INTRODUCTION
Having understood the fundamental laws of circuit theory (Ohm’s law and Kirchhhoff’s laws), we are now prepared to apply to develop two powerful techniques for circuit analysis: nodal analysis, which is based on a systematic application of Kirchhoff’s current law (KCL) and mesh analysis which based on a systematic application of Kirchhoff’s voltage law (KVL). The two techniques are so important that this chapter should be regarded as the most important in the book. Students are therefore encouraged to pay careful attention.
With the two techniques to be developed in this chapter we can analyze any linear circuit by obtaining a set of simultaneous equation that are then solved to obtain the required values of current or voltage. One method of solving simultaneous equations involves Cramer’s rule, which allow us to calculate circuit variables as a quotient of determinants.
Finally, we apply the technique learned in this chapter to analyze transistor circuits.
NODAL ANALYSIS
Nodal analysis provides a general procedure circuits using node voltages as the circuit variables. Choosing node voltages instead of element voltages as circuit variables is convenient and reduces the number of equations one must solve simultaneously.
To simplify matters, we shall assume in this section that circuits do not contain voltage sources. Circuits that contain voltage sources will be analyzed in the next section.
In nodal analysis, we are interested in finding the node voltages. Given a circuit with n nods without voltage sources, the nodal analysis of the circuit involves taking the following three steps.
Steps to determine node voltages:
- Select a node as the reference node. Assign voltages
v1v1 size 12{v rSub { size 8{1} } } {},
v2v2 size 12{v rSub { size 8{2} } } {}, …,
vn−1vn−1 size 12{v rSub { size 8{n - 1} } } {} to remaining n – 1 nodes. The voltages are referenced with respect to the reference node.
2. Apply KCL to each of the n-1 nonreference nodes. Use Ohm’s law to express the branch currents in term of node voltages.
3. Solve the resulting simultaneous equations to obtain the unknown node voltages.
We shall now explain and apply these three steps.
The first step in nodal analysis is selecting a node as the reference or
datum node. The reference node is commonly called the
ground since it is assumed to have zero potential. A reference node is indicated by any of the three symbols in
Figure 1. The type of ground in
Figure 1(b) is called a chassis ground and is used in devices where the case, enclosure, or chassis acts as reference point for all circuits. When the potential of the earth is used as reference, we use the earth ground in
Figure 1(a) or
Figure 1(c). We shall always use the symbol in
Figure 1(b).
Once we have selected a reference node, we assign voltage designations to nonreference nodes. Consider, for example, the circuit in
Figure 2(a). Node 0 is the reference node (v = 0), while nodes 1 and 2 are assigned voltages
v1v1 size 12{v rSub { size 8{1} } } {} and
v2v2 size 12{v rSub { size 8{2} } } {}, respectively. Keep in mind that the node voltages are defined with respect to the reference node. As illustrated in
Figure 1(a) each node voltage is the voltage rise from the reference node to the corresponding nonreference node or simply the voltage of that node with respect to the reference node.
As the second step, we apply KCL to each nonreference node in the circuit. To avoid putting too much information on the same circuit, the circuit in
Figure 2(a) is redrawn in
Figure 2(b), where we now add
i1i1 size 12{i rSub { size 8{1} } } {},
i2i2 size 12{i rSub { size 8{2} } } {}, and
i3i3 size 12{i rSub { size 8{3} } } {} as the circuits through resistors
R1R1 size 12{R rSub { size 8{1} } } {},
R2R2 size 12{R rSub { size 8{2} } } {}, and
R3R3 size 12{R rSub { size 8{3} } } {}, respectively. At node 1, applying KCL gives
I
1
=
I
2
+
i
1
+
i
2
I
1
=
I
2
+
i
1
+
i
2
size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +i rSub { size 8{1} } +i rSub { size 8{2} } } {}
(1)
At node 2,
I
2
+
i
2
=
i
3
I
2
+
i
2
=
i
3
size 12{I rSub { size 8{2} } +i rSub { size 8{2} } =i rSub { size 8{3} } } {}
(2)
We now apply Ohm’s to express the unknown currents
i1i1 size 12{i rSub { size 8{1} } } {},
i2i2 size 12{i rSub { size 8{2} } } {}, and
i3i3 size 12{i rSub { size 8{3} } } {} in term of node voltages. The key idea to bear in mind is that, since resistance is a passive element, by the passive sign convention, current must always flow from a higher potential to a lower potential.
Current flows from a higher potential to a lower potential in a resistor.
We can express this principle as
i
=
v
higher
−
v
lower
R
i
=
v
higher
−
v
lower
R
size 12{i= { {v rSub { size 8{ ital "higher"} } - v rSub { size 8{ ital "lower"} } } over {R} } } {}
(3)
Note that this principle is in agreement with the way we define resistance in chapter 2 (see Figure 2.1). With this in mind, we obtain from
Figure 2(b),
i1=v1−0R1i1=v1−0R1 size 12{i rSub { size 8{1} } = { {v rSub { size 8{1} } - 0} over {R rSub { size 8{1} } } } } {}or
i1=G1v1i1=G1v1 size 12{i rSub { size 8{1} } =G rSub { size 8{1} } v rSub { size 8{1} } } {}
i2=v1−v2R2i2=v1−v2R2 size 12{i rSub { size 8{2} } = { {v rSub { size 8{1} } - v rSub { size 8{2} } } over {R rSub { size 8{2} } } } } {}(4)
or
i2=G2(v1−v2)i2=G2(v1−v2) size 12{i rSub { size 8{2} } =G rSub { size 8{2} } \( v rSub { size 8{1} } - v rSub { size 8{2} } \) } {}
i3=v2−0R3i3=v2−0R3 size 12{i rSub { size 8{3} } = { {v rSub { size 8{2} } - 0} over {R rSub { size 8{3} } } } } {}or
I3=G3v2I3=G3v2 size 12{I rSub { size 8{3} } =G rSub { size 8{3} } v rSub { size 8{2} } } {}
I
1
=
I
2
+
v
1
R
1
+
v
1
−
v
2
R
2
I
1
=
I
2
+
v
1
R
1
+
v
1
−
v
2
R
2
size 12{I rSub { size 8{1} } =I rSub { size 8{2} } + { {v rSub { size 8{1} } } over {R rSub { size 8{1} } } } + { {v rSub { size 8{1} } - v rSub { size 8{2} } } over {R rSub { size 8{2} } } } } {}
(5)
I
2
+
v
1
−
v
2
R
2
=
v
2
R
3
I
2
+
v
1
−
v
2
R
2
=
v
2
R
3
size 12{I rSub { size 8{2} } + { {v rSub { size 8{1} } - v rSub { size 8{2} } } over {R rSub { size 8{2} } } } = { {v rSub { size 8{2} } } over {R rSub { size 8{3} } } } } {}
(6)
I
1
=
I
2
+
G
1
v
1
+
G
2
(
v
1
−
v
2
)
I
1
=
I
2
+
G
1
v
1
+
G
2
(
v
1
−
v
2
)
size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +G rSub { size 8{1} } v rSub { size 8{1} } +G rSub { size 8{2} } \( v rSub { size 8{1} } - v rSub { size 8{2} } \) } {}
(7)
I
2
+
G
2
(
v
1
−
v
2
)
=
G
3
v
2
I
2
+
G
2
(
v
1
−
v
2
)
=
G
3
v
2
size 12{I rSub { size 8{2} } +G rSub { size 8{2} } \( v rSub { size 8{1} } - v rSub { size 8{2} } \) =G rSub { size 8{3} } v rSub { size 8{2} } } {}
(8)
The third step in nodal analysis is to solve for the node voltages. If we apply KCL to n-1 nonreference nodes, we obtain n-1 simultaneous equations such as
Equation 5 and
Equation 6 or
Equation 7 and
Equation 8. For the circuit of
Figure 2, we solve
Equation 5 and
Equation 6 or
Equation 7 and
Equation 8 to obtain the node voltages
v1v1 size 12{v rSub { size 8{1} } } {} and
v2v2 size 12{v rSub { size 8{2} } } {} using any standard method, such as substitution method, elimination method, Cramer’s rule or matrix inversion. To use either of the last two methods, one must cast the simultaneous equations in matrix form. For example,
Equation 7 and
Equation 8 can be cast in matrix form as
[
G
1
+
G
2
−
G
2
...−
G
2
G
2
+
G
3
][
v
1
v
2
]=[
I
1
−
I
2
...
I
2
]
[
G
1
+
G
2
−
G
2
...−
G
2
G
2
+
G
3
][
v
1
v
2
]=[
I
1
−
I
2
...
I
2
]
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@5924@
(9)
which can be solved to get
v1v1 size 12{v rSub { size 8{1} } } {} and
v2v2 size 12{v rSub { size 8{2} } } {}.
(Reference) will be generalized in section 6. The simultaneous equations may also be solved using calculator or with software package such as MATLAB.
NODAL ANALYSIS WITH VOLTAGE SOURCES
We now consider how voltage sources effect nodal analysis. We use the circuit in
Figure 3 for illustration. Consider the following two possibilities.
CASE 1: if a voltage source is connected between the reference node and a nonreference node, we supply set the voltage at the nonreference node equal to the voltage of the voltage source. In
Figure 3, for example,
v
1
=
10
V
v
1
=
10
V
size 12{v rSub { size 8{1} } ="10"V} {}
(10)
Thus our analysis is somewhat simplified by this knowledge of the voltage at this node.
CASE 2: If the voltage source (dependent or independent) is connected between two nonreference nodes, the two nonreference nodes form a generalized node or supernode; we apply both KCL and KVL to determine the node voltages.
A supernode is formed by enclosing a (dependent or independent) voltage source connected between two nonreference nodes and any elements connected in parallel with it.
In
Figure 3, nodes 2 and 3 form a supernode. We analyze a circuit with supernode using the same three steps mentioned in the previous section except that the supernodes are treated differently. Why? Because an essential component of nodal analysis is applying KCL, which requires knowing the current through each element. There is no way of knowing the current through a voltage source in advance. However, KCL must be satisfied at a supernode like any other node. Hence, at the supernode in
Figure 3,
i
1
+
i
4
=
i
2
+
i
3
i
1
+
i
4
=
i
2
+
i
3
size 12{i rSub { size 8{1} } +i rSub { size 8{4} } =i rSub { size 8{2} } +i rSub { size 8{3} } } {}
(11)
or
v
1
−
v
2
2
+
v
1
−
v
3
4
=
v
2
−
0
8
+
v
3
−
0
6
v
1
−
v
2
2
+
v
1
−
v
3
4
=
v
2
−
0
8
+
v
3
−
0
6
size 12{ { {v rSub { size 8{1} } - v rSub { size 8{2} } } over {2} } + { {v rSub { size 8{1} } - v rSub { size 8{3} } } over {4} } = { {v rSub { size 8{2} } - 0} over {8} } + { {v rSub { size 8{3} } - 0} over {6} } } {}
(12)
To apply Kirchhoff’s voltage law to the supernode in
Figure 3, we redraw the circuit as shown in
Figure 4. Going around the loop in the clockwise direction gives
−
v
2
+
5
+
v
3
=
0
⇒
v
2
−
v
3
=
5
−
v
2
+
5
+
v
3
=
0
⇒
v
2
−
v
3
=
5
size 12{ - v rSub { size 8{2} } +5+v rSub { size 8{3} } =0 drarrow v rSub { size 8{2} } - v rSub { size 8{3} } =5} {}
(13)
Note the following properties of a supernode:
1. The voltage source inside the supernode provides a constraint equation needed to solve for the node voltage.
2. A supernode has no voltage of it own.
3. A supernode requires the application of both KCL and KVL.
MESH ANALYSIS
Mesh analysis is also known as loop analysis or the mesh current method. Mesh analysis provides another general procedure for analyzing circuits using mesh currents as the circuit variables. Using mesh currents instead of element currents as circuit variables is convenient and reduces the number of equations that must be solved simultaneously. Recall that a loop is closed path with no node passed more than once. A mesh is a loop that does not contain any other loop within it.
Nodal analysis applies KCL to find unknown voltages in a given circuit, while mesh analysis applies KVL to find unknown currents. Mesh analysis is not quite as general as nodal analysis because it is only applicable to a circuit that is
planar. A planar circuit is one that can be drawn in a plane with no branches crossing one another; otherwise it is
nonplanar. A circuit may have crossing branches and still be planar if it can be redrawn such that it has no crossing branches. For example, the circuit in
Figure 5(a) has two crossing branches, but it can be redrawn as in
Figure 5(b). Hence, the circuit in
Figure 5(a) is planar. However, the circuit in
Figure 6 is nonplanar, because there is no way to redraw it and avoid the branches crossing. Nonplanar circuits can be handled using mesh analysis, but they will not be considered in this text.
To understand mesh analysis, we should first explain more about what we mean by a mesh.
A mesh is a loop which does not contain any other loop within it.
In
Figure 7, for example, paths abefa and bcdeb are meshes, but path abcdefa is not a mesh. The current through a mesh is known as
mesh current. In mesh analysis, we are interested in applying KVL to find the mesh currents in a given circuit.
In this section, we will apply mesh analysis to planar circuits that do not contain current sources. In the next section, we will consider circuits with current sources. In the mesh analysis of a circuit with n meshes, we take the following three steps.
Steps to determine mesh currents:
- Assign mesh currents
i1i1 size 12{i rSub { size 8{1} } } {},
i2i2 size 12{i rSub { size 8{2} } } {}, …, in to the n mesh.
- Apply KVL to each of n meshes. Use Ohm’s law to express the voltages in term of the mesh currents.
- So the resulting n simultaneous equations to get the mesh currents.
To illustrate the steps, consider the circuit in
Figure 7. The first step requires that mesh currents
i1i1 size 12{i rSub { size 8{1} } } {} and
i2i2 size 12{i rSub { size 8{2} } } {} are assigned to meshes 1 and 2. Although a mesh current may be assigned to each mesh in arbitrary direction, it is conventional to assume that each mesh current flows clockwise.
As the second step, we apply KVL to each mesh. Applying KVL to mesh 1, we obtain
−
V
1
+
R
1
I
1
+
R
3
(
i
1
−
i
2
)
=
0
−
V
1
+
R
1
I
1
+
R
3
(
i
1
−
i
2
)
=
0
size 12{ - V rSub { size 8{1} } +R rSub { size 8{1} } I rSub { size 8{1} } +R rSub { size 8{3} } \( i rSub { size 8{1} } - i rSub { size 8{2} } \) =0} {}
Or
(
R
1
+
R
3
)
i
1
−
R
3
i
2
=
V
1
(
R
1
+
R
3
)
i
1
−
R
3
i
2
=
V
1
size 12{ \( R rSub { size 8{1} } +R rSub { size 8{3} } \) i rSub { size 8{1} } - R rSub { size 8{3} } i rSub { size 8{2} } =V rSub { size 8{1} } } {}
(14)
For mesh 2, applying KVL gives
R
2
i
2
+
V
2
+
R
3
(
i
1
−
i
2
)
=
0
R
2
i
2
+
V
2
+
R
3
(
i
1
−
i
2
)
=
0
size 12{R rSub { size 8{2} } i rSub { size 8{2} } +V rSub { size 8{2} } +R rSub { size 8{3} } \( i rSub { size 8{1} } - i rSub { size 8{2} } \) =0} {}
or
−
R
3
i
1
+
(
R
2
+
R
3
)
i
2
=
−
V
2
−
R
3
i
1
+
(
R
2
+
R
3
)
i
2
=
−
V
2
size 12{ - R rSub { size 8{3} } i rSub { size 8{1} } + \( R rSub { size 8{2} } +R rSub { size 8{3} } \) i rSub { size 8{2} } = - V rSub { size 8{2} } } {}
(15)
[
R
1
+
R
2
...−
R
3
−
R
3
R
2
+
R
3
][
i
1
i
2
]=[
V
1
-V
2
]
[
R
1
+
R
2
...−
R
3
−
R
3
R
2
+
R
3
][
i
1
i
2
]=[
V
1
-V
2
]
MathType@MTEF@5@5@+=feaagaart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVCI8FfYJH8YrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfeaY=biLkVcLq=JHqpepeea0=as0Fb9pgeaYRXxe9vr0=vr0=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@5924@
(16)
Notice that the branch currents are different from the mesh current unless the mesh is isolated. To distinguish between the two types of currents, we use i for a mesh current and I for a branch current. The current elements
I1I1 size 12{I rSub { size 8{1} } } {},
I2I2 size 12{I rSub { size 8{2} } } {}, and
I3I3 size 12{I rSub { size 8{3} } } {} are algebraic sums of the mesh currents. It is evident from
Figure 7 that
I1=i1I1=i1 size 12{I rSub { size 8{1} } =i rSub { size 8{1} } } {}(17)
I2=i2I2=i2 size 12{I rSub { size 8{2} } =i rSub { size 8{2} } } {},
I3=i1−i2I3=i1−i2 size 12{I rSub { size 8{3} } =i rSub { size 8{1} } - i rSub { size 8{2} } } {}
MESH ANALYSIS WITH CURRENT SORCES
Applying mesh analysis to circuits containing current sources (dependent or independent) may appear complicated. But it is actually much easier than what we encountered in the previous section, because the presence of the current sources reduces the number of equations. Consider the following two possible cases.
CASE 1. When a current source exists only in one mesh: consider the circuit in
Figure 8, for example. We set
i2=−5Ai2=−5A size 12{i rSub { size 8{2} } = - 5A} {} and write a mesh equation for the other mesh in the usual way, that is,
−
10
+
4i
1
+
6
(
i
1
−
i
2
)
=
0
⇒
i
1
=
−
2A
−
10
+
4i
1
+
6
(
i
1
−
i
2
)
=
0
⇒
i
1
=
−
2A
size 12{ - "10"+4i rSub { size 8{1} } +6 \( i rSub { size 8{1} } - i rSub { size 8{2} } \) =0 drarrow i rSub { size 8{1} } = - 2A} {}
(18)
CASE 2. When a current source exists between two meshes: consider the circuit in
Figure 9(a), for example. We create a
supermesh by excluding the current source and any elements connected in series with it, as shown in
Figure 9(b). Thus,
A supermesh results when two meshes have a (dependent or independent) current source in common.
As shown in
Figure 9(b), we create a supermesh as the periphery of the two meshes and treat it differently. If a circuit has two or more supermeshes that intersect, they should be combined to form a larger supermesh. Why treat the supermesh differently? Because mesh analysis applies KVL-which requires that we know the voltage across each branch-and we do not know the voltage across a current source in advance. However, a supermesh must satisfy KVL like any other mesh. Therefore, applying KVL to the supermesh in
Figure 9(b) gives
−
20
+
6i
1
+