Transcript Review of exponential charging and discharging in RC Circuits

EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
Linear Circuits -Special Properties
• Circuits consisting only of linear elements are linear
circuits.
• There are simple “equivalent circuits” for “one-port”
linear circuits.
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
TWO-TERMINAL LINEAR RESISTIVE NETWORKS
(“One Port” Circuit)
Interconnection of two-terminal linear resistive elements with only two
“accessible” terminals
+

a
b
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
I-V CHARACTERISTICS OF LINEAR TWO-TERMINAL NETWORKS
i
i
5K
+
+
+
5V
5V
v

Associated

v

Unassociated

Apply v, measure i,
or vice versa
Apply v, measure i,
or vice versa
i(mA)
1
+
i(mA)
1
Associated
(i defined in)
.5
Unassociated
(i defined out)
If V = 2.5V
.5
v(V)
5
5
-.5
-.5
-1
-1
v(V)
If R = 2.5K
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
BASIS OF THÉVENIN THEOREM
• All linear one-ports have linear I-V graph
• A voltage source in series with a resistor can
produce any linear I-V graph by suitably adjusting V
and I
THUS
• We define the voltage-source/resistor combination
that replicates the I-V graph of a linear circuit to be the
Thévenin equivalent of the circuit.
• The voltage source VT is called the Thévenin
equivalent voltage
• The resistance RT is called the Thévenin equivalent
resistance
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
FINDING VT, RT BY MEASUREMENT
1) VT is the open-circuit voltage VOC (i.e., i = 0)
RT
+
+
VT


VOC
2a) RT is the ratio of VOC to iSC, the short-circuit current
iSC
V
R T  OC
ISC
2b) If VT = 0, you need to apply test voltage, then RT 
i
Note direction of i !
+

VTEST
i
VTEST
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
FINDING VT, RT BY ANALYSIS
1) Calculate VOC. VT = VOC
2a) If have only resistors and independent voltage/current
sources, turn off independent voltage/current sources and
simplify remaining resistors
OR, if VOC IS NONZERO,
Find ISC, then RT=VOC/ISC
Finish this later when we look at dependent sources…
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
How to turn off voltage/current sources
• Disable a voltage source: make it have zero voltage
Turn voltage source into short circuit (wire)
• Disable a current source: make it have zero current
Turn current source into open circuit (air)
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
NORTON EQUIVALENT CIRCUIT
Corollary to Thévenin: IN  ISC (short - circuit current)
RN is found the same way as for Thévenin equivalent
IN
RN
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
EXAMPLE
Find the Thévenin and Norton equivalents of:
A
25 K
Find VAB = VOC from voltage divider. Left to right:
(4 V rise across 25K + 75K) 
75 K
3 V across 75K, 1 V across 25k, + at bottom.
+

So, VAB = 3 + 6 = 3V = VOC
3
RTH 
 18.8K
3
.16  10
ISC = 6V/75K+2V/25K = 0.16 mA
2V + 6V

B
equivalent to
and equivalent to
A
18.8K A
3V
+
Thévenin
0.16 mA
B
Norton
18.8 K
B
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
Dependent Voltage and Current Sources
A linear dependent source is a voltage or current source that
depends linearly on some other circuit current or voltage.
Example: you watch a certain voltmeter V1 and manually adjust a
voltage source Vs to be 2 times this value. This constitutes a
voltage-dependent voltage source.
Circuit A
+
V1
-
Vs=2V1
+
-
Circuit B
This is just a (silly) manual example, but we can
create such dependent sources electronically. We
will create a new symbol for dependent sources.
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
Dependent Voltage and Current Sources
• A linear dependent source is a voltage or current source that
depends linearly on some other circuit current or voltage.
• We can have voltage or current sources depending on voltages
or currents elsewhere in the circuit.
Here the voltage V is proportional to the voltage across the element c-d .
c
+
Vcd
+
-
V = A v x Vcd
-
d
A diamond-shaped symbol is used for dependent sources, just
as a reminder that it’s a dependent source.
Circuit analysis is performed just as with independent sources.
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
EXAMPLE OF THE USE OF DEPENDENT SOURCE
IN THE MODEL FOR AN AMPLIFIER
AMPLIFIER SYMBOL
Differential Amplifier
V+
V
+
A

V0  A( V  V )
AMPLIFIER MODEL
Circuit Model in linear region
V0
Ri
+

V1
+

AV1
+

V0
V0 depends only on input (V+  V-)
See the utility of this: this Model when used
correctly mimics the behavior of an amplifier but
omits the complication of the many many
transistors and other components.
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
The 4 Basic Linear Dependent Sources
Constant of proportionality
Parameter being sensed
Output
Voltage-controlled voltage source … V = Av Vcd
Current-controlled voltage source … V = Rm Ic
Current-controlled current source … I = Ai Ic
Voltage-controlled current source … I = Gm Vcd
+
Av Vcd
-
+
- Rm Ic
Ai Ic
Gm Vcd
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
NODAL ANALYSIS WITH DEPENDENT SOURCES
Example circuit: Voltage controlled voltage source in a branch
R5
R1
R3
Va
Vb
Vc
R2
+

VAA
+

AvVc
R4
R6
ISS
Write down node equations for nodes a, b, and c.
(Note that the voltage at the bottom of R2 is “known” so current
flowing down from node a is (Va  AvVc)/R2.)
Va  VAA Va  A v Vc Va  Vb


0
R1
R2
R3
Vc  Vb Vc
Vb  Va Vb Vb  Vc

 ISS


0
R5
R6
R3
R4
R5
CONCLUSION:
Standard nodal
analysis works
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
ANOTHER EXAMPLE OF NODAL ANALYSIS WITH
DEPENDENT SOURCES
R1
I
Va
I2
R3
R2
R4
+
-
Rm I2
I2
Standard technique, except an additional equation is needed if the
dependent variable is an unknown current as here.
Dependent voltage
sources also have
I = Va / R2 + (Va - Rm I2)/ R3 and I2 = Va / R2
unknown current—so
no KCL at attached
nodes. Supernode
I = Va (1/R2 + 1/R3 - Rm /R2 R3 )
Solving:
around floating
dependent voltage
So Va = I R2 R3 /(R2 + R3 - Rm )
sources!
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EECS 40
Fall 2000 Lecture 9-10
W. G. Oldham and S. Ross
FINDING VT, RT BY ANALYSIS
1 Calculate VOC. VT = VOC
2a) If have only resistors and independent voltage/current
sources, turn off independent voltage/current sources and
simplify remaining resistors
2b) If only dependent sources and resistors present, you will
need to apply test voltage, calculate i and use RT  v TEST i
as in previous slide
2c) If both independent and dependent sources (with
resistors) present, you may find ISC and let RT=VOC/ISC, or turn
v TEST
off independent sources and use
RT 
i
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