Elec467 Power Machines &
Transformers
Electric Machines by Hubert, Chapter 10
Topic: DC machines
Flux distribution in a DC generator
DC machines
are powered
from a DC
source.
We will first look at
the DC machine as
a generator.
In Fig. 10.1 we
see the flux
distribution thru
the armature
core of a 2-pole
DC machine.
The interpolar
region is the
space between
the poles.
The neutral plane is shown.
In the neutral plane there is no flux.
Plot of flux passing thru surface A-B
In the top diagram a
plane labeled A-B is
designated at 0°
orientated at right angle
to the flux between two
permanent magnet. The
amount of flux flowing is
entirely within the surface
of the plane. As the rotor
rotates, the amount of
flux passing thru the
plane is diagramed in (b).
Adding voltage wave e
generated in the armature
e  Na
d
dt
The Faraday-Lenz formula at the left says that the rate of change of
the flux times the number of coils wrapped around the rotor (surface
A-B represents one coil) generates a voltage. Since the rate of
change is constant, the voltage generated is generally flat.
Rectification by commutation
Equation 10-8 is more useful:
Ea 
nPNa  a
(10-8)
30
Ea is the voltage induced in the armature
P is the number of field poles
n is rotor rpm
Фa is maximum pole flux
Na is the number of rotor conductor turns
Pn
f 
120
(10-7)
f is the frequency of the
induced voltage
How Commutation Works
When a winding is passing
thru the neutral plane (a &
c) the brushes short circuit
the commutator bars but
there is no current being
generator at that time thus
no S/C current will flow.
When the winding is cutting
thru maximum flux the
commutator acts as a
rotary switch to polarize the
external circuit. With more
than one coil on the
armature
a
constant
voltage is created at the
output.
Cutaway view of a DC motor
Interlacing of armature wiring
Shown is the layout of an eight-slot, eight-coil armature
designed for operation with a two pole field. Letter T & B refer
to top and bottom conductors in a single slot.
Physical layout of winding
Here the windings are shown straddling different commutation bars.
DC generator equivalent circuit
In this schematic the field
windings are shown
connected to an electrical
source. Thus in addition to
the prime mover the
generator needs a power
source for the field windings.
Ra and Rf are the DC
resistance of the coils. Ea is
the output DC voltage fed to
the load & controlled by the
rheostat on the field current.
Putting two and two together
z P
kG  a
60 a
Ea  n p kG
p 
Ea 
If 
Nf If

(10-11)
(10-12)
(10-13)
nN f I f kG

(10-14)
za = # armature conductors…=2*Na
a = number of parallel paths in rotor
windings
kG = motor constant
n = rotor rpm
n
Ea
 p kG
( alternate form 10-12
used in Lab, aka 10-19 )
If = field current
Nf = # turns in field coil
R = reluctance of field coil which varies
When equation 10-12 is placed into 10-13 we see that Ea
is dependant on If
Ebat
E
 Rrheo  bat  R f
R f  Rrheo
If
(10-15)
Voltage regulation is % change in terminal voltage when loaded: 
Vnl  Vrated
100
Vrated
Comparison Generator to Motor
When the
prime mover
is causing the
rotor to rotate,
the flux
bunching
created by a
wire moving
through a
magnetic field
cause the
current to flow
in the direction
shown
When the
current in
the rotor coil
is the prime
mover, the
flux
bunching
causes the
mechanical
rotation.
While the
rotor is
rotating in
the same
direction in
both cases,
the current
is reversed.
You should notice which way the current is flowing in both cases.
How a basic DC motor works
There are two branches in this circuit: the shunt field and the armature.
The Φp created by the flow of current in the field branch (formula 10-13)
causes a counter emf in the armature branch that opposes the flow of
current Ia. This force also cause the motor to rotate.
Basic DC motor formulas
VT is the input voltage. When
supplied by an infinite bus the
value doesn’t change.
T D Bp I a kM
(10-17)
Bp = flux density in air gap
produced by shunt field poles
Ia = armature current
kM = motor constant
VT  I a Ra  Ea (10-19b)
Using KVL around
the outside loop
including the
armature branch
Ia 
VT  Ea
Ra
Solving for Ia
(10-19c)
Dynamic behavior with constant load
Speed regulation is the percent change
in speed from no load to rated load with
respect to rated load:
nnl  nrated
100
nrated
If VT in (3) stays constant the rotor rpms will
slow down (1) when the clutch is closed. Ea
decreases due to a smaller n in (2) causing
the numerator in the equation for Ia to increase
and with the increase in Ia the value of TD
increases. Rule is TD (will) = Tload
No load commutation process
(a) Current flow into
commutator bar 3
creates a positive
value. Armature coils
2 and 3 contribute to
the voltage. Coil 3
rotates CCW and coil
2 rotates CW
(c) At this point coil 2 has
reverse its current flow
and both coil 2 & 1 are
now sending current into
the brush keeping the
positive value.
(b) With rotation the
brush shorts commutator
bars 2 & 3 with all the
current going into the
brush. There is no
current flow generated by
coil 2 as it’s in the neutral
plane. The brush is
positive.
Commutation when loaded
When loaded the increased
current takes longer to
discharge thus delaying the
reversal until a spark occurs.
Interpoles
An interpole is a small inductor called a choke used to dampen sparking
in the brushes. They sit right over the brushes and are basically an
inductor used as a low pass filter to lessen high frequency voltage
changes on the DC bus.
Armature reactance transition
Notice that the
armature mmf reverses
direction when the
machine changes from
generator to motoring
and visa versa but
keeps turning in the
same direction.
Compensation windings
corrects armature reactance
Equivalent Circuit
shunt motor
This schematic is nearly the same as Fig. 10-12, can you tell the difference?
If we take formula 10-12’s alternate form and substitute in formula 10-25
we get the General Speed Equation for a DC motor:
V I R
n
T
a
 p kG
acir
(10-26)
Armature control = slowing down
Base speed is the shaft rpm obtained at rated
voltage and load with no extra resistance added.
This slide is the similar to Fig. 10-13 where a constant load was added
except here the motor was running with a load present when Rs is
increased instantaneously. By slowing down n the windage and friction
is decrease allowing the motor to adjust Ia so that TD = Tload
Field control = speeding up
Here resistance is added to the field branch. This
decreases Φp (1) which causes the counter emf in the
armature branch to decrease. The voltage drop Ea
across the armature (2) drops because the field current
decreases to a critical point when it begins to adjust to a
new constant value. Ia & TD will increase (4) & (5). In
(5) Ia is in the numerator and If is in the denominator
(Φp = NfIf) but the denominator drops faster than the
numerator can rise causing shaft rpms to increase. The
motor continues to adjust until TD = Tload
Power flow: Motor & Generator
Interesting aspect of the power flow diagrams is the rotational power losses switch
positions relative to the armature. Ph is the brush contact power loss. The armature
doesn’t actually contribute power or lose power; its function is to convert mechanical power
to electrical power or visa versa. Technically you never get more out than what you put
into the system. The power losses are the cost for converting the energy from one form to
another. The efficiency (η) with which this is accomplished is the ratio of Pout divided by Pin
Power loss formulas
Ploss  Pacir  Ph  Pcore  Pf ,w  Pstray  Pfcl
Power losses are subtracted from the input power. The input power for a
generator is the Pshaft and the input power for a motor is the electrical
power. The voltage across the brush contact used to calculate Ph is either
½ volt for metal-graphic or 2 volts for graphite brushes. In a generator, the
Pfcl are usually not included in the losses as the field is separately excited.
Pacir  I a2 Racir
Pfcl  I 2f R fcl
Pacir is the armature circuit losses (see formula below)
Pcore = hysteresis losses & eddy current losses
for both armature and field iron
Pstray ≈ 1% Pout
Pmech  VT I a  I a2 Racir  Ea I a
Racir  Ra  RIP  RCW
Equivalent Circuit
Locked Rotor
When you start the motor it behaves like a blocked rotor. Large shortcircuit current flows would result burning out the motor were specialize
starters not employed that insert heavy duty resistors into the armature
circuit then switch them out as the motor picks up torque spinning the
shaft faster and faster. When starting the motors in lab you will notice
TD is developed even if the rpm is zero.