Design Procedures
2-8
The sum of the gains of the modulator, the LC filter, and the error amplifier
needs to be 0 dB at the selected unity-gain frequency of 20 kHz. The modulator
and LC filter gain is –14 dB. The two zeroes at 1.87 kHz in the compensation
network that cancels the LC poles will have a total gain of 41.2 dB at 20 kHz.
Therefore, the pole at zero frequency needs to furnish 0–(–14+41.2) =
–27.2 dB (voltage gain = 0.04365) at 20 kHz. R5 and C12 provide this pole.
R6 is already chosen as 4 k
W
. Calculate C12 as:
C12
)
C11
+
1
(2
p
)(f)(R6)(Required Gain)
In practice C12 is much greater than C11, therefore:
C12
+
1
(2
p
)(20 kHz)(4 k
W
)(0.04365)
+
0.045
m
F Use C12
+
0.047
m
F
R4 provides the first zero at the LC break point:
R4
+
1
(2
p
)(1.87 kHz)(C12)
+
1.89 k
W
Use R4
+
1.8 k
W
C13 provides the other zero at the LC break point:
C13
+
1
(1.87 kHz)
*
1
(20 kHz)
2
p
(R6)
+
0.019
m
F
Use C13
+
0.018
m
F
R5 provides the compensation for the ESR zero:
R5
+
1
(2
p
)(26.8 kHz)(C13)
+
330
W
Finally, C11 provides a rolloff filter at high frequency, chosen at 100 kHz:
C11
+
1
(2
p
)(100 kHz)(R4)
+
0.00088
m
F
Use C11
+
1000 pF
