LT3467 LINER [Linear Technology], LT3467 Datasheet - Page 10

• Manufacturer Part Number: LT3467
• Description: 1.1A Step-Up DC/DC Converter with Integrated Soft-Start
• Manufacturer: LINER [Linear Technology]

APPLICATIONS INFORMATION
LT3467/LT3467A
Compensation—Theory
Like all other current mode switching regulators, the
LT3467/LT3467A needs to be compensated for stable
and effi cient 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 re-
placed by the equivalent transconductance amplifi er g
g
proportional to the V
output current of gmp is fi nite due to the current limit
in the IC.
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
ma
mp
C
OUT
PL
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
O
Figure 6. Boost Converter Equivalent Model
–
+
g
mp
g
ma
+
–
C
voltage. Note that the maximum
ma
REFERENCE
1.255V
OUT
C
PL
DIVIDED BY I
3467 F06
R1
R2
LOAD(MAX)
R
ESR
C
OUT
R
L
V
OUT
mp
.
From Figure 6, the DC gain, poles and zeroes can be
calculated as follows:
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.
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
V
1.255
=
OUT
e e P
2
2
:
:
P2=
• •
2 • •
• •
2
π
V
4
4
π
I I N
•
=
π R
=
2
2
V
2
V
R
IN
• •
• •
2
OUT
2
•
ESR
π
π
• •
R
1
• •
2
•
L
π
L
π
g
f
2
3
S S
•
R
R
1
1
ma
•
•
C
O
C
C
C
R C
L
1
OUT
PL
•
OUT
•
•
1
C
C
R
•
1
C C
•
C
O
PL
R
R R
•
1
1
g
+
mp
•
R
2
2
•
R
L
•
3467afd
2 2
1