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LTC3113 数据表(PDF) 17 Page - Linear Technology

部件名 LTC3113
功能描述  3A Low Noise Buck-Boost DC/DC Converter
PDF  24 Pages
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制造商  LINER [Linear Technology]
网页  http://www.linear.com
标志 LINER - Linear Technology

LTC3113 数据表(HTML) 17 Page - Linear Technology

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LTC3113
17
3113f
APPLICATIONS INFORMATION
CLOSING THE FEEDBACK LOOP
The LTC3113 incorporates voltage mode PWM control.
The control-to-output gain varies with the operation region
(buck, buck-boost, boost), but is usually no greater than
15. The output filter exhibits a double pole response, as
given by:
f
FILTER _ POLE =
1
2
π LC
OUT
Hz
()
In Buck Region
()
f
FILTER _ POLE =
1
2
π LC
OUT
V
IN
V
OUT
Hz
()
In Boost Region
()
where L is in Henries and COUT is in Farads. The output
filter zero is given by:
f
FILTER _ ZERO =
1
2
πR
ESRCOUT
Hz
()
where RESR is the equivalent series resistance out the
output capacitor in ohms.
A troublesome feature in the boost and buck-boost region
is the right-half plane (RHP) zero, given by:
f
RHPZ =
V
IN
2
2
πI
OUTLVOUT
Hz
()
The loop gain is typically rolled off before the RHP zero
frequency.
A simple Type I compensation network can be incorporated
to stabilize the loop at the cost of reduced bandwidth and
slower transient response. To ensure proper phase margin
using Type I compensation, the loop must be crossed over
a decade before the LC double pole. Referring to Figure 4,
the unity-gain frequency of the error amplifier with the
Type I compensation is given by:
f
UG =
1
2
πR2C
P1
Hz
()
+
ERROR
AMP
0.6V
R1
R2
FB
VOUT
VC
3113 F04
CP1
Figure 4. Error Amplifier with Type I Compensation
+
ERROR
AMP
0.6V
R1
R2
RP
FB
VOUT
VC
3113 F05
CP2
CZ1
RZ
CP1
Figure 5. Error Amplifier with Type III Compensation
Most applications demand an improved transient response
to allow a smaller output capacitor. To achieve a higher
bandwidth, Type III compensation is required, providing
two zeros to compensate for the double-pole response of
the output filter. Referring to Figure 5, the location of the
poles and zeros are given by:
f
POLE1 =
1
2
π10
5R2C
P1
Hz
()
f
ZERO1 =
1
2
πR
Z CP1
Hz
()
f
ZERO2 =
1
2
πR2C
Z1
Hz
()
f
POLE2 =
1
2
πR
Z CP2
Hz
()
f
POLE3 =
1
2
πR
P CZ1
Hz
()
where resistance is in Ohms and capacitance is in
Farads.



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