AnaPopovic_report2014

FACULTY OF ENGINEERING

RIJEKA

Lab exercise

CONTROL OF ELECTRIC MACHINES

Student: Ana Popović

0069049490

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TABLE OF CONTENT

1. Feedback voltage control ................................................................................................................ 3

1.1. Influence of the offset angle in phase currents and voltage space vector diagram ............... 3

1.2. Decreasing the phase angle to 0° ............................................................................................ 5

1.3. Increasing the phase angle (i.e. to 150°) ................................................................................. 7

2. Field oriented dq-current control .................................................................................................... 9

2.1. Responses for isq=8[A] ............................................................................................................. 9

2.2. Responses for isq=0.4[A] ........................................................................................................ 10



2.5. Responses for isq=2[A] ........................................................................................................... 13

2.6. Calculating the efficiency ...................................................................................................... 14

3. Field oriented speed control with maximum efficiency ................................................................ 15

3.1. Rotational speed jumps (1800 rpm and 2500 rpm) with default parameters ...................... 15

3.2. Definition of parameters (PI controller) and speed jump responses .................................... 16

3.3. Efficiency ............................................................................................................................... 18

4. Field oriented speed control with field weakening operation ...................................................... 19

4.1. Field weakening at 3000[rpm]............................................................................................... 20

4.2. Field weakening at 4000[rpm]............................................................................................... 21

4.3. Efficiency ............................................................................................................................... 22

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Second Lab Exercise – Control

Parameter change for my exercise:

Student Parameter Change

PM

flu

x

Indu

ctiv

ity

Res

ista

nce

Mo

men

t o

f In

erti

a

Bat

tery

Vo

ltag

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Lmd*If Lsd/Lsq Rs Jall UDC

48 Ana Popović 15% -15% -15% 0% 0%

1. Feedback voltage control

1.1. Influence of the offset angle in phase currents and voltage space

vector diagram

Optimal field orientated control with maximum efficiency in steady state is when isd

current is equal to 0. After changing the parameter values, I’ve checked the influence of the

offset angle in the phase currents and managed to find an angle of 90.7° for which isd

component drops to approximately 0.

If we increase or decrease this angle, isq decreases, isd ≠0 and torque drops because it

is proportional with isq and we don’t have maximum efficiency in steady state.

If isd=max and isq =0, torque drops to 0 because magnetic field of permanent magnets

and magnetic field of stator are reversed (𝞿m+ 𝞿s=0). For maximum amplitude of isd current,

voltage has only DC component and the speed is 0.

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1.1 isd=0 for a phase angle of 90.7°

I will show now some responses of currents, voltage, speed and torque, when

increasing/decreasing the phase angle.

E.g., these are the responses for an angle of 0° for which the isq component is

approximately 0 and isd=max.

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1.2. Decreasing the phase angle to 0°

1.2 current response, isq=0 and isd=max

1.3 Torque, for an angle of 0° and isq=0 is approximately 0[Nm]

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1.4 Voltage response, only DC component is left

1.5 Speed is also, like torque, equal to 0, n=0[rpm]

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1.3. Increasing the phase angle (i.e. to 150°)

1.6 Currents response (isq is dropping compared to the one when angle is 90.7° and isd=0)

1.7 Torque is also dropping, because it is proportional to isq

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1.8 Voltage response

1.9 Speed [rpm] for an offset angle of 150°

Changing these angles, I noticed that for every angle of 0°+k*π isq=0. For every odd

'k', isd= -max. For every even k, and for k=0 (shown in example), isd= +max. Same thing

applies in the oposite case, when, isd=0 (the angle is of course, not 0, but 90.7°-k*π).

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2. Field oriented dq-current control

2.1. Responses for isq=8[A]

After changing the parameters to my values, in the input of the controller (iq_set) I set the

value of iq current to 8[A], and I got following responses:

2.1 Current response, isq=8A, isd=0A

2.2 Torque response [mNm]

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2.3 Speed [rpm]

After changing the value of the isq current (0.4, 0.8, 1.2, 2 A) and reading the values of

speeds and torques from following graphs for each of the current value I used formulas to

calculate efficiency.

2.2. Responses for isq=0.4[A]

2.4 Torque [mNm] response for isq=0.4

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2.5 Speed [rpm]


2.6 Torque [mNm] response for isq=0.8[A]

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2.7 Speed [rpm]


2.8 Torque [mNm] response for isq=1.2[A]

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2.9 Speed [rpm]

2.5. Responses for isq=2[A]

2.10 Torque [mNm] response for isq=2[A]

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2.11 Speed [rpm]

2.6. Calculating the efficiency

√

isq [A] Speed [rpm] 𝞰 [%]

8 2937 25.09

2 2937 87.35

1.2 2937 95.1

0.8 2295 93.35

0.4 447 76.54

I read the value of 'U' from the voltage graph.

Cos( )=1

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3. Field oriented speed control with maximum efficiency

3.1. Rotational speed jumps (1800 rpm and 2500 rpm) with default

parameters

3.1 Speed jump to 1800[rpm], overshoot=0[rpm], rise time=0.45s

3.2 Speed jump to 2500[rpm], overshoot=50[rpm], rise time=0.4s

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3.2. Definition of parameters (PI controller) and speed jump responses

To get very fast settling time of the real rotational speed I had to adjust the control

parameters. PI controller is used for control and changing his parameters I managed to

achieve shorter rise up time increasing his integrative parameter, but overshoot increased,

so I had to increase also I parameter to get lower overshoot and to have shorter settling

time.

Following figure shows new parameters of PI controller (e.g., I increased both P and I

parameter 10 times)

3.3 PI regulator with increased parameters P and I

3.4 Speed jump to 1800[rpm], rise time is 0.19[s] and the overshoot is 80[rpm]

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3.5 Speed jump to 2500[rpm], rise time=0.27[s], overshoot=80rpm

3.6 Speed jump to 3000 [rpm] with no overshoot

By changing PI parameters and observing speed jump changes I've come to a conclusion that

it's not possible to get lower rise time, it's only possible to eliminate overshoot and get faster settling

time.

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3.3. Efficiency

To calculate the efficiency we use equations from 2.6.:

For a speed of 3000 rpm, which is 314 rad/s, torque of 60 mNm, isq=1.1 A and cos𝞿=1 we get the

efficiency of

η=

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4. Field oriented speed control with field weakening operation

After changing the parameters to achieve 3000[rpm] in speed control modul, I took

the isq value I got in the current response, and inserted it in current control modul, together

with the isd value that I got. Necessary current is isd=-0.5A, and i got isq current value of

1.1[A].

4.1 Speed response for isd=-0.5

4.2 Torque response for isd=-0.5A

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4.1. Field weakening at 3000[rpm]

I used reduced value of isd current, that was shown on current scope at speed control modul

which was -0.5[A]. With that value, and with isq value of 1.1[A] I got following responses for

speed and currents at current control model:

4.3 Current response

4.4 Speed response

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4.5 Torque response

We can see that speed and torque haven't changed their value.

4.2. Field weakening at 4000[rpm]

In this case, the necessary isd current is -3[A]. I managed to get the speed of 4000[rpm] with

the current values of: isd=-7.2A, isq=1.5A

4.6 Current response, isd=-7.2A, isq=1.5A

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4.7 Speed response for isd=-7.2A, isq=1.5A

4.3. Efficiency

Again, to calculate the efficiency we use formula from 2.6:

=

= 0.7548 = 75.48%

Cosϕ=0.8423

AnaPopovic_report2014

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