1. Lecture Notes 3 Transformers
EE212 Passive AC Circuits
Lecture Notes 3
Transformers EE

2. Magnetic Circuit
What is the relationship between magnetic flux and magnetomotive force, Fm?
What is the relationship between electric current, i, and magnetomotive force?
Magnetic field strength, H = Fm / l EE

3. Magnetic Circuit What is Faraday’s Law?
The voltage induced in an electric circuit is proportional to the rate of change of the magnetic flux linking the circuit EE

4. Magnetic Circuit
Consider a coil around a magnetic core. If a current i flows through the coil, a magnetic flux  is generated in the core. a b i  L
vab = L di dt
 = N i in webers (Wb)
N = number of turns in the coil
Rm = constant known as reluctance
(depends on the magnetic path of the flux)
Direction of flux by
Right-Hand Rule
Rm = in ampere-turns/Wb
l = length of magnetic path
A = cross-section area
m = permeability
Fingers curled around coil
– direction of current
Thumb – direction of flux EE

5. B-H Curve Flux Density, B = in teslas, T
Magnetic field strength, H = in AT/m
B = m H EE

6. Coupled Circuits Circuits that affect each other
by mutual magnetic fields a b i1 f1 L1
vab c d
vcd f2 L2 i2
± depending on whether the fluxes add or oppose each other
L1, L2: self inductance
M: mutual inductance
ratio of induced voltage in one circuit
to the rate of change of current
in another circuit
The flux f2 generated by
current i2 in Coil 2
induces a voltage in Coil 1,
and vice-versa. EE

7. Coupled Circuits in Phasors
If input signals are sinusoidal waveforms,
coupled circuits can be in phasor representation
± depending on flux directions
Left hand side equations imply a four element equivalent circuit with two dependent sources. a b
Vab I1 L1 c d
Vcd I2 L2 M EE

8. Equivalent Circuit with Dependent SourcesEE

9. Dot Convention a b i1 f1 L1 c d f2 L2 i2
Dots are placed at one end of each coil, so that currents entering the dots produce fluxes that add each other.
The dots provide information on how the coils are wound with respect to each other.
A current i entering a dotted terminal in one coil induces a voltage M with a positive polarity
at the dotted terminal of the other coil.
(currents entering the dots
produce upward fluxes)
(+) if both currents enter the dotted terminals (or
the undotted terminals).
(-) if one current enters a dotted terminal and the
other current enters an undotted terminal. a b I1 L1 c d I2 L2
Vab = (jwL1) I1 + (jwM) I2 for the currents as shown EE

10. Coefficient of Coupling, k
k = ≤ k ≤ 1
k depends on the magnetic properties of the flux path.
When k = 0, no coupling
k = 0.01 to 0.1, loosely coupled
k > 0.5, close coupled, e.g. air core
k ≈ 1.0, e.g. power transformer
all the flux generated by one coil is linked to
the other coil (i.e. no leakage flux)
k = 1.0 ideal transformer EE

11. Transformer f if v1 N1 + - e2 N2 e1
primary
winding
secondary
When a voltage V1 is applied to the primary winding, an emf e2 is induced in the secondary winding. The induced emf lags the inducing flux by 900. E1 N2 E2 N1 If + - V1
e2 = N2
V1 = -E1 If f E1 E2
Faraday’s Law EE

12. Transformer Application in Power System
V/I step up/down
Y/D conversion
circuit (dc) isolation
Z matching
(for max power transfer,
min. reflection from load)
Instrument Transformers:
CT (current transformer)
PT or VT (voltage transformer) EE

13. Ideal Transformer No leakage flux
 Coupling Coefficient, k = 1 , i.e. the same flux f goes through both windings =
= a (turns ratio)
e1 = N1
and e2 = N2
=>
Turns ratio (a or n) is also known as the transformation ratio.
No Losses No voltage drops in the windings: V1 = - e1
Instantaneous powers in primary and secondary are equal
(i.e. all the energy from the primary is transferred to the secondary winding).
e1 i1 = e2 i Therefore,
Be careful, the turns ratio can be defined as N2/N1. =
= a EE

14. Transformer Loading a secondary current I2 is drawn by the load
I2 generates a flux that opposes the mutual flux f
(Lenz’s Law: effect opposes the cause)
reduction in f would reduce induced emf e1
since source voltage V1 is constant, and V1 = - e1, f must remain constant
the primary winding must draw an additional current I’1 from the source to neutralize the demagnetizing effect from the secondary e1 N2 e2 N1 I1 V1 + - ZL V2 I2
Primary current I1 = I’1 + If EE

15. Ideal Transformer No leakage flux, i.e., k = 1
Self-inductance, L1 = L2 = ∞ i.e., magnetizing current = 0
Coil losses are negligible
With polarities, dots and currents as shown:
V2 = V1 /a
I2 = a I1
Z1 = a2 Z2 where Z2 is the load impedance
Z1 is the impedance seen by the source a b
Vab I1 L1 c d
Vcd I2 L2 M + + ● ● - - EE

16. Actual Transformer - resistance in primary and secondary windingsv1 N1 + - e2 N2 e1 ZL R1 R2 v2
- resistance in primary and secondary windings
- leakage reactance in pr. and sec. windings
- voltage drops in both windings (leakage impedance)
- losses
- copper loss primary: I12·R1 secondary: I22·R2
- iron loss (core loss)
Core loss depends on voltage and frequency
eddy current loss hysteresis loss EE

17. Iron (Core) Losses Eddy Current Loss:
emf induced in core generates eddy currents which circulate in the core material, generating heat.
laminations (silica sheets between core layers) –
to reduce eddy current, and minimize loss EE

18. Iron (Core) Losses Hysteresis Loss:
The direction of the magnetic flux in the core changes every cycle. Power is consumed to move around the magnetic dipoles in the core material, and energy is dissipated as heat.
Hyst. loss  (vol. of core) x (area of hyst. loop) EE

19. Transformer Construction
Coil Winding
Core Assembly
Core-Coil Assembly
Tank-up
Accessories Mounting and Finishing EE

20. Core Assembly EE

21. Core-Coil Assembly
Core vertical sides – limbs, top horizontal side – yoke
Yoke is removed to insert the coils into the limbs
LV coil is first placed on the insulated core limbs
Insulating blocks are placed at the top and bottom of the LV coil
Cylinder made out of corrugated paper is placed over the LV Coil
HV coil is placed over the cylinder
The top yoke is fixed in position
LV and HV windings are connected as required EE

22. Tank-up EE

23. Accessories Mounting
Connections of LV and HV coil ends to the terminal bushings are made
Transformer tap changer and protection accessories (e.g. Buchholz relay, Conservator, Breather, temperature indicator, etc.) are installed
Tank is closed
Functions of Transformer Oil
Cooling
muffle noise
displace moisture (avoid insulation degradation) EE

24. Transformer Cooling EE

25. Typical Power Transformers
Pole-mounted
Single-phase Transformer
Three-phase Transformer EE

26. Transformer Rating Rated kVA
Rated Voltage primary and secondary: (transformers are normally operated close to their rated voltages)
Rated Current (FL Current) is the maximum continuous current the transformer can withstand
For single phase transformer:
Rated primary current = Rated VA / Rated Pr Voltage
When rated current flows through a transformer, it said to be fully loaded.
The actual current through a transformer varies depending on the load connected at different times of the day. EE

27. Equivalent Circuit
Represent inductively coupled circuits by a conductively connected circuit
Equivalence in terms of loop equations
Assume a load impedance is connected to the secondary.
KVL at Loop 1: V1 + -
L22 R1 R2 V2 M N2 N1
L11 I1 I2
-V1 + R1I1 + jwL11I1 - jwM I2 = 0
KVL at Loop 2:
R2I2 + V2+ jwL22I2 - jwM I1 = 0
inductively coupled circuits
turns ratio,
= a
The loop equations are: EE

28. Equivalent Circuit (continued)
Consider following substitutions: V1 + -
aV2 aM I1 R1 L1
a2R2
a2L2
M  a·M L2  a2·L2 R2  a2·R2
V2  a·V2 I2 
conductively connected circuit
KVL at Loop 1:
-V1 + R1I1 + jwL1I1 + jwaM (I1-I2/a) = 0
KVL at Loop 2:
jwaM (I2/a - I1) + jw a2L2.I2/a + jw a2R2.I2/a + aV1 = 0
Let L11 = L1 + aM
and L22 = L2 + M/a
The loop equations remain the same: V1
- V2
R1 + jwL11 - jwM
- jwM R2 + jwL22 = I1 I2 EE

29. Transformer Equivalent Circuit
a2ZL V1 + - I1 R1 X1
a2R2
a2X2 Ie If Ic Rc Xf
ZL = load impedance
Ie = excitation current
If = magnetizing current
Xf = magnetizing
reactance
equivalent circuit referred to the primary side
R1, R2= primary, secondary winding resistance
X1, X2= primary, secondary leakage reactance
I1, I2= primary, secondary current
Rc = core loss resistance (equivalent resistance contributing to core loss)
Ic = core loss equivalent current EE

30. Example: Transformer Equivalent Circuit
A 50-kVA, 2400/240-V, 60-Hz distribution transformer has a leakage impedance of ( j0.92) W in the high voltage (HV) side and ( j0.0090) W in the low voltage (LV) winding. At rated voltage and frequency, the admittance of the shunt branch of the equivalent circuit is (0.324 – j2.24)x10-2 S [siemens] when viewed from the LV side. Draw the circuit:
a) viewed from the LV side
b) viewed from the HV side
Turns ratio, a = 2400/240 = 10
0.72/a2 W
j0.92/a2 W W
j W
j W W a)
0.72 W
j0.92 W
0.0070a2 W
j0.0090a2 W
a2 W
j a2 W b)
Units of admittance used to be mhos [ohm backwards]. Still see it sometimes, but SI unit is siemen [same as the company]. EE

31. Transformer: Approximate Equivalent CircuitRe Xe Rc Xf
Re = R1 + a2R2
Xe = X1 + a2X2
where, R2 and X2 are referred
to the primary side
Re and Xe  obtained from Short Circuit Test
Rc and Xf  obtained from Open Circuit Test Re Xe
For transformers operating close to full load Xe
For transformers in a large power network analysis EE

32. Short Circuit Test Connect meters on HV side as shown
Short circuit LV side
Energize HV side with a variable
voltage source and increase voltage
gradually to get rated current reading
on the Ammeter
Take the V, I and P readings from meters LV HV
Vsupply + - W A V Re Xe Rc Xf I V + -
P = copper loss = I2 Re  Re =
Ze =  Xe =
Equivalent circuit referred to HV side
X1 ≈ a2 X2 ≈
R1 ≈ a2 R2 ≈ EE

33. Open Circuit Test Connect meters on LV side as shown
Vsupply W A V + - LV HV
Connect meters on LV side as shown
Open circuit HV side
Energize LV side with rated voltage
Take the V, I and P readings from meters Rc Xf I V + - If Ie Ic
P = core loss =  Rc =
Ie = I , Ic =
Equivalent circuit referred to LV side
If =  Xf = EE

34. Transformer Efficiency
Efficiency, h = Power Output / Power Input
Power Output = |VL| |IL| cos  = |IL|2 RL
Power Input = Power Output + Cu losses + Core losses
Cu Loss – varies with load current
Core Loss – depends on voltage (usually a constant for practical purposes)
Transformer efficiency is maximum when
Cu loss = core loss
Transformer efficiency is maximum when Cu loss = core loss EE

35. Transformer Efficiency (continued)
Power transformers: - usually operate at rated capacity,
- designed to have max. h at full load
Distribution transformers: carry a widely varying load
- designed to have max. h at less than full load
- always energized despite load levels – designed to have low core loss
Transformer efficiency is maximum when
Cu loss = core loss EE

36. Example: Transformer
Find the equivalent circuit of a 50-kVA, 2400/240-V transformer. The following readings were obtained from short circuit and open circuit tests:
If rated voltage is available at the load terminals, calculate the transformer efficiency at
full load with 0.8 p.f. lagging
(b) 60% load with 0.8 p.f. lagging.
What is the voltage at the HV terminals of the transformer?
If rated voltage is applied at the primary terminals, calculate the transformer efficiency at
full load with 0.8 p.f. lagging EE

37. Transformer Example 2
The transformer is used to step down the voltage at the load end of a feeder whose impedance is j1.6 ohm. The voltage at the sending end of the feeder is 2400 V. Find the voltage at the load terminals when the connected load is
Zero
60% load at 0.8 p.f. lagging
Full load at 0.8 p.f. laging
Full load at 0.5 p.f. leading
What will be the current on the low voltage side if a short circuit occurs at the load point? EE

38. Voltage Regulation
When a constant rated voltage is applied to the primary,
At no load
- no current, and therefore no voltage drop in transformer
- secondary voltage, Vsec = rated voltage
As load (resistive or inductive) increases
- voltage drop in transformer (Vdrop) increases
- secondary voltage decreases (reverse is the case with capacitive load increase) EE

39. Voltage Regulation (continued)
To maintain rated voltage at the secondary,
At no load, primary voltage required, Vpr = rated voltage
At a certain load, required Vpr = rated voltage + Vdrop in transformer
Voltage regulation is the change in primary voltage required to keep the secondary voltage constant from no load to full load, expressed as a percentage of rated primary voltage.
Load p.f. has a big effect on voltage regulation. EE

40. Example
A 50-kVA, 2400/240-V, 60-Hz transformer has a leakage impedance of (1.42+j1.82) ohms on the HV side. The transformer is operating at full load in all 3 cases below.
1. Calculate the voltage regulation at
0.8 p.f. lagging
unity p.f.
0.5 p.f. leading EE

41. Voltage Control
The allowable voltage variation at the customer load point is usually ± 5% of the rated (nominal) value. Control measures are taken to maintain the voltage within the limits.
Use of transformer tap changers: on-load/off-load
Injecting reactive power (Vars): series/shunt compensation EE

42. Autotransformers
Used where electrical isolation between primary side and secondary side is not required.
Usually relatively low power transformers
Can be readily made to be variable.
Can think of it as two separate windings connected in series.
Usually it is a single winding with a tap point.
Can be used as either step-up or step-down transformer EE

43. Autotransformers (continued)
Schematic diagrams:
Can be connected series opposing, but not terribly useful.
This makes a three-terminal transformer (rather than the typical 4-terminal). Middle terminal is often variable rather than a fixed tap. See next slide.
From J.D. Irwin, “Basic Engineering Circuit Analysis”, 3rd ed. Macmillan EE

44. Autotransformers (continued)
From Jackson et al. “Introduction to Electric Circuits” 8th ed., Oxford EE

45. Autotransformers (continued)
Assume N1 = 200
and N2 = 100
Let Vsource = 120 V
Then what is V2?
I1 = 5 A (from apparent power and source voltage)
Assume I2 is 15 A. Then Sload = ?, and Ssource = ? EE

46. Autotransformers (continued)
What is the node equation at the tap?
Node equation: I1 + IZY = I therefore, IZY =
From transformer action:
N1I1 = N2IZY
therefore,
That is, only part of the load current is due to the magnetomotive force that the current in the primary coil exerts on the secondary coil. The remaining portion is direct conduction of the source current to the load. EE