Electronic Devices and Circuits
EE215
DE-43 MECHATRONICS (B)
Project Report
Group Members:
1. Sajid Ali
Reg # 366087
2. Muhammad Ammar Shabbir
Reg # 382319
3. PC Muhammad Ejaz Ramzan
Reg # 398564
4. Muhammad Farrukh Bin Basheer
Reg # 398565
Submitted to: Muhammad Osama Ali
Background
Transformer is an electrical device that basically works on Faraday’s law of electromagne c induc on
which states that the changing flux induces an emf in the coil of wire. When current flows in the primary
coil it produces a magne c field and the changing flux is produced due to the alterna ng nature of input
current transmi ed through the core of ferromagne c material and induces an emf in the secondary coil.
Transformer flux is unidirec onal in nature. Depending on the turn ra o this input voltage can be stepped
up or down. In an ideal transformer law of conserva on of energy is followed. But in the case of a real
transformer, there are always some losses associated with it.
Objec ve
In this problem-based learning, we must design a transformer depending on the input and required
output parameters.
Parameters
• The input available power and the output desired power are as follows:
Cross
Input Voltage Input Power Output voltage Height Core (cm) sec on
220 V 330 VA 11 V 10 2cm x 2cm
Procedure
The things to be kept in mind while designing a transformer are discussed step by step below:
Wire gauge
In the designing process wire gauge plays a very important role because if a wire of the desired gauge is
not used, the primary and secondary winding may get burnt when a large current flows through it.
Input side: The input/output power of the transformer is 330 VA. The supply voltage at the input side is
220V. Therefore, the current flow through the input coils should be:
P=V*I
I=P/V => 330/220
I=1.5A
So, the current in the input coils of the transformer will be 1.5A.
The following table shows the wire gauge depending on the current in the circuit:
A 20-gauge wire is used at the input side of the transformer.
Output side: As, this is a step-down transformer; therefore, the output current can be calculated from
the rela onship given below:
Np*Ip = Ns*Is
Is=Np*Ip/Ns
The output voltage is stepped down 20 mes the input voltage; therefore, the turn ra o will be 20:1
Is=20*1.5 => 30A
A 30A current flows through the output side and the gauge of wire required to bear such an amount of
current will be:
A 10-gauge wire is used at the output side of the transformer.
Core Geometry
Core has a height of 10cm and a cross-sec onal area of 2*2 cm^2.
The no turns at the primary side are given by:
W = 50/s(cm^2)
Where,
W is no. of turns per volt
S is the cross-sec onal area of the transformer
W= 50/4 => 12.5
According to this 12.5 turns per volt are required and the total number of turns for 220V at the primary
side will be:
Total no. of turns = 220*12.5 => 2750 turns (Primary)
The no. of turns on the secondary side will be:
No. of turns= 2750/20
=137.5 turns
One layer of the winding at the primary side will contain 97.65 turns given by:
Diameter of 18-gauge wire= 0.001024 m
Height of transformer= 10 cm
no. of turns in one layer= 0.001024/0.1
=97.65 turns
Therefore, the total no. of layers on the primary side will be:
Total layers= total no. of turns/turns in one layer
=2750/97.65 => 28.16 layers
Field intensity(H)
Magne c field intensity is given by:
H=N*I /l
Where,
L is the mean path of the transformer
=2750*1.5/0.48 =>8593.75 Wb
Core material
So iron is used as core material due to following reason:
Reluctance
The formula for reluctance is as follows:
R=L/µA
The rela ve permeability of so iron is 200,000:
=0.48/4π*10^-7 * 200,000 *4*10^-4
=4774.6 A turns/Wb
Wire resistance problem
We use copper wire whose resistance can be calculated by using the following formula:
R=ρL/A
Where,
ρ is the resis vity of copper (constant material dependent)
L is the length of the copper wire
A is the cross-sec onal area of the wire
By pu ng the values in the above formula:
𝜌 = 1.77 ∗ 10 Ωm
R= 1.77*10^-8 * 8.8467 / 8.235496
=1.90*10^-8Ω
Where 8.8467 is the length of copper wire which could be found by using the circle circumference length
formula L=2πr, and 8.235496 is the area of the wire which could be found by using formula A=𝜋𝑟
formula.
We can see the copper wire resistance is very low so when a current of 1.5A will flow through it at the
primary side it will burn out. To endure this amount of current resistance of approximatelyb146.7 ohms
is required:
R=V/I => 220/1.5
=146.7ohms
This problem can be solved by adding resistance in series to the primary winding. Although there will be
a drop across the resistor of about 20V but for the sake of safety we have to take this step.
Losses
1. There will be the copper loss at both the input and output side of the transformer which will be
𝐼 𝑅.
2. There will be a hysteresis loss because the material does not magne ze and demagne ze
instantaneously.
3. There will be voltage drop loss across the series resistor connected at the primary side of the
transformer.
4. Eddy current loss in the core of the transformer.
5. Induc ve reactance loss at both primary and secondary side of the transformer.
Equivalent circuit
Conclusion
By using the technique of turn ra o in the primary and secondary side of the transformer voltage at the
input side can be stepped up or down at the output side. But there will be associated some losses in this
whole procedure. Proper wire gauge calcula on should be made before designing a transformer. The
transformer can be represented in the form of an electric circuit which describes all the losses and the
characteris c of the transformer.
Recommenda on
1. Core of the transformer should be cut into pieces of small thickness to avoid the copper losses as
much as possible.
2. Proper lamina on of the core should be done.
3. Use enameled wire for winding to avoid short circuit problem.
4. Use a core of material that can be magne zed and demagne zed easily.
5. Maintain the temperature of the transformer by using a coolant.
6. Avoid the interac on of external electric and magne c field.
