ELECTRICAL DRIVES: An Application of Power Electronics

ELECTRICAL DRIVES: ELECTRICAL DRIVES: An Application of Power ElectronicsAn Application of Power Electronics

T1 conducts va = Vdc

Q1Q2

Va

Ia

T1

T2

D1

+

Va

-

D2

ia

+

Vdc

DC DRIVES

AC-DC-DCAC-DC-DC DC-DC: Two-quadrant Converter

Power Electronic Converters in ED Systems

Jika vdc=110 Volt dan duti cycle=0.75.

Tentukan:

(a). Va (avg, dan V(rms)

(b). Cara Kerja rangkaian untuk operasi 2 kuadran

leg A leg B

+ Va Q1

Q4

Q3

Q2

D1 D3

D2D4

+

Vdc

va = Vdc when Q1 and Q2 are ON

Positive current

Power Electronic Converters in ED SystemsDC DRIVES

AC-DC-DCAC-DC-DC DC-DC: Four-quadrant Converter


Tentukan:




Q1Q2

Va

Ia

T1

T2

D1

+

Va

-

D2

ia

+

Vdc

DC DRIVES




Tentukan:



Q1Q2

Va

Ia

T1

T2

D1

+

Va

-

D2

ia

+

Vdc

D2 conducts va = 0

Va Eb


Quadrant 1 The average voltage is made larger than the back emf

DC DRIVES



Q1Q2

Va

Ia

T1

T2

D1

+

Va

-

D2

ia

+

Vdc

D1 conducts va = Vdc

DC DRIVES



Q1Q2

Va

Ia

T1

T2

D1

+

Va

-

D2

ia

+

Vdc

T2 conducts va = 0

VaEb

D1 conducts va = Vdc

Quadrant 2 The average voltage is made smallerr than the back emf, thus forcing the current to flow in the reverse direction

DC DRIVES



DC DRIVES


+vc

2vtri

vc

+vA

-

Vdc

0


leg A leg B

+ Va Q1

Q4

Q3

Q2

D1 D3

D2D4

+

Vdc


Positive current



leg A leg B

+ Va Q1

Q4

Q3

Q2

D1 D3

D2D4

+

Vdc

va = -Vdc when D3 and D4 are ON


va = 0 when current freewheels through Q and D

Positive current






Positive current

va = Vdc when D1 and D2 are ON

Negative current

leg A leg B

+ Va Q1

Q4

Q3

Q2

D1 D3

D2D4

+

Vdc






Positive current

va = -Vdc when Q3 and Q4 are ON

va = Vdc when D1 and D2 are ON


Negative current

leg A leg B

+ Va Q1

Q4

Q3

Q2

D1 D3

D2D4

+

Vdc




AC-DC-DCAC-DC-DC

vAB

Vdc

-Vdc

Vdc

0vB

vAVdc

0

2vtri

vc

vc

+

_

Vdc+vA

-

+vB

-

Bipolar switching scheme – output swings between VDC and -VDC


AC-DC-DCAC-DC-DCUnipolar switching scheme – output swings between Vdc and -Vdc

Vtri

vc

-vc

vc

+

_

Vdc+vA

-

+vB

-

-vc

vA

Vdc

0

vB

Vdc

0

vAB

Vdc

0


AC-DC-DCAC-DC-DC

Bipolar switching scheme

0.04 0.0405 0.041 0.0415 0.042 0.0425 0.043 0.0435 0.044 0.0445 0.045

-200

-150

-100

-50

0

50

100

150

200

Unipolar switching scheme

0.04 0.0405 0.041 0.0415 0.042 0.0425 0.043 0.0435 0.044 0.0445 0.045

-200

-150

-100

-50

0

50

100

150

200

• Current ripple in unipolar is smaller

• Output frequency in unipolar is effectively doubled

Vdc

Vdc

Vdc

DC-DC: Four-quadrant Converter

Armature current

Armature current

vc

+

Va

−

vtri

Vdc

q

Switching signals obtained by comparing control signal with triangular wave

Va(s)vc(s)DC motor

We want to establish a relation between vc and Va

?

AVERAGE voltage

Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters

dtqT1

dtriTt

ttri

tri

on

Tt

Vdc

0

Ttri

ton

0

1

01

qVc > Vtri

Vc < Vtrivc

dc

dT

0 dctri

a dVdtVT1

Vtri


-Vtri

Vtri

-Vtri

vc

d

vc

0.5

For vc = -Vtri d = 0



0.5

Vtri

Vtri

vc

d

vc

-Vtri-Vtri


For vc = 0 d = 0.5

For vc = Vtri d = 1


0.5

vc

d

-Vtri-Vtri

ctri

vV21

5.0d

Vtri

Vtri

vc


For vc = 0 d = 0.5

For vc = Vtri d = 1

Thus relation between vc and Va is obtained as:

ctri

dcdca v

V2V

V5.0V

Introducing perturbation in vc and Va and separating DC and AC components:

ctri

dcdca v

V2V

V5.0V

ctri

dca v~

V2V

v~

DC:

AC:


Taking Laplace Transform on the AC, the transfer function is obtained as:

tri

dc

c

a

V2V

)s(v)s(v

va(s)vc(s)DC motor

tri

dc

V2V


2vtri

vc

vc

vtri+

Vdc

−

q-Vdc

q

Vdc

+ VAB

vAB

Vdc

-Vdc

ctri

dcABBA v

VV

VVV

tri

cAB V2

v5.0d1d

ctri

dcdcB v

V2V

V5.0V

vB

Vdc

0

tri

cA V2

v5.0d

ctri

dcdcA v

V2V

V5.0V

vA

Vdc

0



tri

dc

c

a

VV

)s(v)s(v

va(s)vc(s)DC motor

tri

dc

VV



+

Vdc

−vc

vtri

qa

Vdc

-vc

vtri

qb

Leg a

Leg b

The same average value we’ve seen for bipolar !

Vtri

vc

-vc

tri

cA V2

v5.0d

ctri

dcdcA v

V2V

V5.0V

vA

tri

cB V2

v5.0d

ctri

dcdcB v

V2V

V5.0V

vB

ctri

dcABBA v

VV

VVV

vAB



tri

dc

c

a

VV

)s(v)s(v

va(s)vc(s)DC motor

tri

dc

VV



DC motor – separately excited or permanent magnet

Extract the dc and ac components by introducing small perturbations in Vt, ia, ea, Te, TL and m

aa

aaat edtdi

LRiv

Te = kt ia ee = kt

dtd

JTT mle

aa

aaat e~dti~

dLRi

~v~

)i~(kT

~aEe

)~(ke~ Ee

dt)~(d

J~BT~

T~

Le

ac components

aaat ERIV

aEe IkT

Ee kE

)(BTT Le

dc components


Perform Laplace Transformation on ac components

aa

aaat e~dti~

dLRi

~v~

)i~(kT

~aEe

)~(ke~ Ee

dt)~(d

J~BT~

T~

Le

Vt(s) = Ia(s)Ra + LasIa + Ea(s)

Te(s) = kEIa(s)

Ea(s) = kE(s)

Te(s) = TL(s) + B(s) + sJ(s)



Tkaa sLR

1

)s(Tl

)s(Te

sJB1

Ek

)s(Ia )s()s(Va

+-

-

+



Tc

vtri

+

Vdc

−

q

q

+

–

kt

Torque controller

Tkaa sLR

1

)s(Tl

)s(Te

sJB1

Ek

)s(Ia )s()s(Va

+-

-

+

Torquecontroller

Converter

peak,tri

dc

VV)s(Te

-+

DC motor


Design procedure in cascade control structure

• Inner loop (current or torque loop) the fastest – largest bandwidth

• The outer most loop (position loop) the slowest – smallest bandwidth

• Design starts from torque loop proceed towards outer loops

Closed-loop speed control – an example


OBJECTIVES:

• Fast response – large bandwidth

• Minimum overshoot good phase margin (>65o)

• Zero steady state error – very large DC gain

BODE PLOTS

• Obtain linear small signal model

METHOD

• Design controllers based on linear small signal model

• Perform large signal simulation for controllers verification



Ra = 2 La = 5.2 mH

J = 152 x 10–6 kg.m2B = 1 x10–4 kg.m2/sec

kt = 0.1 Nm/Ake = 0.1 V/(rad/s)

Vd = 60 V Vtri = 5 V

fs = 33 kHz


• PI controllers • Switching signals from comparison of vc and triangular waveform


Bode Diagram

Frequency (rad/sec)

-50

0

50

100

150From: Input Point To: Output Point

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

103

104

105

-90

-45

0

45

90

Pha

se (

deg)

compensated

compensated

kpT= 90

kiT= 18000


Torque controller design Open-loop gain

Speed controller design

1Speedcontroller sJB

1

* T* T

–

+

Torque loop


Bode Diagram

Frequency (Hz)

-50

0

50

100

150From: Input Point To: Output Point

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

103

104

-180

-135

-90

-45

0

Pha

se (

deg)

Open-loop gain

compensated

kps= 0.2

kis= 0.14

compensated

Speed controller design


Large Signal Simulation results

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45-40

-20

0

20

40

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45-2

-1

0

1

2

Speed

Torque


THANK YOUTHANK YOU

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ELECTRICAL DRIVES: An Application of Power Electronics