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2.004 Dynamics and Control II
Spring 2008
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Massachusetts Institute of Technology
Department of Mechanical Engineering
2.004 Dynamics and Control II
Spring Term 2008
Lecture 81
1
Electrical System Modeling (continued)
Modeling Real Sources: Up to this point we have considered only ideal voltage and
current sources:
v
v
a v o lta g e s o u r c e
a c u rre n t s o u rc e
V
v o lta g e is in d e p e n d e n t o f
th e c u r r e n t s u p p lie d to th e
lo a d .
c u r r e n t is in d e p e n d e n t o f
th e v o lta g e r e q u ir e d to
m a in ta in th e c u r r e n t.
i
0
V (t)
0
+
+
s y s te m
-
V
I
I
i
s y s te m
I(t)
However, practical (real) sources do not have an ideal characteristic – they are power limited.
For example we may think of a battery as an ideal voltage source - say with a 9 volt output,
but if we were to measure the terminal voltage under increasing load currents
i
+
b a tte ry
v
R
L
v
v
R
L
we would find that at high currents the voltage decreases in a nonlinear manner:
1
c D.Rowell 2008
copyright �
8–1
v
o p e n - c ir c u it v o lta g e
w h e n i = 0 (R = ¥ ).
F o r a r e a l s o u r c e , th e v o lta g e
d e c re a s e s a s th e c u rre n t
in c r e a s e s
V
" lin e a r " r e g io n
s h o r t- c ir c u it c u r r e n t
w h e n v = 0 (R = 0 ).
i
0
The voltage–current relationship is determined by the electrochemical reaction within the
battery. It may be possible to define a region of operation at low currents in which there is
an approximately linear relationship between voltage and current
v(i) ≈ voc − Rs i
where voc is the open-circuit voltage (when i = 0), and Ro is a resistance.
How can we model this linear part of the characteristic?
i
+
s
-
R
o
v
v = V
s y s te m
s
- iR
o
s o u rc e m o d e l
One way to do this is to model the real source(the battery) as an ideal voltage source Vs
in series with a resistor Ro (the value of which is found experimentally). The voltage drop
vRo = iRo across the resistor accounts for the “droop” in the source characteristic.
This is known as the Thévenin equivalent source model.
V
v
o c
R
o
=
d v
d i
=
i
v
o c
s c
T h e v e n in s o u r c e m o d e l
a c tu a l s o u rc e
c h a r a c te r is tic
0
i
8–2
s c
i
Note that
(1) The open-circuit (i = 0) voltage is Vs .
(2) The short-circuit (v = 0) current is isc = Vs /R.
Example 1
Electronic instruments are often designed to have a specific output impedance
(resistance) Ro . For example when the Tektronix function generator in the 2.004
laboratory is connected to an arbitrary load resistance RL the measured output
voltage vout will not necessarily be equal to the value set on the front panel:
in d ic a te d v o lta g e
v
i
1 .2 3 4
o u t
R
L
(a) If RL = ∞ (no load connected) the true voltage at the terminals is twice the
indicated value.
(b) If RL = 0 (short circuit) the actual voltage is zero but the indicated voltage
is unchanged.
(c) If RL = 50 Ω the indicated and measured output voltages are the same.
Why?
Answer: The function generator is specified as having an output impedance of
50Ω. In fact, if you look at the circuit schematic, you will find that there is a
50Ω resistor in series with the output terminal:
F u n c tio n G e n e r a to r
a m p lifie r
v1
5 0 W
v
o u t
R
1 .2 3 4
L
a c tu a l v
v o lta g e
v o ltm e te r
T h e m e te r is c a lib r a te d to r e a d
c o r r e c tly o n ly w h e n a 5 0 W lo a d
r e s is to r is c o n n e c te d !
8–3
o u t
= R
L
R
L
+ 5 0 v
1
In the lab we have 50Ω terminator resistors connected across the output terminals
so that the voltage reads correctly.
A second approach to modeling the real source characteristic
v
V
v = V
- iL R
o c
R
o c
o
o
d v
=
=
d i
i
v
o c
s c
0
i
s c
iL
is to use a current source IN in parallel with the resistor Ro
N
(a )
l
R o
N
R
o
v
s y s te m
N o rto n s o u rc e m o d e l
This is known as the Norton equivalent source model.
If no load resistor is connected (open-circuit)
vos = IN Ro .
The short-circuit current is
isc = IN .
If the load is a resistor RL
v=
RL Ro
RL
v/i
=
voc =
RL + R o
RL + R o
vı + Ro
and rearranging
v = voc − iRL .
The Norton and Thévenin source models are equivalent and cannot be dis­
tinguished by measurements at the terminals.
8–4
Example 2
Measurements on an industrial dc power supply showed it to have an open-circuit
voltage of 200 v, but when 10 Ω resistor is connected, the voltage drops to 133v.
Find Thev´enin and Norton equivalent source models.
R
V
o
+
th
R
R
IN
L
-
o
Thevénin Model:
R
V
th
= 2 0 0 v
o
+
R
-
L
= 1 0 W
Vth = voc = 200 v.
Using the voltage divider relationship
vo = vRL =
10
= 133volts
10 + Ro
which gives
Ro = 5Ω.
The Thevénin model is:
R
V
Norton Model:
th
= 2 0 0 v
o
= 5 W
+
-
From above, the short-circuit current is
isc =
200
Vth
=
= 40 amps
Ro
5
and the Norton model is:
8–5
R
L
IN = 4 0 A
R
8–6
N
= 5 W