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Stepper-Motor Operation and Interfacing Fundamentals
Prepared by: P. David Fisher and Diane T. Rover
Ampere’s Law & Biot-Savart Law
An electrical current I in a ware causes (induces) a magnetic
field B. The direction of B is given by the “right-hand rule”.
I
R B
Magnetic Fields for a Long Thin Wire
For a long thin wire, the strength of the magnetic field B a distance R from the
wire is
B = (U0I)/(2R)
(1)
where 0 = permeability of vacuum,
0 = 4x10-7 henry/meter
The Solenoid
A coil of wire with N turns creates a magnetic field B in the direction illustrated,
where
B = kNI,
(2)
with k being a constant. Hence, B is proportional to N and I.
I
B
1
Magnetic
Field
Lines
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The Electromagnet
N (North)
(South) S
A
B
Iron Core
If you place a compass and in the vicinity of the iron core, you would discover that
one end (say A) would be similar to the “South Magnetic Pole” of the earth, while
the other end (say B) would be similar to the Earth’s “North Magnetic Pole”.
Two important properties of electromagnets are the following:
1. All Electromagnets are dipoles; i.e., they have a North Pole (N) and a South
Pole (S).
2. The position of the Poles (at A or B) is determined by the direction of the
current I and the direction of the winding.
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Basic Model for a Stepper Motor
Consider the four electromagnets physically arranged as illustrated.
P1(L11,L12)
A
B
P4(L41,L42)
A
B
A
L31
R31
B
P2(L21,L22)
B
A
P3(L31,L32)
where
R11
L11
i11
i31
va
R12
L12
L32
i12
R32
Windings
for
P1 & P3
i32
vb
L21
R21
L41
R41
1
i21
i41
vc
R22
L22
L42
i22
R42
Windings
for
P2 & P4
i42
vd
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Controlling Magnetic Polarities with Winding Voltages (va, vb, vc and vd)
The magnetic polarities of the electromagnets can be controlled by varying the
winding voltages va, vb, vc and vd. Consider the following two cases.
Case I – P1 and P3
P1(L11,L12)
A
B
P(L31,L32)
A
B
P1(L11,L12)
A
B
P3(L31,L32)
A
B
N
A
B
S
S
A
B
N
S
A
B
N
N
A
B
S
ill > 0
va = 5V
il2 = 0
vb = 0V
i3l > 0
va = 5V
i32 = 0
vb = 0V
ill = 0
va = 0V
il2 > 0
vb = 5V
i3l = 0
va = 0V
i32 > 0
vb = 5V
Note: Winding voltages va and vb control the polarities for electromagnets P1 and
P3.
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Case II – P2 & P4
P2(L21,L22)
A
B
P4(L41,L42)
A
B
P2(L21,L22)
A
B
P4(L41,L42)
A
B
N
A
B
S
S
A
B
N
S
A
B
N
N
A
B
S
i2l > 0
vc = 5V
i22 = 0
vd = 0V
i41 > 0
vc = 5V
i42 = 0
vd = 0V
i2l = 0
vc = 0V
i22 > 0
vd = 5V
i4l = 0
vc = 0V
i42 > 0
vd = 5V
Note: Winding voltages vc and vd control the polarities of electromagnets P2 and
P4.
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Controlling Stepper-Motor State Transitions
The “state” of a stepper motor can be controlled by controlling the winding
voltages of the electromagnets. Consider the following example.
P1
va = 0V
Present State
P1 & P3
vb = 5V
B
N
P4
B
S
N
B
P2
S
B
vc = 5V
C
P2 & P4
vd = 0V
P3
P1
Next State
va = 5V
P1 & P3
vb = 0V
B
S
P4
B
S
N
B
P2
N
B
vc = 5V
C
P2 & P4
vd = 0V
P3
Note: The polarities of electromagnets P1 and P3 can be reversed by
simultaneously changing va from 0V to 5V and vb from 5V to 0V.
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Important Questions and Conclusions
With respect to the previous example, answer the following questions.
1. Assume that the stepper motor is in its “initial state.” If a compass is positioned
with the pivot point of its needle at point C, in what direction would the needle
point?
2. Assume that the stepper motor is in its “next state.” If a compass is positioned
with the pivot point of its needle at point C, in what direction would the needle
point?
3. Did the compass needle move clockwise or counter clockwise?
4. What voltages do we need to change to have the compass needle rotate in the
opposite direction?
5. How many “steps” does it take to make a 360 rotation?
6. How might you add the number of steps for a 360 rotation? Identify two
distinct approaches.
7. Why might you want to add steps?
8. What are the engineering design considerations that must be addressed as a new
stepper-motor assembly is designed for a new commercial application?
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Stepper-Motor Interface Circuit Model
There are a number of significant challenges facing the computer engineer who
must interface a stepper motor to a microcontroller. For example, consider the
following transient circuit response problem.
S1
RS1
VS
RS1 = Shunt
Resistance of
Switch
R11
+ VS1 i11
L11
+
VL11
-
RS1 >> R11
Case 1: Switch closes at t = 0
i11(0-) = i11(0+) = VS/RS  0A
i11(t) = (VS/R11)e-t/, where  = R11/L11
VL11(t) = Vse-t/
(3)
(4)
(5)
Case 2: Switch Opens at t = 0
Because currents through an inductor cannot change discontinuously,
ill(0+) = i11(0-) = VS/R11
(6)
Applying Kirchhoff’s Voltage Law (KVL) around the loop at time t = 0+ yields:
-VS + VS1 + R11i11 + VL11 = 0
-VS + RS1i11 + R11i11 + VL11 = 0
VL11 = VS – (RS1 + R11)i11  VS – (RS1/R11)VS, when RS1 >> R11
VL11  -(RS1/R11)VS  -(1M/0.1k)VS  -104VS
(7)
(8)
(9)
(10)
These large voltages will destroy the solid-state switch.
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Diode Protection
There exists a very standard solution to the problems which arise due to the desire
to rapidly switch electrical currents in circuits containing inductive loads. The
following example illustrates the solution.
The winding of an electromagnet can be modeled as a resistance in series with an
inductance, as illustrated in the figure. Under computer control, current i11 is to be
controlled by controlling voltage v1. As we saw with the stepper-motor example,
i11 will assume one of two steady-state values—i.e., i11 = 0A and i11 = VS/R11. In the
circuit illustrated, diode D1 protects the interface logic from large transient
voltages.
v1
Interface
Computer
Logic
D1
R11
i11
L11
Case 1: v1 = 0V
The voltage drop across the diode is v1 = 0V, and the diode is turned off (an open
circuit). Also, i11 = 0A.
Case 2: v1 = VS, where VS > 0V
The voltage drop across the diode is v1 = VS, and the diode is turned off (an open
circuit). Also, i11 = VS/R11.
Case 3: At t = 0-, v1 = VS, where VS > 0V. Then at time t = 0, the interface logic
switches and presents a high impedance to the rest of the circuit.
At time At t = 0+, the current i11 = VS/R11 and passes through the diode. The diode
is forward biased with v1 = -0.7V. With time, i11 drops to 0A, v1 returns to 0V and
the diode is turned off (an open circuit).
This is the solution to only one interfacing problem. Another common problem is
the fact that actuators, such as the stepper motor, do not operate at standard “logic
voltages.” This problem will be discussed as we investigate the electrical
properties of a specific stepper motor and its computer-interface requirements.
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