Download Lecture 2 Review • Methods of Analysis —Nodal analysis —Mesh analysis

Lecture 2 Review
• Methods of Analysis
—Nodal analysis
• Without Voltage Sources
• With Voltage Sources
—Mesh analysis
• Without Current Sources
• With Coltage Sources
—Analysis by Inspection
Quiz
An open-loop gain: 2105
Input resistance: 2 M
Output resistance: 50  v
Find the closed-loop gain: o
vs
Lecture 3 DC Circuits
Circuit Theorems: to simplify circuit analysis
Lecture Objectives
• Linear circuit
• Superposition
• Equivalent Circuits
—Simple Circuits
—Y- Transforms
—Source transformation
—Thevenin's theorem
—Norton's theorem
• Maximum power transfer
Linear circuit
v  iR
—Homogeneity: i1  ki0 , v1  ki0 R  k (i0 R)  kv0
—Additivity:
i  i1  i2 , v  (i1  i2 ) R  i1 R  i2 R  v1  v2
• Linearity Property
Input(excitation),i
Linear Circuit
Output(response), v
Example: Linearity
3A
6V
2A
5A
14V
2A
3V
1A
8V
Assume I0 = 1 A,… the source current Is =?
5V
5A
Since Is = 15 A = 3 × 5A, according to linearity
(homogeneity), I0 = ? 3 A
Superposition: several sources
• Linear circuit: Additivity
• Voltage or Current = algebraic sum of v or i due to
each independent source acting alone,
respectively.
• Can we use superposition for power? No! Why?
If there are N sources, you’ll need to do similar calculation N times.
Example: Superposition
Use the superposition theorem to find v in the circuit:
Turn off
v = 10 V
Voltage source
0V
Short circuit
Current source
0A
Open circuit
How about dependent sources?
Equivalence of circuits
a
b
•
•
•
•
•
What elements are in the box?
What is the behaviour of the circuit?
Voltage-Current relation
Series-parallel combination
Y- transformation
Source Transformation
Thevenin’s Theorem
Norton’s Theorem
Voltage-current relation at terminals a and b is identical to that of its equivalent circuit.
Simple resistive circuits
• Series circuits
— Current: elements in series carry the same current
— Resistance in series: the sum of the individual resistances
• Parallel circuits
— Voltage: elements in parallel have the same voltage
— Conductance in parallel: the sum of the individual conductances
L1
One path I
L2
Equivalent
L=L1+L2
V
RT  
RT  
A1
two paths
R
L
A
L1  L2
L
L
  1   1  R1  R2
A
A
A
L
,
A1  A2
1 1 A1  A2 1 A1 1 A2 1 1



 
RT  L
 L  L R1 R2
A2
A=A1+A2
GT  G1  G2
 - Y Transformation
R1 
Rb Rc
Ra  Rb  Rc
Rc Ra
R2 
Ra  Rb  Rc


Ra Rb
R3 
Ra  Rb  Rc
T
Y
Y- Transformation
Example
Obtain the equivalent
resistance Rab for
the circuit and use
it to find current i
Source transformation
Replacing a voltage source vs in series with a resistor R by a
current source is in parallel with a resistor R, or vice versa.
A Voltage Source  A CurrentSouce
Check:
Voltage Source
Open Circuit
Short Circuit
vs
vs
R
That’s how this part of circuit acts on
the load as a ’source’.
Current Source
Turn-off source R
is R
R
is
R
That’s what the external sources ‘feel’
this part of the circuit.
Example
Use source transformation to find vo in the circuit
Thevenin's theorem
Load – variable
Other elements - fixed
Vth : Open circuit voltage
RTh : the input resistance at the terminals when the
independent sources are turned off
Example
Find the Thevenin equivalent circuit of the circuit, to the left of the terminals a-b.
Then find the current through RL = 6, 16, and 36 
Norton's theorem
Maximum power transfer
• Minimizing power dissipated in the process
of transmission and distribution
• Maximize the power delivered to a load
p  i 2 RL
i
VTh
RTh  RL
2
 VTh 
2
 RL
p  i RL  
 RTh  RL 
Apendix
Op Amp - typical packages
1.
2.
An active circuit element
Perform mathematical
operations
Pin 1 is always to
the left of the
notch or dot.
For package information:
http://www.intersil.com/design/packages/
Looking from the top
741 General-purpose: Fairchild Semiconductor … Intel
Feedback path - example
Feedback: negative feedback
An open-loop gain: 2105
Input resistance: 2 M
Output resistance: 50  v
o
Find the closed-loop gain:
vs
Nodal analysis
vo
 1.9999699
vs
Insensitive to A
This is tedious.
Ideal Op Amp
Virtual open circuit
i1 = i2 =0
Virtual close circuit
vd = 0; v1 = v2
• Infinite open-loop gain, A  
• Infinite input resistance, Ri  
• Zero output resistance, Ro  0
Idea Op Amp - Example
Rework it using the ideal op amp model
Idea Op Amp: i1  i2  0
v1
i1 =0
v2
i2 =0
Negligibly small error results from
assuming ideal op amp characteristics
Virtual open circuit: i1 = i2 =0
Virtual close circuit: vd = 0; v1 = v2
Node 1:
vs  v1
v1  vo

10 103 20 103
Idea Op Amp: v1  v2  0
vs  0
0  vo

10 103 20 103
vo
 2
vs
Working with non-ideal:
vo
 1.9999699
vs
Lecture 4 DC Circuits
Capacitors and Inductors