Chapter 9: Ideal Transformer - Oakland...

10/9/2003 Electromechanical Dynamics 1

Chapter 9: Ideal Transformer


Introduction

• Transformers are one of the most useful electrical devices– provides a change in voltage and current levels

– provides galvanic isolation between different electrical circuits

– changes the apparent magnitude value of an impedance


Voltage Induction

• For a coil consisting of N turns placed in a time-varying sinusoidal flux, the flux induces a sinusoidal ac voltage

– the rms value of the voltage

where

• f is the sinusoidal frequency

• Φmax is the peak flux as defined by

– the peak flux is useful for working with iron cores and assessing the impact of losses and saturation

max

max

44.4

2

Φ=Φ=

Nf

NfE π

( )φπ +Φ=Φ f2sinmax

coreABmaxmax =Φ

( )dt

tdNte

φ=)(


Applied Voltage

• Consider a coil connected across an AC voltage source– the coil and source resistances are negligible

– the induced voltage E must equal the source voltage; KVL

– a sinusoidal AC flux Φ must exist to generate the induced voltage on the N turns of the coil

• Φmax varies in proportion to Eg

• placing an iron core in the coil will not change the flux Φ

– magnetization current Im drives the AC flux

• the current is 90° out-of-phase and lagging with respect to the voltage

• with an iron core, less current is needed to drive the AC flux

Nf

Eg

44.4max =Φ


Induced Voltages

• Example– a coil, having 4000 turns, links an AC flux with a peak value

of 2 mWb at a frequency of 60 Hz• calculate the rms value of the induced voltage

• what is the frequency of the induced voltage?

• Example– a coil, having 90 turns, is connected to a 120 V, 60 Hz source

– the rms magnetization current is 4 A• calculate the peak value of the flux and the mmf

• find the inductive reactance and the inductance of the coil


Elementary Transformer

• Consider an air-core coil– excited by an AC source Eg

– draws a magnetization current Im

– produces a total flux Φ• A second coil is brought close to

the first– a portion Φm1 of the flux couples

the second coil, the mutual flux

– an AC voltage E2 is induced

– the flux linking only the first coil is called the leakage flux, Φf1

• Improved flux coupling– concentric windings, iron core

– weak coupling causes small E2

• the magnetization current Im

produces both fluxes Φm1 and Φf1

– the fluxes are in-phase

– the voltages Eg and E2 are in-phase

– terminal orientation such that the coil voltages are in-phase are said to possess the same polarity


Ideal Transformer

• An ideal transformer– transformer has no losses

– core is infinitely permeable

– all fluxes link all coils

– there are no leakage fluxes

• Voltage relationship– consider a transformer with two coils

of N1 and N2 turns

– a magnetizing current Im creates a flux Φm

– the flux varies sinusoidally and has a peak value of Φmax

– the induced voltages are

– from these equations, it can be deduced that

– the ratio of the primary and secondary voltages is equal to the ratio of the number of turns

• E1 and E2 are in-phase

• polarity marks show the terminal on each coil that have a peak positive voltage simultaneously

max22

max11

44.4

44.4

Φ=Φ=

NfE

NfE

aN

N

E

E ==2

1

2

1


Ideal Transformer

• Current relationship– let a load be connected across the

secondary of an ideal transformer

– current I2 will immediately flow

– coil voltages E1 and E2 cannot change when connected to a fixed voltage source and hence flux Φm

cannot change

– current I2 produces an mmf

– if mmf2 acts along, it would profoundly change Φm

– Φm can only remain fixed if the primary circuit develops a mmf which exactly counterbalances mmf2

– current I1 must flow such that

• I1 and I2 must be in-phase

• when I1 flows into the positive polarity marking of the primary, I2 flows out of the positive polarity marking of the secondary

ZEI 22 =

222 INmmf =

aN

N

I

I 1

1

2

2

1 ==


Ideal Transformer

• Ideal transformer model– let a = N1/N2

– then E2 = E1 / a and I1 = I2 / a

E1 E2

++

I1 I2Φm


Ideal Transformer

• Example– a not so ideal transformer has 200 turns in the primary coil

and 10 turns in the secondary coil• the mutual coupling is perfect, but the magnetization current is

1 A

– the primary coil is connected to a 480 V, 60 Hz source

– calculate the secondary rms voltage, peak voltage

• Example– for the transformer above, a load is connected to the

secondary coil that draws 80 A of current at a 0.8 lagging pf

– calculate the primary rms current and draw the phasor diagram


Impedance Ratio

• Transformers can also be used to transform an impedance– the source sees the effective

impedance

– on the other side, the secondary winding of the transformer sees the actual impedance

– the effective impedance is related to the actual impedance by

11 IEZ X =

22 IEZ =

ZaI

Ea

aI

Ea

I

EZ X

2

2

22

2

2

1

1 ====


Shifting Impedances

• Impedances located on the secondary side of a transformer can be relocated to the primary side– the circuit configuration remains

the same (series or shunt connected) but the shifted impedance values are multiplied by the turns ratio squared

• Impedance on the primary side can be moved to the secondary side in reverse manner– the impedance values are

divided by the turns ratio squared


Shifting Impedances

• In general, as an impedance is shifted across the transformer– the real voltage across the

impedance increases by the turns ratio

– the actual current through the impedance decreases by the turns ratio

– the required equivalent impedance increases by the square of the turns ratio

• Example– using the shifting of impedances

calculate the voltage E and current I in the circuit, knowing that the turns ratio is 1:100


Ideal Transformer

• Homework– Problems: 9-4, 9-6, 9-10*

Note: problem 9-10 is a design problem

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