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Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
LECTURE 2
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Circuit Analysis Techniques
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Circuit Analysis Techniques
• Voltage Division Principle
• Current Division Principle
• Nodal Analysis with KCL
• Mesh Analysis with KVL
• Superposition
• Thévenin Equivalence• Norton’s Equivalence
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Circuit Analysis using Series/Parallel Equivalents1. Begin by locating a combination of resistances that are in series or
parallel. Often the place to start is farthest from the source.
2. Redraw the circuit with the equivalent resistance for the combination found in step 1.
3. Repeat steps 1 and 2 until the circuit is reduced as far as possible. Often (but not always) we end up with a single source and a single resistance.
4. Solve for the currents and voltages in the final equivalent circuit.
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Working Backward
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Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Voltage Division
total321
111 v
RRR
RiRv
total321
222 v
RRR
RiRv
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Application of the Voltage-Division
Principle
V5.1
156000200010001000
1000
total4321
11
v
RRRR
Rv
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Current Division
total21
2
11 i
RR
R
R
vi
total21
1
22 i
RR
R
R
vi
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
206030
6030
32
32eq RR
RRR
A10152010
20
eq1
eq1
siRR
Ri
Application of the Current-Division Principle
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
•Voltage division •Voltage division and •current division
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
•Current division
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Although they are veryimportant concepts,series/parallel equivalents andthe current/voltage divisionprinciples are not sufficient to solve all circuits.
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Node Voltage (Nodal) Analysis
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Definitions of node and Supernode
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3.1 5
2
211
vvv
(-1.4)- 5
1
122
vvv
At node 1
At node 2
Obtain values for the unknown voltages across the elements in the circuit below.
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Writing KCL Equations in Terms of the
Node Voltages for Figure 2.16
svv 1
03
32
4
2
2
12
R
vv
R
v
R
vv
03
23
5
3
1
13
R
vv
R
v
R
vv
node 1
node 2
node 3
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
02
21
1
1
siR
vv
R
v
04
32
3
2
2
12
R
vv
R
v
R
vv
siR
vv
R
v
4
23
5
3
node 1
node 2
node 3
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
No. of unknown: v1, v2, v3
No. of linear equation : 3
Setting up nodal equation withKCL at Node 1, Node 2, Node 3
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
No. of unknown: v1, v2, v3
No. of linear equation : 3
Setting up nodal equation withKCL at Node 1, Node 2, Node 3
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Problem with node 3, it is rather hard to set the nodal equation at node 3 but still solvable.
Problem with node 3, it is rather hard to set the nodal equation at node 3, but still solvable. Why? As there is no way to determinethe current through the voltage source, but v3=Vs
v3
Same as before.
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May Not Be That Simple
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Circuits with Voltage SourcesWe obtain dependent equations if we use
all of the nodes in a network to write KCL equations.Any branch with a voltage source:
• define SUPERNODE, sum all current either in or out at the supernode with KCL
• use KVL to set up dependent equation involving the voltage source.
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(a) The circuit of Example 4.2 with a 22-V source in place of the 7- resistor. (b) Expanded view of the region defined as a supernode; KCL requires that all currents flowing into the region must sum to zero, or we would pile up or run out of electrons.
4
3 38 3121 vvvv
22154
3
253
23
231312
vv
vvvvvv
At node 1:
At the “supernode:”
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
A B
0)15()15(
3
2
4
2
1
1
2
1
R
v
R
v
R
v
R
v
For supernode A, EXCLUDE THE SOURCE
Why? As the current via the 10V source is equal to the current via R4 plus the current via R3
Summing all the current outfrom the supernode A
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
0
1515
3
2
4
2
1
1
2
1
R
v
R
v
R
v
R
v
For supernode B, EXCLUDE THE SOURCE
Summing all the current into the supernode B
B
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
0)15()15(
3
2
4
2
1
1
2
1
R
v
R
v
R
v
R
v
-v1 -10 + v2 = 0
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Any branch with a voltage source:• define supernode, sum all current either in or out at the supernode with KCL• use KVL to set up dependent equation involving the voltage source.
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010 21 vv
13
32
2
31
1
1
R
vv
R
vv
R
v
04
3
3
23
2
13
R
v
R
vv
R
vv
14
3
1
1 R
v
R
v
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
010 21 vv
13
32
2
31
1
1
R
vv
R
vv
R
v
04
3
3
23
2
13
R
v
R
vv
R
vv
14
3
1
1 R
v
R
v
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Node-Voltage Analysis with a Dependent Source•First, we write KCL equations at each node, including the current of the controlled source just as if it were an ordinary current source.
•Next, we find an expression for the controlling variable ix in terms of the node voltages.
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xs iiR
vv2
1
21
03
32
2
2
1
12
R
vv
R
v
R
vv
024
3
3
23
xiR
v
R
vv
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
xs iiR
vv2
1
21
03
32
2
2
1
12
R
vv
R
v
R
vv
024
3
3
23
xiR
v
R
vv
3
23
R
vvix
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Substitution yields
3
23
1
21 2R
vvi
R
vvs
03
32
2
2
1
12
R
vv
R
v
R
vv
023
23
4
3
3
23
R
vv
R
v
R
vv
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Node-Voltage Analysis1. Select a reference node and assign variables for the unknown node voltages. If the reference node is chosen at one end of an independent voltage source, one node voltage is known at the start, and fewer need to be computed.
2. Write network equations. First, use KCL to write current equations for nodes and supernodes. Write as many current equations as you can without using all of the nodes. Then if you do not have enough equations because of voltage sources connected between nodes, use KVL to write additional equations.3. If the circuit contains dependent sources, find expressions for the controlling variables in terms of the node voltages. Substitute into the network equations, and obtain equations having only the node voltages as unknowns.4. Put the equations into standard form and solve for the node voltages.
5. Use the values found for the node voltages to calculate any other currents or voltages of interest.
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Step 1.Reference node
Step 1. v1
Step 1 v2
Step 2.
10
1510
1
212
v
vvv
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
v1v2
v3
yivv
vvvvvv
vvvv
2
055210
1053
32
331221
2131
refStep 1
Step 1 Step 1
supernode
node 1Step 2
supernode
Step 3
52
5
3132
31
vvvv
vviy
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Mesh Current Analysis
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Definition of a loop
Definition of a mesh
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Choosing the Mesh Currents
When several mesh currents flow through one element, we consider the current in that element to be the algebraic sum of the mesh currents.
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Writing Equations to Solve for Mesh
CurrentsIf a network contains only resistors and independent voltage sources, we can write the required equations by following each current around its mesh and applying KVL.
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
For mesh 1, we have
For mesh 2, we obtain
024123 BviRiiR
For mesh 3, we have
031132 BviRiiR
0213312 AviiRiiR
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Determine the two mesh currents, i1 and i2, in the circuit below.
For the left-hand mesh,
-42 + 6 i1 + 3 ( i1 - i2 ) = 0
For the right-hand mesh,
3 ( i2 - i1 ) + 4 i2 - 10 = 0
Solving, we find that i1 = 6 A and i2 = 4 A.
(The current flowing downward through the 3- resistor is therefore i1 - i2 = 2 A. )
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Mesh Currents in Circuits Containing
Current Sources*A common mistake is to assume the voltages across current sources are zero. Therefore, loop equation cannot be set up at mesh one due to the voltage across the current source is unknown
21 i
0105)(10 212 iii
Anyway, the problem isstill solvable.
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
As the current source common to two mesh, combine meshes 1 and 2 into a supermesh. In other words, we write a KVL equation around the periphery of meshes 1 and 2 combined.
01042 32311 iiiii
Mesh 3: 0243 13233 iiiii
512 ii
It is the supermesh.
Three linear equations and three unknown
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Find the three mesh currents in the circuit below.
Creating a “supermesh” from meshes 1 and 3:
-7 + 1 ( i1 - i2 ) + 3 ( i3 - i2 ) + 1 i3 = 0 [1]
Around mesh 2:
1 ( i2 - i1 ) + 2 i2 + 3 ( i2 - i3 ) = 0 [2]
Rearranging,
i1 - 4 i2 + 4 i3 = 7 [1]
-i1 + 6 i2 - 3 i3 = 0 [2]
i1 - i3 = 7 [3]
Solving,
i1 = 9 A, i2 = 2.5 A, and i3 = 2 A.
Finally, we relate the currents in meshes 1 and 3:
i1 - i3 = 7 [3]
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026420 221 iii
124ii
vx
22ivx
supermesh of mesh1 and mesh2
current source
branch current
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
026420 221 iii
124ii
vx
22ivx
Three equations and three unknown.
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Mesh-Current Analysis1. If necessary, redraw the network without crossing conductors or elements. Then define the mesh currents flowing around each of the open areas defined by the network. For consistency, we usually select a clockwise direction for each of the mesh currents, but this is not a requirement.
2. Write network equations, stopping after the number of equations is equal to the number of mesh currents. First, use KVL to write voltage equations for meshes that do not contain current sources. Next, if any current sources are present, write expressions for their currents in terms of the mesh currents. Finally, if a current source is common to two meshes, write a KVL equation for the supermesh.3. If the circuit contains dependent sources, find expressions for the controlling variables in terms of the mesh currents. Substitute into the network equations, and obtain equations having only the mesh currents as unknowns.4. Put the equations into standard form. Solve for the mesh currents by use of determinants or other means.
5. Use the values found for the mesh currents to calculate any other currents or voltages of interest.
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Superposition• Superposition Theorem – the response of a
circuit to more than one source can be determined by analyzing the circuit’s response to each source (alone) and then combining the results
Insert Figure 7.2
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Superposition
Insert Figure 7.3
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Superposition• Analyze Separately, then Combine Results
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Use superposition to find the current ix.
Current source is zero – open circuit as I = 0 and solve iXv
Voltage source is zero – short circuit as V= 0 and solve iXv
XcXvX iii
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Use superposition to find the current ix.
The controlled voltage source is included in all cases asit is controlled by the current ix.
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Voltage and Current Sources
Insert Figure 7.7
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Voltage and Current Sources
Insert Figure 7.8
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Voltage and Current Sources
Insert Figure 7.9
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Source Transformation
+
VL
_
+_
iLRS
RLVS
iL
RP RLIS
+
VL
_
Under what condition, the voltage and current of the load is the same whenoperating at the two practical sources?For voltage source
L
LS
SL R
RR
VV
,
For current source
L
LP
PSL R
RR
RiV
We have,
LP
PS
LS
S
RR
Ri
RR
V
S
SSSP R
ViRR ,
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Voltage and Current Sources
• Equivalent Voltage and Current Sources – for every voltage source, there exists an equivalent current source, and vice versa
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Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Thevenin’s Theorem• Thevenin’s Theorem – any resistive circuit
or network, no matter how complex, can be represented as a voltage source in series with a source resistance
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Thevenin’s Theorem• Thevenin Voltage (VTH) – the voltage present at the
output terminals of the circuit when the load is removed
Insert Figure 7.18
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Thevenin’s Theorem• Thevenin Resistance (RTH) – the resistance
measured across the output terminals with the load removed
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Thévenin Equivalent Circuits
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Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Thévenin Equivalent Circuits
ocvVt
sc
oc
i
vRt
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Thévenin Equivalent Circuits
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Finding the Thévenin Resistance Directly
When zeroing a voltage source, it becomes a short circuit. When zeroing a current source, it becomes an open circuit.
We can find the Thévenin resistance by zeroing the sources in the original network and then computing the resistance between the terminals.
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Computation of Thévenin resistance
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Equivalence of open-circuit and Thévenin voltage
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A circuit and its Thévenin equivalent
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Superposition
As the voltage source does not contribute any output voltage,Only the current source has the effect.
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Determine the Thévenin and Norton Equivalents of Network A in (a).
Source transformation
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Find the Thévenin equivalent of the circuit shown in (a).
As i = -1, therefore, the controlled voltage source is -1.5V.Use nodal analysis at node v,
v
6.0,123
)5.1(
v
vv Thus, Rth =v/I = 0.6/1 = 0.6 ohms
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Applications of Thevenin’s Theorem
• Load Voltage Ranges – Thevenin’s theorem is most commonly used to predict the change in load voltage that will result from a change in load resistance
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Applications of Thevenin’s Theorem
• Maximum Power Transfer– Maximum power transfer from a circuit to a
variable load occurs when the load resistance equals the source resistance
– For a series-parallel circuit, maximum power occurs when RL = RTH
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Applications of Thevenin’s Theorem
• Multiload Circuits
Insert Figure 7.30
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Norton’s Theorem• Norton’s Theorem – any resistive circuit or network,
no matter how complex, can be represented as a current source in parallel with a source resistance
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Norton’s Theorem• Norton Current (IN) – the current through
the shorted load terminals
Insert Figure 7.35
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Computation of Norton current
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Norton’s Theorem
• Norton Resistance (RN) – the resistance measured across the open load terminals (measured and calculated exactly like RTH)
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Norton’s Theorem• Norton-to-Thevenin and Thevenin-to-Norton
Conversions
Insert Figure 7.39
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Step-by-step Thévenin/Norton-Equivalent-
Circuit Analysis
1. Perform two of these: a. Determine the open-circuit voltage Vt = voc.
b. Determine the short-circuit current In = isc.
c. Zero the sources and find the Thévenin resistance Rt looking back into the terminals.
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2. Use the equation Vt = Rt In to compute the remaining value.
3. The Thévenin equivalent consists of a voltage source Vt in series with Rt .
4. The Norton equivalent consists of a current source In in parallel with Rt .
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Department of Electronic EngineeringBASIC ELECTRONIC ENGINEERING
Maximum Power Transfer
The load resistance that absorbs the maximum power from a two-terminal circuit is equal to the Thévenin resistance.
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Graphical representation of maximum power transfer
Power transfer between source and load
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