Energy Converters – CAD and System Dynamics€¦ · · 2018-04-22saturated iron Air-gap ... Starting time constant of electric machines t J M dt J M M t dt d J M N t M t N N m

TECHNISCHE UNIVERSITÄT

DARMSTADTINSTITUT FÜRELEKTRISCHE ENERGIEWANDLUNG

Prof. Dr. A. Binder, Energy Converters - CAD and System Dynamics

5/1

Energy Converters – CAD and System Dynamics

1. Basic design rules for electrical machines

2. Design of Induction Machines

3. Heat transfer and cooling of electrical machines

4. Dynamics of electrical machines

5. Dynamics of DC machines

6. Space vector theory

7. Dynamics of induction machines

8. Dynamics of synchronous machines

Source:

SPEED program




5/2


Energy Converters - CAD and System Dynamics

Source:

Siemens AG




5/3



5.1 Dynamic system equations of separately excited DC machine

5.2 Dynamic response of electrical and mechanical system of separately

excited DC machine

5.3 Dynamics of coupled electric-mechanical system of separately excited

DC machine

5.4 Linearized model of separately excited DC machine for variable flux

5.5 Transfer function of separately excited DC machine

5.6 Dynamic simulation of separately excited DC machine

5.7 Converter operated separately excited DC machine




5/4


Dynamic system equations of separately excited DC machine

a

pzk

⋅⋅=

π21

2Machine constant:

Induced voltage:

Machine torque:

Flux:

)()()( 2 fmi itktu ΦΩ ⋅⋅=

)()()( 2 fae itiktm Φ⋅⋅=

)()(

)()( tudt

tdiLtiRtu i

aaaaa +⋅+⋅=Armature circuit:

Mechanical acceleration:

Field circuit:

Non-linear differential equations, even if Lf = const.

)()( tmtmdt

dJ se

m −=⋅Ω

dt

tdiLtiRtu

fffff

)()()( ⋅+⋅=

)( fiΦ




5/5


Saturation-dependent field inductance Lf(if)

Main flux per pole:

Field inductance:

)( fiΦ

αΦ tan~/)()( fffff iiNiL ⋅=

α

Lf(if)

if00

unsaturated iron

saturated iron

Air-gap magnetization

iron magnetization

Linearized field inductance:

ϕ

ϕ

kNL

iikNL

fIf

fffIf

fN

fN

⋅=

⋅⋅= /

fNIfL

fNI




5/6

• Armature field: Armature self-inductance La

Main field: Field self-inductance Lf .

Mutual inductance Maf only between

a) commutating armature coils (= short-circuited by brushes) and

b) field coil, otherwise zero.

dttdiLRtitu

titttktu

tudttdiLRtitu

fffff

fmi

iaaaaa

/)()()(

))(()(),()()(

)(/)()()(

2

⋅+⋅=

==

+⋅+⋅=

ΦΦΦΩ

No magnetic coupling between armature and field circuit

coil: Nc

brush

coil

main pole area





5/7


If flux is kept constant, set of differential equations is linear.

ffff IRUtu ⋅==)(

)()(

)()(

)()(

2

2

tmtikdt

dJ

tkdt

tdiLtiRtu

sam

ma

aaaa

−⋅⋅=⋅

⋅⋅+⋅+⋅=

ΦΩ

ΦΩ

One linear differential equation of second order with constant coefficients:

sma

a

aaa

ma

a

a

a mkTTdt

du

RTi

TTdt

di

Tdt

id⋅⋅

⋅+⋅

⋅=⋅

⋅+⋅+

Φ22

2 11111

dt

dm

Jm

JTu

kTTTTdt

d

Tdt

d ss

aa

mam

ma

m

a

m ⋅−⋅⋅−⋅⋅

=⋅⋅

+⋅+1111111

22

2

ΦΩ

ΩΩ

Electrical time constant of armature:

Mechanical time constant of machine and load:

aaa RLT /=2

2 )( Φk

RJT a

m

⋅=




5/8


- Changing of armature current and rotor speed is ruled by armature voltage and is

disturbed by load torque, which both are contained in "right side" of system

differential equation.

- DC machine at constant flux: LINEAR system:

DC machine may be controlled in an easy way.

- Mechanical time constant Tm decreases with the square of increasing flux.

- Machine gets ”weaker” at:

a) Flux weakening

b) Increased stator resistance Ra, which causes voltage drop, thus reducing

internal voltage Ui,

⇒ Mechanical time constant = response to load step: Tm increases !

Features of separately excited DC machine at constant flux




5/9


Starting time constant of electric machines

tJ

Mdt

J

MtM

dt

dJ

M

Nt

tM

NmN

mM ⋅==→=⋅ ∫

=0

)(ΩΩ

N

NM

N

mNMJ

M

nJ

M

JT

⋅⋅=

⋅=

πΩ 2

Relationship between starting time constant and mechanical time

constant is given with per unit resistance ra and flux φ:

:/,/

NNN

aa

IU

Rr ΦΦφ ==

2

2

200

1)/(

⋅⋅=⋅=⋅=

Nam

NN

NNM

N

mMJ

rT

Ik

kUJ

MJT

ΦΦ

ΦΦΩ

Example:

ra = 0.05, Φ = ΦN :

TJ0 = 20.Tm

TJ0 = 10 s, Tm = 0.5 s




5/10


Starting time constant vs. mechanical time constant

:/,/

NNN

aa

IU

Rr ΦΦφ ==

2

2

22

22

22

222

200

11

)(

)(1

)(

)(

1)/(

φΦΦ

ΦΦ

Φ

ΦΦΦΩ

⋅⋅=

⋅⋅=⋅⋅⋅⋅=

=⋅⋅=⋅=⋅=

am

Nam

NN

N

a

aM

NN

NM

NN

NNM

N

mMJ

rT

rT

k

k

I

U

Rk

RJ

kI

UJ

Ik

kUJ

MJT

Starting time to no-load speed n0:

Starting time to rated speed nN:

N

mMJ

MJT 0

0

Ω⋅=

0JN

mNMJ T

MJT <⋅=

Ω




5/11

Summary: Dynamic system equations of separately excited DC machine

- Separately excited DC machine treated in this lecture

- Second order differential electro-mechanical equation for armature circuit

- At constant main flux: Linear differential equation

- Long mechanical and short electrical time constant

- Do not mix mechanical time constant and starting time constant





5/12





excited DC machine


DC machine








5/13


Dynamics of mechanical system of DC machine (1)

Taking ma TT << leads to 0→aT :

a) Steady state condition: d/dt = 0 :

sm

msm

am mJ

Tm

J

Tu

k⋅−=⋅−= 0

2

1Ω

ΦΩ

Steady state condition

sam

mm

m mJ

ukTTdt

d⋅−⋅=⋅+

1111

2ΦΩ

Ωa

amm i

k

R⋅−=

ΦΩΩ

20

aa

mm ik

R⋅−=

ΦΩΩ

20

am uk Φ

Ω2

0

1=

aa

sm

am

aesi

k

Rm

J

T

kRJT

ikmm

ΦΦ

Φ

22

2

2

)/(/=

=

⋅==

No-load speed




5/14


Steady-state characteristic in p.u.

aa

mm ik

R⋅−=

ΦΩΩ

20

xry a ⋅−= 1

am

a

m

m ik

R⋅

⋅−=

020

1ΩΦΩ

Ω

NamNiammN Uukui ==⋅==⇔== 020 :0, ΩΦΩΩΦΦ

N

aa

N

a

NN

a

m

m

I

ir

I

i

IU

R⋅−=⋅−= 1

/1

0ΩΩ

0m

myΩΩ

=

N

a

I

ix =

1

1

•

Example:

Small two-pole PM-excited DC motor:

ra = 0.2

Speed at rated current:

8.012.0110

=⋅−=⋅−=N

aa

m

m

I

ir

ΩΩ

0.8




5/15



b) Dynamic operation:

Example:

Switching on of DC machine at a) rated flux and b) no-load ms = 0:

Initial condition: Zero speed . Armature voltage is switched from zero to

rated value Ua = UN for t > 0:

0)0( =mΩ

( ))/exp(1)( 0 mmm Ttt −−⋅=ΩΩ

sam

mm

m mJ

ukTTdt

d⋅−⋅=⋅+

1111

2ΦΩ

Ω

am

mm

m ukTTdt

d

ΦΩ

Ω

2

111⋅=⋅+

am uk Φ

Ω2

0

1=

ms

ms = 0

Ua = UN

Solution:




5/16


Dynamic speed response of separately excited DC machine to switching with armature voltage step, leading to exponential increase of speed !

)/exp(0m

m

mme Tt

TJ

dt

dJm −⋅⋅=⋅=

ΩΩ yields armature current )/exp( m

a

Na Tt

R

Ui −⋅=


( ))/exp(1)( 0 mmm Ttt −−⋅=ΩΩ t ≥ 0

t ≥ 0

a

a

a

a

m

ma

e

R

u

JR

k

kk

uJ

kT

Ji

k

m=⋅⋅=

⋅⇒=

22

222

0

2

)(1 ΦΦΦΦ

ΩΦ

)0( =sm




5/17

- “Jumping” armature current (due to Ta = 0)

and decreasing armature current due to

speeding up of the rotor to no-load speed.

- At (ideal) no-load the armature current

also in motor operation is zero!

mTt

a

Na e

R

Uti

/)(

−⋅=

t

ia(t)

Tm

Armature currentUN/Ra



NN

a

Na

NN

a II

R

Ui

IU

R⋅===→= 40

025.0ˆ025.0

/

Attention: Starting resistor is necessary:

Too big!




5/18


Taking ma TT << leads to ∞→mT , if electrical system only is investigated.

a

ia

aa

m

aa

aa

a

a

L

uu

RT

k

RT

ui

Tdt

di −=

⋅⋅

−⋅

=⋅+ΩΦ21

a) Steady state condition: d./dt = 0:

a

ia

a

iaaa

R

UU

R

uuIi

−=

−==

Dynamics of electrical system of DC machine (1)

Steady state condition

)( aaia iRuu −⋅−=

-ia > 0: Generator operation

aaiaaia RiuRiuu ⋅−−=⋅+= )(

.0/

:

constdtd

J

mm =→=

∞→

ΩΩ




5/19


b) Dynamic operation:

Example:

Switching on of DC machine rated flux, no-load, zero speed: Ωm = 0

Initial condition: Zero current

Armature voltage is switched from zero to rated value ua = UN for t > 0.

0)0( =ai

( )aTt

a

Na e

R

Uti

/1)(

−−⋅=


aa

Na

a

a

RT

Ui

Tdt

di

⋅=⋅+

1Solution: t ≥ 0

0 ia = UN/Ra

UN

ua

GEN.

MOT.

1mΩ

12 mm ΩΩ <

0=mΩ

ia




5/20


- Dynamic current response of separately excited DC machine at stand still to

switching with armature voltage step of rated voltage, leading to exponential

increase of armature current.

Armature current and so electromagnetic torque react to armature voltage steps

with the short armature time constant Ta.

- So separately excited DC machines are dynamic drives.


( )aTt

a

Na e

R

Uti

/1)(

−−⋅= t ≥ 0




5/21

Summary: Dynamic response of electrical and mechanical system of separately excited DC machine

- Separate solving of electrical and mechanical machine behaviour

- Steady-state solution and step response

- Consumer reference system used

- Steady-state solutions already known from bachelor´s course





5/22





excited DC machine


DC machine








5/23


Homogeneous differential equation:

Characteristic equation for λ:

If , then

dt

tdm

Jtm

JTtu

kTTTTdt

d

Tdt

d ss

aa

mam

ma

m

a

m )(1)(

11)(

1111

22

2

⋅−⋅⋅−⋅⋅

=⋅⋅

+⋅+Φ

ΩΩΩ

011

2

2

=⋅⋅

+⋅+ mma

m

a

m

TTdt

d

Tdt

dΩ

ΩΩ

ttmh eCeCt 21

21)(λλΩ ⋅+⋅=

m

a

aamaa T

T

TTTTT

41

2

1

2

10

112,1

2 −±−=→=⋅

+⋅+ λλλ

am TT 4< 144

1 −=−m

a

m

a

T

Tj

T

T

Dynamics of coupled electric-mechanical system (1)




5/24

a) am TT 4> 21, λλ are real values, so transient speed response contains TWO

time constants 2211 /1,/1 λλ −=−= TT , a short and a long one.

b) am TT 4= λλλ == 21 is real value, so transient speed response contains ONE time constant λ/1−=T (“aperiodic limit”).

c) am TT 4< 21, λλ are complex values dj ωδλ ⋅+−=1 , dj ωδλ ⋅−−=2 , so

transient speed response oscillates with frequency )2/( πωddf = ,

which is damped by damping coefficient δ.

aTT ≥2



a) 21 /2

/1)(

TtTtmh eCeCt

−− ⋅+⋅=Ω

long time constant:

m

a

a

T

T

TT

411

21

−−

= , short time constant

m

a

a

T

T

TT

411

22

−+

=

mTT ≤1




5/25

aTT ≥2


Long and short time constant T1 and T2

mTT ≤1Long time constant:

Short time constant:

am

m

a

m

m

a

m

a

a

m

a

a

TTTT

T

T

T

T

T

T

T

T

T

T

TT

ma−≈

+=

−−−

≈

−−

=<<

12

211

2

411

2

21

+⋅≈

−=

−+

≈

−+

=<<

m

aa

m

a

a

m

a

a

m

a

aTT T

TT

T

T

T

T

T

T

T

T

TT

ma1

1211

2

411

22

• t0

1T mT2TaT




5/26


b) Ttmh eCt /)( −⋅=Ω , aTT 2=

c) )sin()cos()( teBteAt dt

dt

mh ⋅⋅⋅+⋅⋅⋅= ⋅−⋅− ωωΩ δδ A, B: integration constants

- Damping coefficient: aT2

1=δ

- Eigen-frequency: 4

1

2

1

4

1

2

1 2

−⋅⋅

=−⋅

=aJ

a

am

a

ad

rT

T

TT

T

Tf

φππ

, Period: d

df

T1

=


- Number NH of half periods, until oscillations are damped to 5%:

δω

ωπδ

δδ d

ddH

t

T

tNte ≈==→=→=⋅−

/

/3

2//3*05.0

**




5/27


Example: Separately excited DC machine: Inertia TJ = 1 s, electric parameters: Ta = 50 ms, ra = 0.05, NΦΦ = .

- With TJ ≅ TJ0: Mechanical time constant: 501/05.01 2

2=⋅=⋅=

φa

Jm

rTT ms

- As msTmsT am 200504450 =⋅=<= , DC machine oscillates:

- frequency 76.24

1

05.0

05.0

05.02

1=−⋅

⋅=

πdf Hz

- angular frequency /3.172 == dd fπω s,

- oscillation period of 36276.2/1/1 === dd fT ms.

- Damping coefficient: /1005.02

1

2

1=

⋅==

aTδ s

- After 73.110

3.17==≈

δωd

HN half-periods the oscillation is damped

down to 5% of initial value.




5/28


Speed response to mechanical load torque stepData: 1,05.0,50,1 ==== φaaJ rmsTsT , step in shaft torque of 5.0/ =Ns MM∆

Result: msTmsTs mdd 50,363,/3.17 ===ω

MeMN/2 MN

-2.5% -5%

aa

U

aa

a

m

m

aa

mm

irIU

R

Ik

R

N

⋅−=⋅−=

⋅−=

110

20

ΩΩ

ΦΩΩ

Transient speed response:




5/29


Example: Start-up of unloaded, separately excited DC motor (ms = 0, Φ = const., 4.Ta < Tm) (1)

sma

a

aaa

ma

a

a

a mkTTdt

du

RTi

TTdt

di

Tdt

id⋅⋅

⋅+⋅

⋅=⋅

⋅+⋅+

Φ22

211111

dt

du

RTi

TTdt

di

Tdt

id a

aaa

ma

a

a

a ⋅⋅

=⋅⋅

+⋅+111

2

2

)0()0()0()0()0(0)0(,0)0(0)0( aaaiaaaaim uiRuuiLiu =−−=′⋅==⇒=Ω

sUURT

uUs

L

u

TT

I

T

IsIs aa

aa

aa

a

a

ma

a

a

aa /

)0()0(2 =−⋅

=−⋅

+⋅

+(

((((

−−

−⋅

−⋅=

++⋅=⇒=

⋅+

⋅+

21212

2 111

)/(1/

1

λλλλ ssRT

U

TTTssRT

UI

RT

U

TT

I

T

IsIs

aa

a

maaaa

aa

aa

a

ma

a

a

aa

(((

(

( )21 //

/41

2

2)( TtTt

ma

a

aa

aa ee

TT

T

RT

Uti −− −⋅

−⋅=

( )21 //

/41

/)(

TtTt

ma

aaa ee

TT

RUti

−− −⋅−

=

4.Ta < Tm: λ1, λ2 are real numbers!

e.g.: T1 ≈ Tm , T2 ≈ Tama

a

TTT

/411

21 −⋅=− λλ

)()()/(1/ 212 λλ −⋅−=⋅++ ssTTTss maa

aaaa IsiIsiL((⋅=−⋅=′ )0( aaaaaaa LuIsiisIsiL /)0()0()0( 22 −⋅=′−⋅−⋅=′′

((




5/30


Example: Start-up of unloaded, separately excited DC motor (ms = 0, Φ = const.) (2)

Time t

0)0( =ai

t = 0

0)0( =au

0)0( =n )2/()( 20 ΦπkUntn N==∞→

( )21 //

41

/)(

TtTt

m

a

aaa ee

T

T

RUti

−− −⋅

−

=

Na Utu => )0(

0)( =∞→tia

- Current rises approximately with Ta

- Speed rises approximately with Tm

- Current decreases approximately

with Tm

T1 ≈ Tm , T2 ≈ Ta

Qualitative solution:




5/31

Summary: Dynamics of coupled electric-mechanical system of separately excited DC machine

- Mechanical and electrical transients are change of speed and armature current

- DC machine may oscillate, if mechanical time constant is rather short

- In most cases the big load inertia does not allow any oscillation

- Dynamic current overshoot calculated during motor start-up





5/32





excited DC machine


DC machine








5/33

))(()()()(/)()()( 2 titttkdttdiLRtitu fmaaaaa ΦΦΦΩ =+⋅+⋅=


Linearized model for variable flux

dttdiiLRtitu ffffff /)()()()( ⋅+⋅=

)()()(/)( 2 tmtitkdttdJ sam −=⋅ ΦΩNon-linear expressions

For investigations of small disturbances the equations are linearized !

Small transient deviations ∆ua(t), ∆ia(t), ∆Ωm(t), ∆ms(t), ∆Φ(t), ∆if(t), ∆uf(t) from

the steady state operation Ua, Ia, Ωm, Ms, Φ, If, Uf !

ffff iiNL /)(Φ=




5/34


Linearized model variables

For investigations of small disturbances the equations are linearized !

Small transient deviations:

ua(t) = Ua + ∆ua(t)

ia(t) = Ia + ∆ia(t)

Ωm(t) =Ωm + ∆Ωm(t)

ms(t) = Ms + ∆ms(t)

Φ(t) =Φ + ∆Φ(t)

uf(t) = Uf + ∆uf(t)

if(t) = If + ∆if(t)




5/35


Linearized differential equations at variable flux- Small transient deviations from steady state operation: e.g.: from

- ”Small”: Per unit deviation < 10%:

- Neglect product of small deviations in voltage equation:

- Result for induced voltage ui(t) and torque me(t) :

)(tua∆ aU

1/)( <<aa Utu∆

))(

1())(

1())(())(()( 22 Φ∆Φ

Ω∆Ω

ΦΩ∆ΦΦ∆ΩΩtt

kttktum

mmmmi +⋅+⋅⋅⋅=+⋅+⋅=

Φ∆Φ

Ω∆Ω

Φ∆Φ

Ω∆Ω

Φ∆Φ

Ω∆Ω

Φ∆Φ

Ω∆Ω )()(

1)()()()(

1))(

1())(

1(tttttttt

m

m

m

m

m

m

m

m ++≈⋅+++=+⋅+

20.11.01.0101.01.01.011.01.01.01.01)1.01()1.01(21.1 =++≈+++=⋅+++=+⋅+=

)()()()()( ,,222 tutuUtktkktu imiimmmi ΦΩ ∆∆∆ΦΩΦ∆ΩΦΩ ++=⋅⋅+⋅⋅+⋅⋅≅

)()())(())(()( 2222 tIktikIkttiiktm aaaaae ∆ΦΦ∆Φ∆ΦΦ∆ ⋅⋅+⋅⋅+⋅⋅≈+⋅+⋅=




5/36

A) LINEAR differential equations, neglecting products of ∆ :


)()()()(

)()()(

)()(

22

22

tmtIktikdt

tdJ

tktkdt

tidLtiRtu

saam

mma

aaaa

∆∆ΦΦ∆∆Ω

∆ΦΩΦ∆Ω∆

∆∆

−⋅⋅+⋅⋅≈⋅

⋅⋅+⋅⋅+⋅+⋅≈

Linearized dynamic equations (1)

( ) ( )

( )ssaae

mm

mmiaa

aaaaaa

mMIkikMdt

dJ

kkUdt

iIdLiIRuU

∆∆ΦΦ∆∆ΩΩ

∆ΦΩΦ∆Ω∆

∆∆

−−⋅⋅+⋅⋅+≈+

⋅

⋅⋅+⋅⋅+++

⋅++⋅≈+

22

22

( )

ssaaem

mmia

aaaaaa

mMIkikMdt

dJ

kkUdt

idLiIRuU

∆∆ΦΦ∆∆Ω

∆ΦΩΦ∆Ω∆

∆∆

−−⋅⋅+⋅⋅+≈⋅

⋅⋅+⋅⋅++⋅++⋅≈+

22

22

0,0 ==dt

d

dt

dI ma Ω

Stationary equilibrium

cancels !




5/37


)()()()(

)()()(

)()(

22

22

tmtIktikdt

tdJ

tktkdt

tidLtiRtu

saam

mma

aaaa

∆∆ΦΦ∆∆Ω

∆ΦΩΦ∆Ω∆

∆∆

−⋅⋅+⋅⋅≈⋅

⋅⋅+⋅⋅+⋅+⋅≈

B) LINEAR differential equations of deviations :

”Small signal theory”:

Linear differential equation system is only valid within the limits of deviation from

steady state operation of about +/- 10 ... 20 %.

Constant parameters of differential equations depend on steady state values.

C) In case of linear systems NO linearization is necessary: ”Large signal theory”!

Example:

Separately excited DC machines with constant flux operation !

Linearized dynamic equations (2)




5/38

Summary: Linearized model of separately excited DC machine for variable flux

- Method of linearization of non-linear equations shown

- Linearization only possible for small disturbances („perturbation method“)

- Perturbation method leads to small signal theory

- Separately excited DC machine at constant main flux linear also for large signals





5/39





excited DC machine


DC machine








5/40

- Here: Constant flux operation (Φ = const.) = linear system !

- Initial conditions set to zero:

- Laplace transform: and so on ....


Transfer function of separately excited DC machine

0)0( =au∆ , 0)0( =ai∆ , 0)0( =m∆Ω , 0)0( =∆Φ , 0)0( =sm∆ , 0)0( =fi∆

aT/1=γ

)()0()()(

sisisisdt

tidL aaa

a((

∆⋅=∆−∆⋅=

∆

)()()(

)()()()(

2

2

smsikssJ

sksiLssiRsu

sam

maaaaa(((

((((

∆Φ∆Ω∆

ΦΩ∆∆∆∆

−⋅⋅=⋅⋅

⋅⋅+⋅⋅+⋅=

s

Φ = const.:

−⋅⋅

+⋅⋅

+⋅+

+= )()(

1)( 2

2smsu

R

k

sJ

Tss

ss sa

a

m

m(((

∆∆Φ

γγ

γγ

γΩ∆




5/41


Characteristic polynomial of 2nd order differential equation

)()(0)( 212 ssss

TsssP

m

−⋅−=+⋅+==γ

γ

- Real part of roots in the left half plane (Re(s1), Re(s2) < 0): stable operation.

- If imaginary part of roots is zero (Im(s1) = Im(s2) = 0): no oscillations occur.

- Real part of roots Re(s1), Re(s2) small: time constants are very long.- Pairs of conjugate complex roots s1, s2: sine & cosine function as transients.

- Imaginary part of root is far off origin (Im(s1) >> 1): oscillation frequency is high.

- Tangent of angle α = number NH of half periods of transient oscillation,

till damped to 5% of initial value.

Hd N==δω

αtan

s1

s2

djs ωδλ ⋅±−== 2,12,1„Roots“:




5/42


Example: Speed response to mechanical load torque step (1)

Constant armature voltage, load step after no-load operation:

Load torque step: 0)(.;)( === tuconstUtu aaa ∆

)()( tMtm ss ε∆∆ ⋅=

≥

<=

0,1

0,0)(

t

ttε

sJ

M

Tss

s

s

M

R

k

sJ

Tss

ss s

m

s

a

m

m ⋅⋅

+⋅+

+−=

−⋅⋅

+⋅⋅

+⋅+

+=

∆γ

γ

γ∆Φγ

γγ

γ

γΩ∆

2

2

20

1)(

(

Decomposition in single terms for inverse Laplace transformation: for t > 0

s

MtmL s

s

∆∆ =)(

++⋅

⋅−

+++

+−⋅

+=⋅

++

+22

22

222222 )(2)(

121

)(

2

d

d

d

d

ddd ss

s

sss

s

ωδ

ωωδδω

ωδ

δ

ωδ

δ

ωδ

δ

⋅⋅⋅

⋅−

+⋅⋅−⋅ ⋅−⋅− )sin(2

)cos(122

teteT dt

d

dd

tm ω

ωδδω

ω δδ

Laplace

domain:

Time domain:




5/43


Side calculations for inverse Laplace transform

++⋅

⋅−

+++

+−⋅

+=⋅

++

+22

22

222222 )(2)(

121

)(

2

d

d

d

d

ddd ss

s

sss

s

ωδω

ωδδω

ωδδ

ωδδ

ωδδ

⋅⋅⋅

⋅−

+⋅⋅−⋅+

⋅−⋅− )sin(2

)cos(12 22

22tete d

t

d

dd

t

d

ωωδδω

ωωδδ δδ

22222 )(

1

)(

21

dd

m

s

CBs

s

A

ss

s

s

Tss

s

ωδωδδ

γγ

γ++

++=⋅

++

+=⋅

+⋅+

+δωδ 2))(( 222 +=++++⋅ sCsBssA d

δωδδ 2)()2()( 22

2 +=+++⋅++ sACAsBAs d

δωδ

δ

2)(

12

0

22 =+⋅

=+⋅

=+

dA

CA

BA

)/()2(1

)/(2

)/(2

222

22

22

d

d

d

C

B

A

ωδδ

ωδδ

ωδδ

+−=

+−=

+=

++

−+−+⋅

+=⋅

++

+22

22

2222 )(

)2/()3(121

)(

2

d

d

dd s

s

sss

s

ωδδδω

ωδδ

ωδδ

Inverse Laplace transform




5/44


Side calculations for more elegant formula description

⋅⋅

⋅−

−⋅⋅−⋅+

⋅− )sin(2

)cos(12 22

22tte d

d

dd

t

d

ωωδδω

ωωδδ δ

m

m

a

a

m

a

m

adm

ad

a

T

T

T

T

T

T

T

TT

T

T==

−+⋅=

−⋅+

=+

−⋅==4

4

14

1

12

14

22:1

4,

2

1

2222 δ

δδ

δωδδ

δωδ

)cos(sincossin2

cos22

ψωωωωωδδω

ω −⋅=⋅−⋅=⋅⋅−

− tMtLtKtt ddddd

dd

d

dddd MLMKtMtMtM

ωδδω

ψψψωψωψω⋅−

=−===→+=−⋅2

sin,1cossinsincoscos)cos(22

222

22222222

21)sin(cos

−+=+=→+==+

d

dLKMLKMMδωδω

ψψ

d

ma

d

dd

ddd

d

TTM

δωδδ

δωδω

δωδω

δωδωδω 2

)1/4(

2)(

2

1)()2(

2

1 22222222222 +−

=+

=+⋅=−+⋅=

mdd

ma

T

TTM

ωωδ 1

2

/4=

⋅= )arccos(/1cos mdmd TTM ωψωψ =→==




5/45

Data:

1,05.0,50,1 ==== φaaJ rmsTsT , step in shaft torque of 5.0/ =Ns MM∆

Result:

msTmsTs mdd 50,363,/3.17 ===ω , 526.0)05.03.17arccos( =⋅=ψ .

- Transient speed response: (t in seconds)


Speed response to mechanical load torque step (2)

)arccos( mdTωψ =

))526.03.17cos(865.0

11(05.05.0

)( 05.02

0

−⋅⋅⋅−⋅⋅−= ⋅−

tet

t

m

m

Ω∆Ω

−⋅⋅−⋅⋅−=

−)cos(

11)(

2 ψωω

∆Ω∆ te

TT

J

Mt d

T

t

mdm

sm

a

−⋅⋅−⋅⋅−=

−)cos(

11

)( 2

20

ψωωφ

∆Ω

∆Ωte

T

r

M

Mtd

T

t

md

a

N

s

m

m a

Result:

N

a

mN

ma

m

m

M

r

JM

Jr

J

T2

0

02

0

1

φΩΩ

φΩ==




5/46


Speed response to mechanical load torque step (3)

))526.03.17cos(865.0

11(05.05.0

)( 05.02

0

−⋅⋅⋅−⋅⋅−= ⋅−

tet

t

m

m

Ω∆Ω

MeMN/2 MN

-2.5% -5%




5/47

Summary: Transfer function of separately excited DC machine

- (Linearized) differential equation is „transfer function“ in LAPLACE domain

- Step response calculated via transfer function

- Denominator of transfer function is characteristic polynomial of diff. equation

- Zeros (= “roots”) of characteristic polynomial are „poles“ of the system

in the s-plane

- Negative inverse of „poles“ = time constants or natural frequencies of the system

- Positive time constants for stable operation needed

- Hence poles must lie in the negative s-half-plane





5/48





excited DC machine


DC machine








5/49


Input for dynamic simulations

- General model calculation possible via numerical integration

- Here: Only: a) Linearized non-linear magnetization characteristic

„flux vs. field current“

b) Separately excited machine

c) Constant flux operation

ϕϕ ΦψΦ kNLiLNik fffffff ⋅=⇒⋅=⋅=⇒⋅=




5/50

The function uf(t) must be given for t ≥ 0 ! Initial condition:


First order differential equations for RUNGE-KUTTA method, here for Φ = const.

)()(

)()( tudt

tdiLtiRtu i

aaaaa +⋅+⋅=

)()( tmtmdt

dJ se

m −=⋅Ω

)()()()( 2 t

L

k

L

tiR

L

tu

dt

tdim

aa

aa

a

aa ΩΦ

⋅−⋅

−=⇒

J

tmti

J

k

dt

d sa

m )()(2 −⋅=

ΦΩ⇒

Initial conditions:

The functions ua(t), ms(t) must be given for t ≥ 0 !

In case of varying flux add the third differential equation:

)0(),0( mai Ω

dt

tdiLtiRtu

fffff

)()()( ⋅+⋅=

)(

)(

)(

)()(

ff

ff

ff

ff

iL

tiR

iL

tu

dt

tdi ⋅−=⇒

)0(fi




5/51


Example: Separately excited DC machine: motor data, fed from ideal DC voltage (“stiff battery”):

27,5.6,320

min/625,142,460

mkgJAIAI

nkWPVU

MfNN

NNN

⋅===

===

HLRmHLR ffaa 64,25,5.1,05.0 ==== ΩΩ

Two cases investigated:

a) Load step with rated torque at no-load speed, rated armature voltage

and rated flux

b) Armature voltage step of 20% rated voltage at rated motor operation

Load inertia: (i): 8 kgm2 Total inertia: J = 15 kgm2 = JN

(ii): 143 kgm2 Total inertia: J = 150 kgm2




5/52


Nmn

PM

N

NN 6.2169

)60/625(2

142000

2=

⋅=

⋅=

ππRated torque:

VRIUU aNNi 44405.0320460 =⋅−=⋅−=Induced voltage at rated speed and torque:

VsU

kmN

iN 78.6

)60/625(2

4442 =

⋅==

πΩΦMotor constant and flux per pole:

Motor efficiency: %78.955.625320460

14200022=

⋅+⋅=

+⋅=

fNfNN

N

IRIU

Pη

No-load speed at rated armature voltage and main flux: min/9.64778.62

60460

2 20 =

⋅⋅

=⋅

=πΦπ N

N

k

Un

Steady state characteristics of the DC motor at UN

n

1/min

0 M/MN

647.9/min625/min

1

Case a)

Load step with rated torque at no-load

speed, rated armature voltage and fluxLoad step by 100%




5/53


VRIUU aNNi 53605.03205522.1 =⋅−=⋅−⋅=Induced voltage at rated torque:

min/7552

60

78.6

536

2

=⋅=→=πΦ

Ω nk

U

N

imMotor speed at constant flux/pole:

No-load speed at 120% armature voltage & main flux: min/9.77778.62

60552

2

2.1

20 =

⋅⋅

=⋅

=πΦπ N

N

k

Un

Steady state characteristics of the DC motor at 1.2UN

n

1/min

0 M/MN

647.9/min625/min

1

777.9/min

755/minx 1.2

Case b)

Armature voltage step of 20% rated

voltage at rated motor operation

Voltage step by 20%

130/min




5/54


,56.225

64,30

05.0

0015.0s

R

LTms

R

LT

f

ff

a

aa ======

sk

JRT

N

Nam 0163.0

78.6

1505.0

)( 222

=⋅

=⋅

=Φ

(i) at JN = 15 kg.m2

sk

JRT

N

am 163.0

78.6

15005.0

)( 222

=⋅

=⋅

=Φ

(ii) at J = 150 kg.m2

Calculation of time constants

• t0 msTii m 163:)( =msTa 30=

msTi m 3.16:)( =




5/55


Variation of total inertia: (i) J = 15 kgm2, (ii) J = 150 kgm2

(i) Total inertia 15 kgm2: msTmsT am 12043.16 =<= : damped oscillations occur:

Damping coefficient: sTa

/67.1603.02

1

2

1=

⋅==δ .

Natural frequency: HzT

T

Tf

m

a

ad 69.6

4

1

0163.0

03.0

03.02

1

4

1

2

1=−

⋅=−

⋅=

ππ

Period of oscillation is msf

Td

d 5.14969.6

11===

After 5.267.16

69.62=

⋅==

πδωd

HN half periods the oscillation is reduced down to 5%.




5/56


(ii) Total inertia increased by factor 10: 150 kgm2

msTmsT am 1204163 =>= : no oscillations occur:

long time constant: ms

T

T

TT

m

a

a 3.123

163

30411

302

411

21 =

⋅−−

⋅=

−−

= ,

short time constant ms

T

T

TT

m

a

a 6.39

163

30411

302

411

22 =

⋅−+

⋅=

−+

=

Variation of total inertia: (i) 15 kgm2, (ii) 150 kgm2

• t0 msT 1231 = msTm 163=

msT 392 =

msTa 30=




5/57


a) Load step with rated torque at no-load speed, rated armature voltage and flux

At low total inertia DC machine is oscillating

At big total inertia no oscillation occurs

2169.6 Nm

Damping completed after 2.5 half

cycles

msTm 163=

msTd 5.149=

)mkg150( 2⋅=J

)mkg15( 2⋅=J




5/58


a) Load step with rated torque at no-load speed, rated armature voltage and flux

At low total inertia DC machine is oscillating

At big total inertia no oscillation occurs

647.9/min625/min

320 A

Armature current is

proportional to torque me

)mkg150( 2⋅=J

)mkg15( 2⋅=J

)mkg150( 2⋅=J

)mkg15( 2⋅=J




5/59



Big armature current = big torque: to accelerate inertia to higher speed, demanded by higher armature voltage

Big inertia J = 150 kgm2

320 A320 A

460 V 552 V =1.2 x 460 V

625/min755/min

1750 A

msTm 163=

No oscillations




5/60


Small inertia J = 15 kgm2


Smaller peak armature current (1180 A < 1750 A)= smaller peak torque: to accelerate the lower inertia to higher speed

Oscillation occurs320 A320 A

460 V552 V

625/min755/min

1180 A

Damping completed

after 2.5 half cycles

Oscillation occurs

msTd 5.149=




5/61

Summary: Dynamic simulation of separately excited DC machine

- Linear differential equation is solved numerically via RUNGE-KUTTA

- Parameter variation: Big vs. small inertia = without / with oscillations

Usually: Load inertia big, so no oscillations occur!

- Step response to torque as disturbing signal

- Step response to armature voltage as commanding signal

- Further example in the text book with variable series resistor





5/62





excited DC machine


DC machine








5/63


Converter-operated separately excited DC machine

αα cos)( 0 ⋅= dd UU 23

0 ⋅⋅= LLd UUπ

B6C bridge

Rectified voltage from the grid:

rectified voltage

Six-pulse ripple

ULL

T = 1/f0T/6

ft πωω∆α 2, =⋅=

∆t

0)0( dd UU ==α

B: bridge6: 6-pulseC: controlled Thyristor control: Ud > 0, Ud < 0 possible




5/64


Armature current with six-pulse ripple due to B6C converter

Measured armature current

DC motor 40 kW

Separately excited

Fed by B6C armature thyristor bridge

Six-pulse ripple = 6x50 = 300 Hz

Armature inductance is small: La ~ Na

2 ~ z2, so ia-ripple is big!

Six-pulse rippleT/6

Calculation example in “Collection of Exercises”

Iad = 124 A

ia(t)

T = 20 ms, f = 1/T = 50 Hz

T/2 = T =




5/65


B2H-converter (cheap solution)

•

•• •

ugrid(t)

uf(t)

Lf & Rf

T

T

D

D

ugate(t)

α

α

D: diode

T: thyristor

Due to diodes: only Uf > 0 possible

B: bridge2: 2-pulseH: half-controlled




5/66


Field current: Two-pulse ripple due to B2H-converter

Measured field current

DC motor 40 kW

Separately excited

Field current fed by B2H thyristor bridge

Two-pulse ripple = 2x50 = 100 Hz

Field inductance is big:Lf ~ Nf

2, so if-ripple is small!

Measured field voltage

αf

Two-pulse ripple

Six-pulse ripple due to armature reaction

if(t)

uf(t)

T = 20 ms, f = 1/T = 50 Hz

T/2 = T =

T/2




5/67


Magnetic coupling of stator and armature field via iron saturation

Armature and stator field add= increased yoke flux density

= increased saturation

increased saturation

Armature field ~ iaStator field ~ if

Facit: Due to the magnetic common path in the yoke a change of the armature current causes a change

of yoke iron permeability. Hence also the stator field changes and induces a voltages in the field coil,

which causes a (small) six-pulse field current ripple.

ia(t)

(small) six-pulse field current

ripple




5/68


Armature 0,0 >>< aa IU

B6C: One six-pulse, voltage controlled thyristor bridge, 6 thyristors. Voltage and current ripple: 6f = 300 Hz.

0,0 ≥≥≤ Mn , but often only:

One quadrant operation: 0,0 ≥≥ Mn

Armature 0,0 ><>< aa IU

(B6C)A(B6C): Two anti-parallel six-pulse, voltage controlled thyristor bridges, 12 thyristors Voltage and current ripple: 6f = 300 Hz Four quadrants operation:

Field B2H: One two-pulse, voltage controlled thyristor-diode bridge, 2 thyristors, 2 diodes Voltage and current ripple: 2f = 100 Hz

Field

B2C: One two-pulse, voltage controlled thyristor bridge, 4 thyristors Voltage and current ripple: 2f = 100 Hz

Field B6C: One six-pulse, voltage controlled thyristor bridge, 6 thyristors Voltage and current ripple: 6f = 300 Hz

Different configurations of controlled rectifier bridges for armature and

field circuit of separately excited DC machines

0,0 ><>< Mn

0,0,0 ≥≥≥ Φff IU

0,0,0 ≥≥>< Φff IU

0,0,0 ≥≥>< Φff IU




5/69


Four-quadrant operation

0

nU a , 0,0 ≥≥ ΦfI

ea MI ,

Motor operation

Motor operation

Generator operation

Generator operation

1st quadrant2nd quadrant

3rd quadrant 4th quadrant




5/70


Parallel B6C bridge operation

αGR

αWR

°180

tω

tω

0

0

GRdU αcos0

WRdU αcos0−

WRdGRd UU αα coscos 00 −=

GRWRWRGR αααα −°=⇒−= 180coscos

Facit: With the condition both bridges deliver the same average

voltage and can therefore operate in parallel on the DC machine.GRWR αα −°=180

Two anti-

parallel bridges

operate with

the same

average output

voltage in

parallel.




5/71


(B6C)A(B6C)-bridge operation

If both anti-parallel bridges are on at the same time, a

parasitic circular current iK flows via both bridges.

Additional chokes LK are limiting iK

A: anti-parallel

WRGRWRGR uuuu +=−− )(2TK




5/72


Exact calculation of parasitic circular current iK

KKKKKKK TtTTtUdtdiLtu ≤≤−⋅−=⋅= )sin(/)sin(/)( ωω

( )CtTL

ULdttuti

KK

KKKK −⋅

⋅=⋅= ∫ )cos(

)sin(/)()( ω

ωω0)( =−= KK Tti )cos( KTC ω=

)sin(

)cos()cos()(

K

K

K

KK

T

Tt

L

Uti

ωωω

ω−

⋅=

For small values TK the parabolic approximation of iK is derived!

)sin(~)( ttuK ω

)sin(

)cos(1)0(ˆ

K

K

K

KKK

T

T

L

Utii

ωω

ω−

⋅===

)cos(~)( ttiK ω

6/2 TTK ≤UK

2TK




5/73


Parabolic approximation of small parasitic circular current iK

K

KKK

L

TUi

2ˆ =

UK

t

t

T/6

2TK

uK

iK

-TK 0 TK

)sin(

)cos(1ˆ

K

K

K

KK

T

T

L

Ui

ωω

ω−

⋅=

K

KK

K

K

K

K

K

K

K

KK

L

TU

T

T

L

U

T

T

L

Ui

2

)2/)(1(1

)sin(

)cos(1ˆ2

=−−

⋅≈−

⋅=ωω

ωωω

ω

For small values of α: 16/2 <<⇒<< KK TTT ω




5/74


Three-phase AC 400 V grid, B6C-bridge:

maximum voltage: VUUUU LLddd 54024003

23

0cos)0( 00 =⋅⋅=⋅⋅==⋅==ππ

α

rated voltage: 4602

354030cos)30( 0 =⋅=°⋅=°= dd UU α V

voltage margin for voltage control: VVVUd 80460540 =−=∆So thyristor bridge is operated between .

Example:Three-phase AC 400 V grid, usual rated and maximum voltages:

Maximum voltage Rated voltage

B6C 540 V 460 V 1 quadrantoperation

(B6C)A(B6C) 540 V 400 V 4 quadrantoperation

Voltage limits of converter-operated DC machines

°<<° 15030 α




5/75

Φ⋅≈+= nURIUU iaaia ~ nIuIM aRa ⋅⋅ ~~ Φ


Limits of converter-fed DC machine operation

Nnn ≤≤0 Voltage controlled DC machine:

Limits: Maximum armature current ia, maximum main flux Φ

RN nnn ≤≤ Flux controlled DC machine (Field weakening): Φ ↓ Limits: Maximum armature current ia and voltage ua

maxnnnR ≤≤ Flux controlled DC machine (Field weakening): Φ ↓ Limits: Maximum armature voltage ua and maximum reactance voltage of commutation uR,max = 12 V




5/76


Time for reversal of armature current.

a) (B6C)A(B6C) b) (B6C)A(B6C) c) Mechanical switch

Both bridges always active Only one bridge active Polarity changer

< 0.5 ms 5 ... 10 ms 50 ... 1500 ms

The larger numbers correspond with larger drives of several hundreds of kW up to MW range.

Time for reversal of field current.

a) (B6C)A(B6C) b) Mechanical switch

0.5 ... 2 s 1 ... 2.5 s

The larger numbers correspond with larger drives of several hundredsof kW up to MW range.

Typical response times of converter-fed DC machines

aaafff RLTRLT // =>>=

Fast DC machine reaction via change of ia, NOT via change of if!




5/77

Summary: Converter operated separately excited DC machine

- Thyristor converter operation of DC machine is a dynamic operation

- Armature current time signal may be calculated analytically

(see: Collection of Exercises)

- Anti-parallel thyristor rectifier for reversed torque

- Circulating parasitic current between the two thyristor bridges possible

- Limiting operation curves for variable speed DC machine

- Typical dynamic response times of controlled DC machines rise with increased

machine size

- Fast control via change of armature current due to Ta << Tf


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