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ELECTRICAL DRIVES:

An Application of Power Electronics

Power Electronic Converters in ED Systems DC DRIVES

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

V a

+ V dc  T1 D1 T2 D2 +

V a

i a Q2 T1 conducts 

v a = V dc

Jika vdc=110 Volt dan duti cycle=0.75.

Tentukan: (a). Va (avg, dan V(rms) (b). Cara Kerja rangkaian untuk operasi 2 kuadran Q1

I a

Power Electronic Converters in ED Systems DC DRIVES

AC-DC-DC DC-DC: Four-quadrant Converter leg A leg B + V dc  Q1 D1 + V a  D3 Q3 Q4 D4 D2 Q2 Jika vdc=110 Volt dan duti cycle=0.75.

Tentukan: (a). Va (avg, dan V(rms) (b). Cara Kerja rangkaian untuk operasi 4 kuadran

v a = V dc when Q1 and Q2 are ON Positive current

Power Electronic Converters in ED Systems DC DRIVES

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

V a

+ V dc  T1 D1 T2 D2 +

V a

i a Q2 T1 conducts 

v a = V dc

Jika vdc=110 Volt dan duti cycle=0.75.

Tentukan: (a). Va (avg, dan V(rms) (b). Cara Kerja rangkaian untuk operasi 2 kuadran Q1

I a

Power Electronic Converters in ED Systems DC DRIVES

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

V a

T1 D1 + V dc  T2 D2 + V a i a Q2 Q1

I a

D2 conducts 

v a = 0

T1 conducts 

v a = V dc V a E b Quadrant 1

The average voltage is made larger than the back emf

Power Electronic Converters in ED Systems DC DRIVES

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

V a

T1 D1 + V dc  T2 D2 + V a i a Q2 Q1

I a

D1 conducts 

v a = V dc

Power Electronic Converters in ED Systems DC DRIVES

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

V a

T1 D1 + V dc  T2 D2 + V a i a Q2 Q1

I a

T2 conducts 

v a = 0

D1 conducts 

v a = V dc V a E b Quadrant 2

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

Power Electronic Converters in ED Systems DC DRIVES

AC-DC-DC DC-DC: Two-quadrant Converter 2v tri + v A v c + v c V dc 0

Power Electronic Converters in ED Systems DC DRIVES

AC-DC-DC DC-DC: Four-quadrant Converter leg A leg B + V dc  Q1 D1 + V a  D3 Q3 Q4 D4 D2 Q2

Positive current v a = V dc when Q1 and Q2 are ON

Power Electronic Converters in ED Systems DC DRIVES

AC-DC-DC DC-DC: Four-quadrant Converter leg A leg B + V dc  Q1 D1 + V a  D3 Q3 Q4 D4 D2 Q2

Positive current v a v a v a = V dc when Q1 and Q2 are ON = -V dc when D3 and D4 are ON = 0 when current freewheels through Q and D

Power Electronic Converters in ED Systems DC DRIVES

AC-DC-DC DC-DC: Four-quadrant Converter leg A leg B + V dc  Q1 D1 + V a  D3 Q3 Q4 D4 D2 Q2

Positive current v a v a v a = V dc when Q1 and Q2 are ON = -V dc when D3 and D4 are ON = 0 when current freewheels through Q and D Negative current v a = V dc when D1 and D2 are ON

Power Electronic Converters in ED Systems DC DRIVES

AC-DC-DC DC-DC: Four-quadrant Converter leg A leg B + V dc  Q1 D1 + V a  D3 Q3 Q4 D4 D2 Q2

Positive current v a v a v a = V dc when Q1 and Q2 are ON = -V dc when D3 and D4 are ON = 0 when current freewheels through Q and D Negative current v a v a v a = V dc when D1 and D2 are ON = -V dc when Q3 and Q4 are ON = 0 when current freewheels through Q and D

v c + _

Power Electronic Converters in ED Systems DC DRIVES

AC-DC-DC Bipolar switching scheme – output swings between V DC and -V DC 2v tri V dc + v A + v B v A v B v AB v c V dc 0 V dc 0 V dc -V dc

v c + _

Power Electronic Converters in ED Systems DC DRIVES

AC-DC-DC Unipolar switching scheme – output swings between V dc and -V dc

V tri

V dc + v A + v B v A v B v AB -v c

v c -v c

0 V dc 0 V dc 0 V dc

Power Electronic Converters in ED Systems DC DRIVES

AC-DC-DC DC-DC: Four-quadrant Converter V dc V dc 200 150 100 50 0 -50 -100 -150 -200 0.04

0.0405

0.041

0.0415

0.042

0.0425 0.043

0.0435

0.044

0.0445

0.045

Armature current V dc 200 150 100 50 0 -50 -100 -150 -200 0.04

0.0405

0.041

0.0415

0.042

0.0425 0.043

0.0435

0.044

0.0445

0.045

Armature current Bipolar switching scheme Unipolar switching scheme • • Current ripple in unipolar is smaller Output frequency in unipolar is effectively doubled

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters Switching signals obtained by comparing control signal with triangular wave

v c

v tri q We want to establish a relation between

v c

and

V a

+ V a − V dc AVERAGE voltage v c (s)

?

V a (s) DC motor

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters T tri v c q  1  0 V c > V tri V c < V tri t on 1 0 d  1 T tri t  t  T tri q dt  t on T tri

0 V dc

V a  1 T tri 0  dT tri V dc dt  dV dc

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters d 0.5

v c V tri -V tri -V tri v c For v c = -V tri  d = 0

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters d 0.5

V tri -V tri -V tri v c v c V tri For v c = -V tri For v c = 0   d = 0 d = 0.5

For v c = V tri  d = 1

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters d V tri 0.5

-V tri d  0 .

5  1 2 V tri v c -V tri v c v c V tri For v c = -V tri For v c = 0   d = 0 d = 0.5

For v c = V tri  d = 1

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters Thus relation between v c and V a is obtained as: V a  0 .

5 V dc  V dc 2 V tri v c Introducing perturbation in v c and V a and separating DC and AC components:

DC: AC:

V a  0 .

5 V dc  V dc 2 V tri v c v~ a  V dc 2 V tri v~ c

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters Taking Laplace Transform on the AC, the transfer function is obtained as: v a v c ( s ) ( s )  V dc 2 V tri v c (s) V dc 2 V tri v a (s) DC motor

v c

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters

Bipolar switching scheme

V dc -V dc 2v tri v tri q + V dc −

+ V AB

 v A v B q v AB d A  0 .

5  v c 2 V tri V A  0 .

5 V dc  V dc 2 V tri v c d B  1  d A  0 .

5  v c 2 V tri V B  0 .

5 V dc  V dc 2 V tri v c V A  V B  V AB  V dc V tri v c v c V dc 0 V dc 0 V dc -V dc

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters

Bipolar switching scheme

v a v c ( s ( s ) )  V dc V tri v c (s) V dc V tri v a (s) DC motor

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters V dc

Unipolar switching scheme

Leg b

V tri

+ v tri V dc q a v c − v A Leg a v tri -v c q b v B d A  0 .

5  v c 2 V tri d B  0 .

5   v c 2 V tri v AB V A  0 .

5 V dc  V dc 2 V tri v c V B  0 .

5 V dc  V dc 2 V tri v c V A  V B  V AB

The same average value we’ve seen for bipolar !

 V dc V tri v c

v c -v c

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters

Unipolar switching scheme

v a v c ( s ( s ) )  V dc V tri v c (s) V dc V tri v a (s) DC motor

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters DC motor – separately excited or permanent magnet v t  i a R a  L a di dt a  e a T e  T l  J d  m dt T e = k t i a e e = k t  Extract the dc and ac components by introducing small perturbations in V t , i a , e a, T e , T L and  m e v~ t ac components  T ~ a R a e   L a k E ( ~ a d i a dt )  e~ a ~ e  k E ( ~ )  T L  B   J d (  ) dt dc components V t  I a R a  E a T e  k E I a E e  k E  T e  T L  B (  )

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters DC motor – separately excited or permanent magnet Perform Laplace Transformation on ac components v~ t  ~ a R a  L a d i a dt  e~ a V t (s) = I a (s)R a + L a sIa + E a (s) T e  k E ( ~ a ) T e (s) = k E I a (s) ~ e  k E ( ~ ) E a (s) = k E  (s) e  T L  B  J d (  ) dt T e (s) = T L (s) + B  (s) + sJ  (s)

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters DC motor – separately excited or permanent magnet V a ( s ) + R a 1  sL a I a ( s ) k T T e ( s ) + T l ( s ) B 1  sJ  ( s ) k E

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters q v tri Torque controller + T c

+ –

V dc − T e ( s ) + Torque controller Converter V dc V tri , peak V a ( s ) + q k t

DC motor

R a 1  sL a I a ( s ) k T T e ( s ) + T l ( s ) B 1  sJ  ( s ) k E

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters Closed-loop speed control – an example 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

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters Closed-loop speed control – an example OBJECTIVES: • Fast response – large bandwidth • Minimum overshoot good phase margin (>65 o ) • Zero steady state error – very large DC gain BODE PLOTS METHOD • Obtain linear small signal model • Design controllers based on linear small signal model • Perform large signal simulation for controllers verification

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters Closed-loop speed control – an example Ra = 2  B = 1 x10 –4 kg.m

2 /sec k e = 0.1 V/(rad/s) La = 5.2 mH J = 152 x 10 –6 kg.m

2 k t = 0.1 Nm/A V tri = 5 V V d = 60 V f s = 33 kHz • PI controllers • Switching signals from comparison of v c and triangular waveform

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters Torque controller design

Open-loop gain

Bode Diagram From: Input Point To: Output Point 150 100 50 0 -50 90 45 0 -45 -90 10 -2 10 -1 10 0 10 1 10 2 Frequency (rad/sec) 10 3 compensated compensated 10 4 10 5 k pT = 90 k iT = 18000

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters Speed controller design  * + – Speed controller T* 1 T B 1  sJ  Torque loop

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters Speed controller design

Open-loop gain

Bode Diagram From: Input Point To: Output Point 150 100 50 0 compensated k ps = 0.2

k is = 0.14

-50 0 -45 -90 -135 -180 compensated 10 -2 10 -1 10 0 10 1 Frequency (Hz) 10 2 10 3 10 4

Modeling and Control of Electrical Drives

Modeling of the Power Converters: DC drives with SM Converters Large Signal Simulation results Speed 40 20 0 -20 -40 0 Torque 2 1 0 -1 -2 0 0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

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