1. Elements of electrical design
DR. S. & S.S. GHANDHY GOVT. ENGINEERING COLLEGE, SURAT
Elements of electrical design
Design of Small single Phase Transformer & Choke Coils
Guided by -
Prepared by -
: Hemaxi Halpati
: Priyank Hirani
: Manish Jatiya
: Rakesh Joshi
: Piyush Kanani
Prof. M R Vasavada

2. Design of small single phase transformer
Transformer is a machine which changes the voltages level without changing the frequency.
Design procedure of high capacity single phase and three phase transformer is a very tedious job, but small capacity low voltage transformers is very easy.
The general ratings of small single phase transformers are 10 VA to 1kVA.

3. Core design
Primary and secondary voltage (v1 and v2 Given in problem).
From the reference standard tables turn per volt(Te) is taken for further calculation.
VA and Te table

4. Conti… Emf equation, Ep=4.44ΦmfNp So that, Turns/volt = N/E
Te = 1/4.44Φmf
Frequency of transformer is specified and standard frequency is 50 Hz.
Φm is found out from equation of Te .
Flux density Bm is given or assumed(1 to 1.5 Tesla).
Net core area, Ai is find out from equation, Ai = Φm / Bm .
Gross core area is find out by equation, Agi = Ai / Ks.
Where Ks is stacking factor, it is for the core is assumed as 0.9.
Shell type construction is preferred.

5. Conti. For core, square section is normally used.
So, width of central limb is given by, A = [Agi]0.5.
Standard E-I shape, T-U shape or E shape stamping are available and selected for design.
A standard stamping giving a width C nearly equal to the value calculated above may be used.
Prior to start the design procedure for a given problem let us know about the standard core stamping and standard table of copper conductors.
Stamping are available in three categories:
1. E-I type
2. T-U type
3. E-E type

6. E-I Type

7. T-U Type

8. E-E Type

9. Winding Design

10. Round copper wire(Synthetic enamel)

11. Conti.

12. Window area Space required for primary winding = Tpa’p/sfp.
Space required for secondary winding = Tsa’s/sfs.
The space required for insulation and former is estimated as 20% of that required for windings.
Therefore, window area required, Aw = 1.2 (Tpa’p/sfp + Tsa’s/sfs).
It should be noted that the stamping used, gives a higher value of window area than the value calculated above.
Width of window, Ww = A – C – 2D/ 2
Height of window, Hw = B – 2E
Area of window provided, Aw = Ww x Hw

13. Choke
A coil of copper wire wound on laminated iron core has negligible resistance is known as choke coil.
When an ac voltage is applied to the purely inductive coil, an emf known as self-induced emf is induced in the coil due to self-inductance of the coil which opposes the applied voltage.

14. Conti.
In the case of tube light, the sudden stop of current through the choke produces voltage of several hundreds volts (approx. 900 to 1000V) acroos it, because of self inductance of coil.
The high voltage starts flow of electrons from one filament to the other through the gas filled inside the tube.

15. Design Procedure of choke(No air- gap)
The Design Procedure of choke coil is similar to the single phase small transformer design.
Similar terms are use in this case also.
The only different is that , it has only one winding support on the iron core.
Therefore, the Turns per volt is found on the basis of ½ output VA this means if voltage V is applied across choke and I current through the choke, then is taken to find Te from the reference table.

16. Steps in designing the choke
= = ,find Te for this value from the table.
EMF equation
Bm is given. If not take Bm= 1wb/mm2
Stacking factor Ks=0.9 is assumed
Gross area Ai /0.9
E = 4.44ΦmfT

17. Conti. Square section is used. Width of central limb A =
For winding, Turns T=V x Te
Area of the conductor wire =I/δ δ is given = 2.5A/mm2
From this area (a) = π/4 d2
From above equation find d.
Refer standard conductor size table and find the matching size d of conductor and with insertion diameter d1
Space in window for winding = and
Increase the space area by 20% to accommodate former insulation, packing etc.
The total area required = 1.2

18. Conti.

19. Design of variable air-gap single phase and 3 phase choke coil
In this section we have study design procedure of choke operating on (I) single phase (II)Three phase with adjustable air-gap between limbs of iron core.

20. Construction
Iron core of one part variable by rotating wheel and one part is fixed.
Copper coils with N/2 turns.
Rotating wheel to adjust air-gap.

21. Conti. Reluctance (s) is the opposition to the magnetic flux (Φ).
Reluctance (s) of air gap is very high as compared with reluctance of iron material.
To produce the same flux more ampere turns (NI) i.e. m.m.f is required in air compared with ampere turns for iron parts.
Generally for the choke coil made copper material R<<L and R is neglected.

22. Conti. and In circuit R= Resistance of coil L= Inductance of coil
N=Number of turns of coil
V=voltage with frequency f
According to basic equation,
and

23. Step in Designing a Variable choke
For Magnetic part
For Electric part
Mechanical dimension
To find R,L,Z of the coil

24. For Magnetic part To find constant k where µ0=Permeability
f= frequency
Vph=phase voltage in r.m.s
Iph=phase voltage current in r.m.s

25. Conti.
Knowing the relation

26. Conti.
3. Assume staking factor Ks=0.9

27. Conti. AT is Amp.Turns (m.m.f) for the air gap lg.
lg is total length of each air gap
if lg’ is the length of each air gap
then lg= 2 lg’ for single phase choke
and lg= 1.5 lg’ for 3 phase values
AT required for iron part = ATi
ATi << ATg
ATi = 10 to 20 % of ATg
so that ATtotal = ATi + ATg
Attotal =1.1 to 1.2[Atg]

28. For Electric part Number of Turn per coil = AT/I where I= current
There are 2 coil of N/2 turns for single phase variable choke
There are 3 coil each of N turns for a 3-phase variable choke.
thus number of turns for the coil are decided.

29. Conti. Current density (δ)
Generally enameled copper conductors are used for the choke winding.
δ=2.3 to 2.5 Amp/mm2
for single phase choke current is I
for 3-phase choke current is taken i.e. Iph

30. Conti.
3. To find diameter (d) of bare conductor C.S. area of bare conductor a= hence

31. Conti. Use of standard size conductor tables
from these table exact or nearest size of conductor is selected.( d and a )
d’ = diameter of insulated conductor
So, area of insulated conductor
a’ =

32. Mechanical dimension Window size average value of space factor sf=0.8
sf= active area/gross area
these is for single phase choke
Window area for single phase choke
Window area for 3-phase choke
gross area of window Aw=1.2 to 1.25 Aw’

33. Conti.
2. Depth df and height hf of coil to be accommodated in the window space. hf =actual height of coil hf’ =available height df =actual depth of coil df’ =depth of coil Window area Aw=Hw . Ww

34. Conti.
Some clearance on both side 10+10=20mmbe kept for formar etc. as shown in the figure.
Available height for winding hf’=Hw-20mm
Height wise turns per larger Nn=
Depthwise turns per layer Nd=
Actual depath of coil,df=df’+5mm
Actual height of coil,hf=hf’+(2^5)

35. Conti..
3. To find and distance between the two coils in the window (dc) and overall dimension. dc=Ww-2df Distance (D) between of two line. D = D= Ww + A Total width = D+A = Ww+2A

36. To find R,L,Z of the coil where ρ =resistivity of copper=1.73 x 10-8
a=C.S area of conductor
Impedance of coil with maximum air-gap
Reactance XL of coil = Ω
AND XL =2πfL Ω
L= Henry

37. Choke main dimensions Single phase variable choke

38. Thank you
Thank you