1. EET 306 ELECTRICAL MACHINES
TRANSFORMER
Syafruddin Hasan
Syafruddin Hasan

2. INTRODUCTION Transformer Fundamentals Loaded Transformers
Transformer Efficiency
Parallel Operation
Three-phase Transformers
Impedance Matching
Syafruddin Hasan

3. Introduction Transformer is an ac static machine that
(i) transfers electrical energy from one electric circuit to another
(ii) does so without a change of frequency
(iii) does so by the principle of electromagnetic induction
(iv) has electric circuits that are linked by a common magnetic circuit.
(v) the energy transfer usually take place with a change of voltage level,
althought this is not alaway necessary
Caution
Transformer must not be connected to a dc source. If the primary winding of a transformer is connected to a dc supply, the flux produced will not vary but remain constant in magnitude and therefore no emf will be induced in the secondary winding, except at the moment of switching on.
There will be no back induced emf in the primary winding and therefore a heavy current will be drawn from the supply which may result in the burning out of the winding.
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4. FACTS ABOUT TRANSFORMERS
Transformers operate on mutual inductance.
A transformer has a primary winding and a secondary winding.
The coefficient of coupling is the portion of primary flux that links the secondary.
With 100% coupling, the turns-per-volt ratio is the same for all windings.
Transformers can have hysteresis, eddy current, and copper (I2R) losses.
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5. MORE FACTS ABOUT TRANSFORMERS
Transformer losses can be reduced by using silicon steel cores, laminated cores, and small gage wires.
As the load on a transformer increases, angle theta decreases.
Three-phase transformers use a three-legged core.
Transformer windings can be connected in series or parallel.
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6. Transformer Fundamentals
Types of transformer
Single phase transformer construction
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7. Ideal transformer Assumptions:
Winding resistances are zero (perfect conductivity)
No leakage flux (inductance)
No iron loss
Core material’s permeability is constant
Magnetization current generates a flux that induces voltage in both windings
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8. TRANSFORMER FUNDAMENTALS
Primary
Secondary
Load
A transformer has a primary coil
and a secondary coil.
The primary is connected to a source.
The secondary is connected to a load.
During the first half-cycle,
the flux builds up and collapses.
This creates a half-cycle of induced voltage in the secondary.
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9. TRANSFORMER FUNDAMENTALS
Primary
Secondary
Load
During the next half-cycle,
the flux again builds up and collapses.
This creates another half-cycle of induced voltage in the secondary.
Notice that the primary and secondary voltages are out-of-phase.
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10. PRIMARY-SECONDARY TERMINOLOGY
Transformers are bidirectional devices.
120 V
Primary
Source
90 V
Secondary
Load
This transformer was designed to step 120 V down to 90 V.
120 V
Secondary
Load
Source
90 V
Primary
However, either winding can be used as the primary.
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11. Mutual Inductance
Transformers are constructed of two coils placed so that the charging flux developed by one will link the other.
The coil to which the source is applied is called the primary coil.
The coil to which the load is applied is called the secondary coil.
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12. TURNS-PER-VOLT RATIO + _ 1.0 V + _ 0.25 V _ + 0.75 V
The 4-turn primary with a 1 volt source provides 4 turns-per-volt;
therefore, each turn has 0.25 V across it.
Each turn will produces a specific amount of flux in the core.
This same flux will, in turn, produce 0.25 V in each secondary turn.
and a 3 turn secondary provides 0.75 V.
Thus, a 1 turn secondary provides 0.25 V,
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13. Un-loadedTransformers
Current, voltages and flux in an unloaded ideal transformer Ep
ImNp Vp Im Φ
Closed loop Es
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14. PRIMARY CURRENT Load Iref Ipri With no load, Ipri is the
Vpri
Vpri
Ipri
Ien
Ipri
With no load, Ipri is the
energizing current (Ien).
With a load, Ipri
is composed of Ien
and Iref .
The transformer acts like an inductor.
Iref is the resistive load current reflected from the secondary to the primary.
Theta is large; power is low.
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15. Loaded Transformers Ep ImNp Vp Im Φm Es Loaded Es Is IsNs Φs Φp
Closed loop
Currents and fluxes in a loaded ideal transformer Ep
ImNp Vp Im Φm Es
Loaded Es Is
IsNs Φs Φp
Φs = Φp
I1 = Im + Ip Ip
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16. Transformers Ideal transformer Turn ratio is:
The input and output complex powers are equal or
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17. Real Transformer
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18. EDDY CURRENT ILLUSTRATION
If an eddy current is induced in the core, one will also be induced
in the aluminum disk placed on top of the core. The eddy current in
the aluminum disk will produce a magnetic field that opposes the
field that created the eddy current.
Eddy current
Aluminum disk
Coil connected to
a 60-Hz supply
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19. TRANSFORMER LOSSES
Copper (I2R) loss B H
Eddy current loss
(I2R in the core)
Hysteresis loss (green area of the hysteresis loop)
These losses are minimized by a narrow hystersis loop, thin laminations, and large diameter wire.
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20. Transformers Real transformer
Figure Leakage flux in the real transformer
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21. No-load Transformer Ic = Io cos φo Im = Io sin φo
No-load circuit equivalent
Ic = Io cos φo
Im = Io sin φo
The exciting current (Io) magnitude about 1% - 5% of the rated current of the primary for power transformer, but may be much higher for small transformers
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22. primary current Io has two components:
Active or working or iron (core) loss component Ic (in phase with V1), because it mainly supplies the iron loss plus small quantity of primary copper loss
Ic = Io cos φo
Magnetizing component Im (in quadrature with V1), because its function to sustain the alternating flux in the core
Im = Io sin φo
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23. Loaded Transformers Equivalent circuit of a real transformer
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24. Primary side I1 = Io + I2’ with, Secondary side with, and
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25. (c) Capacitive load (a) Resistive load (b) Inductive load
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26. Equivalent circuit refer to primary side
R2’ = a2 R2 = equivalent secondary resistance as referred to primary
X2’ = a2 X2 = equivalent secondary reactance as referred to primary
ZL’ = a2 ZL = equivalent load impedance as referred to primary
E2’ = a E V2’ = aV I2’ = I2 / a
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27. R01 = R1 + R2’ = R1 + a2 R2
= equivalent or effective resistance of transformer as referred to primary
X01 = X1 + X2’= X1 + a2X2
= equivalent or effective reactance of transformer as referred to primary
Z01 = √(R012 + X012)
= equivalent or effective impedance of transformer as referred to primary
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28. Equivalent circuit refer to secondary side
R02 = R2 + R1’ = R02 = R2 + R1/a2
= equivalent or effective resistance of transformer as referred to seconadry
X02 = X2 + X1’= X2 + X1/a2
= equivalent or effective reactance of transformer as referred to secondary
Z02 = √(R022 + X022)
= equivalent or effective impedance of transformer as referred to secondary
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29. Total Approximate Voltage Drop in a Transformer
= AN = AD + DN
= I2R02 cosφ + I2X02 sinφ
In general, approximate voltage drop is
The approximate voltage drop as referred to primary is
Note: (+) sign for inductive load
(-) sign for capacitive load
% voltage drop in secondary is
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30. Determination of transformer parameters by measurement
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31. Transformer Test
No-Load Test
Objectives: or
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32. Short Circuit Test
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33. Regulation of a Transformer
Regulation of a transformer is the change in secondary terminal voltage from no-load to full-load and divided it by no-load terminal voltage at a primary input voltage and power factor are constant.
and
In further treatment, unless stated otherwise, regulation is to be taken as regulation ‘down’.
The total voltage drop that was explained above can be used as % voltage regulation.
The regulation may also be expressed in terms of primary values.
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34. Losses in Transformer Pcu = I12 R1 + I22R2 = I12 R01 = I22 R02
Core or Iron loss. It includes both hysteresis and eddy current. Because the flux in a
transformer remains practically constant for all loads ( its variation being 1 to 3%
from no load to full load), the core loss is practically the same at all loads.
Hysteresis loss,
Eddy current loss,
Copper Loss. This loss is due to the ohmic resistance of the transformer windings.
Total Cu loss
Pcu = I12 R1 + I22R2 = I12 R01 = I22 R02
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35. Efficiency Iron loss = Cu loss Condition for Maximum Efficiency
Variation of Efficiency with Power Factor
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36. Differentiating both sides with respect to I1, we get
Primary input = V1 I1 cos φ1
Differentiating both sides with respect to I1, we get
For efficiency (η) to be maximum, dη / dI1 = 0
Hence, the above equation becomes or
Wi = I12 R01 or = I22 R02
Iron loss = Cu loss
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37. TRANSFORMER EFFICIENCY
Source
provides
1640 W
Transformer
90 W loss
(heat loss)
Load
1550 W
consumed
The transformer core and copper losses cause the transformer to heat up as electric energy is converted to heat energy.
Efficiency = 1550 W / 1640 W =
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38. to increase power capability of transformer for maintenance
Parallel Operation of Single-phase Transformer
Objectives:
to increase power capability of transformer
for maintenance
The conditions that must be satisfied
Primary windings of the transformers should suitable for the supply
system voltage and frequency.
The transformers should be properly connected with regard to polarity.
The voltage rating of both primaries and secondaries should be
identical. In other words, the transformers should have the same
transformation ratio.
The percentage impedances should be equal.
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39. Parallel Operation of Single-phase Transformer
The current carries by each transformer
The VA carries by each transformer
and
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40. Polarity Test 1. additive polarity 2. subtractive polarity
Connect the high voltage winding to a
low voltage ac source Eg (say 120V)
2. Connect a jumper J between any two
adjacent HV and LV terminals
3. Connect a voltmeter Ex between the other two adjacent HV
and LV terminals
4. Connect another voltmeter Ep across the HV winding
If Ex >> Ep, the polarity is additive.
H1 and X1 are diagonally opposite.
If Ex << Ep, the polarity is subtractive.
H1 and X1 are adjacent.
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41. SERIES-OPPOSING WINDINGS
120 V + _
8 V
2 A
6 V
3 A
2 V _ +
2 A
Connect two terminals with the same instantaneous polarities.
Take the output from the other two terminals.
The output voltage equals the difference between the two voltages.
The current is limited to the lesser of the two winding currents.
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42. SERIES-AIDING WINDINGS
120 V + _
8 V
2 A
6 V
3 A
14 V _ +
2 A
Connect two terminals that have opposite instantaneous polarities.
Take the output from the other two terminals.
The output voltage equals the sum of the two winding voltages.
The current is limited to the lesser of the two winding currents.
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43. PARALLEL SECONDARY WINDINGS
120 V + _
9 V
4 A
9 V _ +
8 A
The two windings must have equal voltage ratings.
The two windings should have equal current ratings.
Connect the negative terminals
and the positive terminals.
Take the output from the negative and positive terminals.
Voutput = Vwinding and
Ioutput = 2 x Iwinding
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44. IMPEDANCE MATCHING 10:1 ratio 100 W 1 W 0.1 A 1 A 10 V 1 V 20 V 100 W
Notice the source in both circuits
provides 0.1 A at 10 V. Thus, the
transformer makes the 1-W resistor
act like a 100-W resistor in terms
of the load on the source.
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