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PDF AN021 Data sheet ( Hoja de datos )
| Número de pieza | AN021 | |
| Descripción | A Handy Method to Obtain Satisfactory Response of Buck Converter | |
| Fabricantes | AIC | |
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AN021
A Handy Method to Obtain Satisfactory Response of
Buck Converter
Introduction
The focus of this application note is to help users to
select a good set of components of the compensation
section of the Buck converter. Traditionally, engineers
first concern about the power circuit for a circuit design.
After meeting the power requirement, what follows is
control loop design. The first priority of control design is
the stability. And after that, the choice of components
can substantially effect the transient response. Here
shows a simple method to obtain a satisfactory
transient response with an acceptable steady state
error.
Analysis:
Fig.1 shows the unity feedback system for Buck
converter. We first identify transfer functions for each
of the corresponding block.
As to the modeling of the low-frequency behavior of
power switches in square-wave power converters,
please refer to appendix [1]. The circuit of Buck
converter is shown in Fig.2 and the model of its power
switches is shown in Fig.3. Please note that the circuit
in Fig.3 is linear. Fig.4 shows the circuit for small signal
analysis.
+
dVOUT(s)=0
_
Gc(s)
PWM
Gp(s) dVOUT(s)
dVc(s)
dD(s)
Fig.1 The Unity Feedback Control Loop for Buck Converter
R1
VIN
R2
VOUT
+
Fig. 2 Buck Converter
October, 2001
1
1 page  
AN021
Design Example
Design example:
VIN=8V,VOUT=5V, the operation point of output current
is 1A
L=27µH, rL=38.5mΩ
C=1000µH, rC=52mΩ
MOSFET=IR3103, RDS-ON=14 mΩ
RL=5Ω
VM=1.3V
Design Procedure:
1. Construct the model of Buck converter and obtain
the transfer function.
27µH 38.5m
dVOUT(s)
14m
S
VIN *
52m
+
5
1000µF
Fig. 9 The Model of the Given Buck Converter
response are shown in Fig. 10 and Fig. 11, respectively.
Obviously, the rising time is too long and the overshoot
is too large. To increase the speed of response and to
have more phase margin, we try to extend the
crossover frequency.
Fig. 10 The Transient Response (Ch2 is Current
Curve, 1A/div)
dVOUT (s)
dD(s)
=
8
×
1.93k(s + 19.2k )
s 2 + 4.07ks + 37M
…
…
…
…
…
..(6)
From equation (6), the location of zero and the
complex conjugate poles can be easily obtained.
2. Locate the zero and the second pole
compensation network.
Since ωn= 6.08k,
zero of Buck= 19.2k,
fSwitching= 200kHz= 1.25M(rad/s)
Let zero of compensation network=2.04k
Pole of compensation network=250k
of
3. Choose the components:
According to the equation (4),
Let R2=75k then C2=6800p, and C3=56p
The transient response and the simulation of frequency
Fig. 11
The Corresponding Simulation of
Frequency Response
According to equation (4), we can add the extra gain
which is provided by the compensation network at the
middle frequency without changing the location of zero
and pole of the original loop transfer function.
Let R2=560k, then
R2C2=
1
2.04k
C2=875 ≈ 820p
October, 2001
5
5 Page | ||
| Páginas | Total 7 Páginas | |
| PDF Descargar | [ Datasheet AN021.PDF ] | |
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