rev.1.00

Page 13 of 25

Jan 26, 2011

equation (3),

=

13700 / 820

Ω × 470 nH × 5 V × 500 kHz

− 8 × 470 nH × 5 V × 500 kHz × (0.134 V × 13700 / 820 Ω) }

19.63

=

SQRT {3.955}

= 9.871

A0 =

……(3)

Nt/RCS

× L × Vin × Fsw

− 8 × L × Vin × Fsw × (VCS0 × Nt / RCS) }

The frequency of the pole established by the power stage and modulator is

F0 = Nt / (2

  Co  RCS  A0) ......(5)

Thus,

F0 = 13700 / (2

  600 F  820   9.871) = 448.967 Hz

Thus,

Fzero = 10

 F0 = 4.489 kHz

Cf = (2

Therefore, we select 1500 pF for Cf.

Basically, the transient response is faster when Cf is smaller, but too small a value will make the system-loop

unstable.

F0

Fzero

BW/Af

Error amp. unity gain

frequency

BW

Af

A0

Fsw

Converter open loop

Compensated error amp.

Power and modulator

Gain (dB)

Freq. (Hz)

−20 dB/dec

−40 dB/dec

−20 dB/dec

Asw

Figure 4