Design of small transformer and chock coil.

1. Guided by -
Prof. Raj Patel
Prepared by
-14024010950 7: Anish Singh
14024010950 8: Manish Singh
150243109011 : Krunal Panchal
150243109501: Shubham
Champaneri

2. Introduction of 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.

3.  In the case of tube light, the
sudden stop of current through
the choke produces voltage of
several hundreds volts (approx.
900 to 1000V) across 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.

4. 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.
2
VI

5. Steps in designing the choke
 = = ,find Te for this value from the table.
 EMF equation E = 4.44Φmf N
 Bm is given. If not take Bm= 1Wb/m2
 .
 Stacking factor Ks=0.9 is assumed
 Gross area Ai /0.9
2
VI
2
VA
2
Q
m
m
i
B
A
φ
=

6.  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
gA
fS
aT ×
2
1
8.0 





=
d
d
Sf
fS
aT ×

7. 2

8. 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.

9. 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.

10.  Reluctance is the opposition to the magnetic flux (Φ).
 Reluctance 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.

11.  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,
dt
di
LiRV +=
dt
d
NL
φ
=and
( )φNdtiRv +=
dt
d
NiR
φ
+=

12. Step in Designing a Variable choke
A. For Magnetic part
B. For Electric part
C. Mechanical dimension
D. To find R,L,Z of the coil

13. For Magnetic part
1. To find constant k
where µ0=Permeability
f= frequency
Vph=phase voltage in r.m.s
Iph=phase voltage current in r.m.s
g
phph
ig
fL
IV
ABK
π
µ
2
.
0
==

14. 2. Knowing the relation
2
2
.
g
i
ig
B
K
A
KAB
=
=
g
i
B
K
A =

15. 3. Assume staking factor Ks=0.9
9.0≅=
gi
i
s
A
A
K
igi
i
s
i
gi AA
A
K
A
A ≥== ;
9.0

16. 4. 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
0
gl.
µ
g
g
B
AT =

17. For Electric part
1. 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.

18. 2. 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

19. 3. To find diameter (d) of bare conductor
C.S. area of bare conductor
a =
Hence
2
4
d
π
π
a
d
.4
=

20. 4. 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’ = mm
2
1
4
d
π 2

21. Mechanical dimension
1. 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
2
1






d
d
'
'
2
2
wA
a
N
××
=
sf
aN
Aw
'
'
×
=
sf
aN
Aw
'2
'
××
=

22. 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
2
w
w
A
W =
2
ww 2WA =

23.  Some clearance on both side 10+10=20mm be kept
for formar etc. as shown in the figure.
 Available height for winding hf’=Hw-20mm
 Height wise turns per larger Nh=
 Depth wise turns per layer Nd=
 Actual depth of coil, df=df’+5mm
 Actual height of coil, hf= hf’+ (2×5)
1
'
d
hf
hN
N

24. 3. To find and distance between the two coils in the
window (dc) and overall dimension.
dc=Ww-2df
Distance (D) between of two lines.
D =
D = Ww+ A
Total width = D+A
= W +2A
22
AA
Ww ++

25. To find R,L,Z of the coil
•
where = 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
Ω=
a
l
R
ρ
Ω=
ph
ph
I
V
Z
22
RZ −
f
Xl
π2
ρ

26. Choke main dimensions
Single phase variable choke 3-phase variable choke

27. Thank You….