Rainbow Electronics MAX5098A - Manual (Page 22)

MAX5098A
Dual, 2.2MHz, Automotive Buck or Boost
Converter with 80V Load-Dump Protection
22 ______________________________________________________________________________________
bandwidth and phase margin. Use a simple pole-zero
pair (Type II) compensation if the output capacitor ESR
zero frequency is below the unity-gain crossover fre-
quency (f
C ). Type III compensation is necessary when
the ESR zero frequency is higher than f
C
or when com-
pensating for a continuous mode boost converter that
has a right-half-plane zero.
Use procedure 1 to calculate the compensation net-
work components when f
ZERO,ESR
< f
C
.
Buck Converter Compensation
Procedure 1 (See Figure 4)
1) Calculate the f ZERO,ESR and LC double-pole fre-
quencies:
2) Select the unity-gain crossover frequency:
If the f
ZERO,ESR
is lower than f
C
and close to f
LC
, use a
Type II compensation network where R F C F provides a
midband zero f MID,ZERO , and R F C CF provides a high-
frequency pole.
3) Calculate modulator gain G
M
at the crossover fre-
quency.
where V OSC is a peak-to-peak ramp amplitude equal to
1V.
The transconductance error amplifier gain is:
G E/A = g M x R F
The total loop gain at f C should be equal to 1:
G M x G E/A = 1
or
4) Place a zero at or below the LC double pole:
5) Place a high-frequency pole at f P = 0.5 x f SW .
Procedure 2 (See Figure 5)
If the output capacitor used is a low-ESR ceramic type,
the ESR frequency is usually far away from the targeted
unity crossover frequency (f
C ). In this case, Type III
compensation is recommended. Type III compensation
provides two-pole zero pairs. The locations of the zero
and poles should be such that the phase margin peaks
around f
C . It is also important to place the two zeros at
or below the double pole to avoid the conditional stabil-
ity issue.
1) Select a crossover frequency:
2) Calculate the LC double-pole frequency, f LC :
f
L C
LC
OUT OUT
=
Ч Ч
1
2 π
f
f
C
SW ≤
20
C
C
f R C
CF
F
SW F F
=
Ч Ч Ч
( )
− 2 0 5 1
π
.
C
R f
F
F LC
=
Ч Ч
1
2
π
R
V ESR f L V
V g ESR
F
OSC C OUT OUT
IN M
=
+ Ч Ч
( )
Ч
Ч Ч Ч
2
0 8
π
.
G
V
V
ESR
ESR f L V
M
IN
OSC C OUT OUT
= Ч
+ Ч Ч
( )
Ч
2
0 8
π
.
f
f
C
SW ≤
20
f
ESR C
f
L C
ZERO ESR
OUT
LC
OUT OUT
,
=
Ч Ч
=
Ч
1
2
1
2
π
π
R
1
FB_
R F
COMP_
V OUT
V REF
C
CF
C
F
R
2
-
+
g M
Figure 4. Type II Compensation Network