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Huang-Jen Chiu

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F<br />

1<br />

Fco = Fs<br />

5<br />

z<br />

F<br />

F<br />

co<br />

esr<br />

EA<br />

=<br />

Compensator Design Example<br />

20 k<br />

= = 8 ⇒ lag = 97°<br />

2.<br />

5k<br />

F<br />

K<br />

lag<br />

co<br />

=<br />

F p = K × Fco<br />

=<br />

20 kHz<br />

= 360 − 45 − 97 = 218 ° ⇒ K =<br />

20<br />

4<br />

=<br />

= 5kHz<br />

⇒<br />

C1<br />

=<br />

4<br />

1<br />

2π<br />

( 100 k )( 5k<br />

)<br />

=<br />

318<br />

pF<br />

1<br />

4 × 20 = 80 kHz ⇒ C 2 =<br />

=<br />

2π<br />

( 100 k )( 80 k )<br />

20<br />

pF

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