NX2124/2124A
Case 1:
F
LC
<F
O
<F
ESR
R
1
=
R
2
×
V
REF
10k
Ω ×
0.8V
=
=
8k
Ω
V
OUT
-V
REF
1.8V-0.8V
Choose R
1
=8kΩ.
Gain(db)
power stage
3. Set zero F
Z2
= F
LC
and F
p1
=F
ESR
.
4. Calculate R
4
and C
3
with the crossover
F
LC
40dB/decade
frequency at 1/10~ 1/5 of the switching frequency. Set
F
O
=30kHz.
C
3
=
loop gain
1
1 1
×
(
-
)
2
× π ×
R
2
F
z2
F
p1
F
ESR
20dB/decade
compensator
1
1
1
×
(
-
)
2
× π ×
10k
Ω
6.2kHz 60.3kHz
=2.3nF
=
R
4
=
=
V
OSC
2
× π ×
F
O
×
L
×
×
C
out
V
in
C
3
1.5V 2
× π ×
30kHz
×
1.5uH
×
×
440uF
5V
2.2nF
=16.9k
Ω
F
Z1
F
Z2
F
O
F
P1
F
P2
Choose C
3
=2.2nF, R
4
=16.9kΩ.
5. Calculate C
2
with zero F
z1
at 75% of the LC
double pole by equation (11).
Figure 4 - Bode plot of Type III compensator
Design example for type III compensator are in
order. The crossover frequency has to be selected as
F
LC
<F
O
<F
ESR
and F
O
<=1/10~1/5F
s.
Here two POSCAP
2R5TPE220MC(220uF,12 mΩ) are chosen as output
capacitor.
1.Calculate the location of LC double pole F
LC
and ESR zero F
ESR
.
C
2
=
=
1
2
× π ×
F
Z1
×
R
4
1
2
× π ×
0.75
×
6.2kHz
×
16.9k
Ω
=
2nF
Choose C
2
=2.2nF.
6. Calculate C
1
by equation (14) with pole F
p2
at
half the switching frequency.
C
1
=
1
1
2
× π ×
R
4
×
F
P2
F
LC
=
=
2
× π ×
L
OUT
×
C
OUT
1
2
× π ×
1.5uH
×
440uF
=
6.2kHz
1
2
× π ×
16.9k
Ω ×
150kHz
=
63pF
=
Choose C
1
=68pF.
7. Calculate R
3
by equation (13).
F
ESR
=
=
1
2
× π ×
ESR
×
C
OUT
R
3
=
=
1
2
× π ×
F
P1
×
C
3
1
2
× π ×
6m
Ω ×
440uF
=
60.3kHz
2. Set R
2
equal to 10kΩ.
Rev.1.8
02/28/08
1
2
× π ×
60.3kHz
×
2.2nF
=
1.2k
Ω
Choose R
3
=1.2kΩ.
9
MICROSEMI [ MICROSEMI CORPORATION ]
