Datasheet

LT3579/LT3579-1
30
35791fa
For more information www.linear.com/LT3579
As with any feedback loop, identifying the gain and phase
contribution of the various elements in the loop is critical.
Figure 18 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power stage of the IC, inductor and diode have been replaced
by a combination of the equivalent transconductance
amplifier g
mp
and the current controlled current source
(which converts I
VIN
to
ηV
IN
V
OUT
I
VIN
). G
mp
acts as a current
source where the peak input current, I
VIN
, is proportional
to the V
C
voltage.
APPENDIX
1.215V
REFERENCE
I
VIN
η
•
V
IN
V
OUT
•
I
VIN
V
OUT
C
OUT
C
PL
R
ESR
R
L
R
O
R
C
C
C
C
F
R1
FB
R2
R2
–
+
–
+
35791 F18
g
mp
g
ma
C
C
: COMPENSATION CAPACITOR
C
OUT
: OUTPUT CAPACITOR
C
PL
: PHASE LEAD CAPACITOR
C
F
: HIGH FREQUENCY FILTER CAPACITOR
g
ma
: TRANSCONDUCTOR AMPLIFIER INSIDE IC
g
mp
: POWER STAGE TRANSCONDUCTANCE AMPLIFIER
R
C
: COMPENSATION RESISTOR
R
L
: OUTPUT RESISTANCE DEFINED AS V
OUT
/I
LOADMAX
R
O
: OUTPUT RESISTANCE OF gma
R1, R2; FEEDBACK RESISTOR DIVIDER NETWORK
R
ESR
: OUTPUT CAPACITOR ESR
η
: CONVERTER EFFICIENCY (~90% AT HIGHER CURRENTS)
Figure 18. Boost Converter Equivalent Model
Note that the maximum output currents of g
mp
and g
ma
are finite. The output of the g
mp
stage is limited by the
minimum switch current limit (see Electrical Specifications)
and g
ma
is nominally limited to about ±12μA.
From Figure 18, the DC gain, poles and zeros can be
calculated as follows:
DC Gain:
A
DC
= g
ma
•R
O
•g
mp
• η•
V
IN
V
OUT
•
R
L
2
•
0.5R
2
R
1
+ 0.5R
2
Output Pole:P1=
2
2• π•R
L
•C
OUT
Error AmpPole:P2 =
1
2• π• R
O
+R
C
( )
•C
C
Error Amp Zero: Z1=
1
2• π•R
C
•C
C
ESR Zero: Z2 =
1
2• π•R
ESR
•C
OUT
RHP Zero: Z3 =
V
IN
2
•R
L
2• π• V
OUT
2
•L
High Frequency Pole:P3>
f
S
3
Phase Lead Zero: Z4 =
1
2• π•R
1
•C
PL
Phase LeadPole:P4 =
1
2• π•
R
1
•0.5R
2
R
1
+ 0.5R
2
•C
PL
Error AmpFilter Pole:
P5 =
1
2• π•
R
C
•R
O
R
C
+R
O
C
F
,C
F
<
C
C
10