Md. Shahadat Hasan Sohel
#0606028
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Introduction:
Transformer is one of the most used electrical equipments of todays world. It is
widely used in the power transmission. It has made possible the transmission
of power, from where it is generated, to where it will be used, with very little
loss.
Without transformer, the transmission of power would have been quite
impossible.
Basic principle:
Transformer uses magnetic usk to transfer electrical energy.
There are two
types of transformers:
1) Step up: Changes alternating low voltage to alternating high voltage
2) Step down: Changes alternating high voltage to alternating low voltage
If the number of turns in the primary and secondary windings are
ns
Vp
np
and
respectively and the voltage across the primary and secondary windings are
Vs respectively, then :
np
Vp
=
Vs
ns
and
Figure 1: Transformer
2
Theories and Corresponding Calculation:
Core Area:
If the core area of the transformer is taken as a and va is the corresponding
volt-ampere, thena=c
√
va
Where c is a constant.
If a is taken in sq. inch then-
√
a=0.16
va
Calculation for core area:
Required output voltage for my transformer is 18 V and its current capacity in
the secondary winding is 1 A.
So,Volt-ampere, va = 18×1=18
√
Now, core area, a = 0.16
18
= 0.6788
inch2
Selection of the core:
For a shell type transformer, which I planned to make, the core area is equal
to the area of the bobbin.
But in the market, the available bobbins are of
some xed areas. So, I had to choose a bobbin of area which is not less than
the calculated core area and close to the calculated value. The most suitable
bobbin that I found for this purpose, which fullls the required conditions, had
an area of 1inch
2
.
Number of turns per volt:
Number of turns per volt can be calculated from the following equation-
N
108
V = 4F f aB
Here,
V = Voltage across the winding in volts in rms
F = Form factor of the ac waveform
f = Frequency of the wave in Hz
a = Cross section area of the core in sq. cm
N = Number of turns on the considered
B = Flux density in maxwells per unit area
Calculation for number of turns per volt:
Here,
F = 1.1
f = 50 Hz
a = 6.4516
cm2
B = 9, 600 maxwells per
So,
cm2
3
N
108
80
V = 4×1.1×50×6.4516×9600 = 11
Voltage in the primary, V p = 220 V
Number of turns in the primary,
Voltage in the secondary
Vs =
np
=
80
11 ×220 = 1600
18 V
Number of turns in the secondary,
ns =
80
11 ×18= 130.91
Number of turns given:
In the primary winding, calculated turns i.e. 1600 turns are given. But, as to
counter leakage eect while loaded, the number of turns in secondary is increased
a bit. The turns given in the secondary winding are 140.
Selection of wire:
The selection of wire is done on the basis of how much current will be drawn.
In the secondary winding, maximum current that will be drawn is 1A. So, wire
of gauge 21 is used in the secondary, which has a current carrying capacity of
1.0377 A.
The maximum current in the primary can be calculated from the following
formula:
Vs
18
Vp ×Is = 220 × 1 = 0.0818A
As a result, the wire used in the secondary is of 34 gauge, which has a current
Ip =
carrying capacity of 0.0858 A.
Determination of Transformer Parameters:
Open Circuit Test:
The purpose of the open circuit test is to determine the magnetizing reactance
(XM )and
the equivalnet core-loss resistance
(Rf e )
In the open cercuit test, the secondary was kept open and 220 volt ac was
applied to the primary. A power-factor meter and an ammeter were connected
with the primary coil to measure the angle between applied voltage and current. Then the open circuit voltage
(Voc ),
between input current and input voltage,
primary current,
(θoc )
(Ioc )
and the angle
were measured.
Data from the open-circuit test:
The data is given beow:
Voc = 220V
Ioc = 0.025A
θoc = 14◦
The no-load voltage found in the secondary was 19.6. As there is very little
leakage at no load, the voltage was a bit higher than as it should be.
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Calculation of magnetizing branch:
From the open circuit test data:
Poc = Voc × Ioc × cosθoc = 220 × .025 × cos14
Now, P = Voc × If e
⇒ If e = VPoc = 5.33665
220 = 0.02426 A
= 5.3366 W
Again,
2
2
Ioc
= If2e + IM
q
2 − I2
⇒ IM = Ioc
fe
√
2
⇒ IM = 0.025 − 0.024262
⇒ IM = 0.006A = 6mA
Then,
Rf e = VIfoce
220
⇒ Rf e = 0.025
⇒ Rf e = 8800Ω
And,
oc
XM = VIM
220
⇒ XM = 0.006
⇒ XM = 36666.67Ω
Short Circuit Test:
The purpose of the open circuit test is to determine the equivalent resistance
(Req ) ,equivalent leakage reactance (Xeq ) and equivalen impedance (Zeq ).
In the short circuit test, the secondary was shorted. An ammeter was connected in series with the secondary coil. 220 volt ac was applied to the primary.
A power-factor meter and an ammeter were connected with the primary coil
to measure the angle between applied voltage and current. Rated current (1A)
was allowed to ow through the secondary and primary current (Isc ), primary
voltage (Vsc ), and the angle between voltage and current (θsc ) were measured.
Data from the open-circuit test:
The data is given beow:
Isc = 0.1A
Vsc = 19.5V
θsc = 3◦
Calculation of equivalent circuit:
From the open circuit test data:
Psc = Vsc × Isc × cosθsc = 220 × .1 × cos3=
Now,
sc
Zeq,HS = VIsc
⇒ Zeq,HS = 19.5
0.1
1.9473 W
5
⇒ Zeq,HS = 195Ω
And,
2
Psc = Isc
Req,HS
⇒ Req,HS = PI 2sc
sc
⇒ Req,HS = 1.9473
0.12
⇒ Req,HS = 194.73Ω
Again,
2
2
2
Zeq,HS
= ReqHS
+ XeqHS
q
2
2
⇒ Xeq,HS = ZeqHS
− ReqHS
√
⇒ Xeq,HS = 1952 − 194.732
⇒ Xeq,HS = 10.258Ω
Eciency Test:
For eciency test, 220 Volt was applied in the primary and a load was connected
that draws a current of 1A through it. For this load, primary current and powerfactor angle was measured.
The data is given below:
Vp = 220V
Ip = 0.12A
voltage, Vs = 18.2V
current, Is = 1A
Primary voltage,
Primary current,
Secondary
Secondary
Angle between primary voltage and primarey current,
θ = 14◦
Calculation of eciency:
From the data,
Pin = Vp Ip cosθ = 220 × 0.11 × cos14◦ = 23.481W att
Output power, pout = Vs Is = 18.2 × 1 = 18.2W att
P
18.2
Eciency, η = out × 100% =
Pin
23.481 × 100% = 77.51%
Input power,
Equivalent Circuit:
From the parameters calculated from short circuit and open circuit tests, we
can draw the equivalent circuit as:
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Figure 2: Equivalent circuit
5V DC Output:
The output of the transformer is applied to the input of a bridge rectier consisting of four diodes (1N4007). Then the rectied voltage is supplied to a 7805 IC
through a capacitor(1000µF ). And from the output of the IC, through another
capacitor(1000µF ), 5V constant dc output is taken. The circuit is as following:
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Figure 3: 5V DC Output Circuit
Discussion:
Eciency:
The eciency of the transformer is about 78%. The probable reasons behind
this are1) Presence of core losses such as: Eddy current loss and hysteresis loss
2) As the exact B-H curve of the core was not available, the maximum B
that was taken in designing the transformer, may not be at the highest point of
the linear region of the B-H curve.
Output Voltage:
The output voltage required was 18 for my transformer. I designed the transformer to give 18 V at rated load which draws 1A current. That was made by
taking in consideration the leakage eect.
To counter the leakage eect, few
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additional turns were given in the secondary winding as mentioned in the calculation part. Accordingly, at loaded conditon the secondary voltage is 18.2. But
at no load, the leakage eect is suciently low. So, the output voltage becomes
around 20 (19.8 to be exact). On the day of submission, the input voltage was
also higher (230V instead of 220).
And, the output was checked at no load
condition. So, the output voltage got even higher (around 21V).
Comment:
The transformers that are available in the market, which give similar voltage,
found to have eciencies between 65% to 70%. So, I think, it is quite a success
to have a transformer with eciency around 80%.
And, as it was my rst
practical project, I enjoyed a lot, doing every bit of it. It was really great to do
something in practical and acquire rst hand experience.