Download Network Analysis Superposition

Network Analysis
Aims:
• Consolidate use of KCL in circuit analysis.
• Use Principle of Superposition.
• Learn basics of Node Voltage Analysis (uses KCL)
• Learn basics of Mesh Current Analysis (uses KVL)
Lecture 6
1
Superposition
• How can we deal with a network branch which is part of two networks
each with a source?
R1
V1
R2
R3
V2
R3 is carrying current
supplied by each battery
Principle of superposition: The current in a branch of a linear circuit
supplied from several sources is equal to the sum of the currents in that
component provided by each source alone with the other sources reduced
to zero
“Reduced to zero” means:
Voltage sources short circuit (V=0, R=0)
Current sources open circuit (I=0, R=∞)
Lecture 6
2
1
2Ω
Example#1
2A
8V
14 Ω
I2I1I?
Lecture 6
3
2Ω
Example #1
2A
14 Ω
8V
I?
2Ω
Stage 1:
Current source to zero (open circuit):
I1 = 8V/16 Ω = 0.5 A
14 Ω
8V
I1
Stage 2:
Voltage source to zero (short circuit):
RAB = 14 || 2 = 28/16 Ω
VAB = RAB x 2 A = 56/16 V
I2 = VAB/ 14 = 0.25 A
A
2A
2Ω
14 Ω
I2 B
Stage 3: Superposition: I = I1 + I2 = 0.75 A
Lecture 6
5
2
Applications
Superposition tells us that branch currents and voltages are always
proportional to the ‘driving’ currents or voltages. This means that resistor
networks are LINEAR CIRCUITS
Network analysis
We know that we can replace complex networks of sources and resistors by
single sources (Thévenin, Norton), so we can use superposition to calculate
the effects of networks (or sub-networks) feeding the same load
Sensitivity analysis
How much does a branch current or node voltage change when one of the
sources changes? (This becomes very important when you have AC signals
in your network)
“Working back”
Do a calculation with an “easy” number and then scale the result to get the
actual answer you want for a ‘difficult’ number
Lecture 6
6
Working back (or the “unit” method)
5
A simple example: Calculate the
voltage required to give 17 mA in the
1Ω resistor
A
2
V?
17 mA
3
1
The ‘working back” approach: calculate the voltage required to
give 1 A in the resistor:
VA = 1 x (2 + 1) = 3V
Current in 5Ω = 1 + VA/3 = 2 A
V = VA + 5 x 2 = 13V
So 13V gives 1 A
To get 17mA, the superposition principle then tells us that we
need 13V x (17 mA / 1 A) = 221 mV.
Lecture 6
7
3
Superposition Example#2I
1
2
A
Find the current flowing through AB
B
6
3
42 V
42V
3
12
12
12
3
5A
3Ω
4Ω
Lecture 6
8
Superposition Example#2
A
Useful when there are multiple sources:
B
6
3
Find the current flowing through AB
42 V
12
12
1 1 1 1+ 2 + 3 1
+ + =
=
12 6 4
12
2
3
12
3
5A
I n Req 2
=
=
I
Rn 6
I1
A
6
1: Reduce voltage source to zero:
I1 = -5/3 A
B
0V
12
3
12
3
12
3
5A
Lecture 6
10
4
Network analysis using superposition
I2
A
2: Reduce current source to zero:
B
6
42 V
I2 = 14/3 A
12
3
12
3
12
0A
3
So the total current (from superposition) is
I = I1 + I2 = -5/3 + 14/3 = 9/3 = 3 A
Lecture 6
11
General network analysis
So far, most of the methods we have considered are ‘tricks’ to simplify the
analysis of a network (e.g. by bundling up resistors and sources, turning sources
off one by one)
Sometimes we wish to know the current and voltage at every branch/node in any
general network.
For this we fall back on KCL and KVL.
There are two different approaches:
Node voltage analysis: Get an equation for each node voltage in the network
and use these to solve for the branch currents
Mesh current analysis: Get an equation for each branch current in the network
and use these to solve for node voltages
Lecture 6
12
5
Node Voltage Analysis
Most appropriate when
• There are only current sources in the network OR
• There are fewer nodes than meshes
Method:
• A systematic use of KCL at all but one of the nodes in a network. (The node
left out is the reference or ground node)
• The currents for KCL are calculated in terms of the node voltages and branch
resistances
• If there are N nodes, you get N-1 equations for node voltage, which allows you
to obtain all the unknown voltages
As always, it is essential to get the signs and polarities right
Lecture 6
13
An example:
• 3 nodes, 4 meshes, current sources
only:
• Use node voltage analysis
4A
v1
• Label nodes 0, 1 and 2. Node 0 is
the reference (ground), so we will
try to find v1 and v2
v2
1A
2A
8
2
8
• Choose a sign convention for KCL
(e.g. currents leaving are positive)
and mark this on the diagram
v0 = 0
• Branch currents are then given
either by the current source value
or Ohm’s law on the resistors
(voltage difference / resistance)
Lecture 6
14
6
An example:
KCL at node 1:
v1 − v0 v1 − v 2
+
− 4 −1 = 0
2
8
4A
v1
v2
1A
2A
8
2
KCL at node 2:
v 2 − v0 v 2 − v1
+
+4−2 = 0
8
8
8
v0 = 0
5
1
v1 − v 2 = 5
8
8
1
1
− v1 + v 2 = −2
8
4
Lecture 6
15
Matrix formulation
5
1
v1 − v 2 = 5
8
8
1
1
These simultaneous equations can be solved to give:
− v1 + v 2 = −2
v1 = 64/9 V and v2 = -40/9 V
8
4
And the unknown currents can then be found
The equations can also be written in MATRIX form:
I is the matrix of the sum of source currents at each node
V is the matrix of node voltages
G is the conductance matrix
• Diagonal terms are the sum of conductances at each node
• Non-diagonal terms are the conductance joining each pair
of nodes
⎛ 5
⎜ 8
⎜
⎜−1
⎜
⎝ 8
or
1⎞
− ⎟
8 ⎛ v1 ⎞ ⎛ 5 ⎞
⎟⎜ ⎟ = ⎜ ⎟
1 ⎟ ⎝ v 2 ⎠ ⎝ −2 ⎠
⎟
4 ⎠
GV = I
Matrix algebra is a very powerful tool for handling large numbers of simultaneous
equations, and this gives us a route to scale up the analysis to a general number
of nodes – and to write computer programs to do this!
Lecture 6
16
7
Mesh current analysis
Most appropriate when
• There are only voltage sources in the network OR
• There are fewer meshes or loops than nodes
Method:
• A systematic use of KVL to obtain the current in each mesh in a network.
• Identify each mesh (1 to N) and mark clockwise mesh currents (in)
• Identify each branch (1 to M)
• Branch currents of shared branches are the difference of the mesh currents
1Ω
4Ω
5Ω
63
V
I1
6Ω
3Ω
I2
I3
2Ω
Lecture 6
17
Example
1Ω
5 nodes, 3 meshes, only voltage
sources: Use mesh current
analysis.
63
V
I1
I2
3Ω
I3
2Ω
−6i1 + 10i2 − 3i3 = 0
mesh 2 :
0i1 − 3i2 + 9i3 = 0
6(i2 − i1 ) + i2 + 3(i2 − i3 ) = 0
3(i3 − i2 ) + 4i3 + 2i3 = 0
6Ω
11i1 − 6i2 + 0i3 = 63
KVL in mesh 1 :
−63 + 5i1 + 6(i1 − i2 ) = 0
mesh 3 :
4Ω
5Ω
This can be solved by substitution to give
i1=9 A, i2 = 6 A and i3 = 2 A
The branch currents and node voltages then follow
Lecture 6
18
8
Another matrix:
⎛ 11 −6 0 ⎞ ⎛ i1 ⎞ ⎛ 63 ⎞
⎜
⎟⎜ ⎟ ⎜ ⎟
⎜ −6 10 −3 ⎟ ⎜ i2 ⎟ = ⎜ 0 ⎟
⎜ 0 −3 9 ⎟ ⎜ i ⎟ ⎜ 0 ⎟
⎝
⎠⎝ 3 ⎠ ⎝ ⎠
RI = V
11i1 − 6i2 + 0i3 = 63
−6i1 + 10i2 − 3i3 = 0
0i1 − 3i2 + 9i3 = 0
V is the matrix of the sum of voltage sources in each mesh
I is the matrix of mesh currents
R is the resistance matrix
• Diagonal elements are sum of resistance in each mesh
• Off-diagonal elements are ‘coupling’ resistance between meshes
(e.g. meshes 1 and 2 are coupled by 6Ω resistor,
meshes 1 and 3 have no coupling)
Once again, matrix techniques form the basis of a general approach to
solving any arbitrary network
Lecture 6
19
Network analysis software
These programs allow you to define a network of nodes with components
linking them.
Not just resistors and sources, but capacitors, inductors, dependent
sources or any other components that can be defined as an equivalent
circuit (e.g. FETs, transistors).
The programs build the conductance, resistance and source matrices for the
network and use these to determine mesh currents and node voltages.
Many features such as time and frequency dependence, graphs of everything
against everything else, etc…
Lecture 6
20
9
Network analysis software
SPICE : The industry
standard professional package
used for modelling everything
from complex integrated
circuits to telephone networks.
The full version is massively
expensive, but it has spawned
a whole family of SPICE-like
packages, some of which have
limited student or demo
versions.
Microcap PSPICE: This
may be available on the
Physics workstations.
Otherwise try:
http://www.cadencepcb.com/products/downloads/PSpicestudent
http://www.5spice.com/download.htm
Lecture 6
21
Summary of techniques for network analysis
Technique
Typical application
Ohm’s law
Current and voltage in simple loops
Series and parallel
Simplifying networks with ONLY resistors
Source transformation Calculating the current/voltage at only one branch/node of
(Norton, Thévenin)
a complex network when you are not interested in what is
happening in the rest
Superposition
Calculating current/voltage in a branch fed with multiple
sources
Node voltage analysis
Calculate the current and voltage at every node and
branch in a network
Mesh current analysis
Loop current analysis
Calculate the current and voltage at every node and
branch in a network
Lecture 6
22
10