LTC3785-1 Linear Technology, LTC3785-1 Datasheet - Page 14

• Manufacturer Part Number: LTC3785-1
• Description: Buck-Boost Controller
• Manufacturer: Linear Technology

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APPLICATIONS INFORMATION
LTC3785-1
CLOSING THE FEEDBACK LOOP
The LTC3785-1 incorporates voltage mode control. The
control to output gain is given by:
The output fi lter exhibits a double-pole response and is
given by:
where C
The output fi lter zero is given by:
where R
A troublesome feature in boost mode is the right half plane
zero (RHP), and is given by:
The loop gain is typically rolled off before the RHP zero
frequency.
A simple type I compensation network (Figure 5) can be
incorporated to stabilize the loop but at a cost of reduced
bandwidth and slower transient response. To ensure proper
phase margin, the loop must cross over almost a decade
before the L-C double pole.
14
G
G
f
f
f
FILTER POLE
RHPZ
FILTER ZERO
Buck
BOOST
Figure 5. Error Amplifi er with Type I Compensation
ESR
OUT
=
=
_
_
=
2 • •
1 6
is the capacitor equivalent series resistance.
is the output fi lter capacitor.
. •
1 6
π
. •
=
=
V
V
I
IN
OUT
2 π
V
2 π
IN
ERROR
V
AMP
,
OUT
• •
• •
IN
Buck Mode
• •
2
+
–
2
L V
, ,
R
1
Boost Mode
ESR
L C
1
1.225V
OUT
FB
V
•
C
•
OUT
C
C
P1
OUT
V
OUT
R1
R2
37851 F05
The unity gain frequency of the error amplifi er with the
type 1 compensation is given by:
Most applications demand an improved transient response
to allow a smaller output fi lter capacitor. To achieve a higher
bandwidth, type III compensation is required as shown
in Figure 6. Two zeros are required to compensate for the
double pole response.
EFFICIENCY CONSIDERATIONS
The percentage effi ciency of a switching regulator is
equal to the output power divided by the input power
times 100%.
It is often useful to analyze individual losses to determine
what is limiting the effi ciency and which change would
produce the most improvement. Although all dissipative
elements in circuits produce losses, four main sources
account for most of the losses in LTC3785-1 application
circuits:
f
f
f
f
POLE
PO
ZERO
ZERO
f
L L E
UG
Figure 6. Error Amplifi er with Type III Compensation
1
2
2
1
=
≈
=
≈
=
2
2
2
2
2
• •
• •
• •
• •
• •
ERROR
π
π
π
AMP
π
π
R C
1
32 3
R
R
+
–
R C
1
1
1
1
Z
Z
1
e
•
•
1
•
•
C
C
P
•
1.225V
FB
V
P
Z
C
P
1
C
1
1
2
P
1
•
R
R
Z
1
C
(
P2
a very low frequen
C
P1
37851 F06
V
OUT
R1
R2
C
Z1
37851f
c c y
)