lt3467aes67 Linear Technology Corporation, lt3467aes67 Datasheet - Page 10

• Manufacturer Part Number: lt3467aes67
• Description: 1.1a Step-up Dc/dc Converter With Integrated Soft-start
• Manufacturer: Linear Technology Corporation

Compensation—Theory
Like all other current mode switching regulators, the
LT3467/LT3467A needs to be compensated for stable and
efficient operation. Two feedback loops are used in the
LT3467/LT3467A: a fast current loop which does not
require compensation, and a slower voltage loop which
does. Standard Bode plot analysis can be used to under-
stand and adjust the voltage feedback loop.
As with any feedback loop, identifying the gain and phase
contribution of the various elements in the loop is critical.
Figure 6 shows the key equivalent elements of a boost
converter. Because of the fast current control loop, the
power stage of the IC, inductor and diode have been
replaced by the equivalent transconductance amplifier g
g
proportional to the V
current of g
LT3467/LT3467A
APPLICATIONS
10
mp
R
V
acts as a current source where the output current is
C
C
C
C
C
C
g
g
R
R
R
R1, R2: FEEDBACK RESISTOR DIVIDER NETWORK
R
C
C
OUT
PL
ma
mp
C
L
O
ESR
: COMPENSATION CAPACITOR
: COMPENSATION RESISTOR
: OUTPUT RESISTANCE DEFINED AS V
: OUTPUT RESISTANCE OF g
: PHASE LEAD CAPACITOR
: TRANSCONDUCTANCE AMPLIFIER INSIDE IC
: POWER STAGE TRANSCONDUCTANCE AMPLIFIER
R
: OUTPUT CAPACITOR
: OUTPUT CAPACITOR ESR
Figure 6. Boost Converter Equivalent Model
O
mp
–
+
is finite due to the current limit in the IC.
g
mp
g
ma
C
U
voltage. Note that the maximum output
+
–
INFORMATION
ma
REFERENCE
U
1.255V
OUT
C
PL
DIVIDED BY I
W
3467 F06
R1
R2
LOAD(MAX)
R
ESR
C
OUT
U
R
L
V
OUT
mp
.
The Current Mode zero is a right half plane zero which can
be an issue in feedback control design, but is manageable
with proper external component selection.
From Figure 6, the DC gain, poles and zeroes can be
calculated as follows:
Output Pole: P1=
Error Amp Pole:
Error Amp Zero: Z1=
DC GAIN: A=
ESR Zero:
RHP Zero: Z3=
High Frequency Pole: P3>
Phase Lead Zero Z
Phase Lead Pol
Z
2
1.255
V
=
OUT
e e P
2
2
:
P2=
:
• •
2 • •
• •
2
π
V
4
4
π
I I N
•
=
π R
2
2
=
V
2
V
IN
• •
• •
R
2
OUT
2
•
π
π
ESR
R
• •
•
1
2
• •
L
L
f
g
π
3
2
π
S S
•
R
R
1
1
ma
•
•
C
O
C
C
C
L
R C
1
OUT
•
•
•
PL
OUT
1
R
C
C
1
•
C
C C
O
•
PL
R
•
R R
1
g
1
mp
+
•
R
2
•
2
R
L
•
3467afc
2 2
1