Mar12-07 RevD

SP7663 Evaluation Board Manual

©2007 Sipex Corp.

Page 10 of 16

As a particular example, consider for the following SP7663EB with a Type II

Voltage Loop Compensation component selections:

Input requirements and inductor selection

Vin = 5 to 13.5V

Vout = 3.30V @ 0 to 6A load

Select L = 1.5uH => yield ≈ 48% of maximum 6A output current ripple.

Ω

Ω

Ω

)

fs = 600kHz SP7663 Internal Oscillator Frequency

Step-by-step design procedures:

Choose

fs/10

fco =

60Khz

600Khz/10

fco

=

=

Calculate fp_LC, the double pole frequency of the filter

)

C

L

(

2

1

fp_LC

⋅

=

π

Khz

2

.

7

153

.

7

330

5

.

1

2π

1

fp_LC

≈

=

⋅

⋅

=

Khz

uF

uH

Calculate fz_ESR, the ESR zero frequency

Cout

Cesr

2π

1

fz_ESR

⋅

⋅

=

hz

Khz

µF

m

π

ESR

fz

Κ

≈

=

⋅

⋅

=

14

77

.

13

)

330

(

)

35

(

2

1

_

Select R1 component value such that 50kΩ ≤ R1 ≤ 100kΩ

R1 = 68.1kΩ, 1%

Note: Type II Loop Compensation component calculations discussed in this

These calculations shown here can be quickly iterated with the Type III

Loop Compensation Calculator on the web at:

http://www.sipex.com/files/ApplicationNotes/Copy%20of%20Type%20II%20calculator5.xls