# Slide 1 - ENCON

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```							Simulation of Power Electronic
Systems Using PSpice

1
Presentation Outlines
In order to use Pspice for power electronic
systems, we have to:
 Know background of SPICE
 Understand Power Electronics Circuits/Systems
 Know how to use VPULSE to generate useful
waveforms
 Know how to make simple models using ABM

2
Scope

This presentation covers:
 PSpice
 System/Circuit Level Simulation
 Power Electronic Circuits/Systems
 Simulation

3
SPICE/PSpice
Did you know?
 SPICE turns 38 years old this year
 I Knew SPICE when she was 17 years old
 I love PSpice because she can do almost
anything I need with FOC.
 I like to talk about her.

4
Why simulation?
Simulations are essential ingredients of the
analysis and design process in power
electronics:

 Saving of development time
 Saving of costs („burnt power circuits tend
to be expensive‟)
 Better understanding of the function

5
… continued
 Testing and finding of critical states and
regions of operation (Worst Case Analysis)
 Stress test (Smoke Analysis)
 Optimization of system
 Testing new ideas

6
Overview

Simulation of analog circuits normally
uses three basic tools:

 SPICE simulator,
 Mathematical analysis package,
 and Microsoft Excel.

7
SPICE
 Simulation Program for Integrated Circuit
Emphasis

 Intended for ICs, not for power electronics.

 Uses iterative Newton-Raphson Algorithm
to solve a set of nonlinear equations.

8
SPICE LIMITATIONS
 The Newton-Raphson algorithm is guaranteed
to converge if the equations is continuous.

 The transient analysis has the additional
possibility of unable to converge because of the
discontinuity in time.

9
SPICE LIMITATIONS

Computer Hardware Limitation:
• Voltage and currents are limited to +/-1e10.
• Derivatives in PSpice are limited to 1e14.
• The arithmetic used in PSpice is double
precision and has 15 digits of accuracy.

10
Power Electronic Circuit
 Power electronic circuits
are characterized by switching
on and off of power
semiconductor switches; the
generated waveform is
passed through inductors and
capacitors for filtering.

11
Power Electronic Circuit

 Due to switching action of the switch,
discontinuity (in circuit variables and in
time) can easily occur during simulation,

“Avoid discontinuity”

12
Discontinuity Analogy:
Unacceptable Bump

Acceptable Bump

“Whole car shakes when I hit a bump on the road”
PSpice doesn’t like discontinuity as we don’t like
13
Avoid Discontinuity
S

G

VGS                      VGS

t                          t
 All signals must be made „less discontinuous‟
 All relationships must be continuous
14
VPULSE
Waveform generator
 PULSE
 SAWTOOTH
 TRIANGULAR

15
VPULSE
Waveform generator
In order to use PSpice for power electronic
circuits, the first thing you have to know is to
program VPULSE to produce these waveforms:

 PULSE
 Sawtooth
 Triangular

16
VPULSE
Waveform Generator Part
 has 7 parameters to set
 TD can be zero, others can not!

V1=                        PW
V2=           V2
TD=
TR=
TF=
V1 TD
PW=
PER
PER=
 know   what parameters to adjust and to fix.
17
VPULSE
To Generate Pulse Waveform
 Very small values for TR and TF
 Duty cycle = PW/PER

PW
V1=0
V2=12      V2
TD=0
TR=10n
TF=10n
PW=10u     V1
TR ≈ 0        TF ≈ 0
PER=20u
PER
18
A Typical application
Buck Converter (Open Loop)
M2
IR F15 0

10 0uH

V2                                                       68 0uF   10
V3                          MU R1 520
20 V
V+                V-
TD = 0
TF = 1 0n
PW = 10u
PE R = 20u        0
V1 = 0
TR = 1 0n
V2 = 1 2V

A Pulse waveform is used to drive a MOSFET
ON and OFF.
19
Its Pulse (I)

V1=0
V2=12
TD=0
TR=10n
TF=10n
PW=10u
PER=20u

PW 10
Duty Cycle, D          50%
PER 20

20
Its Pulse (II)
V1=0
V2=12
TD=0
TR=10n
TF=10n
PW=5u
PER=20u

PW   5
D         25%
PER 20
21
VPULSE
To Generate Sawtooth
 Very small values for TF and PW
 TR≈PER
V1=0
V2=12                            PW
TD=0
TR={20u-20n}
TF=10n                              TF
PW=10n
PER=20u
PER

22
A Typical application
Buck Converter (Closed Loop)
M2
IR F15 0

10 0uH

V2                                                   68 0uF   10
E1
Gate

+
-
MU R1 520

+
-
20 V                         E
Driver        GA IN = 4

Comparator                        0

-
Sawtooth   V4       +                  For Closed-loop, the control signal
Gen.
is compared with a sawtooth
Control                   waveform to produce the pulse
0 Signal
waveform.

23
PSpice Implementation
M2
IR F15 0

10 0uH

V2                                                          68 0uF   10
Gate                 E1

+
-
MU R1 520

+
-
20 V                         E
Driver                GA IN = 1

0

V5
Comparator
E2
Control                    IN +OU T+
IN - OU T-
Signal
2.5 Vdc

Gate Driver  E
ETABL E
V3
0                                           0

Comparator  ETABLE
TD = 0       TA BLE = (0 ,0 ( 200u ,12 )
TF = 1 0n      V( %IN +, % IN-)
PW = 10n
0PE R = 20u

Sawtooth VPULSE
V1 = 1 V
TR = { 20u- 20n}
V2 = 4 V

Vpulse                           Control  VDC

24
Its Waveform (I)

Control

Sawtooth
D = 50 %

Pulse

25
Its Waveform (II)
Control

Sawtooth

D = 33%

Pulse

Control Signal.
26
VPULSE
To Generate Triangular wave
 Very small value for PW
 TR≈TF ≈ PER/2

V1= -1               PW
V2= +1
TD=0
TR= {10u-10n}
TF= {10u-10n}
PW=20n
PER=20u             PER

27
VPULSE
Its Triangular Wave

28
Triangular Wave
Typical applications
Bipolar SPWM

TR I                   SI NE

V                        V
V1 = -1                     V1                               VD C*( V(SI NE) -V(TRI))/ ABS (V(S INE )-V( TRI) )
V2 = +1
TD = 0                                              V2
TR = { (1/(F TRI *2))- 10n}    VO FF = 0                                                 SP WM
TF = {( 1/(F TRI* 2)-1 0n)}    VA MPL = { Ma}
PW = 20n                       FR EQ = {F SIN E}                                                  V
PE R = {1/F TRI }              PH ASE = { -90/ Mf }
Comparator
0                          0
PARAMETERS:
Ma = 0 .8
Mf = 2 1
FTRI = {FS INE *Mf }
FS INE = 5 0
VD C = 100

29
Triangular Wave
Typical applications
Bipolar SPWM
1.0V

0V

-1.0V
V(TRI)   V(SINE)   0
100

0

-100
40ms     42ms          44ms   46ms   48ms      50ms   52ms   54ms   56ms   58ms   60ms
V(SPWM)    0
Time [ms]

30
Triangular Wave
Typical applications
Unipolar SPWM
TR I                   SI NE1

V                        V
V1 = -1                      V1                                         0.5 *VD C*( V(SI NE1 )-V( TRI) )/AB S(V (SIN E1) -V(TRI))
V2 = +1
TD = 0                                                   V2
TR = { (1/(F TRI *2))- 10n }      VO FF = 0                                                   A
TF = {( 1/(F TRI *2)-1 0n)}       VA MPL = { Ma}
PW = 20n                          FR EQ = {F SIN E}                                                       V
PE R = {1/F TRI }                 PH ASE = { -90/ Mf }
0                            0
Comparator 1
SI NE2

V
PARAMETERS:                                                       0.5 *VD C*( V(SI NE2 )-V( TRI) )/AB S(V (SIN E2) -V(TRI))
Ma = 0 .8
Mf = 2 1                                  V2 a
FTRI = {FS INE *Mf } VO FF = 0                                                                B
FS INE = 5 0         VA MPL = { Ma}
VD C = 100           FR EQ = {F SIN E}                                                                  V
PH ASE = { -90/ Mf +180}
0
Comparator 2

31
Triangular Wave
Typical applications
Unipolar SPWM
1.0V

0V

-1.0V
V(SINE1)   V(SINE2)   V(TRI)
100V

0V

-100V
40ms        42ms      44ms         46ms   48ms    50ms   52ms   54ms   56ms   58ms   60ms
V(A)-V(B)
Time [ms]

32
Analog Behavior Model (ABM)
Makes the Circuit Simpler
Use equations to model circuits
 Comparator
 Single Phase Rectifier
 Three Phase Rectifier
 Buck Converter in CCM
 Single Phase Inverter

33
ABM
Behavior Model of Comparator
V(-)   IF the voltage at the terminal V(+) is
V(out)
-          greater than the voltage at terminal
V(+)
+          V(-) the output V(out) is HIgh,
otherwise the output is LOw.

(1) Using IF-Then-Else function
IF(V(+)>V(-),HI, LO)

(2) Using signum function
(V(+)-V(-))/ABS(V(+)-V(-))

34
ABM
Behavior Model of Comparator
V(-)
V(out)
-
V(+)
+

(3) Using I/O graph           (4) Using Op-amp alike
V(out)                            V(+)      V(out)

+
0      V(+)-V(-)                       - A*(V(+)-V(-))
V(-)
0

35
ABM
Comparator in PSpice
IF (V(S INE )>V( TRI) ,10, -10)
TR I                   SI NE

ou t1
V1 = -1
V2 = +1
V1
V                       V
1
V
TD = 0                                                   V2
TR = { (1/(F TRI *2))- 10n }      VO FF = 0                                 SI NE E1
TF = {( 1/(F TRI *2)-1 0n)}       VA MPL = { Ma}                                                        ou t2
IN +OU T+
PW = 20n
PE R = {1/F TRI }
FR EQ = {F SIN E}
PH ASE = { -90/ Mf }               2               IN - OU T-
ETABL E                              V
TR I
0                            0                          V( %IN +, % IN-)            0
PARAMETERS:                                  TA BLE = (- 100 u,-10 ) (1 00u, 10)
Ma = 0 .8
Mf = 2 1                           VD C*( V(SI NE) -V(TRI)) /ABS (V(S INE )-V( TRI ))
FTRI = {FS INE *Mf }
FS INE = 5 0
ou t3
VD C = 10
3
 NO 2 is implemented using ETABLE                                                                                            V
LIMIT( 10k* (V(S INE )-V( TRI) ),10 ,-10)
 Others are implemented using ABM part
 NO 2 & NO 4 are suitable for Op-amp                                                   4
ou t4

(Error Amplifier)                                                                                                             V

36
ABM
Behavior Model of Comparator
1.0V

0V

-1.0V
V(TRI)   V(SINE)
10V

0V
These waveforms come from the outputs of four comparators

-10V
40ms      42ms    44ms     46ms      48ms     50ms   52ms   54ms   56ms   58ms   60ms
V(OUT3) V(OUT2) V(OUT1) V(OUT4)

Time [ms]

37
ABM
Behavior Model of Rectifier (I)
D3                D4
Db reak           Db reak
V1 a
VO FF = 0
VA MPL = 3 40                                                               R1 a
FR EQ = 50
1k

D5                D6
Db reak           Db reak

0

in
V(out)=ABS(V(IN))
V1                E1
VO FF = 0                                                        R1 b
VA MPL = 3 40                           IN +OU T+
FR EQ = 50                            IN - OU T-                  1k
EV ALU E
ABS(V(IN))

0                               0
38
ABM
Behavior Model of Rectifier (II)
+

Van
V(out) = 0.5*(ABS(V(an)-V(bn)
Vbn
+ABS(V(bn)-V(cn))
Vcn                +ABS(V(cn)-V(an)))

-

39
ABM
Behavior Model of Buck in CCM
IR F54 0

10 0uH
+
V2
68 0uF            RL
20 Vdc                   V3
MU R1 520       Vd

TD = 0                -                               Vd = d*Vin
TF = 1 0n
PW = 10u
PE R = 20u        0
V1 = 0
TR = 1 0n
V2 = 1 2V
10 0uH
+
 d is a PWM signal                     d
E1
IN +OU T+
68 0uF          RL

with 1V amplitude.                    Vin     IN - OU T-
EV ALU E
Vd
V(%IN+)*V( %IN-)
-
0
40
ABM
Behavior Model of Inverter

a
+
VDC               Vab
-
b                 Bipolar SPWM

E1          +
SINE
IN+ OUT+
IN- OUT-
Vab
TRI
EVALUE      -
VDC*(V(%IN+)-V( %IN-))/ABS(V(%IN+)-V( %IN-))
0
41
#TIPS
 There are many different ways
to model the same thing. So, be
creative!
 Use a simple model wherever
possible to reduce modeling time
and make simulation run faster
and converge better!
42

“Models are like shoes; there is no one-size-
fits-all model.”

43
Our Case Study
A Buck Converter with VMC
 A Simple PWM Controller IC Model
 A PWM IC Controller IC Model including
Soft-start
 A PWM IC Controller IC Model Including
Soft-start, Duty Cycle Max and Current
Limiter

44
Our Case Study
A Buck Converter with VMC

+
-
+
-

0

SG3525
PWM Controller IC      0

45
SG3525
PWM Controller IC
Key Functions:
 Oscillator
(Sawtooth Generator)
 PWM Comparator
and SR Flip-flop
 Error Amplifier
 5.1 V Reference
 Pulse Steering
Logic
 Shutdown and
Soft-start Circuitry

46
SG3525
 We do not need to have SG3525 model in
PSpice‟s library to simulate buck converter with
VMC.
 To verify the controller design, all we need are
functional models of these:
 Error Amplifier
 Comparator
 Sawtooth generator

47
SG3525
A Simple Model

Sawtooth

-
+                          To MOSFET
+            Driver
-
Comparator
Error Amp.

48
A Buck Converter with VMC

Buck Converter

+
-
+
-
 Consider we know all             0
circuit parameters.

 Our interest is to       Comparator         Error Amp.
0
-
simulate the system.                +
-                        +

 The controller is used                           VP ULSE
Vref

to regulate the output                         0
Sawtooth      0
voltage at 5 V.
49
A Buck Converter with VMC
 The controller is a linear controller and the
design is based on a small-signal model.

 So, the controller can not cope with large
signal scenario such as start-up.

 Initial values, which are equal to their steady
state values, for the inductor current and the
capacitor voltage must be set.

50
How to set a load disturbance ?

3A

1A
0A
8 ms            8.5 ms

R = 1.666 W
R=5W
R=5W

 R is changed from 5 W to 1.666 W
51
Our Case Study
How to set load disturbance ?
Using IPULSE

I1 = 1       I1

1
I2 = 3
TD = 8m
TR = 0.1u
TF = 0.1u
PW = 0.5m
PER = 1m

 Allocate enough times for TR and TF

52
How to set load disturbance ?
Using SW_tclose and SW_topen

N
TOPE = 8.5m
1         2

E
TCLOS = 8m

2               5                          2 .5

5//2.5 =1.666
53
M1
IRF150                               R1av            L1
i nput                                                                                        out
{L}
I
I
IC = 1A               R2av
V2
15Vdc                                               D5                                                 {Resr}
E1               Dbreak                                                         I1 = 1        I1
+
-
GA IN = 3   +                                                                                    I2 = 3
-     E                                                               C1av           TD = 8m
TR = 0.1u
{C}        TF = 0.1u
IC = 5V    PW = 0.5m
PE R = 1m
0

C3av

R6av{C3}                C2av        R7av            C4av

{R2}                    {C2}        {R1}            {C1}
E2av                                                      E1av                                      R4av
OUT+ IN+                                             OUT+ IN+
OUT- IN-                                             OUT- IN-             V2av                     {R3}
(0,0) (250u,5) V1 = 0                 V1                 ET ABLE
ET ABLE          V2 = 3                         -V(%IN+, %IN-)                        R2
V(%IN+, %IN-)    TD = 0                           (0,0) (250u ,6)
0                    TR = {10u-20n}                                                       {Rbias}
TF = 10n                                  {Vref }
PW = 10n         0        0
PE R = 10u                                                  0    0

54
5.2V

Output Voltage
5.0V

4.8V
V(OUT)
4.0A

2.0A                            Inductor Current

0A
7.8ms    7.9ms    8.0ms   8.1ms   8.2ms    8.3ms   8.4ms   8.5ms   8.6ms   8.7ms 8.8ms
Time [ms]

55
Input Disturbance
How to set an input disturbance ?
Let the input disturbance is:

25 V

15 V

0V
8 ms        8.5 ms

56
Input Disturbance
How to set an input disturbance ?
Use VPWL (Piece-Wise Linear Voltage Source)

25 V

15 V

0V
8 ms            9 ms

PWL(T1,V1)(T2,V2)(T3,V3)(T4,V4)(T5,V5)
PWL (0,15) (8m,15) (8.0001m,25) (9m,25) (9.0001m,15)
57
Input Disturbance
Responses

30V
25V
20V
Input Voltage
10V
V(INPUT)

5.1V
5.0V
4.9V                  Output Voltage
4.8V
V(OUT)
2.0A
Inductor Current
1.0A

0A
7.8ms      8.0ms   8.2ms   8.4ms    8.6ms   8.8ms   9.0ms   9.2ms   9.4ms   9.6ms   9.8ms 10.0ms
I(L1)
Time [ms]

58
Start-up Scenario
 Previous simulation skips start-up scenario.
 To know how the controller handles start-up, set the initial
values for iL and vc to zero.
20

15

Inductor Current
10

Output Voltage
5

0
0s        100us    200us   300us   400us   500us   600us   700us   800us
I(L1)   V(OUT)
Time [s]
59
Start-up Scenario
 A very large overshoot and undershoot occur in inductor current.
 The duty cycle is at first at 1 for a long time and later at 0 for a long
time too, then after that it gradually increases.

 Convergence problem can easily occurs at this extreme condition.

5.0V

2.5V

Gate Signal
0V
V(E1:1)
20

15

10

5

0
0s                  100us     200us   300us   400us   500us   600us   700us   800us
I(L1)     V(OUT)

Time

60
Start-up
 In practical circuit, another auxiliary controller is
required to handle start-up.
 This circuit is known as soft-start.
Soft start                            VMC
5.0V
Controller                           Controller

2.5V

Gate Signal
0V
V(E1:1)
20

15

10

5

0
0s                  100us    200us      300us     400us   500us   600us    700us   800us
I(L1)     V(OUT)

Time [s]
 Soft-start circuit works by gradually increasing the duty
cycle. So do the inductor current and capacitor voltage.
61
Soft-start
 The previous PWM IC model is very useful and it is
simple to set-up in PSpice.

 It is enough to verify the design of controller based on
small signal model.

 However, to add soft-start controller and other
protection circuits, we need a more flexible PWM IC
model.

62
A Modified PWM IC Model

Oscillator    Clock
Sawtooth

Comparator           SR Flip-flop

+               S
+                              R
-                   Q
-

Error Amp.

 The output of SR flip-flop is set by the Clock.
 The output of SR flip-flop is reset by Comparator.

63
A Modified PWM IC Model
Oscillator        Clock
Sawtooth
Error Amp.
Comparator       SR Flip-flop
+
+                                           S
-                 R
-                                              Q
-
Analog                             R
-                     R
Signals                  Digital
Signals
 Analog signals can be added at minus terminals of the
comparator.
 Digital signals can be added at the input Resets of FF.
64
Soft-start
Sawtooth
+
Error Amp.                 To R of SR
-
Control     Output                     Flip-Flop
Signal
Soft-start   -

 Sawtooth is still compared with the control signal.

 But, Control Signal can be either Error Amp. output
(EAO) or Soft-start signal (SS), whichever is lower.

65
Soft-start
Sawtooth
50 A                 +
Soft-start (SS)              To R of SR
-
Flip-Flop
Error Amp.
C
Output (EAO) -

 The soft-start voltage is the capacitor voltage.
 The capacitor C is charged by a constant current
source of 50 A. The result is a ramp voltage.
 C determines the duration of soft-start.
66
Soft-start
How Soft-start works?
Soft-start
V
I C
Voltage

t
4V
50 
Slope =
C

C = 125 nF

10 ms   t

 Use PWL to emulate soft-start voltage
 For the graph, PWL(0,0)(10ms,4V)
67
Soft-start
Sawtooth
50 A                        +
SS                              To R of SR
Control   -   Flip-Flop
EAO                Signal
C

Selector

 We need a selector to select either SS or EAO,
whichever is lower, to be Control Signal.
 We can use IF-Then-Else function
 IF(SS < EAO, SS, EAO)
68
Soft-start
In PSpice
C3av           Error Amplifier
SELECTOR
IF-Then-Else                       R6av{C3}          C2av        R7av        C4av

-V(%IN+, %IN-)                   IF( V(%IN2)<V(%IN1),
(0,0) (250u,5)                     V(%IN2),V(%IN1) )              {R2}              {C2}        {R1}        {C1}
E1av                            R4av
R       E2av   ET ABLE                                 1 err_out
OUT+ IN+
Vout
control   3
OUT+ IN+                                      2                OUT- IN-        V2av                 {R3}
OUT- IN-         Sawtooth                                      ET ABLE
-V(%IN+, %IN-)                  R2
Comparator                                 SoftS                 (0,0) (500u ,5)               {Rbias}
V1 = 0                    V1
V2 = 3                                                           {Vref }
0            TD = 0                               V3         0
TR = {10u-20n}                                                                 0    0
TF = 10n                        TRAN = PWL(0,0)(10m,4)
PW = 10n                  0
PE R = 10u                       0
Sawtooth
Generator

69
Soft-start
Start-up Signals
Control = IF(SS < EAO, SS, EAO)
5.0V

2.5V     Error Amplifier Output
0V

5.0V

2.5V
Soft-Start Signal

0V

2.0V

Control Signal
1.0V

0V
0s     1.0ms        2.0ms      3.0ms    4.0ms   5.0ms   6.0ms

Time [ms]

70
Soft-start
C = 125 nF (Too Small!)
7.5V

5.0V

2.5V
V(OUT)          tstart-up = 1ms
0V

4.0A

2.0A
I(L1)

0A
0s     1.0ms           2.0ms       3.0ms     4.0ms   5.0ms   6.0ms

Time [ms]

 Soft-start        signal ramps up too fast
71
Soft-start
Start-up Current and Voltage
7.5V

5.0V
C = 25 nF
2.5V
V(OUT)                 tstart-up = 3.2 ms
0V

2.0A

1.0A
I(L1)

0A
0s         1.0ms        2.0ms     3.0ms     4.0ms    5.0ms   6.0ms

Time [ms]
 Still has a small overshoot and undershoot in
inductor current
 has a room for improvement by increasing C.
72
Soft-start
Start-up Current and Voltage
6.0V

4.0V

2.0V              V(OUT)
0V
V(OUT)
2.0A

1.0A

I(L1)
SEL>>
0A
0s            5ms           10ms   15ms          20ms   25ms   30ms   35ms
I(L1)
Time

 C = 125 nF ; Start-up time is 30 ms.

73
A Modified PWM IC Model
Oscillator        Clock
Sawtooth
Error Amp.
Comparator       SR Flip-flop
+
+                                           S
-                 R
-                                              Q
-
Analog                             R
-                     R
Signals                  Digital
Signals
 To add digital signals for protection.
 For examples, Maximum Duty Cycle and Current Limiter
 Flip-flop can be reset either by PWM comparator, or
Maximum duty cycle, or Current Limiter.
74
DutyMax and CurrentLimit
 Maximum duty cycle limiter is in digital form. It can be
applied directly to the Reset of FF.
The switch current (or inductor current) must be compared
with its limit value to produce a digital signal.
Dutymax
V1 = 0           Vdutymax
V2 = 5V
RESET 3 (DMax)            TD = {10u*0.85 }
TR = 10n
1                TF = 10n
2   3     U12A                        PW = {(10u-10u*0.85)-20n}
Q   1   U10A             7432       2                PE R = 10u        0                        I(L1)
7402   3                                                                 Ecurr_l imit

OUT+ IN+
RESET 2 (CL)                     OUT- IN-
8A
1                   ET ABLE
2                             3     U16A                    0 +V(%IN+, %IN-)
1   U11A                                       7432      2     R
7402   3       SET
S
(0,0) (250u,5)

V1 = 0       VClock
RESET 1 (EAO)
V2 = 5V
TD = 0                 Set only by one i. e. the clock
 Reset can be done by three, whichever
TR = 1n
TF = 1n
PW = 0.1u
PE R = 10u   0
comes first.
75
DutyMax and CurrentLimit
5.0V
CLOCK
2.5V

0V
5.0V
V(S)           DUTYMAX
2.5V

0V
V(DUTYMAX)
4.0V
SAWTOOTH
2.0V

0V
0s          20us   40us   60us   80us   100us   120us   140us   160us   180us   200us
V(SAWTOOTH)
Time

 DUTYMAX signal will only reset FF if the duty cycle is more than 0.85
 This DUTYMAX is to make sure that the MOSFET always turns-off for
each cycle
 CurrentLimit signal will only appear and reset FF if the peak switch is
greater than pre-specified value.
76
DutyMax and CurrentLimit
10

Output Voltage
5

Inductor Current
0
5.6ms          5.7ms   5.8ms   5.9ms   6.0ms   6.1ms   6.2ms   6.3ms   6.4ms   6.5ms   6.6ms
V(OUT)   I(L1)
Time

We want to limit this current at 8A

77
DutyMax and CurrentLimit
What do we expect ?
10
8A Limiter

Output Voltage
5

Inductor Current

0
5.6ms          5.7ms      5.8ms   5.9ms   6.0ms    6.1ms    6.2ms   6.3ms    6.4ms   6.5ms   6.6ms
V(OUT)   I(L1)
Time

Reset by EAO                       Reset by        Reset by         Reset by EAO
DutyMax         CurrentLimit

78
DutyMax and CurrentLimit
5.0V

2.5V

0V
V(CLOCK)
5.0V

2.5V

0V
V(PWMCOMP)      V(Q)
5.0V

2.5V

0V
V(CURRENTLIM)     V(Q)
5.0V

2.5V

0V
5.90ms          5.95ms           6.00ms     6.05ms    6.10ms   6.15ms   6.20ms
V(DUTYMAX)     V(Q)
Time [ms]
at 6.0 ms
79
Knowing

“There is no substitute for knowing
what we are doing”

80
CONCLUSION
In order to simulate power electronic circuit:
 Know how to program VPULSE for Pulse,
Sawtooth, and Triangular waveforms.
 Avoid discontinuity at any cost
 Use the simplest model possible
 Use a simple model first, and add
complexity in stages.
 No replacement for good understanding

81
Q&A

82

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