Magnetic circuit
Yoke
Armature Core
Flux Path
• The path of magnetic flux is
called magnetic circuit
• Magnetic circuit of dc machine
comprises of yoke , poles,
airgap, armature teeth and
armature core
• Flux produced by field coils
emerges from N pole and cross
the air gap to enter the
armature tooth. Then it flows
through armature core and
again cross the air gap to enter
the S pole
N
S
Pole Body
S
N
Magnetic Circuit of 4-Pole DC Machine
ly
hpl
lc
hpl
Magnetic circuit
Bg – Max. flux density in the core
Kg- Gap contraction factor
lc – Length of magnetic path in the core
l y – Length of magnetic path in the yoke
ds - Depth of the slot
dc - Depth of core
hpl - Height of field pole
Dm – Mean diameter of armature
When the leakage flux is neglected magnetic circuit of a DC machine consists
of following:
i. Yoke
ii. Pole and pole shoe
iii. Air gap
iv. Armature teeth
v. Armature core
Let
Total MMF to be developed by each pole is given by the sum of
MMF required for the above five sections.
MMF for air gap ATg=800000 Bg Kg lg
MMF for teeth ATt=att X ds
MMF for core ATc=atc X lc/2
MMF for pole ATp = atp X hpl
MMF for yoke ATy= aty X ly/2
att , atc , atp , aty - are determined B-H curves
lc = πDm/P = π(D – 2ds – dc)/P
ly = πDmy/P = π(D+ 2lg + 2hpl +dy)/P
AT total =ATg + ATt + ATc + ATp +ATy
Design of field system
 Consists of poles, pole shoe and field winding.
 Types:
 Shunt field
 Series field
 Shunt field winding – have large no of turns made of thin
conductors ,because current carried by them is very low
 Series field winding is designed to carry heavy current
and so it is made of thick conductors/strips
 Field coils are formed, insulated and fixed over the field
poles
Design of field system
Factors to be considered in design:
 MMF/pole &flux density
 Losses dissipated from the surface of field coil
 Resistance of the field coil
 Current density in the field conductors
Design of field system
Tentative design of field winding
Let ,
ATfl -MMF developed by field winding at full load
Qf
- Copper loss in each field coil(W)
qf
- Permissible loss per unit winding surface for normal temperature rise(W/m2 )
Sf
- Copper space factor
ρ
- Resistivity ( –m)
hf
- Height of winding(m)
df
- Depth of winding(m)
S
- Cooling surface of field coil(m2 )
Lmt - Length of mean turn of field winding(m)
Rf
- Resistance of each field coil (ohms)
Tf
- Number of turns in each field coil
Af
- Area of each conductor of field winding(m2)
If
- Current in the field winding (A)
δf
- Current density in the field winding(A/mm2 )
Design of field system
Cooling surface of the field winding, S=2Lmthf -- (1)
Permissible copper loss in each field coil, Sqf=2Lmthfqf -- (2)
Area of X-section of field coil=hfdf -- (3)
Area of copper in each section=Sfhfdf -- (4)
i.e, T f a f =S f h f d f -- (5)
Copper loss in each field coil, Qf=If2 Rf=If2 (TfLmt)/af
 T f  L mt
Qf  I 
 a
f

2
f

  a 
f
f



2
 T f  L mt

 a
f

Q f   f  T f a f  L mt

Q f   f   Volume
of Copper
2
2
 ( 6 )
i.e., Copper loss  f2 (Square of the current density)




Design of field system
To have temperature rise within the limit, the copper loss should be equal to the
permissible loss.
Using Eqns. (2) & (6),
2q f
(7)
2Lmt hf qf =f2 Lmt (Sfhf df ) =>  f 
S f d f
MMF per metre height of field winding

AT fl

hf

I fTf
hf
 f afTf

hf
hf
 fSfdf 

 f S f d f hf
2q f S f d f

2q f
S f d f
S fd f
[    2  10
MMF per meter height
 10 
4
8
 .m ]
qfSfdf
- - (8)
Design of field system
 Normal values:
 Permissible loss, qf -700W/m2
 Copper Space factor, Sf :
 Small wires: 0.4
 Large round wires: 0.65
 Large rectangular conductors: 0.75
 Depth of the field winding, df :
Armature Dia (m)
Winding Depth (mm)
0.2
30
0.35
35
0.5
40
0.65
45
1.00
50
1.00 and above
55
Design of field system
Height of field, h f
AT fl

Ampere Turns per meter height
Using Eqn (8),
hf 
AT fl  10
4
qfSfdf
Total height of the pole,
hpl=hf+hs+ height for insulation and curvature of yoke
where,
hs - Height of the pole shoe (≈0.1 to 0.2 of the pole height)
Design of shunt field winding
 Involves
the determination of the following information
regarding the pole and shunt field winding







Dimensions of the main field pole ,
Dimensions of the field coil ,
Current in shunt field winding,
Resistance of coil,
Dimensions of field conductor,
Number of turns in the field coil ,
Losses in field coil.
 Dimensions of the main field pole
 For rectangular field poles
o Cross sectional area, length, width , height of the body
 For cylindrical pole
o Cross sectional area, diameter, height of the body
Design of shunt field winding
 Area of the pole body can be estimated from the knowledge
of flux per pole , leakage coefficient and flux density in the
pole
 Leakage coefficient (Cl) depends on power output of the
DC machine
 Bp in the pole 1.2 to 1.7 wb/m2
 Фp = Cl. Ф
 Ap = Фp/Bp
 When circular poles are employed, C.S.A will be a circle
 Ap = πdp2 /4
dp 
4 Ap / 
Design of shunt field winding
 When rectangular poles employed, length of pole is chosen
as 10 to15 mm less than the length of armature
 Lp=L –(0.001 to 0.015)
 Net iron length Lpi = 0.9 Lp
 Width of pole, bp = Ap/Lpi
 Height of pole body hp = hf + thickness of insulation and
clearance
 Total height of the pole hpl = hp + hs
Design of shunt field winding
 Field coils are former wound and placed on the poles
 They may be of rectangular or circular cross section
depends on the type of poles
 Dimensions – Lmt, depth, height, diameter
 Depth(df) – depends on armature
 Height (hf) - depends on surface required for cooling the
coil and no. of turns(Tf)
 hf, Tf – cannot be independently designed
Design of shunt field winding
 Lmt - Calculated using the dimensions of pole and depth of
the coil
 For rectangular coils
 Lmt =2(Lp + bp + 2df) or (Lo +Li)/2
 Where Lo – length of outer most turn & Li – length of inner most turn
 For cylindrical coils
 Lmt = π(dp +df)
 No of turns in field coil: When the ampere turns to be
developed by the field coil is known, the turns can be
estimated
 Field ampere turns on load, ATfl= If. Tf
 Turns in field coil, Tf = ATfl/If
Design of shunt field winding
Power Loss in the field coil:
• Power loss in the field coil is copper loss, depends on
Resistance and current
• Heat is developed in the field coil due to this loss and it is
dissipated through the surface of the coil
• In field coil design , loss dissipated per unit surface area is
specified and from which the required surface area can be
estimated.
• Surface area of field coil – depends on Lmt, depth and height of
the coil
Design of shunt field winding
• Lmt – estimated from dimensions of pole
• Depth – assumed (depends on diameter of armature)
• Height – estimated in order to provide required surface area
Heat can be dissipated from all the four sides of a coil. i.e,
inner , outer, top and bottom surface of the coil
Inner surface area= Lmt (hf – df)
Outer surface area = Lmt (hf + df)
Top and bottom surface area = Lmt df
Total surface area of field coil, S= Lmt (hf – df)+ = Lmt (hf + df)+
Lmt df + Lmt df
S= 2Lmt hf +Lmt df = 2Lmt (hf +df)
Permissible copper loss, Qf=S.qf
[qf -Loss dissipated/ unit area]
Design of shunt field winding
Substitute S in Qf,
Qf= 2Lmt (hf +df).qf
Actual Cu loss in field coil=If2Rf=Ef2/Rf
Substituting Rf=(Lmt Tf)/ af ,
Actual Cu loss in field 2coil=Ef2 .af /(Lmt Tf)
Ef af
 2L mt q f (h f  d f ) 
ρL mt Tf
Conductor
field coil
area in 
 Area of X - section of 

  No.of turns X 
field
conductor



T f af
Conductor
Field coil
area in 
 Area of X - section of 

  Copper space factor X 

 field coil

 S fh f d f
Procedure for shunt field design
Step1 : determine the dimensions of the pole. Assume a
suitable value of leakage coefficient and B = 1.2 to 1.7 T
Фp= Cl. Ф
Ap = Фp/Bp
When circular poles are employed, C.S.A will be a circle
Ap = πdp2 /4 : dp =Ѵ(4Ap/π) When rectangular poles
employed, length of pole is chosen as 10 to15 mm less than
the length of armature
Lp=L –(0.001 to 0.015)
Net iron length Lpi = 0.9 Lp
Width of pole = Ap/Lpi
Procedure for shunt field design
Step 2 : Determine Lmt of field coil
Assume suitable depth of field winding
For rectangular coils
Lmt =2(Lp + bp + 2df) or (Lo +Li)/2
For cylindrical coils Lmt = π(dp +df)
Step 3: Calculate the voltage across each shunt field coil
Ef = (0.8 to 0.85) V/P
Step 4 : Calculate C.S.A of filed conductor
Af = ρLmt ATfl/Ef
Step 5:Calcualate diameter of field conductor
dfc =Ѵ(4af/π)
Diameter including thickness dfci = dfc + insulation thickness
Copper space factor Sf = 0.75(dfc/dfci)2
Procedure for shunt field design
Step 6 : Determine no. of turns (Tf) and height of coil (hf)
They can be determined by solving the following two
equations 2Lmt(hf + df) = Ef2 af/ρLmt Tf
Tf.af = Sf.hf.df
Step 7 : Calculate Rf and If : Rf = Tf. ρLmt /af
If = Ef/Rf
Step 8 : Check for δf
δf = If / af
δf – not to exceed 3.5A/mm2 .
If it exceeds then increase af by 5% and then proceed
again
Procedure for shunt field design
Step 9 : Check for desired value of AT
ATactual= If.Tf
ATdesired- 1.1 to 1.25 times armature MMF at full load
When ATactual exceeds the desired value then increase the
depth of field winding by 5% and proceed again.
Check for temp rise:
Actual copper loss = If2 Rf
Surface area = S = 2Lmt (hf + df)
Cooling coefficient C = (0.14 to 0.16)/(1 + 0.1 Va)
m = Actual copper loss X (C/S)
If temperature rise exceeds the limit , then increase the depth
of field winding by 5% and proceed again.
Design of Series Field Winding
Step 1: Estimate the AT to be developed by series field coil,
AT /pole = (Iz . (Z/2))/P
For compound m/c, ATse = (0.15 to .25) (Iz . Z)/2P
For series m/c, ATse = (1.15 to 1.25) (Iz . Z)/2P
Step 2: Calculate the no. of turns in the series field coil,
Tse = ATse/Ise (Corrected to an integer)
Step 3: Determine cross sectional area of series field conductor,
ase = Ise /δse
Normally, δse - 2 to 2.3 A /mm2
Design of Series Field Winding
Step 4 : Estimate the dimension of the field coil
Conductor area of field coil = Tse.ase
Also Conductor area of field coil = Sfse.hse.dse
When circular conductors are used
Sfse = 0.6 to 0.7
For rectangular conductors, Sfse – depends on thickness and
type of insulation
On equating above two expressions,
Tse.ase = Sfse.hse.dse
hse= (Tse.ase )/(Sfse.dse)
Design of commutator and brushes
 Commutator and brush arrangement are used to convert the
bidirectional current to unidirectional current
 Brushes are located at the magnetic neutral axis ( mid way
between two adjacent poles)
 The phenomenon of commutation is affected by resistance of
the brush , reactance emf induced by leakage flux, emf induced
by armature flux.
Design of Commutator and brushes
Classification of commutation process
1.
2.
3.
4.




Resistance commutation
Retarded commutation
Accelerated commutation
Sinusoidal commutation
Commutator is of cylindrical in shape and placed at one end of the
armature
Consists of number of copper bars or segments separated from one
another by a suitable insulating material of thickness of 0.5 to 1mm
Number of commutator segments = no. of coils in the armature
Materials used :



Commutator segments: Hard Drawn Copper or Aluminum Copper
Insulation :Mica, Resin Bonded Asbestos
Brushes :Natural Graphite, Hard Carbon , Electro Graphite, Metal Graphite
Design of Commutator and brushes
Design formulae
1. No. of commutator segments, C = ½ u.Sa
where, u – coils sides/slot
Sa – no. of armature slots
2. Minimum no. of segments = Ep/15
3. Commutator segment pitch = βc = πDc/C
where,
Commutator Diameter Dc – 60% to 80% of diameter of armature
βc ≥ 4mm
4. Current carried by each brush Ib= 2Ia/P for lap winding
Ib= Ia
 for wave winding
5. Total brush contact area/spindle Ab= Ib/δb
6. Number of brush locations are decided by the type of winding
Lap winding: No of brush location = no. of poles
Wave winding : No of brush location =2
Design of Commutator and brushes
7.
Area of each individual brush should be chosen such that , it does not carry
more than 70A
Let ,
ab – Contact area of each brush
nb – Number of brushes / spindle
 Contact area of brushes in a spindle, Ab = nb. ab
also ab = wb.tb
Ab = nb. wb.tb
Usually, tb = (1 to 3) βc
wb = Ab/ nb. Tb = ab/tb
8. Lc – depends on space required for mounting the brushes and to dissipate the
heat generated by commutator losses
Lc = nb(wb + Cb) + C1 + C2
where, Cb - Clearnace between brushes (5mm)
C1 - Clearance allowed for staggering of brushes (10mm, 30mm)
C2 – Clearance for allowing end play (10 to 25 mm)
Design of Commutator and brushes
9.
Losses :
 Brush contact losses: depends on material, condition, quality of
commutation
 Brush friction losses
Brush friction loss Pbf = μ pb AB.Vc
μ – Coefficient of friction
pb-Brush contact pressure on commutator (N/m2)
AB - Total contact area of all brushes (m2)
AB =P Ab (for lap winding)
= 2 Ab (for wave winding)
Vc – Peripheral speed of commutator (m/s)
Design of Interpoles
 Interpoles: Small poles placed between main poles
 Materials Used: Cast steel (or) Punched from sheet steel
without pole shoes
 Purposes:
 To neutralize cross magnetizing armature MMF
 To produce flux density required to generate rotational voltage in the
coil undergoing commutation to cancel the reactance voltage.
 Since both effects related to armature current, interpole
winding should be connected in series with armature
winding
 Average reactance voltage of coil by Pitchelmayer’s Equation
is, Erav = 2Tc ac Va.L .λ
Inductance of a coil in armature =2Tc2 .L .λ
Design of Interpoles
Normally, Length of interpole = length of main pole
Flux density under interpole, Bgi = ac. λ .(L/Lip)
where, Lip- length of interpole
In general,
Bgi = 2 Iz. Zs. (L/Lip). (1/Va.Tc).λ
 mmf required to   mmf required to over come 
ATi  


establish
B
armature
reaction
gi




MMF required to establish Bgi = 800000Bgi.Kgi.lgi
MMF required to

 I z .Z
overcome
 
2P
armature reaction 

 ( without compensati
 (1 -  ) 
I z .Z
2P
ng winding)
 ( with compensati
ng winding)
Design of Interpoles
No.of turns 
AT i
Ia
Current density in 
 , δ i  2.5 to 4 A/mm
interpole winding 
Area of X - section of 
 A ip
interpole conductor, 

2
Ia
δi
Losses and efficiency :
1.
2.
3.
Iron Loss - i)Eddy current loss ii) Hysteresis loss
Rotational losses - Windage and friction losses
Variable or copper loss
Condition for maximum efficiency :
Constant Loss= Variable Loss