Simulation of Power Electronic Systems Using PSpice

Presented by Nik Din Muhamad

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

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

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.

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

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

Overview
Simulation of analog circuits normally uses three basic tools:
SPICE simulator, Mathematical analysis package, and Microsoft Excel.

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.

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.

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, which leads to convergence problem. Avoid discontinuity

12

Discontinuity Analogy: A Bump on the Road

Unacceptable Bump

Acceptable Bump

Whole car shakes when I hit a bump on the road PSpice doesnt like discontinuity as we dont like a bump on the road.
13

Avoid Discontinuity
S G

VGS

VGS

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= V2= TD= TR= TF= PW= PER=
know

V2

PW

V1

TD PER

what parameters to adjust and to fix.

17

VPULSE
To Generate Pulse Waveform
Very small values for TR and TF Duty cycle = PW/PER

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

PW
V2

V1

TR 0

TF 0

PER
18

A Typical application Buck Converter (Open Loop)

M2 IR F15 0 10 0uH

V2 V3 20 V
V+ V-

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

10

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

Duty cycle of the waveform is adjusted by adjusting PW

PW 5 D 25% PER 20
21

VPULSE To Generate Sawtooth

Very small values for TF and PW TRPER V1=0 V2=12 TD=0 TR={20u-20n} TF=10n PW=10n PER=20u PER

PW

TF

22

A Typical application Buck Converter (Closed Loop)

M2 IR F15 0 10 0uH V2 20 V 68 0uF 10

Gate Driver Comparator +

E1 E GA IN = 4 MU R1 520

+ -

+ -

Sawtooth Gen.

V4

Control 0 Signal

For Closed-loop, the control signal is compared with a sawtooth waveform to produce the pulse waveform.
23

PSpice Implementation
M2 IR F15 0 10 0uH V2 20 V 68 0uF 10

Gate Driver

E1 E GA IN = 1 MU R1 520

+ -

+ -

V5

Control 2.5 Vdc Signal

V3

Comparator E2
IN +OU T+ IN - OU TETABL E

TD = 0 TA BLE = (0 ,0 ( 200u ,12 ) TF = 1 0n V( %IN +, % IN-) PW = 10n 0PE R = 20u V1 = 1 V TR = { 20u- 20n} V2 = 4 V

Gate Driver E Comparator ETABLE

Vpulse

Sawtooth VPULSE Control VDC

24

Its Waveform (I)

Control
Sawtooth

D = 50 %

Pulse

25

Its Waveform (II)

Control Sawtooth

D = 33%
Pulse

Duty Cycle of the Pulse is adjusted by adjusting Control Signal.

26

VPULSE To Generate Triangular wave

Very small value for PW TRTF PER/2 V1= -1 V2= +1 TD=0 TR= {10u-10n} TF= {10u-10n} PW=20n PER=20u PW

PER

27

VPULSE Its Triangular Wave

28

Triangular Wave Typical applications

Bipolar SPWM
TR I
V

SI NE
V

V1 = -1 V1 V2 = +1 TD = 0 V2 TR = { (1/(F TRI *2))- 10n} VO FF = 0 TF = {( 1/(F TRI* 2)-1 0n)} VA MPL = { Ma} PW = 20n FR EQ = {F SIN E} PE R = {1/F TRI } PH ASE = { -90/ Mf }

VD C*( V(SI NE) -V(TRI))/ ABS (V(S INE )-V( TRI) ) SP WM

V

0
PARAMETERS:
Ma = 0 .8 Mf = 2 1 FTRI = {FS INE *Mf } FS INE = 5 0 VD C = 100

Comparator

29

Triangular Wave Typical applications

Bipolar SPWM
1.0V

0V

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

-100 40ms 42ms V(SPWM) 0

44ms

46ms

48ms

50ms

52ms

54ms

56ms

58ms

60ms

Time [ms]
30

Triangular Wave Typical applications

Unipolar SPWM
TR I
V

SI NE1
V

V1 = -1 V1 V2 = +1 TD = 0 TR = { (1/(F TRI *2))- 10n } TF = {( 1/(F TRI *2)-1 0n)} PW = 20n PE R = {1/F TRI }

0.5 *VD C*( V(SI NE1 )-V( TRI) )/AB S(V (SIN E1) -V(TRI)) A
V

V2 VO FF = 0 VA MPL = { Ma} FR EQ = {F SIN E} PH ASE = { -90/ Mf }

0
SI NE2
V

Comparator 1

PARAMETERS:
Ma = 0 .8 V2 a Mf = 2 1 FTRI = {FS INE *Mf } VO FF = 0 VA MPL = { Ma} FS INE = 5 0 FR EQ = {F SIN E} VD C = 100 PH ASE = { -90/ Mf +180}

0.5 *VD C*( V(SI NE2 )-V( TRI) )/AB S(V (SIN E2) -V(TRI)) B
V

Comparator 2
31

Triangular Wave Typical applications

Unipolar SPWM
1.0V

0V

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

0V

-100V 40ms 42ms V(A)-V(B)

44ms

46ms

48ms

50ms

52ms

54ms

56ms

58ms

60ms

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(out)

V(-) V(+)

IF the voltage at the terminal V(+) is greater than the voltage at terminal 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(out)

V(-) V(+)

+
(3) Using I/O graph V(out)

(4) Using Op-amp alike V(+) V(out)

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

V(+)-V(-) V(-)

0
35

ABM Comparator in PSpice

IF (V(S INE )>V( TRI) ,10, -10) TR I
V

SI NE
V

V1 = -1 V1 V2 = +1 TD = 0 TR = { (1/(F TRI *2))- 10n } TF = {( 1/(F TRI *2)-1 0n)} PW = 20n PE R = {1/F TRI }

1
SI NE E1

ou t1
V

V2 VO FF = 0 VA MPL = { Ma} FR EQ = {F SIN E} PH ASE = { -90/ Mf }

2
0
TR I

0
PARAMETERS:
Ma = 0 .8 Mf = 2 1 FTRI = {FS INE *Mf } FS INE = 5 0 VD C = 10

IN +OU T+ IN - OU TETABL E V( %IN +, % IN-)

ou t2
V

TA BLE = (- 100 u,-10 ) (1 00u, 10) VD C*( V(SI NE) -V(TRI)) /ABS (V(S INE )-V( TRI ))

ou t3
V

NO 2 is implemented using ETABLE Others are implemented using ABM part NO 2 & NO 4 are suitable for Op-amp (Error Amplifier)

LIMIT( 10k* (V(S INE )-V( TRI) ),10 ,-10) ou t4

V

36

ABM Behavior Model of Comparator

1.0V

0V

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

0V

These waveforms come from the outputs of four comparators

-10V 40ms 42ms 44ms 46ms V(OUT3) V(OUT2) V(OUT1) V(OUT4)

48ms

50ms

52ms

54ms

56ms

58ms

60ms

Time [ms]

37

ABM Behavior Model of Rectifier (I)

D3 Db reak V1 a VO FF = 0 VA MPL = 3 40 FR EQ = 50 R1 a 1k D4 Db reak

D5 Db reak

D6 Db reak

0
in

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

ABS(V(IN))
0 0
38

ABM Behavior Model of Rectifier (II)

+
Van Vbn Vcn

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

39

ABM Behavior Model of Buck in CCM

IR F54 0

+
V2

10 0uH

20 Vdc

V3

MU R1 520

Vd
0

68 0uF

RL

TD = 0 TF = 1 0n PW = 10u PE R = 20u V1 = 0 TR = 1 0n V2 = 1 2V

Vd = d*Vin

d is a PWM signal with 1V amplitude.

+
d Vin
E1 IN +OU T+ IN - OU TEV ALU E

10 0uH

Vd
-

68 0uF

RL

V(%IN+)*V( %IN-)

0
40

ABM Behavior Model of Inverter

a

VDC

+ Vab b

Bipolar SPWM
E1 IN+ OUT+ IN- OUTEVALUE

SINE TRI

+ Vab 0
41

VDC*(V(%IN+)-V( %IN-))/ABS(V(%IN+)-V( %IN-))

#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

Quote about Model !

Models are like shoes; there is no one-sizefits-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

+ -

+ -

SG3525 PWM Controller IC

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 PSpices 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

+ Error Amp.

+
Comparator

To MOSFET Driver

48

A Buck Converter with VMC

Buck Converter
+ + -

Consider we know all circuit parameters. Our interest is to simulate the system.

Comparator
+ -

Error Amp.
+
Vref VP ULSE

The controller is used to regulate the output voltage at 5 V.

Sawtooth

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

Load Disturbance
How to set a load disturbance ? Let the load disturbance is:
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 I2 = 3 TD = 8m TR = 0.1u TF = 0.1u PW = 0.5m PER = 1m

I1

ILOAD

Allocate enough times for TR and TF

52

Load Disturbance
How to set load disturbance ? Using SW_tclose and SW_topen
TOPE = 8.5m N 1 2

TCLOS = 8m E

2 .5

5//2.5 =1.666
53

Load Disturbance: PSpice

i nput M1 IRF150 R1av 50m V2 15Vdc E1 E D5 Dbreak
+ -

L1 out {L}
I V 0Vdc

ILOAD
I

IC = 1A

R2av {Resr} I1 = 1 I2 = 3 TD = 8m TR = 0.1u TF = 0.1u PW = 0.5m PE R = 1m I1

+ -

GA IN = 3

C1av {C} IC = 5V

C3av R6av{C3} C2av R7av C4av

{R2}
E2av OUT+ IN+ OUT- INE1av OUT+ IN+ OUT- INV1 V2 = 3 TD = 0 TR = {10u-20n} TF = 10n PW = 10n PE R = 10u ET ABLE -V(%IN+, %IN-)

{C2}

{R1}

{C1}
R4av

V2av R2 {Rbias}

{R3}

(0,0) (250u,5) V1 = 0 0
ET ABLE V(%IN+, %IN-)

(0,0) (250u ,6) 0 0 {Vref } 0 0

54

Load Disturbance: Results

5.2V

Output Voltage
5.0V

4.8V 4.0A V(OUT)

2.0A

Inductor Current

0A 7.8ms 7.9ms 8.0ms I(L1) I(ILOAD)

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 10V V(INPUT) 5.1V 5.0V 4.9V 4.8V V(OUT) 2.0A 1.0A 0A 7.8ms I(L1)

Input Voltage

Output Voltage Inductor Current

8.0ms

8.2ms

8.4ms

8.6ms

8.8ms

9.0ms

9.2ms

9.4ms

9.6ms

9.8ms 10.0ms

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 I(L1)

100us V(OUT)

200us

300us

400us

500us

600us

700us

800us

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

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 Controller
Gate Signal
0V V(E1:1) 20 15 10 5 0 0s I(L1) 100us V(OUT) 200us 300us 400us 500us 600us 700us 800us

5.0V

VMC Controller

2.5V

Time [s]

Soft-start circuit works by gradually increasing the duty cycle. So do the inductor current and capacitor voltage.
61

Soft-start To add 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 + + Analog Signals Digital Signals S SR Flip-flop

R
R R

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 To add Soft-start Signal

Sawtooth +
To R of SR Flip-Flop Error Amp. Output Soft-start

Control Signal

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 To add Soft-start Signal

Sawtooth
50 A

+
To R of SR Flip-Flop

Soft-start (SS)
C

Error Amp. 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 Voltage 4V Slope =

V I C t
50 C

C = 125 nF

10 ms

Use PWL to emulate soft-start voltage For the graph, PWL(0,0)(10ms,4V)

67

Soft-start To add Soft-start Signal

Sawtooth
50 A

+
To R of SR Flip-Flop

SS
C

EAO
Selector

Control Signal

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
SELECTOR IF-Then-Else
-V(%IN+, %IN-) C3av R6av{C3}

Error Amplifier
C2av R7av C4av

(0,0) (250u,5)
R E2av ET ABLE control OUT+ IN+ OUT- IN-

IF( V(%IN2)<V(%IN1), V(%IN2),V(%IN1) )

1 err_out 3 2 Sawtooth SoftS V1 V3

{R2}
E1av OUT+ IN+ OUT- INET ABLE -V(%IN+, %IN-)

{C2}

{R1}

{C1}
R4av

Vout

V2av R2 {Rbias}

{R3}

Comparator
0
V1 = 0 V2 = 3 TD = 0 TR = {10u-20n} TF = 10n PW = 10n PE R = 10u

(0,0) (500u ,5) 0 {Vref } 0 0

TRAN = PWL(0,0)(10m,4)

Sawtooth Generator

69

Soft-start Start-up Signals

Control = IF(SS < EAO, SS, EAO)
5.0V 2.5V 0V 5.0V 2.5V 0V 2.0V 1.0V 0V 0s

Error Amplifier Output

Soft-Start Signal

Control Signal
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
V(OUT)

2.5V

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)
V(OUT)

0V 2.0A

1.0A

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

I(L1)

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

73

A Modified PWM IC Model

Oscillator Clock

Sawtooth
Error Amp. Comparator + + Analog Signals Digital Signals S SR Flip-flop

R
R R

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

To Add Digital Signals 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.
RESET 3 (DMax)
1 2 Q 1 U10A 7402 3 3 U12A 7432 2 Dutymax Vdutymax V1 = 0 V2 = 5V TD = {10u*0.85 } TR = 10n TF = 10n PW = {(10u-10u*0.85)-20n} PE R = 10u 0

Ecurr_l imit OUT+ IN+ OUT- IN-

I(L1)

RESET 2 (CL)
1 2 1 U11A 7402 3 3

8A

SET S
V1 = 0 V2 = 5V TD = 0 TR = 1n TF = 1n PW = 0.1u PE R = 10u VClock

U16A 7432

0 +V(%IN+, %IN-) (0,0) (250u,5)

ET ABLE

Set only by one i. e. the clock Reset can be done by three, whichever comes first.
75

RESET 1 (EAO)

To Add Digital Signals DutyMax and CurrentLimit

5.0V 2.5V 0V V(S) 5.0V 2.5V 0V V(DUTYMAX) 4.0V 2.0V 0V 0s 20us V(SAWTOOTH) 40us 60us 80us 100us Time 120us 140us 160us 180us 200us

CLOCK

DUTYMAX

SAWTOOTH

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

To Add Digital Signals DutyMax and CurrentLimit

10

Output Voltage
5

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

We want to limit this current at 8A

77

To Add Digital Signals DutyMax and CurrentLimit

What do we expect ?
10

8A Limiter
Output Voltage

Inductor Current
0 5.6ms V(OUT)

5.7ms I(L1)

5.8ms

5.9ms

6.0ms

6.1ms
Time

6.2ms

6.3ms

6.4ms

6.5ms

6.6ms

Reset by EAO

Reset by DutyMax

Reset by CurrentLimit

Reset by EAO

78

To Add Digital Signals DutyMax and CurrentLimit

5.0V 2.5V 0V

V(CLOCK)
5.0V 2.5V 0V

5.0V
2.5V 0V 5.0V 2.5V

V(PWMCOMP)

V(Q)

V(CURRENTLIM)

V(Q)

0V 5.90ms

V(DUTYMAX)

5.95ms V(Q)

6.00ms

6.05ms

6.10ms

6.15ms

6.20ms

Time [ms]

A Load disturbance 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