TRANSFORMERS
3.1 Introduction
A transformer is a static electrical device that changes A.C
electric power at one voltage level to A.C electric power at anther voltage
level of the same frequency through the action of a magnetic field. It
consists of two or more coils of wire wrapped around a common
ferromagnetic core, these coils are (usually) not directly connected. One
of the transformer winding is connected to a source of A.C electric
power, and the second transformer winding supplies electric power to
loads. The transformer winding connected to the power source is called
the primary winding or input winding, and the winding connected to
the loads is called the secondary winding or output winding. Power can
flow in either direction, as either winding can be used as the primary or
the secondary. Constructionally, the transformers are of two general
types, distinguished from each other merely by the manner in which the
primary and secondary coils are placed around the laminated core. The
two types are known as is core-type and shell-type. In the so-called core
type transformers, the windings surround a considerable part of the core
where as in shell-type transformers, the core surrounds a considerable
portion of the winding as shown in fig.(3.1).
Core-Type
Shell-Type
Fig.(3.1)
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3.2 The Ideal Transformer
An ideal transformer is a lossless device with an input winding and
an output winding. The relationships between the input voltage and
output voltage and between the input current and the output current, are
given by two simple equations. Fig.(3.2) shows an ideal transformer.
Flux φm
VP
VS
NP
NS
Fig.(3.2)
When a sinusoidal voltage is applied to primary winding, the same
magnetic flux (φm) goes through both windings. According to Faraday's
law, the voltage across the primary winding is
vp = NP
dφ m
dt
While that across the secondary winding is
vS = N S
dφ m
dt
Dividing, we get
vp
vS
=
Np
NS
=a
Where the subscript "m" indicates mutual flux linking and,
N P : number of turns of primary
N S : number of turns of secondary
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a : is the turn ratio or transformer ratio
dφ m
: is the time rate of change of flux linking the coil
dt
For the reason of power conservation, the energy supplied to the
primary must equal the energy absorbed by the secondary, since there
are no losses in an ideal transformer. This implies that
v p .i p = v S .i S
vp
vS
iS
ip
=
Combining equations gives
vp
vS
Np
iS
=
=a
ip NS
=
In terms of phase quantities, these equation are
VP I S N P
=
=
=a
VS I P N S
Where
VP : The voltage across the primary winding
VS : The voltage across the secondary winding
I P : The primary current
I S : The secondary current
If (a>1), we have a step-down transformer, since the voltage is
decreased from primary to secondary (vs< vp) on the other hand, if (a< 1),
the transformer is a step-up transformer, as the voltage is increased
from primary to secondary (vs> vp).
Note that: The rating of a transformer is stated in terms of the voltamperes that it can transform without overheating. The transformer
rating is either (VpIP) or (VSIS), where (IS) is the full-load secondary
current.
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3.3 The Transformer (E.M.F) Equation
The induced e.m.f. due to mutual flux, N
dφ m
dt
Since the flux varies sinusoidally, φ m = φˆm sin ωt , where ( φˆm ) is the
maximum value reached in the cycle. The induced e.m.f. is thus.
e=N
e.m.f
(
)
d ˆ
φ m sin ωt = ωN .φˆm cos ωt
dt
The maximum value of e.m.f. occurs when ( cos ωt = 1 ), i.e. Eˆ m = ω.N .φˆm
The r.m.s value is
E=
Eˆ m
2
=
ω.N .φˆm
2
, now ω = 2.π . f hence E =
1
2
(2πf .N .φˆ ) = 4.44 f .N .φˆ
m
m
Applying equation to two coil
E P = 4.44 f .N p .φˆm = The r.m.s primary e.m.f due to mutual flux.
E S = 4.44 f .N S .φˆm =
The r.m.s secondary e.m.f due to mutual flux.
φm
φ m = φˆm sin ωt
Fig.(3.3)
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volt
3.4 The real Transformer
In section (3.2), we idealized the transformer. We now add back
the effects that we ignored.
(a) Leakage flux
While most flux is confined to the core, a small amount (called
leakage flux) passes outside the core and through air at each winding as
in fig.(3.4). The effect of this leakage can be modeled by inductances (Lp)
and (Ls) as indicated in figure. The remaining flux, the mutual flux ( φ m )
links both windings and is accounted for by the ideal transformer as
previously.
Fig.(3.4)
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(b) Winding Resistance
The effect of coil resistance can be approximated by resistances (Rp)
and (Rs) as shown in fig.(3.5). The effect of these resistances is to cause
as light power loss and hence a reduction in efficiency as well as a small
voltage drop. (The power loss associated with coil resistance is called
copper loss and varies as the square of the load current).
Fig.(3.5)
(c ) Core Loss
Losses occur in the core because of eddy current and hysteresis.
First, consider eddy currents. Since iron is a conductor, voltage is induced
in the core as flux varies. This voltage creates current that circulate as
"eddies" within the core itself. One way to reduce these currents is to
break their path of circulation by constructing the core from thin
laminations of steel rather than using a solid back of iron.
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Laminations are insulated from each other by a coat of ceramic,
varnish, or similar insulating material. (Although this does not eliminate
eddy current, it greatly reduces them).
Another way to reduce eddy current is to use powered iron held
together by an insulating binder. Ferrite core are made like this.
Now consider hysteriesis. Because flux constantly reverses,
magnrtic domains in the core steel constantly reverse as well. This takes
energy. However, this energy is minimized by using special
grain-oriented transformer steel.
The sum of hysteresis and eddy current loss is called core loss or
iron loss. The core-loss current (Ic) is a current proportional to the
applied to the core that is in phase with the applied voltage, so it can be
modeled by a resistance (Rc) connected across the primary voltage
source. As long as voltage is constant (which it normally is), core losses
remain constant.
IP
IS
IP
Io
VP
EP
IC
Im
Fig.(3.6)
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ES
VS
(e) magnetizing current
We have also neglected magnetizing current. In a real transformer,
however, some current is required to magnetize the core. The
magnetization current(Im) is a current proportional (in the unsaturated
region) to the voltage applied to the core and lagging the applied voltage
by (900), so it can be modeled by inductance (Lm) connected across the
primary voltage source. Although fig.(3.6) is an accurate model of a
transformer, it is not a very useful one. To analyze practical circuit
containing transformers, it is normally necessary to convert the entire
circuit to an equivalent circuit at a single voltage level. Therefore, the
equivalent circuit must be referred either to its primary side or to its
secondary side in problem solutions.
Fig.(3.7) Exact transformer models.
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Resistance (RS) in fig.(3.6) can be replaced by inserting and
additional resistance (a2 RS) in the primary circuit such that the power
absorbed in (a2 RS) when carrying the primary current is equal to that in
(RS) due to the secondary current, i.e
⎛I
From which RS = (a .RS ) = ⎜⎜ S
⎝ IP
2
I P2 .(a 2 .R S ) = I S2 .R S
2
⎞
⎟⎟ .RS
⎠
The simplified equivalent circuit of a transformer is shown in fig.(3.8).
Then the total equivalent resistance in the primary circuit ( Req ) is equal to
the primary and secondary resistances of the actual transformer.
Hence
ReqP = RP + RS
i.e
Req P = RP + a 2 .RS
By similar reasoning, the equivalent reactance in the primary circuit is
given by
X eqP = X P + X S ,
i.e
X eqP = X P + a 2 . X S
.
Fig.(3.8) Approximate transformer models.
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The excitation branch has a very small current compared to the
load current of the transformer. The excitation branch is simply moved to
the front of the transformer, and the primary and secondary impedances
are left in series with each other. These impedances are just added,
creating the approximate equivalent circuit in fig.(3.8).
3.5 Determining the value of components in the transformer
It is possible to experimentally determine the values of the
inductances and resistances in the transformer model. An adequate
approximation of these values can be obtained with only two tests. The
open-circuit test and the short-circuit test.
3.5.1 Open-Circuit or no-load test
The purpose here is to measure the iron losses (core loss) and the
components of the no-load current which in turn will give the relevant
components of the equivalent circuit. One winding is open-circuited and
rated voltage at rated frequency is applied to the other winding. Quite
often the low-voltage winding is supplied, to reduce the test voltage
required. Providing that the applied voltage per turn is normal, either
winding may be used, since the flux and iron losses will then be normal.
The primary winding is connected to a full-rated line voltage, and the
input voltage, input current, and input power to the transformer are
measured. Look at the equivalent circuit in fig.(3.8). under the conditions
described, all the input current must be flowing through the excitation
branch of the transformer. The series element (Req) and (Xeq) are too
small in comparison to (RC) and (Xm) to cause a significant voltage drop,
so essentially all the input voltage is dropped across the excitation
branch. Iron loss and will be so taken.
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The equivalent circuit and the phasor diagram of the no-load
current, from fig.(3.9) are
IC
V
φ
Io
Im
Fig.(3.9)
I C = I o cos(φ )
I m = I o sin(φ )
And the power-factor angle (φ ) is given by
⎛
Po.c
⎝ Vo.c × I o.c
φ = cos −1 ⎜⎜
⎞
⎟⎟
⎠
The magnetizing circuit resistance is RC =
reactance is X m =
Vo.c
IC
and the magnetizing
Vo.c
Im
3.5.2 Short-Circuit Test
This test used to determine the leakage impedance and the
effective current loss (I2Req) . In this test, the secondary terminals of the
transformer are short-circuited. And the primary terminals are connected
to a fairly low-voltage source. The input voltage is adjusted until the
current in short-circuited windings is equal to its rated value.
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(Be sure to keep the primary voltage at a safe level. It would not be a
good idea to burn out the transformer's windings while trying to test it).
The input voltage, current, and power are again measured. Since
the input voltage is so low during the short-circuit test, negligible
current flows through the excitation branch. Since the windings are
coupled, any current in one will be opposed magnetically by a current in
the other giving the same m.m.f the mutual flux will be very small, just
sufficient to provide the secondary leakage impedance drop, so the net
magnetizing m.m.f. is negligible. If the excitation current is ignored, then
all the voltage drop in the transformer can be attributed to series elements
in the circuit.
Req
I S .C
X eq
VS .C
Fig.(3.10)
The magnitude of the series impedances referred to the primary side of
the transformer is the leakage impedance is
Z eq =
VS .C
I S .C
The power input reading (PS.C), which is due to the loss
PS .C = I S2.C × Req
The effective resistance
Req =
PS .C
I S2.C
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The total leakage reactance is
X eq =
(Z ) − (R )
2
eq
2
eq
If (RP) can be measured, then knowing (Req), we can find
RS = Req − RP
3.6 Transformer Voltage Regulation and Efficiency
Because a real transformer has series impedances within it, the
output voltage of a transformer varies with the load even if the input
voltage remains constant. To conveniently compare transformers in this
respect, it is customary to define a quantity called voltage regulation
(VR). Full-load voltage regulation is a quantity that compares the output
voltage of the transformer at no load with the output voltage at full load,
it is defined by the equation.
V .R =
VS , N .L − VS , F .L
VS , F .L
× 100 0
0
Usually it a good practice to have as small a voltage regulation as
possible. For an ideal transformer, V .R = 0 percent.
The transformers are also compared and judged on their
efficiencies. The efficiency of a device is defined by the equation.
η=
Pout
× 100 0
0
Pin
To calculate the efficiency of a transformer at a given load, just add the
losses from each resistance and apply equation. Since the output power is
given by
Pout = VS I S cos(φ S )
The efficiency of the transformer can be expressed by
η=
VS I S cos(φ s )
× 100 0
0
Pcu + p core + VS I S cos(φ S )
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η=
VS I S cos(φ )
× 100
I R1 + I R2 + PCore + VS I S cos(φ )
2
1
2
2
Greater accuracy is possible by expressing the efficiency thus,
η=
=
output. power
× 100
input. power
input. power − losses
× 100 ,
input. power
η = 1−
losses
× 100
input. power
Example 3.1: The primary and secondary windings of a 500 KVA
transformer have resistance of 0.42 Ω and 0.0011 Ω respectively. The
primary and secondary voltages are 6600V and 400 V respectively and
iron loss is 2.9 KW. Calculate the efficiency on (i) Full-load and,
(ii) Half-load, assuming the power factor of the load to be 0.8.
Solution: (i) Full-load secondary current=
And, full-load primary current=
500 × 1000
= 1250 A
400
500 × 1000
= 75.8 A
6600
∴ Secondary copper loss on full-load= (1250) 2 × 0.0011 = 1720 W
And, Primary copper loss on full-load= (75.8) 2 × 0.42 = 2415 W
∴ Total copper loss on full-load= 4135 W = 4.135 KW
And, Total loss on full-load = 4.135 + 2.9 = 7.035 KW
Output power on full-load= 500 × 0.8 = 400 KW
∴ Input power on full-load = 400 + 7.035 = 407.035 KW
Efficiency on full-load= ⎛⎜1 −
⎝
7.035 ⎞
⎟ × 100 = 98.2 0 0
407.0 ⎠
(ii) Since the copper loss varies as the square of the current.
∴ Total copper loss on half-load= 4.135 × (0.5) 2 = 1.034 KW
And, total loss on half-load= 1.034 + 2.9 = 3.934 KW
∴ Efficiency on half-load= ⎛⎜1 −
⎝
3.934 ⎞
⎟ × 100 = 98.07 0 0
203.9 ⎠
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