Chapter 9: Ideal Transformer
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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
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Voltage Induction
• For a coil consisting of N turns placed
in a time-varying sinusoidal flux, the
flux induces a sinusoidal ac voltage
dφ (t )
e( t ) = N
dt
– the rms value of the voltage
E = 2π f N Φ max
= 4.44 f N Φ max
where
– the peak flux is useful for
• f is the sinusoidal frequency
working with iron cores and
• Φmax is the peak flux as defined by
assessing the impact of losses and
saturation
Φ = Φ max sin(2π f + φ )
Φ max = Bmax Acore
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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
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Φ max =
Eg
4.44 f N
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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
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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
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• the magnetization current Im
produces both fluxes Φm1 and Φf1
– the fluxes are in-phase
– the voltages Eg and E2 are inphase
– terminal orientation such that the
coil voltages are in-phase are
said to possess the same polarity
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Ideal Transformer
– the induced voltages are
• An ideal transformer
–
–
–
–
transformer has no losses
core is infinitely permeable
all fluxes link all coils
there are no leakage fluxes
E1 = 4.44 f N1 Φ max
E2 = 4.44 f N 2 Φ max
– from these equations, it can be
deduced that
• Voltage relationship
E1 N1
=
=a
E2 N 2
– consider a transformer with two coils
of N1 and N2 turns
– the ratio of the primary and secondary
– a magnetizing current Im creates a
voltages is equal to the ratio of the
flux Φm
number of turns
– the flux varies sinusoidally and has a
• E1 and E2 are in-phase
peak value of Φmax
• polarity marks show the terminal
on each coil that have a peak
positive voltage simultaneously
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Ideal Transformer
• Current relationship
– let a load be connected across the
secondary of an ideal transformer
– current I2 will immediately flow
I 2 = E2 Z
– Φm can only remain fixed if the
primary circuit develops a mmf
which exactly counterbalances
mmf2
– current I1 must flow such that
– 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
mmf 2 = N 2 I 2
I1 N 2 1
=
=
I 2 N1 a
• 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
– if mmf2 acts along, it would
profoundly change Φm
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Ideal Transformer
• Ideal transformer model
– let a = N1/N2
– then E2 = E1 / a and I1 = I2 / a
I1
+
Φm
I2
+
E1
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E2
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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
1A
– 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
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Impedance Ratio
• Transformers can also be used to
transform an impedance
– the source sees the effective
impedance
Z X = E1 I1
– on the other side, the secondary
winding of the transformer sees
the actual impedance
Z = E2 I 2
– the effective impedance is related
to the actual impedance by
E1 a E2 a 2 E2
ZX =
=
=
= a 2Z
I1 I 2 a
I2
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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
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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
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• Example
– using the shifting of impedances
calculate the voltage E and
current I in the circuit, knowing
that the turns ratio is 1:100
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Ideal Transformer
• Homework
– Problems: 9-4, 9-6, 9-10*
Note: problem 9-10 is a design problem
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