1. BASIC ELECTRICAL TECHNOLOGY Chapter 6: Single Phase Transformer
DET 211/3
Chapter 6: Single Phase Transformer

2. Introduction to Transformer
A transformer is a device that changes ac electric energy at one voltage level to ac electric energy at another voltage level through the action of a magnetic field.

3. The most important tasks performed by transformers are:
changing voltage and current levels in electric power systems.
matching source and load impedances for maximum power transfer in electronic and control circuitry.
electrical isolation (isolating one circuit from another or isolating dc while maintaining ac continuity between two circuits).
It consists of two or more coils of wire wrapped around a common ferromagnetic core. One of the transformer windings is connected to a source of ac electric power – is called primary winding and the second transformer winding supplies electric power to loads – is called secondary winding.

5. Ideal Transformer
An ideal transformer is a lossless device with an input winding and output winding.
a = turns ratio of the transformer
lossless

6. Power in ideal transformer
Where q is the angle between voltage and current

7. Impedance transformation through the transformer
The impedance of a device – the ratio of the phasor voltage across it in the phasor current flowing through it:

8. The equivalent circuit of a transformer
The major items to be considered in the construction of such a model are:
Copper (I2R) losses: Copper losses are the resistive heating in the primary and secondary windings of the transformer. They are proportional to the square of the current in the windings.
Eddy current losses: Eddy current losses are resistive heating losses in the core of the transformer. They are proportional to the square of the voltage applied to the transformer.
Hysteresis losses: Hysteresis losses are associated with the arrangement of the magnetic domain in the core during each half cycle. They are complex, nonlinear function of the voltage applied to the transformer.
Leakage flux: The fluxes ΦLP and ΦLS which escape the core and pass through only one of the transformer windings are leakage fluxes. These escaped fluxes produce a self inductance in the primary and secondary coils, and the effects of this inductance must be accounted for.

9. Nonideal or actual transformer
Mutual flux

10. Nonideal or actual transformer

11. Nonideal or actual transformer
Transformer equivalent circuit, with secondary impedances referred to the primary side
Ep = primary induced voltage Es = secondary induced voltage
Vp = primary terminal voltage Vs = secondary terminal voltage
Ip = primary current Is = secondary current
Ie = excitation current IM = magnetizing current
XM = magnetizing reactance IC = core current
RC = core resistance Rp = resistance of primary winding
Rs = resistance of the secondary winding Xp = primary leakage reactance
Xs = secondary leakage reactance

12. Nonideal or actual transformer Transformer equivalent circuit

13. Dot convention
If the primary voltage is positive at the dotted end of the winding with respect to the undotted end, then the secondary voltage will be positive at the dotted end also. Voltage polarities are the same with respect to the dots on each side of the core.
If the primary current of the transformer flows into the dotted end of the primary winding, the secondary current will flow out of the dotted end of the secondary winding.

14. Exact equivalent circuit the actual transformer
The transformer model referred to primary side
The transformer model referred to secondary side

15. Approximate equivalent circuit the actual transformer
The transformer model referred to primary side
The transformer model referred to secondary side

16. Exact equivalent circuit of a transformer
Ep = primary induced voltage Es = secondary induced voltage
Vp = primary terminal voltage Vs = secondary terminal voltage
Ip = primary current Is = secondary current
Ie = excitation current IM = magnetizing current
XM = magnetizing reactance IC = core current
RC = core resistance Rp = resistance of primary winding
Rs = resistance of the secondary winding Xp = primary leakage reactance
Xs = secondary leakage reactance

17. Primary side
Secondary side

18. Exact equivalent circuit of a transformer referred to primary side
Exact equivalent circuit of a transformer referred to secondary side

19. Approximate equivalent circuit of a transformer referred to primary sideVp Ip + -
Reqp
Is/a
jXeqp Rc
jXM
aVs
Reqp=Rp+a2Rs
Xeqp=Xp+a2Xs
Approximate equivalent circuit of a transformer referred to secondary side
Vp/a
aIp + -
Reqs Is
jXeqs
Rc/a2
jXM/a2 Vs
Reqs=Rp/a2+Rs
Xeqs=Xp/a2+Xs

20. Example
A single phase power system consists of a 480V 60Hz generator supplying a load Zload=4+j3W through a transmission line ZLine=0.18+j0.24W. Answer the following question about the system.
If the power system is exactly as described below (figure 1(a)), what will be the voltage at the load be? What will the transmission line losses be?
Suppose a 1:10 step-up transformer is placed at the generator end of the transmission line and a 10:1 step down transformer is placed at the load end of the line (figure 1(b)). What will the load voltage be now? What will the transmission line losses be now?

21. Example ZLoad=4+j3 V=48000V IG + - VLoad ILine ILoad Figure 1 (a)T1 T2
1:10
10:1
ZLine=0.18+j0.24
ILoad
ILine
Figure 1 (b)

22. Solution Example
(a) From figure 1 (a) shows the power system without transformers. Hence IG = ILINE = ILoad. The line current in this system is given by:

23. Therefore the load voltage is:
Solution Example
Therefore the load voltage is:
and the line losses are

24. Solution Example
(b) From figure 1 (b) shows the power system with the transformers. To analyze the system, it is necessary to convert it to a common voltage level. This is done in two steps:
i) Eliminate transformer T2 by referring the load over to the transmission’s line voltage level.
ii) Eliminate transformer T1 by referring the transmission line’s elements and the equivalent load at the transmission line’s voltage over to the source side.
The value of the load’s impedance when reflected to the transmission system’s voltage is

25. The total impedance at the transmission line level is now:
Solution Example
The total impedance at the transmission line level is now:
The total impedance at the transmission line level (Zline+Z’load) is now reflected across T1 to the source’s voltage level:

26. Solution Example
Notice that Z’’load = 4+j3  and Z’line = j . The resulting equivalent circuit is shown below. The generator’s current is:
a) System with the load referred to the transmission system voltage level
b) System with the load and transmission line referred to the generator’s voltage level

27. Solution Example
Knowing the current IG, we can now work back and find Iline and ILoad. Working back through T1, we get:

28. Working back through T2 gives:
Solution Example
Working back through T2 gives:
It is now possible to answer the questions. The load voltage is given by

29. and the line losses are given by:
Solution Example
and the line losses are given by:
Notice that raising the transmission voltage of the power system reduced transmission losses by a factor of nearly 90. Also, the voltage at the load dropped much less in the system with transformers compared to the system without transformers.