2. Transformers Modified

8/12/2019 2. Transformers Modified

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CHAPTER 2 TRANSFORMERS

2.1 Single-phase Transformer

A transformer is a device that couples two or more windings through a

common magnetic core.

Usages:

In power systems to step-up or step-down the voltage.

In low-power low-current electronic circuits.

A 1:1 transformer is used to isolate grounds of two electronic circuits

2.1.1 No-Loa Coni!ions

Assume ideal transformer.

i

21NN

+

e

+

v Open

20 1010

10

20

10

2.5All dimensions are in cms

Core type transformar

Figure 2-1 wo-winding transformer

For ideal transformer! v " e

1#


2/26

Faraday$s law

mrms

mrms

mm

m

m

m

fNE

fNE

NE

where

tEte

or

tNte

Then

tt

Ifdt

de

1

1

1

1

%%%

2

2

.

"os#$

"os#$

sin#$

=

=

=

=

=

=

=

mwb

turnsN

Hzf

VV

ge

fN

V

Therefore

VE

m

m

m

rmsrms

.

%%.

&'e(ill"ore!heinfl)*!heThen

..

.

+

&'

&'''('%%%

2%'

&'''

('

2%'

%%%

1

1

=

=

=

=

=

=

=

E*er"ise: )etermine the relative permea*ility of the a*ove core if the

current is 1'mA.

2.1.2 ,o! Conen!ion

1. )etermination of the dots on a two-winding transformer

2'


3/26

2

1

1i

1

1

b

a

2

2

b

a

21NN

2i

Figure 2-2 wo-winding transformer dot determination

a. Assign a dot to one of the windings. +et a1*e the dotted terminal

of winding N1.

*. Assume that current i1is into the dotted terminal a1of winding N1.

c. +et a current i2 into a2terminal of winding N2.

d. ,hec the direction of the flues that are provided *y the two

windings using the right-hand rule.e. If the flues 1and 2are supporting one another! i.e. in the same

direction! a2is the dotted terminal of winding N2. If not! then *2is

the dotted terminal.

2. )etermination of the direction of the currents in the two dotted windings.

A current i1entering a dotted terminal of winding N1induces a voltage

dt

diMe 1

2 = with the positive polarity at the dotted terminal of winding N2

which is *2.If a load is connected to N2 winding! it will draw a current i2that is leaving the

dotted terminal *2.

Figure 2-& wo-winding ideal transformer with given dots

2.1. /eal Transformer

21


4/26

1i

21NN Load

2i

+

1v

+

2v

+

1e

+

2e

Figure 2-% Ideal transformer

For the ideal transformer! we have the following assumptions:

1. he leaage flu of the two windings is neglected.

2. he two windings resistances are neglected.

&. he relative permea*ility of the core is so high that negligi*le mmf isre/uired to esta*lish the flu in the core.

%. ,ore losses are negligi*le.

Using Fraday$s law!

0dt

dN

dt

de

1

1

1 ==

Using assumptions 1 and 2!

011 ve =

Also!

0dt

dN

dt

de

2

2

2 =

and!

022

ve =

therefore!

Ne

Ne

2

2

1

1==

dtd

or

N

N

e

e

2

1

2

1

2

1==

he net mmf acting on the core at any instant is

22


5/26

!)rn- A1 221 iNiNF =

Using assumption &! i.e. " !

'==A

lR

therefore0

1

'221 === RiNiNF

and1 221 iNiN =

or

1

2

2

1

N

N

i

i=

21

2211

1

2

2

1

pp

iviv

or

N

N

i

i

=

=

==

1

2

For ideal transformer! instantaneous power in"instantaneous power

out.

2.1. Sin)soial E*"i!a!ion

ince the sources used for transformers are usually sinusoidal! phasor

representation can *e used for the ideal transformer.

1I 21

NN

2I

+

1V

+

2V

+

1E

+

2E L

LoadIdeal

Transforme

Source

Figure 2- Ideal transformer with sinusoidal ecitation

2&


6/26

sie.primar!he!oreferreimpean"eloa

sieprimar!he!oreferre")rren!Se"onarN

N/

Also+

000

NN

00

sieprimar!he!oreferre-ol!ageSe"onar

N

N

E

E

3

31

1

3

1

21

3

221

2

1

2

1

3

2

1

2

1

=

=

=

==

==

==

=

==

=

L

LL

N

N

IN

N

VN

N

I

V

II

N

N

EEN

NE

2

2

1

2

2

2

2

1

1

1

1

22

2

1

22

2

1

1

he ideal transformer with the load impedance circuit will reduce to:

1I

+1V

3

L

LoadSource

Figure 2-( +oad impedance of an ideal transformer referred to the primary

side

E*ample 2-1

An ideal transformer rated 2%'412' 5 is supplying a load impedance of

6"&78% 9hms at 12' 5. )etermine the secondary current! primary current!

impedance seen *y the source! the load real and reactive power! the load

power factor! and the supply real power .

2%


7/26

2.1.4 E5)ialen! Cir")i! of a Real Transformer

Assumptions for a eal ransformer:

1. ;ach winding resistance may *e represented *y a lumped parameter at

the terminal of the winding.

2. Flu produced *y the mmf of one of the windings may *e divided into

two distinct parts:

2


8/26

a. +eaage flu lining all the turns of the winding producing it! *ut

none of the turns of the other winding.

*. ), model of the magnetic core

2(


9/26

?rimary @inding N1 B econdary @inding N2B

1. Flu +inage

111 N=

1. Flu +inage

222 N=

2. C,+ Flues

11

lm

+= 2. C,+ Flues

22

lm

=

&. Faraday$s +aw

dt

dN

dt

de 1

1

1

11

==

dt

dN

dt

dNe

lm 11111

+=

&. Faraday$s +aw

dt

dN

dt

de 2

2

2

22

==

dt

dN

dt

dNe

lm 22222

=

%. +eaage Flu induced voltage

1111

iLN ll =

dt

diL

dt

dNe l

l

l1

11

1

1

==

%. +eaage Flu induced voltage

2222

iLN ll =

dt

diL

dt

dNe

l

l

l2

22

2

2

==

.


10/26

+et im*e the current in N1re/uired to produce the mutual flu Dm. his

current is nown as the magnetiEing current of the transformer.

im1 1221

NRiNiNF!mnet

===

herefore0

im1 1221 NiNiN +=

and

m1 ii

N

Ni +=

2

1

2

or m3

1 iii +=2

Figure 2-1' ;/uivalent circuit with the ideal transformer *etween primary

and secondary circuits

+mis the magnetiEation inductance of the transformer.

m

iLNmmm

==1

m

mm

i

NL

1

=

m"m RiN =1

1

"mmN

Ri =

"

2

1

R

N

N

R

NL

!m

mm ==

1

1

2.1.6 Real Transformer (i!h Sin)soial E*"i!a!ion

If v1tB is a sinusoidal voltage and the load is linear! we get the following

e/uivalent circuit:

2>


11/26

Figure 2-11 ;/uivalent circuit in the fre/uency domain

@here0

1111

2 "$fL%L" lll ===

2222

2 "$fL%L" lll ===

N

NII

1

2

22 =

3

eferring all secondary parameters to the primary side! we get the followinge/uivalent circuit:

Figure 2-12 ;/uivalent circuit referred to the primary side

;act e/uivalent circuit referred to the primary side.

EN

NEE

1

2

1

22 ==3

N

NVV

2

1

22 =3

N

NII

1

2

22 =

3

N

NRR

2

2

1

22

=3

N

N""

2

2

1

22

=3

2#


12/26

N

N LL

2

2

1

=3

2222 "osIV&&Load ==

2

1

2

2

2

1

22 "osN

NI

N

NV&&Load ==

2222 "os33 IV&&

Load ==

2.1.7 Phasor ,iagram

#$

#$

3

33333

11111

21

1

222212

'"RIEV

III

'"EI

'"RIVEE

m

m

m

++=

+=

=

++==

Figure 2-1& ;/uivalent circuit phasor diagram

2.1.8 Transformer Core Losses

@ith no load connected to the secondary side of the transformer! the

secondary current is Eero. his means that the primary current supplied to

the transformer is primarily to magnetiEe the core and to supply the core

losses.

&'


13/26

Figure 2-1% ingle-phase transformer with no load connected

i.e. II e(=1ince Ie( is flowing in an inductive circuit! it lags the applied voltage *y a

certain angle NL .

Figure 2-1 No-load transformer phasor diagram.Ie( has two components! magnetiEing component! Im ! that produces the

flu in the core and core loss component! !I .

m!e( III +=

IINLe(! "os= and II NLe(m sin=

III m!e(22

+= and!

m

NLI

I

1= !an

he flu will *e modeled *y a magnetiEing reactance mand the core loss

will *e modeled *y a resistance c.

he following is the complete eact e/uivalent circuit referred to the

primary side:

&1


14/26

Figure 2-1( ;act e/uivalent circuit.

[ ]

#$

#$

3

33333

11111

21

11

1

1

222212

11

'"RIEV

III

')*E"'

REIII

R

EI

'"

EI

'"RIVEE

e(

m!

m!

m!e(

!

!

m

m

++=

+=

=

=+=

=

=

++==

E*ample 2-2

A 12 pf lagging.

)etermine the following:

1B Input 5oltage 2B Input current &B Input power factor %B 5oltage

egulationB ;fficiency (B All-day efficiency for the following load cycle:

12 hours full-load at '.> pf and 12 hours no-load .

&2


15/26

2.1.= Appro*ima!e E5)ialen! Cir")i!s

1. he magnetiEing impedance moved across the source

Figure 2-1= Approimate circuit with magnetiEing impedance across the

source.

3

3

33

3

#$

21

21

221

211

"""

RRR

'"RIVV

EEV

e+

e+

e+e+

+=

+=

++=

==

he approimation is that the ecitation current is not flowing in the primary

impedance 1781B.

&&


16/26

2. he magnetiEing impedance is across the load

Figure 2-1> Approimate circuit with magnetiEing impedance across the

load.

#$3

33

e+e+ '"RIVV

VEE

++=

==

121

221

he approimation is that the ecitation current is flowing in the secondaryimpedance $278$2B.

&. Neglecting the ecitation current

Figure 2-1# ;/uivalent circuit with ecitation current neglected.

&%


17/26

#$3

3

e+e+ '"RIVV

II

++=

=

121

21

he approimation is that the ecitation current is too small compared to

rated currentand can *e neglected.

%. Neglecting the e/uivalent resistance

Figure 2-2' ;/uivalent circuit with windings$ resistances neglected.

#$3

3

e+'"IVV

II

121

21

+=

=

he e/uivalent resistance of large transformers several


18/26

and since0

m!

m!

'"'"R

therefore

'"R'"R

>>R!o"omparenegle"!eis

#>>$#$

11

11

+


19/26

Figure 2-2& hort circuit test

he primary and secondary impedances are connected in series and can

*e com*ined together to form an e/uivalent impedance.

Figure 2-2% ;/uivalent circuit for short circuit test.

3

3

21

21

"""

RRR

e+

e+

+=

+=


20/26

Open-"ir")i! !es!


21/26

1B )etermine the parameters of the approimate circuit referred to the

low voltage side +5B!

2B ;press the ecitation current as a percentage of the rated current!

&B )etermine the efficiency of the transformer at rated-load and '.> pf

lagging!

%B )etermine the voltage regulation of the transformer at rated-load and

'.> pf lagging!

B )etermine the all-day efficiency of the transformer for the following

load cycle:

> hours full-load and '.> pf lagging

> hours 'H full-load and '.> pf lagging

> hours H full-load and '.> pf lagging.

2.2 Three-phase Transformer Conne"!ions

2.2.1 ?e-(e "onne"!ion


22/26

AI21

NN

aI

+

ANV

)I bI

-I !I

+

)NV

+

-NV

+

A)V

+

)-V

+

-AV

+

abV

+

b!V

+

!aV

+

anV

+

bnV

+

!nV

A

)

-

N n

a

b

!

Figure 2-2 hree-phase transformer - connection

Figure 2-2( ?hasor diagram for - connection.

ab

A)

A

a

an

AN

V

V

I

I

N

N

V

V===

2

1

anabANA) VVVV && == an

For a wye-connected circuit0

phLphL IVV == /an&

here is no phase shift *etween the line voltages of the primary and

secondary windings of a wye4wye connected three-phase transformer. i.e.phase-inarean abA) VV .

%'


23/26

&&&

&

&

&

&

&

&&

&

'.&/

IVIV.

IVIV

IVIV

IVIV

IVIV&

AANLL

AANLL

AANAA)

AANAA)

AANAAN

+=

=

=

=

=

=

#sin$

#"os$

#"os$

#"os$

#"os$

2.2.2 ,el!a>el!a "onne"!ion

A)I21

NN

abI

)-I b!I

-AI!aI

+

A)V

+

-AV

+

abV

+

b!V

+

!aV

A

)

-

a

b

!

AI

)I

-I

aI

bI

!I

Figure 2-2= hree-phase transformer delta-delta connection

Figure 2-2> ?hasor diagram for delta-delta connection.

A

a

A)

ab

ab

A)

I

I

I

I

N

N

V

V===

2

1

=== =


24/26

here is no phase shift *etween the line voltages of the primary and

secondary windings of a )elta4delta connected three-phase transformer.

i.e. phase-inarean abA) VV .

2.2. ?e>el!a Conne"!ion

21NN

abI

b!I

!aI

+

A)V

+

-AV

+

abV

+

b!V

+

!aV

A

)

-

a

b

!

AI

)I

-I

aI

bI

!I

+

)V

+

ANV

+

)NV

+

-NV

Figure 2-2# hree-phase transformer -J connection.

Figure 2-&' ?hasor diagram for -J connection.

here is a &'K phase shift *etween the line voltages of the primary and

secondary windings of a @ye4delta connected three-phase transformer.

i.e.


25/26

ANA)

ab

AN VVN

N

V

V&

2

1 == an

2

1

& N

N

V

V

ab

A) =

2

1&N

N

V

V

ab

A) =

=

ph

ph,

L

L,

N

N

V

V&

aba

ab

A IN

N

I

I&

1

2== /an

1

2

&N

N

I

I

a

A=

ph,

ph

L

L,

N

N

I

I

&

=

2.2. Open ,el!a

21NN

+

A)V

+

)-V

+

abV

+

b!V

+

!aV

A

)

-

a

b

!

AI

)I

-I

aI

bI

!I

+

-AV

Figure 2-&1 9pen delta connection.

In a delta4delta connection! if one of the three transformers is removed!

the remaining two transformers can still provide >H of the delta4delta

three-phase power.

)elta4delta 9pen delta

phL VV =

phL II &=

pfIV& LL&& =

phL VV =

phL II =

pfIV&LL

&& =

%&


26/26

pfIV& phph&& = pfIV& phph&& =

>'&

1.

P

P

>-

open- ==

E*ample 2-

A three-phase transformer supplies a load of = C5A at %%' 5 line-to-

line and unity power factor. he primary to secondary phase turns ratio is

2:1. )etermine the phase and line voltages and phase and line currents of

the primary and secondary of the transformer for the following

connections:

aB )elta4delta!

*B @ye4delta!cB @ye4wye. Assume ideal transformers.

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2. Transformers Modified

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