Summary from previous lecture
•
Z2(s)
Electrical dynamical variables and elements
Vi(s)
Z1(s)
I1(s)
Charge q(t), Q(s).
Current i(t) = q̇(t), I(s) = sQ(s).
Voltage v(t), V (s) = Z(s)I(s).
Ia(s)
···
ik
P
KVL −
+
−
v1
P
···
P
P
Z1
KCL
Vk (s) = 0
2.004 Fall ’07
Z2
Z = Z1 + Z2
1
1
1
=
+
.
G
G1
G2
+
vk (t) = 0
+
Vo (s)
Z2 (s)
=−
.
Vi (s)
Z1 (s)
Impedances in series
Ik (s) = 0
vk
Vo(s)
Z1
ik (t) = 0
+−
I2(s)
Op—Amp in feedback
configuration:
Electrical networks
i1
-
Figure by MIT OpenCourseWare.
Resistor v(t) = Ri(t), ZR (s) = R.
Capacitor i(t) = C v̇(t), ZC (s) = 1/Cs.
Inductor v(t) = Ldi(t)/dt, ZL (s) = Ls.
•
V1(s)
Z1
Z2
+
Vi
−
+
−
Vi
Impedances in parallel
1
1
1
+
=
Z
Z1
Z2
G = G1 + G2 .
Lecture 05 – Friday, Sept. 14
I
Z2
+V
2
−
Z2
Z1 + Z2
Voltage divider
V2
Goals for today
•
•
The DC motor:
– basic physics & modeling,
– equation of motion,
– transfer function.
Next week:
– Properties of 1st and 2nd order systems
– Working in the s-domain: poles, zeros, and their significance
2.004 Fall ’07
Lecture 05 – Friday, Sept. 14
Power dissipation in electrical systems
i(t)
+
v(t)
–
Instantaneous power dissipation
P (t) = i(t) · v(t).
Unit of power: 1 Watt=1 A · 1 V.
NOTE: P (s) 6= I(s) · V (s). Why?
2.004 Fall ’07
Lecture 05 – Friday, Sept. 14
DC Motor as a system
Transducer:
Converts energy from one domain (electrical)
to another (mechanical)
2.004 Fall ’07
Lecture 05 – Friday, Sept. 14
Physical laws applicable to the DC motor
Lorentz law:
Faraday law:
magnetic field applies force to a current
(Lorentz force)
moving in a magnetic field results
in potential (back EMF)
F = (i × B) · l = iBl
2.004 Fall ’07
(i ⊥ B)
ve = V × B · l = V B l
Lecture 05 – Friday, Sept. 14
(V ⊥ B)
DC motor: principle and simplified equations of motion
multiple windings N:
continuity of torque
T = 2F r = 2(iBN l)r
(Lorentz law)
ve = 2V BN l = 2(ωr)BN l
(Faraday law)
or
where
T = Km i
• Km ≡ 2BN lr torque constant
ve = K v ω
• Kv ≡ 2BN lr back-emf constant
2.004 Fall ’07
Lecture 05 – Friday, Sept. 14
DC motor: equations of motion in matrix form
·
ve
i
¸
=
"
2BN lr
0
0
1
2BN lr
#·
ω
T
¸
or
·
2.004 Fall ’07
ve
i
¸
=
"
Kv
0
0
1
Km
#·
ω
T
¸
Lecture 05 – Friday, Sept. 14
multiple windings N:
continuity of torque
DC motor: why is Km=Kv?
Pin = Pout
(power conservation)
⇒ ive = T ω
⇒ Kv iω = Km iω
⇒ Kv = Km
2.004 Fall ’07
QED.
Lecture 05 – Friday, Sept. 14
DC motor with mechanical load and
realistic electrical properties (R, L)
inductance
(due to windings)
dissipation
(resistance of windings)
Equation of motion — Electrical
vs − vL − vR − ve = 0
KCL:
di
− Ri − Kv ω = 0
dt
Equation of motion — Mechanical
⇒ vs − L
Torque Balance:
T = Tb + TJ
⇒ Km i − bω = J
dω
dt
Combined equations of motion
dissipation
(viscous friction
in motor bearings)
2.004 Fall ’07
L
load
(inertia)
di
+ Ri + Kv ω = vs
dt
Lecture 05 – Friday, Sept. 14
J
dω
+ bω = Km i
dt
DC motor with mechanical load and
realistic electrical properties (R, L)
inductance
(due to windings)
dissipation
(resistance of windings)
Equation of motion — Electrical
KCL:
Vs (s) − VL (s) − VR (s) − Ve (s) = 0
Vs (s) − LsI(s) − RI(s) − Kv Ω(s) = 0
Equation of motion — Mechanical
Torque Balance:
T (s) = Tb (s) + TJ (s)
Km I(s) − bΩ(s) = JsΩ(s)
Combined equations of motion
LsI(s) + RI(s) + Kv Ω(s) = Vs (s)
dissipation
(viscous friction
in motor bearings)
2.004 Fall ’07
JsΩ(s) + bΩ(s) = Km I(s)
·
µ
¶
¸
Js + b
⇒ (Ls + R)
+ Kv Ω(s) = Vs (s)
Km
¶
µ
¶¸
µ
·
Km Kv
Km
Lb
LJ 2
s +
+J s+ b+
Ω(s) =
Vs (s)
⇒
R
R
R
R
load
(inertia)
Lecture 05 – Friday, Sept. 14
DC motor with mechanical load and
realistic electrical properties (R, L)
inductance
(due to windings)
dissipation
(resistance of windings)
Neglecting the impedance
L≈0
·
µ
¶¸
Km Kv
Km
⇒ Js + b +
Ω(s) =
Vs (s)
R
R
This is our familiar 1st —order system!
If we are given step input vs (t) = V0 u(t)
⇒ we already know the step response
´
Km ³
−t/τ
u(t),
V0 1 − e
ω(t) =
R
where now the time constant is
dissipation
(viscous friction
in motor bearings)
2.004 Fall ’07
load
(inertia)
τ= Ã
Lecture 05 – Friday, Sept. 14
b+
J
Km Kv
R
!.
Review: step response of 1st order systems we’ve seen
•
Inertia with bearings (viscous friction)
Step input Ts (t) = T0 u(t) ⇒ Step response
•
•
´
T0 ³
−t/τ
ω(t) =
,
1−e
b
where τ =
J
.
b
RC circuit (charging of a capacitor)
Step input vi (t) = V0 u(t) ⇒ Step response
³
´
−t/τ
vC (t) = V0 1 − e
, where τ = RC.
R
+
vi
−
+
C
vC
−
DC motor with inertia load, bearings and negligible inductance
Step input vs (t) = V0 u(t) ⇒ Step response
´
Km ³
−t/τ
ω(t) =
,
V0 1 − e
R
where
τ= Ã
2.004 Fall ’07
J
b+
Km Kv
R
!.
Lecture 05 – Friday, Sept. 14