pwm buck converter using average model
DESCRIPTION
This document is Conceptkit of PWM Buck Converter using Average Model using PSpice. pre-version, Bee Technologies prepare to products now.TRANSCRIPT
Concept Kit:PWM Buck Converter Average Model
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1
Power Switches Filter & LoadPWM Controller (Voltage Mode Control)
VREF
VOUT
REF
PWM
1 / V p
-
+
U ?P W M _ C TR L
V P = 2 . 5V R E F = 1 . 2 3
D
U ?B U C K _ S W
L1 2
C
R lo a d
V o
E S R
Pre Version
Contents
1. Concept of Simulation
2. Buck Converter Circuit
3. Switches
4. Filter & Load
4.1 Inductor
4.2 Capacitor
5. PWM Controller
5.1 Error Amp.
5.2 PWM
6. Stabilizing the Converter (Example)
7. Load Transient Response Simulation (Example)
Type 2 Compensator Calculator
Simulation Index
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Power Switches
Averaged Buck Switch Model
Filter & Load
Parameter:• L• C• ESR• Rload
PWM Controller (Voltage Mode Control)
Parameter:• VP
• VREF
Models:
Block Diagram:
1.Concept of Simulation
VREF
VOUT
D
U ?B U C K _ S W
REF
PWM
1 / V p
-
+
U ?P W M _ C TR L
V P = 2 . 5V R E F = 1 . 2 3
L1 2
C
R lo a d
V o
E S R
L1 2
C
R lo a d
0
C o m p
C 2
R 2 C 1
F B
Type 2 Compensator
R u p p e r
R lo we r
0
d
V inD
U 2B U C K _ S W
REF
PWM
1 / V p
-
+
U 3P W M _ C TR L
V P = 2 . 5V R E F = 1 . 2 3
V o
E S R
2.Buck Converter Circuit
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4
Filter & Load
PWM Controller
Power Switches
3.Switches
• The Averaged Buck Switch Model represents relation between input and output of the switch that is controlled by duty cycle – d (value between 0 and 1).
• Transfer function of the model is
vout = d vin
• The current flow into the switch is
iin = d iout
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D
U 2B U C K _ S W
vin
+
-
vout
+
-D
iin iout
4.1 Filter & Load: Inductor
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6
Inductor Value• The output inductor value is selected to set the
converter to work in CCM (Continuous Current Mode) or DCM (Discontinuous Current Mode).
• Calculated by
Where
• LCCM is the inductor that make the converter to work in CCM.
• VI,max is input maximum voltage
• RL(max) is load resistance at the minimum output current (IOUT)
• fosc is switching frequency
L1 2
C
R lo a d
V o
E S R
max,
(max)max,
2 Iosc
LOICCM
Vf
RVVL
(1)
4.2 Filter & Load: Capacitor
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7
Capacitor Value• The minimum allowable output capacitor value should
be determined by
Where
• IL, RIPPLE is an inductor ripple current, chosen to be 25% of IOUT.
• VO,RIPPLE is an output ripple voltage.
• fosc is switching frequency
• In addition, the output ripple voltage due to the capacitor ESR must be considered as the following equation.
L1 2
C
R lo a d
V o
E S R
RIPPLEOosc
RIPPLEL
Vf
IC
,
,
8(min)
RIPPLEL
RIPPLEO
I
VESR
,
,
(2)
(3)
• The Error Amp. compares the feedback voltage ( FB ) to the reference voltage ( Parameter: VREF ), the output signal will be fed back to the controller to regulate the converter output voltage as the above equation.
5.1 PWM Controller: Error Amp.
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C o m p
C 2
R 2 C 1
Type 2 Compensator
F B
R u p p e r
R lo we r
0
d
REF
PWM
1 / V p
-
+
U 3P W M _ C TR L
V P = 2 . 5V R E F = 1 . 2 3
Error Amp.
Vo
lower
upperREFOUT
R
RVV 1 (4)
TimeV(PWM)
V(osc) V(comp)0V
2.0V
3.0V
SEL>>
5.2 PWM Controller: PWM
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• The PWM block is used to transfer the error voltage (between FB and REF) to be the duty cycle.
• The error voltage (vcomp) will be compared with
sawtooth signal ( amplitude = VP ) to create the
pulse that the duty cycle depends on the vcomp
• Transfer function of the PWM block is
VP
Duty cycle (d) is a value from 0 to 1
d = vcomp / VP
GPWM = 1/VP
REF
PWM
1 / V p
-
+
U ?P W M _ C TR L
V P = 2 . 5V R E F = 1 . 2 3
vcomp
d
Error Amp.
FB
• Loop gain for this configuration is
L1 2
R lo a d
C
0
C o m p
C 2
R 2 C 1
Type 2 Compensator
F B
R u p p e r3 . 0 6 6 k
R lo we r1 . 0 k
0
d
V in1 2 V d c
D
U 2B U C K _ S W
REF
PWM
1 / V p
-
+
U 3P W M _ C TR L
V P = 2 . 5V R E F = 1 . 2 3
V o
E S R
• The purpose of the compensator G(s) is to tailor the converter loop gain (frequency response) to make it stable when operated in closed-loop conditions.
5.3 PWM Controller: Compensator
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PWMGsGsHsT )()()( GPWM
G(s)
H(s)
6.Stablilizing the Converter (Example)
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11
Specification:VOUT = 5VVIN = 7 ~ 40VILOAD = 0.2 ~ 1A
L = 330uH, C = 330uF (ESR = 100m),Rupper = 3.1k,Rlower = 1k,
PWM Controller:fOSC = 52kHzVP = 2.5VVREF = 1.23V
Task: to find out the element of the Type 2 compensator ( R2, C1, and C2 )
L3 3 0 u H
1 2
C3 3 0 u F
R lo a d5
0
0
C O L1 k F
L O L
1 k H
C 2
R 2 C 1
F B
R u p p e r3 . 1 k
Type 2 Compensator
R lo we r1 . 0 k
0
d
V 31 V a c0 V d c
V in1 2 V d c
D
U 2B U C K _ S W
REF
PWM
1 / V p
-
+
U 3P W M _ C TR L
V P = 2 . 5V R E F = 1 . 2 3
V o
E S R1 0 0 m
G(s)
e.g. Characteristics from National Semiconductor Corp. IC: LM2575
L3 3 0 u H
1 2
C3 3 0 u F
R lo a d5
0
0
C O L1 k F
L O L
1 k H
R 20 . 7 5 6 k
F B
R u p p e r3 . 1 k
Type 2 Compensator
R lo we r1 k
0
d
V 31 V a c0 V d c
V in1 2 V d c
D
U 2B U C K _ S W
REF
PWM
1 / V p
-
+
U 3P W M _ C TR L
V P = 2 . 5V R E F = 1 . 2 3
V o
E S R1 0 0 m
C 21 f
C 11 k
6.Stablilizing the Converter (Example)
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Step2 Set C1=1kF, C2=1fF, and R2= calculated value (Rupper//Rlower) as the initial values.
Step1 Open the loop with LoL=1kH and CoL=1kF then inject the ac signal to generate Bode plot.
The element of the Type 2 compensator ( R2, C1, and C2 ), that stabilize the converter, can be extracted by using Type 2 Compensator Calculator (Excel sheet) and open-loop simulation with the average models (ac models).
6.Stablilizing the Converter (Example)
Type 2 Compensator Calculator
Switching frequency, fosc : 52.00 kHzCross-over frequency, fc (<fosc/4) : 10.00 kHzRupper : 3.1 kOhmRlower : 1 kOhmR2 (Rupper//Rlower) : 0.756 kOhm (automatically calculated)
PWMVref : 1.230 VVp (Approximate) : 2.5 V
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Step3 Select a crossover frequency (fc < fosc/4), for this example, 10kHz is selected. Then complete the table.
Calcuted value of the Rupper//Rlower
from
d = vcomp / VP
Suppose that the error amp. gain is 100.
vcomp = gain (-vFB) then
d = (100 (-vFB) ) / VP
From the graph on the left, vFB = -25mV
VP = (100 (-vFB) ) / dVP ≈ (100 (-(-25mV)) ) / 1
≈ 2.5V
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If the VP ( sawtooth signal amplitude ) does not informed by the datasheet, It can be approximate from the characteristics below.
LM2575: Feedback Voltage vs. Duty Cycle
6.Stablilizing the Converter (Example)
vFB = -25mV
d = 1 (100%)
Parameter extracted from simulationSet: R2=R1, C1=1k, C2=1fGain (PWM) at foc ( - or + ) : -44.211Phase (PWM) at foc : 65.068
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Frequency
100Hz 1.0KHz 10KHz 100KHzP(v(d))
0d
90d
180d
SEL>>
(10.000K,65.068)
DB(v(d))
-80
-40
0
40
80
(10.000K,-44.211)
Step4 Read the Gain and Phase value at the crossover frequency (10kHz) from the Bode plot, Then put the values to the table .
6.Stablilizing the Converter (Example)
Tip: To bring cursor to the fc = 10kHz type “ sfxv(10k) ” in Search Command.
Cursor Search
Gain: T(s) = H(s)GPWM
Phase at fc
K-factor (Choose K and from the table)K 6q -199 ° (automatically calculated)
Phase margin : 46 (automatically calculated)
R2 : 122.780 kOhm (automatically calculated)C1 : 0.778 nF (automatically calculated)C2 : 21.600 pF (automatically calculated)
6.Stablilizing the Converter (Example)
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Step5 Select the desired amount of phase margin you need at fc ( > 45 ). Then change the K value until it gives the satisfied phase margin, for this example K=25 is chosen for Phase margin = 46.
R2, C1, and C2 are calculated
K Factor, introduce by Dean Venable, enable the circuit designer to choose a loop cross-over frequency and phase margin, and then determine the necessary component values to achieve these results from a few straight-forward algebraic equations.
R 21 2 2 . 7 8 0 k
Type 2 Compensator
C 22 1 . 6 p
C 10 . 7 7 8 n
L3 3 0 u H
1 2
C3 3 0 u F
R lo a d5
0
0
C O L1 k F
L O L
1 k H
F B
R u p p e r3 . 1 k
R lo we r1 k
0
d
V 31 V a c0 V d c
V in1 2 V d c
D
U 2B U C K _ S W
REF
PWM
1 / V p
-
+
U 3P W M _ C TR L
V P = 2 . 5V R E F = 1 . 2 3
V o
E S R1 0 0 m
6.Stablilizing the Converter (Example)
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The element of the Type 2 compensator ( R2, C1, and C2 ) extraction can be completed by Type 2 Compensator Calculator (Excel sheet) with the converter average models (ac models) and open-loop simulation.
The calculated values of the type 2 elements are, C1=0.778nF, C2=21.6pF, and R2=122.780k.
*Analysis directives: .AC DEC 100 0.1 10MEG
Frequency
100Hz 1.0KHz 10KHz 100KHzP(v(d))
0d
90d
180d
(9.778K,45.930)
DB(v(d))
-40
0
40
80
-100SEL>>
(9.778K,0.000)
• Phase margin = 45.930 at the cross-over frequency - fc = 9.778kHz.
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6.Stablilizing the Converter (Example)
Tip: To bring cursor to the cross-over point (gain = 0dB) type “ sfle(0) ” in Search Command.
Cursor Search
Phase at fc
Gain: T(s) = H(s) G(s)GPWM
7. Load Transient Response Simulation (Example)
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R 21 2 2 . 7 8 0 k
C 22 1 . 6 p
Type 2 Compensator
C 10 . 7 7 8 n
L o a d
V o
I 1
TD = 1 0 mTF = 2 5 u
P W = 0 . 4 3 mP E R = 1
I 1 = 0I 2 = 0 . 8
TR = 2 0 u
R lo a d2 5
0
F B
R u p p e r3 . 1 k
R lo we r1 k
0
d
V in2 0 V d c
D
U 2B U C K _ S W
REF
PWM
1 / V p
-
+
U 3P W M _ C TR L
V P = 2 . 5V R E F = 1 . 2 3
L3 3 0 u H
1 2
C3 3 0 u F
E S R1 0 0 m
The converter, that have been stabilized, are connected with step-load to perform load transient response simulation.
5V/2.5 = 0.2A step to 0.2+0.8=1.0A load
*Analysis directives: .TRAN 0 20ms 0 1u
Simulation Measurement
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Output Voltage Change
Load Current
• The simulation results are compared with the measurement data (National Semiconductor Corp. IC LM2575 datasheet).
Time
9.9ms 10.1ms 10.3ms 10.5ms 10.7ms 10.9ms1 V(vo) 2 I(load)
4.4V
4.5V
4.6V
4.7V
4.8V
4.9V
5.0V
5.1V
5.2V1
0A
0.5A
1.0A
1.5A
2.0A
2.5A
3.0A
3.5A
4.0A2
>>
7. Load Transient Response Simulation (Example)
Type 2 Compensator Calculator
Switching frequency, fosc : 52.00 kHz Spec, datasheetCross-over frequency, fc (<fosc/4) : 10.00 kHz Input the chosen valueRupper : 3.1 kOhm Spec, datasheetRlower : 1 kOhm Spec, datasheetR2 (Rupper//Rlower) : 0.756 kOhm (automatically calculated)
PWMVref : 1.230 V Spec, datasheetVp (Approximate) : 2.5 V Spec, datasheet + calculating, (or leave default 2.5V)
Parameter extracted from simulationSet: R2=R2, C1=1k, C2=1fGain (PWM) at foc ( - or + ) : -44.211 dB Read from simulation resultPhase (PWM) at foc : 65.068 ° Read from simulation result
K-factor (Choos K and q from the table)K 6 Input the chosen valueq -199 ° (automatically calculated)
Phase margin : 46 (automatically calculated)
R2 : 122.780 kOhm (automatically calculated)C1 : 0.778 nF (automatically calculated)C2 : 21.60 pF (automatically calculated)
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Simulation Index
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Simulations Folder name
1. Stabilizing the
Converter....................................................
2. Load Transient Response..................................................
ac
stepload
Libraries :1. ..\bucksw.lib2. ..\pwm_ctr.lib
Tool :• Type 2 Compensator Calculator (Excel sheet)