NX2124/2124A
Case 2:
F
LC
<F
ESR
<F
O
2. Set R
2
equal to 10kΩ.
R
1
=
R
2
×
V
REF
10k
Ω ×
0.8V
=
=
8k
Ω
V
OUT
-V
REF
1.8V-0.8V
Gain(db)
power stage
F
LC
40dB/decade
Choose R
1
=8.06kΩ.
3. Set zero F
Z2
= F
LC
and F
p1
=F
ESR
.
4. Calculate C
3
.
C
3
=
1
1 1
×
(
-
)
2
× π ×
R
2
F
z2
F
p1
F
ESR
loop gain
1
1
1
×
(
-
)
2
× π ×
10k
Ω
2.3kHz 8.2kHz
=4.76nF
=
20dB/decade
compensator
Choose C
3
=4.7nF.
5. Calculate R
3
.
R
3
=
=
1
2
× π ×
F
P1
×
C
3
1
2
× π ×
8.2kHz
×
4.7nF
=
4.1k
Ω
F
Z1
F
Z2
F
P1
F
O
F
P2
Choose R
3
=4kΩ.
6. Calculate R
4
with F
O
=30kHz.
R
4
=
V
OSC
2
× π ×
F
O
×
L R
2
×
R
3
×
×
V
in
ESR
R
2
+
R
3
Figure 5 - Bode plot of Type III compensator
(F
LC
<F
ESR
<F
O
)
If electrolytic capacitors are used as output
capacitors, typical design example of type III
compensator in which the crossover frequency is
selected as F
LC
<F
ESR
<F
O
and F
O
<=1/10~1/5F
s
is shown
as the following steps. Here two SANYO 16MV-WG1500
with 13 mΩ is chosen as output capacitor.
1. Calculate the location of LC double pole F
LC
and ESR zero F
ESR
.
1.5V 2
× π ×
30kHz
×
1.5uH 10k
Ω ×
4k
Ω
×
×
5V
6.5m
Ω
10k
Ω +
4k
Ω
=37.3k
Ω
=
Choose R
4
=37.4kΩ.
7. Calculate C
2
with zero F
z1
at 75% of the LC
double pole by equation (11).
C
2
=
=
1
2
× π ×
F
Z1
×
R
4
1
2
× π ×
0.75
×
2.3kHz
×
37.4k
Ω
=
2.4nF
F
LC
=
=
1
2
× π ×
L
OUT
×
C
OUT
1
Choose C
2
=2.2nF.
8. Calculate C
1
by equation (14) with pole F
p2
at
half the switching frequency.
2
× π ×
1.5uH
×
3000uF
=
2.3kHz
C
1
=
=
1
2
× π ×
R
4
×
F
P2
F
ESR
1
=
2
× π ×
ESR
×
C
OUT
1
2
× π ×
6.5m
Ω ×
3000uF
=
8.2kHz
=
1
2
× π ×
37.4k
Ω ×
150kHz
=
28pF
Choose C
1
=27pF.
Rev.1.8
02/28/08
10
MICROSEMI [ MICROSEMI CORPORATION ]
