PSIM tutorial 2

7/27/2019 PSIM tutorial 2

1/12

Announcements

H5CPE2 module homepage will soon change

http://hermes.eee.nott.ac.uk/teaching/h5cpe2

Simulation packages:

PSIM: http://www.powersimtech.com/download.html

PSPICE: http://hermes.eee.nott.ac.uk/teaching/h54pqe/

LIVE Power Electronics Course (prof. Mohan)

http://www.ece.umn.edu/groups/PowerElectronics_Drives/

User: [email protected]

Password: powerelectronics

Free Power Electronics Literature on Internet

Power Modules - Application Manual (Semikron)

http://www.semikron.com/skcweb/e/applica/applica_help.html

Switched-Mode Power Supplies (Philips)

http://www.semiconductors.philips.com/acrobat/applicationnotes/APPCHP2.pdf

Power Control with Thyristors and Triacs (Philips)http://www.semiconductors.philips.com/acrobat/applicationnotes/APPCHP6.pdf


2/12

Ideal DiodeRectifier (1)

0

Neutral

A

BC

Line A

Phase

voltage

Vxy

D1 D3 D5

D4 D6 D2

Id=constx

y

Vx

VxyVy

D1, D3, D5 select the most positive input line potential to x

D4, D6, D2 select the most negative input line potential to y

IA

D4 (-Id)

D1 (+Id)

IA

30

150

Six-pulse diode rectifier

Highly inductive load(excitation of synchronous generator)

D1 D3 D5

D6D4D2


3/12


Mean (average) rectified voltage

0

3 35

cos cos 2 2 3 36 6

5 5 2

6 6 6 6

x

EE

V E

0 0

3 3

2

y xV V E

ypotential= xpotential mirrored

0

3 32

xy x

EV V

If E= 240 then2 561.4xyV V

0 0

030 150x ANV Volt time Area V from

Or it could be determined from integrating the line-

to-line voltage over 60

120

60

13 sin

3

3 2 3 3cos cos

3 3 3

xyV E d

E E


4/12

Measures ofPower Quality (1)

VA

IA

Lets consider a nonlinear load that draws a nonlinear

and displaced current from a sinusoidal supply VA

IA

Iah= IA -IA1

IA1

1. Extract the fundamental component IA1(Amp&angle)

IA=1*sin(100t)+1/3*sin(300t)+1/5*sin(500t)+1/7*sin(700t)

FFT of IA


5/12

Displacement Power Factor (DPF) DPF= Cosine of phase shift between fundamental

component of line current and the associated phase

voltage (ie IA and VAN)

Accounts for proportion of fundamental current that

does something useful. DPF = 1 is good, DPF = 0.5 isbad (50% of fundamental current does nothing useful)


VA

IA1

IA1-reactive

IA1-active2 2 2

1 1 1A A active A reactiveI I I

1 1

1 1

cos

sin

A active A

A reactive A

I I

I I

2. Decompose the fundamental component into an

active IA1-active in phase with VA and a reactive IA1-reactive

which is 90-deg displaced to VA

IA1

IA1-active IA1-reactive


6/12


RMS (root-mean square) current = the value of an equivalent

DC current that produces the same amount of heat into a given

resistor as the current in question

2 2

0

1T

RMSt

P R I i dtT

2

0

1T

RMStI i dtT

2 2 2

1

2

RMS RMS k RMS

k

I I I

Distortion Factor (DF) DF = RMS fundamental current/total RMS current (


7/12

Relationship between quantities

Assume voltage is undistorted compared to current

normally reasonable for a decent power system

Then total RMS voltage = fundamental RMS voltage1

1

1

Power delivered

where fund RMS current

Hence:

RMS RMS

RMS

RMS RMS

Tot

RMS RMS

P V I DPF

I

V I DPF PF DF DPF

V I

Power factor = Displacement factor * Distortion factor

Note if the current is undistorted then DF = 1 and Power

factor = Displacement factor = Cosine of phase shift (as

given in 1st year notes)

Power Factor (PF)

PF= active power delivered/total RMS volts * total RMS amps

Tells us what proportion of the total apparent power (VA)

represents useful power delivered to the load (Watts)

PF=1 is good, PF = 0.5 is bad (have to supply 1000VA to get500W for example)



8/12


+Id

-Id

T/2 T0

Half-wave symmetric: f(t+T/2) = -f(t)

4sin

2

dn

II n

n

where n=1,3,5,7,9

In =0 for n =2,4,6because of half-wave symmetry

Analyze Input Current Waveform=general case

Decomposition in Fourier series results in:

In =0 for the situations where n* = 2

As for 3-phase diode rectifier (ideal case) = 2/3

Harmonics current component:In =0 for n=3, 9, 15 etc

1 2 3 dII

155

II

177

II

11111

II

(peak)

RMS input current: 2 3RMS dI I

12 3 3

0.952 3

d

RMS d

II

I I

Distortion factor DF =

>Id

Displacement factor DPF = 1

1

0.95RMS RMS

Tot

RMS RMS

V I DF PF DPF DF

V I

Power factor


9/12

Non-ideal DiodeRectifier (1)

0

Neutral

A

B

C

Line A

Phasevoltage

Vxy

D1 D3 D5

D4 D6 D2

Id=constx

y

L

L

L

ab

c

Effect of supply inductance = this causes overlap since current

cant transfer instantaneously from one device to another

VA VB VC

D1

a

D3

b

x

ia

; 0aa

dii const L

dt 0 aa d

dii I L

dt

Instant commutation between D1 and D3 is not possible !

ia

ib


10/12


'

aa AN

diV V L

dt

'

bb BN

diV V L

dt

a b di i I const a bdi di

dt dt

Outgoing diode (D1) Incoming diode (D3)

Both diodes conduct simultaneously (overlap) ' 'a bV V

a aAN BN

di diV L V L

dt dt

2

a AN BN di V V Ldt

'2 2

AN BN AN BNa AN x

V V V V

V V V

y CNV V

2 2

AN BN AC BCxy CN

V V V V V V

Loss in Volt-time area!! smaller Vxy


11/12


The missing VTA in Vxy is required to commutate the load

current Id from D1 to D3

Loss in mean Vxy

Total VTA loss from Vx in one cycle = 3*L*Id

aL

diV L

dt

6 32

d dI L I L

Effect on the line current waveform

lost L a d VTA V dt L di L I

Similar for Vy hence total VTA loss in Vxy = 6*L*Id

3 33xy d

EV I L

Causes volt-drop with Id

known as regulation

Negligible effect on

the harmonics unless is very large

ia ibId

0 a b di i I


12/12


Calculation of

0

A

B

D1 D3

Id=constx

L

L

ab

VBAia

ib

Easiest way to find it is by considering the following loop:

3 E

VBA

3 cos 0 cos 2 dE I L

3

1 cos2

dI L E

Area =

2 Id

*L

During this time, Vba causes the change of :

ia from Id to 0

ib from 0 to Id

Detrimental effect on other users

0

A

BC

Id=const

L1 L2

L1 L2

To other users

Current in L2 is

distorted due to

overlap

Voltage in PCC =

partially distorted

PCC

1

1 2

L

L L

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PSIM tutorial 2