School of Electrical and Computer Engineering
Applied Electronics and Electrical
Machines
(ELEC 365)
Fall 2015
Transformer
1
Transformer
 Key educational goals:
Formulate basic voltage, current and impedance transformation in a single phase
transformer from Faraday’s law and evaluate its performance.
 Reading/Preparatory activities for class
i) Textbook:
Chapter 15: Sections 15.5-15.6.
2
Transformer
Questions to guide your reading and to think about ahead of time.
1. Why are transformers required and in what circuits (AC &/ DC) do they work ?
3. What are the basic voltage, current and impedance relationships of a transformer
with respect to the turns ratio?
5. How is the exact and approximate equivalent circuit of the transformer?
4. What are the necessity of open and short circuit tests?
5. What are the main losses in a transformer?
Introduction: Basics of Magnetic circuits.
3
Review of 1- AC Circuit Fundamentals
I    ICos  jISin 
Power factor = Cos  = Real Power/ Apparent power
• Apparent Power =VI (multiply the rms value of input voltage and current
(ignore phase angle))
• Real Power =I2 R (square of the rms current flowing through the reristor
times the resistor (ignore phase angle))
4
Transformer
• A transformer consists of a magnetic system in which a time varying flux links
two or more coils.
i1(t)
M
i2(t)
e2(t)
e1(t)
Coil 1
(Primary has N1 turns)
Coil 2
(Secondary has N2 turns)
5
Transformer classifications
 Step-up
The secondary has more turns than the primary.
 Step-down
The secondary has fewer turns than the primary.
 Isolating
Intended to transform from one voltage to the same voltage. The two coils have
approximately equal numbers of turns, although often there is a slight difference
in the number of turns, in order to compensate for losses (otherwise the output
voltage would be a little less than, rather than the same as, the input voltage).
 Variable
The primary and secondary have an adjustable number of turns which can be
selected without reconnecting the transformer.
6
Polarity marking of windings
7
Transformer core construction
8
Ideal transformer
ASSUMPTIONS:
1.
Winding resistances are negligible.
2.
All magnetic flux is confined to the Ferromagnetic core.
3.
Permeability (μr) of core material is so high that negligible mmf is required to
establish the flux in the core.
4.
Core losses are negligible.
10
Ideal transformer
1) Now we prove that the voltage induced across each coil is proportional to its
number of turns.
2) The current induced in each coil is inversely proportional to its number of
turns.
3) Instantaneous input power to the transformer is equal to instantaneous output
power from the transformer.
4) Impedance seen by the source.
11
Ideal transformer
TURNS AND VOLTAGE RATIOS
According to Faraday’s law, we have
which can be rearranged and integrated to yield
Assuming that all of the flux links all of the turns, the secondary voltage is given
By
12
Ideal transformer
Current Ratio
Ideally, the reluctance is zero, and the mmf required to establish the flux in the
core is zero.
Power in an Ideal Transformer
13
Ideal transformer
Impedance Transformations
Using previous equations to substitute for I2 and V2, we have
Rearranging this, we get
In which Z’L is the impedance seen by the source. We say that the load
impedance is reflected to the primary side by the square of the turns ratio.
14
Magnetizing current
Observation: It was shown that the flux in the core is m Sin(t). Since the
permeability of the core is infinite ideally zero current can produce this flux!!! In
actuality, a current Im, known as magnetizing current is required to setup the flux
in the transformer. This current is within 5% of the full load current in a well
designed transformer.
Im 
V 1rms
Lm
2
N1
Lm 

Lm is the primary side self inductance.
15
Example 1
 Working from the circuit , find the values of V2 and the power delivered to
the load.
16
Example 1
First, we reflect the load impedance ZL to the primary side:
The total impedance seen by the source is:
17
Example 1
18
Practical transformer
19
Practical transformer
20
Transformer with sinusoidal excitation
We use phasor equivalent circuit; replace all inductances and capacitances by their
reactances at the supply frequency, voltages and currents by their phasor values.
All quantities are sinusoidal functions of time. Reactance’s of leakage inductances Ll1
and Ll2 are:
Xl1 = (2πf)Ll1 = ωLl1 Ω,
Xl2 = (2πf)Ll2 = ωLl2 Ω.
Reactance of magnetizing inductance L’m is X’m = (2πf)L’m = ωL’m
21
Equivalent circuit to primary side
• Referring all the variables and parameters to primary side
where ZL’ is the load impedance, ZL, referred to primary-side;
Xl2’ is the leakage reactance of secondary, Xl2, referred to primary-side; and
R2’ is the secondary winding resistance, R2, referred to primary-side.
22
Equivalent circuit to primary side
Also, eliminate ideal transformer from the equivalent circuit by shorting secondary.
Complete (or exact) linear equivalent phasor circuit referred to the primary-side.
23
Approximate equivalent circuits
Calculations involved in using the exact equivalent circuit are much reduced if
further approximations are made (acceptable in regulation calculations). We can
assume that V1 ≅ E1.
Neglect parallel branch (Ie is very small compared to I1).
Approximate phasor equivalent circuit (neglecting parallel branch).
24
Approximate equivalent circuits
Approximate
phasor equivalent
circuit (referred to
primary-side).
Approximate
phasor equivalent
circuit (referred to
primary-side).
25
Transformer under no-load conditions – Phasor diagram
Under no-load conditions, the exciting current is made up of two components:
1- i’m, the magnetizing component
2- i’c, the core-loss (iron-loss) component.
Phasor circuit and phasor diagram for the exciting current branch. Reactance of L’m
is X’m = ωL’m, ω is the supply frequency in rads/s.
26
Approximate equivalent circuits & phasor diagrams for different loads
Approximate
phasor equivalent
circuit (referred to
primary-side).
𝑉1
𝑉𝐿′
𝐼2′
𝜃2
90̊
𝑉1
′ 𝐼′
𝑋𝑒𝑞
2
𝐼2′
′ 𝐼′
𝑅𝑒𝑞
2
𝜃2
𝑉𝐿′
′ 𝐼′
𝑋𝑒𝑞
2
90̊
′ ′
𝑅𝑒𝑞
𝐼2
27
Phasor diagram
Phasor diagram for the loaded transformer using exact equivalent circuit referred
to primary side.
28
Transformer under no-load conditions – Phasor diagram steps
1) Load current referred primary side :
I 2 
V2
V 0
V    2
 2
 2
Z L
Z L  2
Z L
Amps.
2) Find E1
E1  E2  V2  R2  jX l2 I 2
3) Find Ie
I e  I c  I m
4) Find I1
I1  I 2  I e
Where
E1
E1
Ic 
, Im 
Rc
jX m
5) Find V1
V1  E1  R1  jX 'l1 I1
29
Complete equivalent circuit referred to secondary side
Referring all the variables and parameters to secondary side:
where Xm” is the magnetizing reactance Xm’ referred to secondary-side;
Xl1” is the leakage reactance of primary, Xl1, referred to secondary-side;
And R1” is the primary winding resistance, R1, referred to secondary-side.
30
Complete equivalent circuit referred to secondary side
Equivalent phasor circuit referred to the secondary-side.
31
Open circuit Test
In this test the rated primary voltage V1 = Voc at rated frequency is applied as
shown in next slide. with the secondary open circuit. The secondary voltage V2, the
no-load current Ioc and power Poc are measured. Poc gives the core losses. The
copper losses during this test can be neglected.
Now, the turn’s ratio
k
N1 V1

N 2 V2
And power factor is
cos θ oc 
Poc
Voc I oc
Therefore, θ oc can be calculated. Now
Xm

V
Voc
 oc 
 I oc sin( oc )
Im
And
Voc
 V
Rc  oc 
 I oc cos( oc )
Ic
Hence the values of Rc , X m and Poc can be obtained.
32
Open circuit Test
 OPEN-CIRCUIT (O.C.) Test (usually done on LP side)
 Apply rated voltage to the primary.
 Measure Voc, Ioc & Poc. V1 = Voc
33
Short circuit Test
The secondary is short-circuited using a suitable ammeter A2, as shown in next slide,
and a reduced primary voltage (Vsc) is applied such that a rated current flows in the
primary and secondary circuits. For this condition, the copper losses in the windings are
the same as that on a full load. On the other hand, the core losses are negligible, since
the applied voltage is only a fraction of the rated voltage. Hence the wattmeter reading
(Psc) gives the copper loss. We have:
Z sc 
R  k R   X
Vsc

I sc
2
2
1
2
2

k
X l2
L1

2
Psc  I sc ( R1  k 2 R2 ) Watts
2
Req  R1  R2  R1  k R2 
2
and
  Z sc  ( Req ) 2  X l1  X l2  X l1  k 2 X l 2 
X leq
2
It is usually assumed that
 /2
R1  R2  Req
and
 /2
X l1  k 2 X l 2  X leq
Hence R 1 , R 2 , X l1 and X l2 can be calculated
34
Short circuit Test
Short-circuit the secondary using an ammeter. Increase the input voltage slowly
until rated current flows. Measure Vsc, Isc & Psc.
P

Req  sc2
I sc
Z sc
Vsc

I sc
Total leakage reactance:
 
X leq   Z sc


2
2 1/ 2
 
  Req  

 
Divide this equally :

X l1  X l 2 
X leq

2
35
Transformer losses
I.
CORE LOSSES (IRON LOSSES) (Pc) = Hysteresis Loss + Eddy Current Loss.
𝐸12
𝑉12
𝑃𝑐 = ′ ≅ ′
𝑅𝑐
𝑅𝑐
Watts.
• Pc is almost constant and almost independent of load (only changes if supply
voltage changes).
II.
COPPER LOSSES (due to winding resistances) (Pw).
𝑃𝑤 = 𝑅1 𝐼12 + 𝑅2 𝐼22
Watts.
36
Transformer efficiency
Power Efficiency =
Output (active) power
Input (active) power
Or
Output (active) power
Power Efficiency =
Output(active) power + Losses

V2 I 2 Cos 
V1 / Rc  I 1 R1  I 2 R2  V2 I 2 Cos 
2
2
2
* 100
• where V2 and I2 are the load voltage and load current, respectively; Cos is
the power factor of the load.
• Unless specified, we refer efficiency of a transformer as “Power Efficiency”.
37
Transformer Regulation
The voltage regulation of a transformer is the change in the secondary terminal
voltage from no-load to full-load and is usually expressed as a percentage of the
full value.
Voltage Regulation

| Vno load |  | Vload |
*100%
| Vload |
Vno-load = RMS voltage across the load terminals without load.
V load = RMS voltage across the load terminals with a specified load.
38
Example 2
 The following measurements were obtained from tests carried out on a 10 kVA,
2300:230-V, 60 Hz distribution transformer:
OPEN-CIRCUIT TEST, with LOW-POTENTIAL winding excited:
Voc = 230 V, Ioc = 0.45 A, Poc = 70 W.
SHORT-CIRCUIT TEST, with HIGH-POTENTIAL winding excited:
Vsc = 120 V, Isc = 4.5 A, Psc = 240 W.
Winding resistances measured by a bridge: RHP = 5.8 Ω, RLP = 0.0605 Ω.
(a) Determine the equivalent circuit of the transformer referred to the low-potential
(LP) side.
39
Example 2.a
SOLUTION:
𝑇𝑢𝑟𝑛𝑠 𝑟𝑎𝑡𝑖𝑜 = 𝐾 =
𝑉1 𝑁1 2300 (𝑣)
=
=
= 10.
𝑉2 𝑁2
230 (𝑣)
OPEN-CIRCUIT (O.C.) Test, LP side:
cos(𝜃𝑜𝑐 ) =
𝑃𝑜𝑐
𝑉𝑜𝑐 𝐼𝑜𝑐
=
70
230×0.45
= 0.676
Core loss resistance referred to LP side:
∴ 𝜃𝑜𝑐 = 47.44 °
𝑅𝑐 ′′ =
𝑉𝑜𝑐
230
=
= 756Ω
𝐼𝑜𝑐 cos 𝜃𝑜𝑐 0.45 × 0.676
Magnetizing reactance referred to LP side: 𝑋𝑚′′ =
𝑉𝑜𝑐
230
=
= 694Ω
𝐼𝑜𝑐 sin 𝜃𝑜𝑐 0.45 × 0.763
40
Example 2.a
SHORT-CIRCUIT (S.C.) TEST, HP SIDE EXCITED:
𝑅𝑒𝑞 ′ =
Z sc
𝑃𝑠𝑐
𝐼𝑠𝑐 2
=
240
= 11.85Ω
4.52
V sc 120


 26.66
I sc
4 .5
Total leakage reactance referred to HP side:
X leq


  Z sc


2
 
  Req  

 
2
1
2

 26.66  11.85
2
2

1
2
 23.88
Total leakage reactance referred to low-potential (LP) side:
2
  230 

X leq  
 . X leq  0.2388
 2300 
41
Example 2.a


X lLP (or X l 2 )  X lHP (or X l1 ) 
X leq
2


0.2388
 0.1194 
2
Equivalent total winding resistance referred to LP side:

Req 
Req
k2

11.85

 0.1185
(10) 2
From dc resistance measurement,
RLP  0.0605
Referred to LP side:
RHP" 
RHP
5 .8

 0.058
2
2
k
(10)
Note that ac resistance Req” = R1dc” + R2dc (= Reqdc”) is satisfied.
[Note: Usually, ac resistances are higher than dc resistances especially at high
frequencies – due to skin effect].
42
Example 2.b
(b) Find the supply voltage if the transformer is supplying a load of 0.8 pf lag at
230V; at rated current. Find regulation: (i) Using exact equivalent circuit. (ii) Using
approximate equivalent circuit.
43
Example 2.b
Solution: Let load voltage ∴V L = 230∠0° V.
Rated load (secondary-side) current,
I2 
10,000(VA)
 43.47 A
230(V )
-1

Load pf angle,  2   cos (0.8)   36.87 , lagging; ∴ I 2  43.47   36.87 A

(Note:
V L 2300 )
I2 

Z L Z L  2 
(i) LOAD REGULATION USING EXACT EQUIVALENT CIRCUIT:
From equivalent circuit,
44
Example 2.b
E2 = V L + (R2 + jXl2 )I 2
= (230 + j0) + (0.0605 + j0.1194)(43.47∠-36.86)
= 230 + (0.13349∠63.05)(43.47∠-36.86)
= 230 + (5.8028∠26.19)
= 230 + (5.2071 + j2.561)
= 235.2071 + j2.561
= 235.22∠0.6238 V
 E
I c  2  235.220.6238  /7550  0.311550.6238  A

Rc
E2

Im 
 [235.22∠0.6238°]/[694∠90°] = 0.3389∠-89.376° A.

jX m
45
Example 2.b



Ie  Ic  Im 
[0.31155∠0.6238] + [0.3389∠-89.376]
=
(0.31153 + j0.0033918) + (0.00369 - j0.33888)
=
0.31522 – j0.33549
=
0.4603∠-46.78o A.


I1  I 2  I e 
[43.47∠-36.86] + [0.4603∠-46.78]
= (34.7805 – j26.076) + (0.31522 – j0.33549)
= 35.0957 – j26.4115
= 43.924∠-36.96o A.
46
Example 2.b
Supply voltage on secondary-side of ideal transformer,


 
V p  E2   R1  jX l1  I1


= (235.2071 + j2.561) + (0.058 + j0.1194)(43.924∠-36.96°)
= (235.2071 + j2.561) + (0.13238∠64.02°)( 43.924∠-36.96°)
= (235.2071 + j2.561) + (5.8147∠27.056°)
= (235.2071 + j2.561) + (5.17837 + j2.64488)
= 240.385 + j5.20588
= 240.44∠1.2406° V
47
Example 2.b
∴ Voltage (rms) required on the primary side,
V1 = Vp = 240.44 × 10 = 2404.4 V.
VNL  VFL
VNO LOAD  VFULL LOAD
Regulation 
 100% 
VFULL LOAD
VFL

240.44 - 230
 100  4.539%
230
48
Example 2.b (ii)
(ii) LOAD REGULATION USING APPROXIMATE EQUIVALENT CIRCUIT:
Approximate equivalent
circuit referred to LP (or
secondary side).
49
Example 2.b
Take load voltage as reference, ∴VL = 230∠0 V.


 
V P  V L   Req  jX leq  I 1


= (230 + j0) + (0.1185 + j0.2388)(43.47∠-36.86)
= 230 + (0.266585∠63.608)( 43.47∠-36.86)
= 230 + 11.58845∠26.748
= 230 + (10.3484 + j 5.21558)
= 240.3484 + j5.21558
= 240.405∠1.243 V
∴ V1 =V p = 10 (V p") = 2404.05∠1.243 V
VNL  VFL
VNO LOAD  VFULL LOAD
Regulation 
 100% 
VFULL LOAD
VFL
240.405 - 230

 100  4.524%
230
50