e1 e2 I1 R1
Lecture 4 DC Circuit Analysis III 1. Topics Covered (a) The slightly tricky problem of a floating voltage source (b) The general principle of linearity (c) The general principle of superposition (d) I-V characteristics (e) The Thevenin equivalent circuit (f) The Norton equivalent circuit (g) Examples (h) Summary of circuit analysis 2. The slightly tricky problem of a floating voltage source (a) We will consider a simple sequence of circuits that shows that under certain conditions it becomes tricky using nodal analysis the way it has been previously described. (b) Once the problem has been defined, we present a simple method of dealing with it. (c) The trickiness arises because of the presence of a “floating voltage” source, which will be defined shortly. (d) To begin consider a sub-section of a general circuit as shown below.
e1
e2 I1
R1
Figure 4.1 1
(e) Let us now write down the current I1 flowing out of node 1. I1 =
e1 − e2 R1
(1)
(f) This current is used in the standard way in the application of nodal analysis. (g) Consider next a slightly more complicated circuit with voltage source and resistor as shown below e1
+
e2
V1
R1
I1
-
Figure 4.2 (h) The current I1 is now given by I1 =
e1 − V1 − e2 R1
(2)
(i) This current can also be used in the standard way in nodal analysis (j) Finally consider the circuit below in which the branch resistor has been eliminated. All that remains is a voltage source connected between two non-grounded nodes. This is a floating voltage source. e3
e1
+
R3
I1
-
R4
V1
R1
Figure 4.3 2
e4
e2
R2
(k) Note that now there is no simple way to calculate I1 in terms of the nodal voltages. We know that the nodal voltages are related by e1 − V1 − e2 = 0 but so what. (l) The resolution to the problem is found by writing down the full nodal equations for nodes #1 and #2 in addition to the voltage relation above. In writing down these equations the quantity I1 is treated as an unknown. e1 e3 − e1 − − I1 = 0 R3 R1 e2 e2 − e4 − + I1 = 0 − R4 R2 e1 − V1 − e2 = 0
(3)
(m) Note that rather than having two independent nodal equations as would be the usual case, we now have three independent equations. However, there is now an additional unknown, I1 , so the number of equations and unknowns remains the same. (n) With a floating voltage source one can always reduce the number of equations by one by simply adding the nodal equations thereby eliminating the unknown I1 . The result is a “standard” set of nodal equations where the only unknowns are the nodal voltages. However, the matrix symmetry properties are no longer preserved with respect to e1 and e2 in the presence of a floating voltage source. e2 e2 − e4 e2 e3 − e1 − − − =0 (4) R3 R1 R4 R2 e1 − e2 = V1
(5)
3. The general principle of linearity (a) The circuit equations obtained from nodal analysis are in general a set of linear, inhomogeneous algebraic equations. (b) The linearity refers to the fact that the unknown nodal voltages all appear in the equations as quantities raised to the first power. Similarly, all the sources appear as inhomogeneous terms, also raised to the first power.
3
(c) This would not be true if the circuit element relations were not linear. For instance, if the value of a resistor depended upon the value of current flowing through it (e.g. R = R0 + aI 2 ) the circuit would no longer be described by a linear set of equations. (d) Why is linearity important? One important consequence of linearity is as follows. Consider a general circuit consisting of various resistors, voltage sources and current sources. If the values of the resistors are held fixed and all the sources are multiplied by a factor f , then all the resulting voltages and currents in the circuit are also multiplied by the same factor f . One can write the answer by inspection without having to re-analyze the circuit. (e) This “scaling result follows trivially from examining the form of the circuit equations. If every inhomogeneous source term in the equations is multiplied by f , then multiplying every unknown nodal voltage by f leaves the equations unchanged; that is the set of equations for the unknowns f ej in terms of the sources f Vj and f Ij is the same as the original set of equations. (f) The example below is an explicit demonstration of linearity where the factor f has been chosen to be f = 3. e 4
2 4
+
-
e'
1
4
2 6
+
-
4−e 6−e e + − =0 2 4 1 or e=2
12
+
-
1
18
+
-
12 − e 18 − e e + − =0 2 4 1
(6)
e = 6 = 3e
(7)
Figure 4.4 4. The general principle of superposition (a) Another important consequence of linearity is the principle of superposition. (b) Superposition is usually applied to circuits with more than one source. 4
(c) Consider such a circuit where the goal is to calculate, for instance a certain current I. (d) The principle of superposition states that the current I can be calculated as the sum of several separate terms, one for each source. (e) Each term is calculated by setting all source values to zero except the one denoted by the subscript j. One then solves the reduced single source problem for the current Ij due to the source j. This process is then repeated for each of the other sources. The total current is obtained by summing the individual contributions over j. I= Ij (8) j
(f) Superposition states that the value of I calculated by summing over the individual sources is identical to the value that would be obtained by solving the equations a single time simultaneously including all the sources. (g) Although it may seem to require more effort to use superposition since several circuit calculations need to be performed versus a single calculation for simultaneous sources, this may not be true. (h) Often, the reduced circuits obtained by setting all but one source to zero are so simple that they can be solved by inspection. (i) The fastest method depends upon the particular problem of interest. In this sense superposition can be viewed as another tool for simplifying the analysis of circuits. (j) Why does superposition work? The reason is that in a linear circuit the source terms appear as a sum of individual, linear, uncoupled terms driving the circuit. There are no terms for instance of the form Vj Ik or Vj2 . (k) It then follows from the theory of linear equations that one can calculate the unknown nodal voltages by simply adding the contributions from each source, setting all others to zero. (l) A technical point to keep in mind. Note that setting a voltage source to zero implies that the voltage across its terminals is equal to zero. Thus, when setting source voltages to zero in superposition, each voltage must be replaced by a short circuit. 5
(m) Similarly, setting a current source to zero implies that zero current flows through the source. Thus, setting a current source to zero in superposition requires replacing it with an open circuit. (n) We demonstrate superposition by a simple example. Consider the circuit shown below. The goal is to calculate the nodal voltage e.
e R1 V
+
I
R2
-
Figure 4.5 (o) Using standard nodal analysis we find e V −e +I − =0 R1 R2 or e=
R2 R1 R2 V + I R1 + R2 R1 + R2
(9)
(10)
(p) Note that the answer is the sum of two terms, one due to the voltage source and one due to the current source. (q) Now lets solve the problem by superposition. First set the current source to zero. The circuit reduces to a simple voltage divider e1 R1 V
+
R2
-
e1 =
R2 V R1 + R2
Figure 4.6
6
(11)
(r) Next set the voltage source to zero. This leads to a simple parallel resistor circuit. e2 R1 I
R2
e2 = R|| I =
R1 R 2 I R1 + R2
(12)
Figure 4.7 (s) Clearly, adding e1 + e2 by superposition yields the same answer as using standard nodal analysis. 5. I-V characteristics (a) The I-V characteristics are a convenient way to describe the properties of a given circuit. (b) In many applications a circuit containing a large number of elements may just be a subset of a larger more complex network. (c) Knowing the I-V characteristics of the subset circuit is often the only information we require in order to design and assemble a more complex system. (d) To understand the concept of the I-V characteristics consider a general, multi-element linear circuit as shown below +
-
+
-
I
+ V +
-
+
-
-
Figure 4.8 7
(e) This circuit may be a module that would be connected to other circuits through the output terminals as shown above. A circuit, regardless of its complexity, in which there are only two output terminals is referred to as a single port network. (f) Now, assume that all the resistors and sources are known. (g) Imagine that we apply a voltage V across the output and measure the current I noting the sign convention as shown. We then repeat this procedure for a number of different values of V . (h) We could then make a plot of I vs V as shown below. I
V
Figure 4.9 (i) This curve, or equivalently the mathematical relation between I and V are known as the I-V characteristics. (j) Note that in principle we could apply a current source I across the terminals and measure the resulting voltage V . Would we obtain the same I-V characteristic? (k) We will prove later that for a general linear resistive network the I-V characteristic turns out to be a linear relation I = C1 V + C2
(13)
(l) Below are some simple examples of I-V characteristics. Consider first a resistor.
8
I +
I R
V V
V R Figure 4.10 I=
(14)
(m) In general the I-V characteristic does not have to pass through the origin. To see this we calculate the I-V characteristics for a voltage source in series with a resistor I Rs +
-
Vs
I
+ Vs
V
-V s /R
I=
V − VS RS
V
(15)
Figure 4.11 (n) A similar result holds for a current source in parallel with a resistor.
9
I +
I IS
Rs
R s Is
V
V
-I s
V = (I + IS )RS
(16)
I = −IS + V /RS
(17)
or Figure 4.12 6. The Thevenin equivalent circuit (a) If we are interested only in the I-V terminal characteristics of a complicated network, we can take two simple measurements (or make two simple calculations) and replace the complicated network by a simple equivalent circuit. (b) This circuit is known as the Thevenin equivalent circuit. (c) Thevenin circuits are often very helpful in making circuit analysis much simpler (d) The idea of the Thevenin equivalent circuit is based on the fact that across any two terminals of a linear network with ideal sources, the relation between voltage and current is a linear relation of the form I + Vj
+
Rj
V
-
Ij
-
I = C1 V + C2 Figure 4.13 10
(18)
(e) This result follows from linearity and superposition. To see this assume a voltage source V is applied across the output terminals and use nodal analysis to calculate I. (f) Using the principle of superposition the solution for the current I will be of the form I = α0 V +
αj Vj +
j
βj Ij
(19)
j
(g) In this expression the Vj and Ij are known internal sources of the network. The coefficients αj and βj are known functions of the values of the known resistors in the circuit. Thus, for a general network we can write I = C1 V + C2 C1 = α0 C2 = αj Vj + βj Ij j
(20)
j
(h) We now focus on interpreting and calculating the coefficients C1 and C2 . (i) Physically, C2 represents the negative of the short circuit current as shown below. This is the current that flows through the output terminals with the polarity as shown below, when the output voltage is short circuited: V = 0. I
I sc
I = −Isc → C2 = −Isc
(21)
Figure 4.14 (j) Physically, C1 is related to the open circuit voltage-the voltage measured when the output is an open circuit: I = 0. 11
+ V oc -
C2 = −Voc C1 Figure 4.15
V = Voc →
(22)
(k) Using these relations we can rewrite and plot the I-V relationship as follows. I Voc
V
-Isc
Isc V − Isc Voc Figure 4.16
I=
(23)
(l) In terms of measurements, the open circuit voltage and short circuit current are measurements that can be made solely across the output terminals. One does not have to interfere with any of the internal components of the circuit. (m) In terms of circuit calculations, we can calculate the open circuit voltage Voc using nodal analysis. In principle we could redo the circuit analysis and calculate the short circuit current Isc by shorting the output terminals. (n) An often easier procedure to calculate Isc than redoing the circuit analysis is as follows. Set all the current and voltage sources in the circuit to zero and then calculate the input resistance to the circuit denoted by Rt . The short circuit current then turns out to be given by the relation Isc = Voc /Rt 12
(24)
(o) The I-V relation can now be rewritten as I Voc V - Voc /Rt
I=
V Voc − Rt Rt
(25)
Figure 4.17 (p) The relation between Rt and Voc , Isc given above may not be completely obvious. It follows from superposition. (q) If all the sources are set to zero and a current source I is applied across the output terminals then this generates an output voltage V given by V = Rt I (26) (r) If the sources are now turned on, the output voltage, by superposition becomes V
= Rt I +
j
= Rt I + Voc
αj Vj +
βj Ij
j
(27)
(s) The key point is that in a linear circuit the contributions from the internal sources are independent of the external source I. Thus, the slope of the I-V characteristic is just 1/Rt as stated above in step (o). (t) The last step is to give a simple equivalent circuit that satisfies the I-V characteristics of a general circuit. This is the Thevenin equivalent circuit and is given by
13
+
-
+
I
Rt V oc
V
I=
V − Voc Rt
(28)
Figure 4.18 7. The Norton circuit (a) A Norton equivalent circuit plays the same role as a Thevenin equivalent circuit except that it uses a current source rather than a voltage source. (b) The Norton equivalent circuit is given by I In
Rn
+ V
Figure 4.19 (c) Using KCL we can easily derive the I-V relationship for the Norton circuit V I + In − =0 (29) Rn or V − In (30) I= Rn (d) Note that the open circuit voltage is given by Voc = Rn In . Hence, if we choose Rn = Rt and In = Voc /Rn the Thevenin and Norton circuits are equivalent to each other.
14
+ V oc /R t
V
Rt
Rt
+
V oc
V
-
Figure 4.20
(e) When analyzing circuits, it does not matter whether you use Thevenin or Norton equivalents. Use the one that is most convenient for the problem at hand. 8. Example #1 (a) Find the Thevenin equivalent circuit by straightforward nodal analysis e 1 I1 50 Ω
+
-
100 Ω
0.5A
I
25V 300 Ω
+ V -
KCL: e1 e1 − 25 − = 0 → e1 = 44.4V 100 350 e1 − 25 = = 0.0555A 350 = 300I1 = 16.7V
.5 − I1 V0c
Figure 4.21.1 Rt :
15
(31)
50 Ω 100 Ω
300 Ω
Rt
1 1 1 = + = 0.01 → Rt = 100Ω Rt 300 150
(32)
Figure 4.21.2 SOLUTION: 100 Ω
+
-
16.6 V
Figure 4.21.3 (b) A slightly smarter procedure 100 Ω 0.5 A
100 Ω
50 V
Figure 4.22.1
16
+
-
Now
+ 100 Ω
+
50 V
50 Ω
-
25 V
-
300 Ω
300 × 25 = 16.7V 450 = 300 150 = 100Ω
Voc = Rt
(33)
Figure 4.22.2 9. Example #2 (a) Consider the circuit shown below 3V
e2
e1 3Ω
+
-
I
6Ω
6Ω
3Ω
2A
Figure 4.23 (b) Find the current I by three methods: (1) straightforward nodal analysis, (2) superposition and (3) Thevenin/Norton equivalents (c) Nodal analysis: KCL: e1 − 3 e1 e1 − e2 1 1 2 + + =0 → e1 − e2 = 6 6 3 3 3 2 e2 e1 − e2 2 1 − − 2 = 0 → − e1 + e2 = 2 3 3 3 3 Solve for e1 :
(34)
4 1 e1 = 1 + 2 → e1 = 3 − 3 3 17
(35)
Solve for I: I=
e1 1 = 6 2
(36)
(d) Superposition: I1 : 6
e1 3 I1 :
e1
I1
+
3
-
3
6 6
+
3
-
3
3 ×3=1 9 e1 1 = = 6 6
e1 = I1
(37)
Figure 4.24.1 I2 :
3 I2 :
6
6
Ia 2
3
6
3
2
I2
3 2 ×2= 9 3 Ia 1 = = 2 3
Ia = I2
1 1 1 + = 6 3 2 Figure 4.24.2
I = I1 + I2 =
(e) Thevenin/Norton equivalents 18
(38) (39)
+
6
I
3
3 6
1 2
2
3
I=
6
6
6
1 1 1 1+ = 3 2 2 Figure 4.25
(40)
10. Summary of circuit analysis (a) Combine parallel and series resistors (b) Use Thevenin/Norton equivalents when convenient (c) Use superposition for multi-source problems when convenient (d) Apply nodal analysis
19
1
e1
e2 I1
R1
Figure 4.1 1
(e) Let us now write down the current I1 flowing out of node 1. I1 =
e1 − e2 R1
(1)
(f) This current is used in the standard way in the application of nodal analysis. (g) Consider next a slightly more complicated circuit with voltage source and resistor as shown below e1
+
e2
V1
R1
I1
-
Figure 4.2 (h) The current I1 is now given by I1 =
e1 − V1 − e2 R1
(2)
(i) This current can also be used in the standard way in nodal analysis (j) Finally consider the circuit below in which the branch resistor has been eliminated. All that remains is a voltage source connected between two non-grounded nodes. This is a floating voltage source. e3
e1
+
R3
I1
-
R4
V1
R1
Figure 4.3 2
e4
e2
R2
(k) Note that now there is no simple way to calculate I1 in terms of the nodal voltages. We know that the nodal voltages are related by e1 − V1 − e2 = 0 but so what. (l) The resolution to the problem is found by writing down the full nodal equations for nodes #1 and #2 in addition to the voltage relation above. In writing down these equations the quantity I1 is treated as an unknown. e1 e3 − e1 − − I1 = 0 R3 R1 e2 e2 − e4 − + I1 = 0 − R4 R2 e1 − V1 − e2 = 0
(3)
(m) Note that rather than having two independent nodal equations as would be the usual case, we now have three independent equations. However, there is now an additional unknown, I1 , so the number of equations and unknowns remains the same. (n) With a floating voltage source one can always reduce the number of equations by one by simply adding the nodal equations thereby eliminating the unknown I1 . The result is a “standard” set of nodal equations where the only unknowns are the nodal voltages. However, the matrix symmetry properties are no longer preserved with respect to e1 and e2 in the presence of a floating voltage source. e2 e2 − e4 e2 e3 − e1 − − − =0 (4) R3 R1 R4 R2 e1 − e2 = V1
(5)
3. The general principle of linearity (a) The circuit equations obtained from nodal analysis are in general a set of linear, inhomogeneous algebraic equations. (b) The linearity refers to the fact that the unknown nodal voltages all appear in the equations as quantities raised to the first power. Similarly, all the sources appear as inhomogeneous terms, also raised to the first power.
3
(c) This would not be true if the circuit element relations were not linear. For instance, if the value of a resistor depended upon the value of current flowing through it (e.g. R = R0 + aI 2 ) the circuit would no longer be described by a linear set of equations. (d) Why is linearity important? One important consequence of linearity is as follows. Consider a general circuit consisting of various resistors, voltage sources and current sources. If the values of the resistors are held fixed and all the sources are multiplied by a factor f , then all the resulting voltages and currents in the circuit are also multiplied by the same factor f . One can write the answer by inspection without having to re-analyze the circuit. (e) This “scaling result follows trivially from examining the form of the circuit equations. If every inhomogeneous source term in the equations is multiplied by f , then multiplying every unknown nodal voltage by f leaves the equations unchanged; that is the set of equations for the unknowns f ej in terms of the sources f Vj and f Ij is the same as the original set of equations. (f) The example below is an explicit demonstration of linearity where the factor f has been chosen to be f = 3. e 4
2 4
+
-
e'
1
4
2 6
+
-
4−e 6−e e + − =0 2 4 1 or e=2
12
+
-
1
18
+
-
12 − e 18 − e e + − =0 2 4 1
(6)
e = 6 = 3e
(7)
Figure 4.4 4. The general principle of superposition (a) Another important consequence of linearity is the principle of superposition. (b) Superposition is usually applied to circuits with more than one source. 4
(c) Consider such a circuit where the goal is to calculate, for instance a certain current I. (d) The principle of superposition states that the current I can be calculated as the sum of several separate terms, one for each source. (e) Each term is calculated by setting all source values to zero except the one denoted by the subscript j. One then solves the reduced single source problem for the current Ij due to the source j. This process is then repeated for each of the other sources. The total current is obtained by summing the individual contributions over j. I= Ij (8) j
(f) Superposition states that the value of I calculated by summing over the individual sources is identical to the value that would be obtained by solving the equations a single time simultaneously including all the sources. (g) Although it may seem to require more effort to use superposition since several circuit calculations need to be performed versus a single calculation for simultaneous sources, this may not be true. (h) Often, the reduced circuits obtained by setting all but one source to zero are so simple that they can be solved by inspection. (i) The fastest method depends upon the particular problem of interest. In this sense superposition can be viewed as another tool for simplifying the analysis of circuits. (j) Why does superposition work? The reason is that in a linear circuit the source terms appear as a sum of individual, linear, uncoupled terms driving the circuit. There are no terms for instance of the form Vj Ik or Vj2 . (k) It then follows from the theory of linear equations that one can calculate the unknown nodal voltages by simply adding the contributions from each source, setting all others to zero. (l) A technical point to keep in mind. Note that setting a voltage source to zero implies that the voltage across its terminals is equal to zero. Thus, when setting source voltages to zero in superposition, each voltage must be replaced by a short circuit. 5
(m) Similarly, setting a current source to zero implies that zero current flows through the source. Thus, setting a current source to zero in superposition requires replacing it with an open circuit. (n) We demonstrate superposition by a simple example. Consider the circuit shown below. The goal is to calculate the nodal voltage e.
e R1 V
+
I
R2
-
Figure 4.5 (o) Using standard nodal analysis we find e V −e +I − =0 R1 R2 or e=
R2 R1 R2 V + I R1 + R2 R1 + R2
(9)
(10)
(p) Note that the answer is the sum of two terms, one due to the voltage source and one due to the current source. (q) Now lets solve the problem by superposition. First set the current source to zero. The circuit reduces to a simple voltage divider e1 R1 V
+
R2
-
e1 =
R2 V R1 + R2
Figure 4.6
6
(11)
(r) Next set the voltage source to zero. This leads to a simple parallel resistor circuit. e2 R1 I
R2
e2 = R|| I =
R1 R 2 I R1 + R2
(12)
Figure 4.7 (s) Clearly, adding e1 + e2 by superposition yields the same answer as using standard nodal analysis. 5. I-V characteristics (a) The I-V characteristics are a convenient way to describe the properties of a given circuit. (b) In many applications a circuit containing a large number of elements may just be a subset of a larger more complex network. (c) Knowing the I-V characteristics of the subset circuit is often the only information we require in order to design and assemble a more complex system. (d) To understand the concept of the I-V characteristics consider a general, multi-element linear circuit as shown below +
-
+
-
I
+ V +
-
+
-
-
Figure 4.8 7
(e) This circuit may be a module that would be connected to other circuits through the output terminals as shown above. A circuit, regardless of its complexity, in which there are only two output terminals is referred to as a single port network. (f) Now, assume that all the resistors and sources are known. (g) Imagine that we apply a voltage V across the output and measure the current I noting the sign convention as shown. We then repeat this procedure for a number of different values of V . (h) We could then make a plot of I vs V as shown below. I
V
Figure 4.9 (i) This curve, or equivalently the mathematical relation between I and V are known as the I-V characteristics. (j) Note that in principle we could apply a current source I across the terminals and measure the resulting voltage V . Would we obtain the same I-V characteristic? (k) We will prove later that for a general linear resistive network the I-V characteristic turns out to be a linear relation I = C1 V + C2
(13)
(l) Below are some simple examples of I-V characteristics. Consider first a resistor.
8
I +
I R
V V
V R Figure 4.10 I=
(14)
(m) In general the I-V characteristic does not have to pass through the origin. To see this we calculate the I-V characteristics for a voltage source in series with a resistor I Rs +
-
Vs
I
+ Vs
V
-V s /R
I=
V − VS RS
V
(15)
Figure 4.11 (n) A similar result holds for a current source in parallel with a resistor.
9
I +
I IS
Rs
R s Is
V
V
-I s
V = (I + IS )RS
(16)
I = −IS + V /RS
(17)
or Figure 4.12 6. The Thevenin equivalent circuit (a) If we are interested only in the I-V terminal characteristics of a complicated network, we can take two simple measurements (or make two simple calculations) and replace the complicated network by a simple equivalent circuit. (b) This circuit is known as the Thevenin equivalent circuit. (c) Thevenin circuits are often very helpful in making circuit analysis much simpler (d) The idea of the Thevenin equivalent circuit is based on the fact that across any two terminals of a linear network with ideal sources, the relation between voltage and current is a linear relation of the form I + Vj
+
Rj
V
-
Ij
-
I = C1 V + C2 Figure 4.13 10
(18)
(e) This result follows from linearity and superposition. To see this assume a voltage source V is applied across the output terminals and use nodal analysis to calculate I. (f) Using the principle of superposition the solution for the current I will be of the form I = α0 V +
αj Vj +
j
βj Ij
(19)
j
(g) In this expression the Vj and Ij are known internal sources of the network. The coefficients αj and βj are known functions of the values of the known resistors in the circuit. Thus, for a general network we can write I = C1 V + C2 C1 = α0 C2 = αj Vj + βj Ij j
(20)
j
(h) We now focus on interpreting and calculating the coefficients C1 and C2 . (i) Physically, C2 represents the negative of the short circuit current as shown below. This is the current that flows through the output terminals with the polarity as shown below, when the output voltage is short circuited: V = 0. I
I sc
I = −Isc → C2 = −Isc
(21)
Figure 4.14 (j) Physically, C1 is related to the open circuit voltage-the voltage measured when the output is an open circuit: I = 0. 11
+ V oc -
C2 = −Voc C1 Figure 4.15
V = Voc →
(22)
(k) Using these relations we can rewrite and plot the I-V relationship as follows. I Voc
V
-Isc
Isc V − Isc Voc Figure 4.16
I=
(23)
(l) In terms of measurements, the open circuit voltage and short circuit current are measurements that can be made solely across the output terminals. One does not have to interfere with any of the internal components of the circuit. (m) In terms of circuit calculations, we can calculate the open circuit voltage Voc using nodal analysis. In principle we could redo the circuit analysis and calculate the short circuit current Isc by shorting the output terminals. (n) An often easier procedure to calculate Isc than redoing the circuit analysis is as follows. Set all the current and voltage sources in the circuit to zero and then calculate the input resistance to the circuit denoted by Rt . The short circuit current then turns out to be given by the relation Isc = Voc /Rt 12
(24)
(o) The I-V relation can now be rewritten as I Voc V - Voc /Rt
I=
V Voc − Rt Rt
(25)
Figure 4.17 (p) The relation between Rt and Voc , Isc given above may not be completely obvious. It follows from superposition. (q) If all the sources are set to zero and a current source I is applied across the output terminals then this generates an output voltage V given by V = Rt I (26) (r) If the sources are now turned on, the output voltage, by superposition becomes V
= Rt I +
j
= Rt I + Voc
αj Vj +
βj Ij
j
(27)
(s) The key point is that in a linear circuit the contributions from the internal sources are independent of the external source I. Thus, the slope of the I-V characteristic is just 1/Rt as stated above in step (o). (t) The last step is to give a simple equivalent circuit that satisfies the I-V characteristics of a general circuit. This is the Thevenin equivalent circuit and is given by
13
+
-
+
I
Rt V oc
V
I=
V − Voc Rt
(28)
Figure 4.18 7. The Norton circuit (a) A Norton equivalent circuit plays the same role as a Thevenin equivalent circuit except that it uses a current source rather than a voltage source. (b) The Norton equivalent circuit is given by I In
Rn
+ V
Figure 4.19 (c) Using KCL we can easily derive the I-V relationship for the Norton circuit V I + In − =0 (29) Rn or V − In (30) I= Rn (d) Note that the open circuit voltage is given by Voc = Rn In . Hence, if we choose Rn = Rt and In = Voc /Rn the Thevenin and Norton circuits are equivalent to each other.
14
+ V oc /R t
V
Rt
Rt
+
V oc
V
-
Figure 4.20
(e) When analyzing circuits, it does not matter whether you use Thevenin or Norton equivalents. Use the one that is most convenient for the problem at hand. 8. Example #1 (a) Find the Thevenin equivalent circuit by straightforward nodal analysis e 1 I1 50 Ω
+
-
100 Ω
0.5A
I
25V 300 Ω
+ V -
KCL: e1 e1 − 25 − = 0 → e1 = 44.4V 100 350 e1 − 25 = = 0.0555A 350 = 300I1 = 16.7V
.5 − I1 V0c
Figure 4.21.1 Rt :
15
(31)
50 Ω 100 Ω
300 Ω
Rt
1 1 1 = + = 0.01 → Rt = 100Ω Rt 300 150
(32)
Figure 4.21.2 SOLUTION: 100 Ω
+
-
16.6 V
Figure 4.21.3 (b) A slightly smarter procedure 100 Ω 0.5 A
100 Ω
50 V
Figure 4.22.1
16
+
-
Now
+ 100 Ω
+
50 V
50 Ω
-
25 V
-
300 Ω
300 × 25 = 16.7V 450 = 300 150 = 100Ω
Voc = Rt
(33)
Figure 4.22.2 9. Example #2 (a) Consider the circuit shown below 3V
e2
e1 3Ω
+
-
I
6Ω
6Ω
3Ω
2A
Figure 4.23 (b) Find the current I by three methods: (1) straightforward nodal analysis, (2) superposition and (3) Thevenin/Norton equivalents (c) Nodal analysis: KCL: e1 − 3 e1 e1 − e2 1 1 2 + + =0 → e1 − e2 = 6 6 3 3 3 2 e2 e1 − e2 2 1 − − 2 = 0 → − e1 + e2 = 2 3 3 3 3 Solve for e1 :
(34)
4 1 e1 = 1 + 2 → e1 = 3 − 3 3 17
(35)
Solve for I: I=
e1 1 = 6 2
(36)
(d) Superposition: I1 : 6
e1 3 I1 :
e1
I1
+
3
-
3
6 6
+
3
-
3
3 ×3=1 9 e1 1 = = 6 6
e1 = I1
(37)
Figure 4.24.1 I2 :
3 I2 :
6
6
Ia 2
3
6
3
2
I2
3 2 ×2= 9 3 Ia 1 = = 2 3
Ia = I2
1 1 1 + = 6 3 2 Figure 4.24.2
I = I1 + I2 =
(e) Thevenin/Norton equivalents 18
(38) (39)
+
6
I
3
3 6
1 2
2
3
I=
6
6
6
1 1 1 1+ = 3 2 2 Figure 4.25
(40)
10. Summary of circuit analysis (a) Combine parallel and series resistors (b) Use Thevenin/Norton equivalents when convenient (c) Use superposition for multi-source problems when convenient (d) Apply nodal analysis
19
1
