Rainbow Electronics MAX15046B - Manual (Page 17)

40V, High-Performance, Synchronous
Buck Controller
MAX15046
______________________________________________________________________________________ 17
The total loop gain, which is the product of the modulator
gain and the error-amplifier gain at f
O , is:
( )
MOD EA
IN FB
M F
OSC O OUT OUT
F
OSC O OUT OUT
F
FB IN M
1) GAIN GAIN 1
So :
V V ESR
g R 1
V (2 f L ) V
Solving for R :
V 2 f L V
R
V V g ESR
Ч =
Ч Ч Ч Ч =
π Ч Ч
Ч π Ч Ч Ч
=
Ч Ч Ч
2) Set a midband zero (f Z1 ) at 0.75 x f PO (to cancel one
of the LC poles):
= = ×
π Ч Ч
Z1 PO
F F
1
f 0.75 f
2 R C
Solving for C F :
=
π Ч Ч Ч
F
F PO
1
C
2 R f 0.75
3) Place a high-frequency pole at f P1 = 0.5 x f SW (to
attenuate the ripple at the switching frequency f SW )
and calculate C CF using the following equation:
=
π Ч Ч
CF
F SW
F
1
C
1
R f -
C
Type III Compensation Network
(See Figure 4)
When using a low-ESR tantalum or ceramic type, the
ESR-induced zero frequency is usually above the tar-
geted zero crossover frequency (f
O ). Use Type III com-
pensation. Type III compensation provides two zeros
and three poles at the following frequencies:
=
π Ч Ч
=
π Ч Ч +
Z1
F F
Z2
I 1 I
1
f
2 R C
1
f
2 C (R R )
Two midband zeros (f Z1 and f Z2 ) cancel the pair of com-
plex poles introduced by the LC filter:
f P1 = 0
f P1 introduces a pole at zero frequency (integrator) for
nulling DC output-voltage errors:
=
π Ч Ч
P2
I I
1
f
2 R C
Depending on the location of the ESR zero (f ZO ), use f P2
to cancel f ZO , or to provide additional attenuation of the
high-frequency output ripple:
=
×
π Ч Ч
+
P3
F CF
F
F CF
1
f
C C
2 R
C C
f P3 attenuates the high-frequency output ripple.
Place the zeros and poles such that the phase margin
peaks around f
O .
Ensure that R F >> 2/g M and the parallel resistance of
R 1 , R 2 , and R I is greater than 1/g M . Otherwise, a 180N
phase shift is introduced to the response making the
loop unstable.
Use the following compensation procedures:
1) With R
F >> 10kI, place the first zero (f Z1 ) at 0.8 x
f PO :
= = ×
π Ч Ч
Z1 PO
F F
1
f 0.8 f
2 R C
Figure 3. Type II Compensation Network
V REF
R 1
V OUT
R 2
g M
R F
COMP
C F C CF