# AN012 A Handy Method to Obtain Satisfactory Response of Buck Converter

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```					                                                                                                                     AN012

A Handy Method to Obtain Satisfactory Response of
Buck Converter

Introduction                                                        Analysis:

The focus of this application note is to help users to
Fig.1 shows the unity feedback system for Buck
select a good set of components of the compensation
converter. We first         identify transfer functions for each
section of the Buck converter. Traditionally, engineers
of the corresponding block.
first concern about the power circuit for a circuit design.
After meeting the power requirement, what follows is
As to the modeling of the low-frequency behavior of
control loop design. The first priority of control design is
power switches in square-wave power converters,
the stability. And after that, the choice of components
please refer to appendix [1]. The circuit of Buck
can substantially effect the transient response. Here
converter is shown in Fig.2 and the model of its power
shows a simple method to obtain a satisfactory
switches is shown in Fig.3. Please note that the circuit
transient response with an acceptable steady state
in Fig.3 is linear. Fig.4 shows the circuit for small signal
error.
analysis.

+                                                   dVOUT(s)
Gc(s)            PWM            Gp(s)
dVOUT(s)=0   _              dVc(s)          dD(s)

Fig.1 The Unity Feedback Control Loop for Buck Converter

R1                                           VOUT

+
VIN
R2

Fig. 2    Buck Converter

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VOUT

+
R
VIN

D* IL                          D*

Fig.3 R = DR1 + (1 − D)R 2

dVOUT(s)

L                rL
rc
R                                          RL
+
S
VIN *

Fig. 4        Small Signal Circuit

The transfer function of output with respect to duty ratio is:
1
1            1
+
RL                 1
rC +
dVOUT (s)                                       sc
= VIN ×
dD(s)                                               1
sL + (R + rL ) +
1              1
+
RL                   1
rC +
sc
… … ..(1)
(CRL rC )s + R L
= VIN ×
L(R L C + rC C)s + [L + C(RR L + rL R L + RrC + rCrL + R L rC )]s + (R L + R + rL )
2

where:                                                                              when RL is >> rL , rC and R
VIN: input voltage                                                                  Equation 1 can be simplified as below:
VOUT: output voltage                                                                dVOUT (s)                           rC Cs + 1
= VIN ×
R: the equivalent resistance of power switches.                                      dD(s)                           L
LCs 2 + [    + C(R + rL + rC )]s + 1
L: the inductor                                                                                                      RL
1
C: the output capacitor                                                                                           s+
V ×r                       rC C
rL: the DC resistance of inductor.                                                  = IN C ×
L                1    R + rL + rC       1
rC: the ESR of output capacitor.                                                                 s2 + [     +             ]s +
CRL        L           LC
… … … … … … … … … … … … ..(2)

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1                           According to its frequency response in Fig. 7, the
where     the zero ZP is at
rC C                         network has high gain at low frequency, which helps to

1                                   reduce the steady state error. The attenuation of gain
the ω n =                                        at high frequency helps to weaken the noise
LC
disturbance.
C3
R2        C2

dVout(s)   R1                _
dVc(s)
+

Fig. 6     The Compensation Network

Fig. 5   The    Simulation      of   Typical   Frequency
REsponse of Buck Converter

Fig. 7    The       Simulation        of   Typical    Frequency
Fig. 5 represents the simulation of typical frequency                   Response of the Compensation Circuit
response of Buck converter. Note that the effect of
complex conjugate poles of LC will make the gain           Its transfer function is:
curve at -40dB/decade and phase curve towards -180°.          dVC (s)             1+ R 2C2s
And after the region around ωn, we will meet the zero,                =                                       ×β … …
dVOUT (s) sR1 (C 2 + C 3 )[1 + sR 2 (C 2 //C 3 )
which is donated by the E.S.R of output capacitor. The
… … … … … … … … … (4)
gain curve becomes -20dB/decade and the phase is
Where:
towards -90°.
the first pole P0 is at zero frequency (ω=0).
1
As to the PWM (Pulse Width Modulation), its transfer              one zero Z1 is at
R 2C 2
function is
dD(s)   1                                                        the 2nd pole P2 is at
=   … … … … … … … … … … ..(3)
dVC (s) VM                                                       1             1
≈      ( when C3<<C2 )
R 2 (C 2 //C 3 ) R 2C3
where VM is the amplitude of ramp in PWM.
1
gain at low frequency: GC (s ) =
About the compensation network, we choose the type                                                            sR1(C 2 + C 3 )
that is shown in Fig. 6.                                          gain at middle frequency:

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R 2C 2                                                               the origin pole is already set. Then here comes the
GC (s ) =                  … ..(Constant)
R1 (C 2 + C3 )                                                           strategy for the location of the zero and the second

1                         pole.
gain at high frequency: GC (s ) =
sR1C 3
For the reason of stability, the zero of compensation
β is the gain of the feedback resistor divider.
network must be set lower than the zero of Buck
Note that due to the origin pole, the phase at low
frequency is -90° and the gain curve is at -20dB/                                    converter. Without compensation zero, the zero of

decade. When the frequency approaches to zero, the                                   Buck converter would face three poles before it. The
phase increases toward 0° with a flat gain curve. When                               phase of the loop transfer function falls down toward
it moves toward the second pole, the frequency will be                               -270° and passes through -180° between the poles of
towards -90° and the gain curve goes back to -20dB/                                  LC and the zero of Buck converter. This situation can
lead to an unstable condition. To avoid that, the
compensation circuit zero is advised to be lower than
Control Strategy
that of Buck converter. However the location of the
The loop transfer function of Fig.1 is shown as below:                               compensation zero will affect the phase margin around
T(s) = G C (s) × PWM × G P (s)                                                       the crossover frequency. To improve the phase margin
1
R 2 C 2 (s +       )                                              of the loop transfer function and keep high gain at low
R 2C2
=                                                                                    frequency, the compensation zero is suggested
1
sR 1 (C 2 + C 3 )R 2 (C 2 //C 3 )(s +
R s (C 2 //C 3 )                                                                  1
between ωn and 0.1 × ωn. (ωn=                ).
1                                                                                      LC
rC (s +      )
1                              rC C
×ß ×    × VIN ×                                                                         The second pole of compensation network is
VM                    1      R + rC + rL       1
L[s 2 + (      +              )s +    ]                              suggested far from the zero of Buck converter. In this
CR L         L           LC
way, we can obtain the –20dB/decade slope around
.… … … (5)
the crossover frequency. According to the sampling
After meeting the power circuit requirement, we start
theorem, the second pole of compensation network is
to design the control loop. Therefore the LC poles and
suggested to be under one fifth of switching frequency
output capacitor zero are already allocated. What we
can do is to distribute the location of pole and zero of                                                        1
of Buck converter (          f Switching ) to reduce the noise at
5
compensation         network       to     achieve             the   desired
response demand. To obtain high gain at low frequency,                               high frequency.

+                                                           dVOUT(s)
Gc(s)            PWM             Gp(s)
dVOUT(s)=0        _                 dVc(s)           dD(s)

Fig. 8      The Unity Feedback Control Loop for Buck Converter

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Design Example                                                      response are shown in Fig. 10 and Fig. 11, respectively.
Obviously, the rising time is too long and the overshoot
Design example:
is too large. To increase the speed of response and to
VIN=8V,VOUT=5V, the operation point of output current
have more phase margin, we try to extend the
is 1A
crossover frequency.
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.

dVOUT(s)
27µH    38.5m

14m                         52m                     Fig. 10    The Transient Response (Ch2 is Current
5
+
S                                                                 Curve, 1A/div)
VIN *                   1000µF

Fig. 9      The Model of the Given Buck Converter

dVOUT (s)      1.93k (s + 19.2k )
= 8× 2                  … … … … … ..(6)
dD(s)        s + 4.07ks + 37M
From equation (6), the location of zero and the
complex conjugate poles can be easily obtained.

2. Locate        the    zero   and     the    second    pole   of
compensation network.                                                Fig. 11     The Corresponding Simulation of

Since ωn= 6.08k,                                                                  Frequency Response
zero of Buck= 19.2k,
fSwitching= 200kHz= 1.25M(rad/s)                                 According to equation (4), we can add the extra gain

Let        zero of compensation network=2.04k                     which is provided by the compensation network at the
Pole of compensation network=250k                           middle frequency without changing the location of zero
and pole of the original loop transfer function.

3. Choose the components:                                               Let R2=560k, then
1
According to the equation (4),                                     R2C2=                       C2=875 ≈ 820p
Let R2=75k then C2=6800p, and C3=56p                                       2.04k
The transient response and the simulation of frequency

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1                                        idea to verify the steady state error by the load
R2C3=         … … … … … … . C3=7.14 ≈ 8p
250k                                       regulation. From Fig.14, it shows that the steady state
error of two settings is small, and the two curves are
almost identical.

5.1

5.05

Vout(V)
5

C2=820p R2=560k
4.95                                     C3=8p
C2=6800p R2=75k
C3=56p

4.9
0     2      4        6       8       10       12
Iout(A)

Fig.12      The Transient Response
(Ch2 is current curve, 1A/div)                    Fig.14           The Comparison of Load Regulation of
Different Components (VIN=8V, VOUT=5V)

From the results of transient response and the steady
state error, the second set of components can achieve
a satisfactory response.

Summary
By taking the advantage of a high gain at low
frequency with a low gain at high frequency, the
best performance of the Buck Converter can be
achieved. The objectives are to improve the steady
Fig.13       The Corresponding Simulation of
state error at low frequency and reduce noise
Frequency Response
disturbance at high frequency.
The key points of the compensation are:
Referring to Fig. 13, the crossover frequency of open
1) Locate the zero around the LC resonant
loop transfer function is raised. And we could also
frequency for the issue of stability.
expect the bandwidth of the closed loop transfer
function being increased. Indeed, in Fig 12, the set of                                             1
2) Locate the pole around           f Switching for the noise
5
control components, which we just used, shows high
speed of response and less overshoot.                                    reduction at high frequency.
3) The combination of the above 2 procedures can
Although the pole at origin (according to equation (5))                  make the crossover frequency at the -20dB/
means the infinitive gain at DC (ω=0), it is still a good                decade situation. Thus a satisfactory phase

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margin is achieved.

Reference:
[1] Yim-Shu Lee, Computer-Aided Analysis and Design
of Switch-Mode Power Supplies., Marcel Dekker, Inc.
Hong Kong,1993

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