Transcript Lab 7
Motors
Discussion D10.2
Chapter 15
Hans Christian Oersted (1777 – 1851)
X
1822
In 1820 he showed that a current
produces a magnetic field.
Ref: http://chem.ch.huji.ac.il/~eugeniik/history/oersted.htm
André-Marie Ampère (1775 – 1836)
French mathematics professor who only
a week after learning of Oersted’s
discoveries in Sept. 1820 demonstrated
that parallel wires carrying currents
attract and repel each other.
attract
A moving charge of 1 coulomb
per second is a current of
1 ampere (amp).
repel
What do Maxwell’s Eqs. Predict?
B = 0
B = magnetic flux density
(magnetic induction)
B = mH
m = magnetic permeability
Magnetic field lines
must be closed loops
Force on moving charge q
Lorentz force
F = q ( v B)
B
Vector Multiplication
Scalar (Dot) Product
Vector (Cross) Product
A
A
q
q
B
B
C=A B
C = AB
C = A B cosq
C = A B sin q
Direction of C given by "Righthand rule" (into the screen)
Force on current in a magnetic field
Force on moving charge q -- Lorentz force
F = q ( v B)
Current density, j, is the amount of charge passing per unit area
per unit time. N = number of charges, q, per unit volume moving
with mean velocity, v.
L
S
j
F = N V q(v B)
F = ( j B)V
F = (i B)L
vt
V = S L
dQ NqSvt
i=
=
= j S
dt
t
j = Nqv
Force per unit length on a wire is
iB
Rotating Machine
B
Force out
i
+
i
X
Force in
Rotating Machine
B
Force in
X
i
+
i
Force out
Rotating Machine
B
Force out
commutator
i
+
i
brushes
X
Force in
What do Maxwell’s Eqs. Predict?
E =
B
t
Corresponds to Faraday’s law of
electromagnetic induction
A changing magnetic flux B density induces a curl of E
B
nda
S t
E ds = E nda =
S
E ds = t S B nda = t flux through S
Back emf
B
B
dq
=
dt
Force out
+-
i
a
b
r
q
l
i
X
X
Force in
area = A = lw = l 2r cos q
w = 2r cos q
flux = BA = Β2rl cos q
emf = E ds = flux = B2rl cos q
t
t
q
emf = eab = B2rl sin q
= 2Brl sin q
t
B
Back emf
eab = 2Brl sin q
Force out
+
-
i
a
b
i
X
eab
Force in
q
B
Back emf
Force out
commutator
i
+
-
eab = 2Brl sin q
i
brushes
X
Force in
eab
q
Armature with four coil loops
eab
S
X
X
X
X
N
q
Motor Circuit
Ia
Ra
Vt
Ea
Vt = Ea I a Ra
Ia = Vt Ea / Ra
Ia = Vt Ka / Ra
Ea = Ka
Power and Torque
Pd = Ea I a = d
d = Ka I a
Vt K a K a
d =
Ra
Ra
2
Armature Current, Speed, Power and Torque
1
Ia =
d
K a
d = Ka I a
Vt K a K a
d =
Ra
Ra
Vt
Ra
=
2 d
Ka Ka
Pd = Ea I a = d
Vt
Ra
2
Pd =
d
2 d
Ka
Ka
2
FF-130 Series Motors
Vt
Ra
2
Pd =
d
2 d
Ka
Ka
1
Ia =
d
K a
Vt
Ra
=
2 d
Ka Ka
Vt
Ra
=
2 d
Ka Ka
Vt
Ra
=
=0
2 d
Ka Ka
dω=0
Vt Ka
=
= 6.4 mN-m
Ra
Vt
Ra
=
2 d
Ka Ka
d = 0
d
=0
Vt
=
= 7100 rpm
K a
Ia =
1
d
K a
I aω=0
Vt
=
Ra
Vt
4.8
Ra = =
= 4.0
I a 1.21
At = 0
Vt K a
d =
= 6.4 mN-m
Ra
Pd max
Vt 2
=
4 Ra
Vt
Ra
2
Pd =
d
2 d
Ka
Ka
Maximum Power
dPd
V
2Ra
= t
=0
2 d
d d Ka Ka
d max
Vt K a
=
2 Ra
d =0
Vt K a
=
Ra
d
=0
Vt
=
= 7100 rpm
K a
K a =
I aω=0
Vt
4.8
=
= 0.00646 volt-sec
7100 rpm 7100(2 / 60)
Vt
=
Ra
Vt
4.8
Ra = =
= 4.0
I a 1.21
Pd max
Vt 2
4.82
=
=
= 1.4 W
4Ra 4 4
dω=0
Vt K a 4.8 0.00646
=
=
= 7.6 mN-m
Ra
4
1.18 W
6.4 mN-m
Generator Circuit
Ia
Ra
Vt
Ea
Vt = Ea I a Ra
Ea = Ka
Ea = 0.00646 7100 2 / 60 = 4.8 V
Ka = 0.00646 volt-sec
Vt = 4.8 0.15 4 = 4.2 V
+5v
V
Motor Driver Circuit
470
1
SSR
V
PWM
p31
6
2
14 13 12 11 10
9
4
8
MOTOR
7406
1
2
3
4
5
6
7
GND
MOS FET Relays
G3VM-61B1
Pulse-Width Modulation
count[3]
set = ~| count
pwm
duty
period
reset = (count == duty)
Vpwm
duty
=
Vref
period
pwm.v
module pwm(clk,clr,duty,pwm);
input clk;
count[3]
input clr;
input [3:0] duty;
set = &count
output pwm;
pwm
reg [3:0] count;
reg pwm;
wire set, reset;
//
duty
period
reset = (count == duty)
4-bit counter
always @(posedge clk or posedge clr)
if(clr == 1)
count <= 0;
else
count <= count + 1;
pwm.v
assign set = ~| count;
assign reset = (count == duty);
always @(posedge clk)
begin
if(set == 1)
pwm <= 1;
if(reset == 1)
pwm <= 0;
end
endmodule
count[3]
set = &count
pwm
duty
period
reset = (count == duty)
Lab 11
Lab 11
+5V +5V
+5V
470 Ω
14
6
7406
1
2
7
1
4
G3VM-61B1
Solid-State
2
Relay
pwm
1N4004
Motor
Generator
VOUT