Page 565 - The Mechatronics Handbook

0066_Frame_C20 Page 35 Wednesday, January 9, 2002 5:49 PM

The induced emf or back emf will oppose the applied voltage. Since the emf is directly proportional
to the motor speed then on startup, there will be no back emf generated. This will have consequences
on the current, which will be drawn by the coils, and some measures will have to be taken to counteract
this effect. This topic will be considered later.

Armature Torque
The force on a current carrying conductor is given as

F = BLI (20.17)
where B is the magnetic flux density under a pole, I is the current flowing in the conductor, and L is the
axial length of the conductor.
The torque on one armature conductor is, therefore,
T = Fr = B av LI a r (20.18)

where r is the radius of the armature conductor about the center of rotation, I a is the current flowing in
the armature conductor, L is the axial length of the conductor, and B av is the average flux density under
a pole.
Given that B av = Φ/(2πrL)/2p], the resultant torque per conductor is
[
Φ2pLI a r ΦpI a
T = --------------------- = ------------ (20.19)
2prL π

For Z s armature conductors connected in series, the total torque (in Nm) on the armature is given by

T = ΦpI a Z s (20.20)
-----------------
π
Terminal Voltage

Denoting the terminal voltage by V, in normal running conditions we have a balanced electrical system
where:

V = E + I a R a (20.21)

Since the number of poles and number of armature conductors are fixed, then from Eq. (20.16) we
have a proportionality relationship between the speed, the induced emf, and the magnetic flux, i.e.,

E
N ∝ ---- (20.22)
Φ
Using Eq. (20.21)
( VI a R a )
–
N ∝ ------------------------ (20.23)
Φ
Since the value of I a R a is normally less than about 5% of the terminal voltage then to a reasonable
approximation
V
N ∝ ---- (20.24)
Φ

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