Notes for course EE1.1 Circuit Analysis TOPIC 3 CIRCUIT ANALYSIS USING SUBCIRCUITS


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1 Notes for course EE1.1 Circuit Analysis TOPIC 3 CIRCUIT ANALYSIS USING SUBCIRCUITS OBJECTIVES 1) To introduce the Source Transformation 2) To consider the concepts of Linearity and Superposition 3) To introduce Thevenin and Norton Equivalent Subcircuits 4) To consider Practical Sources and Matching 5) To introduce Voltage and Current Source Transportation 6) To introduce Voltage and Current Source Substitution 1 SOURCE TRANSFORMATION. Consider a voltage source connected in series with a resistor: We can test its vi characteristic by attaching an independent source to its terminals, varying that source to all possible values, and recording the resulting value of the response variable; we have arbitrarily chosen a current source above but a voltage source would have worked just as well. We can compute v very simply using KCL and Ohm's law: v = v s + ir s Let's invert this relation and express i in terms of v: i = v s R s + v R s This can be interpreted as a KCL equation for the following subcircuit: This subcircuit can be tested also with a current source as shown or a voltage source to obtain the i v characteristic. Since the equations describing the voltage source with series resistor and current source with parallel resistor are equivalent, it follows that these subcircuits are equivalent subcircuits. No test at the terminals can distinguish one from the other.
2 The source equivalence is a powerful tool for analyzing circuits. Consider the following example: Example 3.1 Find the voltage v and the current i in the following circuit using an equivalent subcircuit. Solution We see that there are two subcircuits consisting of a voltage source with a series resistor, as follows: We can replace the two subcircuits with equivalent subcircuits: Notice that we have lost the branch that current i flowed in but the branch corresponding to voltage v is retained. We proceed to determine v: We apply KCL equation at the top node (noting that this node voltage is v): I out = I in Resistors Sources v 6 + v 3 + v 6 = 8 4 v = 4 v 2 3 = 4 v = 6 V To determine i, we redraw the original circuit with the voltage v = 6 V shown explicitly: We can now write KCL at the top node to obtain: 2
3 I out = I in Resistors Sources ) = i 6 i = 2 5 = 7 A Using the source transformation, we can analyze circuits that do not consist merely of a single loop or a single pair of nodes and that are neither series nor parallel. Example 3.2 Find the current i in the following circuit: Solution We apply the source transformation to the twoterminal subcircuit on the left : Analysis of this simple series circuit gives i = v effective r effective = 24 ( 12 ) = = 6 A 2 LINEARITY AND SUPERPOSITION Consider a simple circuit whose independent source values are variables, i s and v s : Using the source transformation on the parallel subcircuit consisting of the current source and its nearest resistor, we derive the equivalent circuit shown: The desired response is the current in one of those elements. We easily analyze this single loop (series) circuit to obtain 3
4 i = v effective r effective = 2i s + v s = 1 3 i s v s Thus, i is a linear combination of the two source values, i s and v s. This is a property that holds in general for all voltage and current responses in circuits with only independent sources and resistors. It will be proved from first principles in the next topic. Now we derive an important result; we can define current parameters i 1 and i 2 as follows: i 1 = i vs =0 = 1 3 i s i 2 = i is =0 = 1 6 v s We call i 1 and i 2 the partial responses. We can now write the equation for i in the form: i = i 1 + i 2 In words, i 1 is the response with the voltage source reduced to zero and i 2 the response with the current source reduced to zero. When a source value, voltage or current, is set to zero, we say the the source is deactivated. However, a deactivated voltage source is a short circuit and a deactivated current source is an open circuit. Hence the partial response expressions describe the circuits shown: These two circuits are referred to as partial circuits. We can now generalize. Consider a circuit having only a number of resistors and n independent sources with values x 1, x 2,..., x n. Any response y, voltage or current, has the form: y = a 1 x 1 + a 2 x a n x n where a i are constants determined by the resistive portion of the circuit. Thus, we can define n partial responses: We can write: y i = a i x i for i = 1, 2,..., n y = y 1 + y y n where y i = y xk =0 k = 1,2,...,n with k i 4
5 Thus, we can compute each partial response y i from the circuit that results when all the independent sources other than the i th are deactivated. This is known as the principle of superposition. It is often a useful analysis tool. By its means, we can split a complicated circuit analysis problem into a number of simpler ones. Example 3.3 Find the indicated response current the following circuit using superposition: Deactivation of the 12A isource and the 8A isource gives the following partial circuit: Using parallel and series equivalents, Ohm's law, and voltage and current division, one finds the partial response to be: i 1 = 2 A Deactivation of the 12A isource and the vsource gives the second partial circuit: We can determine that: i 2 = 5 A We then allow the 12A source to remain active and deactivate the 8A isource and the vsource: This single source circuit can be analyzed to determine that: i 3 = 3 A Finally, we add each of the three partial response currents to obtain the actual current in the original circuit: i = i 1 + i 2 + i 3 = = 0 A This surprising result can be verified by checking that KCL and KVL hold for all loops and nodes. 5
6 3 THEVENIN AND NORTON EQUIVALENT SUBCIRCUITS 3.1 The Thevenin Equivalent As an example, consider the following subcircuit: We assume that this subcircuit is connected into a larger circuit; that is, there are elements external to this subcircuit connected to it through the two terminals shown. Therefore, both the terminal voltage v and the terminal current i have nonzero values in general. If we apply a source transformation to the current source and its parallel 4 Ω resistor, we obtain the equivalent shown: The two voltage sources and the two 4 Ω resistors are series connected and can be combined. Elements in series, such as the two voltage sources may be moved around in the series loop because this does not change the KVL equation. Now we apply one last source transformation to the elements in the shaded box: We can now combine the two 8 Ω parallel resistors to obtain the final equivalent subcircuit: Notice that all of our transformations have left i and v unaffected. 6
7 We can now write one KCL equation at the top node to get: v 4 = i i s v s v = 2i s v s + 4i We see that the voltage is a linear combination of the internal independent sources and the terminal current i. The first two terms on the right side of the equation for v are independent of i and that the third is directly proportional to the terminal current i. Thus the equation may be represented as follows: where v = v oc + ir eq v oc = 2i s v s R eq = 4 The simpler equation describes the series combination of a vsource and a resistor: This subcircuit is, therefore, equivalent to the original because it has the same vi characteristic. It is called the Thevenin equivalent (subcircuit) after the French telegraph engineer Charles Thevenin. Since the superposition property applies to any twoterminal subcircuit composed of resistors and independent sources, then any such subcircuit, however complex, has a valid Thevenin equivalent circuit in this form. We now seek to understand the significance of the two parameters v oc and R eq. Consider the parameter v oc : Although only a part of the total terminal voltage v, it becomes the actual terminal voltage under the condition that i = 0. v = v oc + ir eq v = v oc The condition i = 0 is imposed by simply removing the subcircuit from the elements to which it is connected: It is important to notice that this is precisely the same as the original subcircuit with only two small notational changes: we have explicitly specified that the terminal current is zero, and we have labelled the terminal voltage with the symbol v oc to denote that the voltage obtained is the voltage 7
8 source in the Thevenin equivalent circuit. Note that v oc depends on all the sources and also on the resistors. Using superposition to analyse the subcircuit for determining v oc gives: which is correct. v oc = 2i s v s In order to consider the significance of the term R eq, consider again the equation for v in our example: v = 2i s v s + 4i Note that if we deactivate the internal sources (set i s and v s to zero in our example) the v oc term will be forced to zero, leaving only the term proportional to i: v = 4i Deactivating the sources will make v oc = 0 in the Thevenin equivalent circuit. The equation describing the Thevenin equivalent circuit then becomes: v = v oc + ir eq v = ir eq Thus R eq is v/i for the subcircuit with all its sources deactivated, which is as follows: v/i is the equivalent resistance of the deactivated circuit. Analysis of the deactivated circuit yields R eq = 4 Ω, which agrees with the previous figure. In general, all independent vsources are replaced by short circuits and all isources by open circuits. Example 3.4 Find the Thevenin equivalent subcircuit for the subcircuit shown: Solution We first determine v oc and set i = 0: 8
9 Since i = 0, the current in the resistor must equal the current in the current source, as shown. Then KVL gives: v oc = = 25 V To compute R eq we deactivate the subcircuit, thus replacing the 2V source with an equivalent short circuit and the 3A source with an equivalent open circuit: Clearly, R eq = 9 Ω. The Thevenin equivalent subcircuit is as follows: Example 3.5 Check the Thevenin equivalent circuit we have derived by determining the vi equation for the subcircuit. Solution KCL at the left upper node tells us that the current in the 9 Ω resistor is i 3 A (downwards). KVL around the loop gives us: v = 9( i 3) + 2 = i This confirms the values of v oc = 25 V and R eq = 4 Ω obtained previously. 1.2 The Norton Equivalent The Thevenin equivalent circuit is equivalent to any 2terminal subcircuit containing resistors and independent voltage and current sources. The Norton equivalent circuit is an alternative general equivalent circuit which can be derived independently or it can be obtained from the Thevenin equivalent circuit by applying the source transformation. Application of the source transformation to the Thevenin subcircuit leads to the Norton equivalent circuit as follows: In order for this subcircuit to be equivalent to the original circuit, we must have 9
10 i sc = v oc R eq Notice that the terminal vi characteristic for the Norton equivalent is: i = v R eq i sc Combining these two equations leads to the equation describing the Thevenin equivalent: i = v R eq v oc R eq v = v oc + ir eq Hence, the Thevenin and Norton equivalent circuits are equivalent to each other. To understand the significance of the parameter i sc, suppose the subcircuit is shorted  that is, a short circuit is placed across its terminals. This is the same as shorting the Norton equivalent: (Arrows on the short denote that it is merely a test connection.) The terminal voltage and therefore the resistor voltage are both zero, so the resistor current is zero also. Thus, all of the source current flows out of the top terminal and down through the short circuit, as shown above. Thus i sc, or i shortcircuit, is the current that flows in the subcircuit when the terminals are shorted together. The Thevenin and Norton equivalents involve three parameters: v oc, R eq and i sc. R eq is common to both equivalents and in addition all three parameters are related via the source transformation equation: i sc = v oc R eq Hence, one need only compute two of the parameters from the circuit in order to determine al three. We can see the connection graphically starting from the vi characteristic for the Thevenin equivalent: We can plot a graph of this equation as follows: v = v oc + ir eq We have sketched this graph under the assumption that v oc (and hence i sc ) is positive. 10
11 It is clear that the vertical intercept is v oc, the horizontal intercept is i sc and the gradient is R eq. Example 3.6 Find v oc, i sc, and R eq by direct tests on the 2terminal subcircuit shown: Solution When we determine Thevenin or Norton equivalent circuits, we are free to define the reference directions as we wish. The open circuit voltage can be computed by analyzing the following circuit where we have imposed the condition i = 0: We first deactivate the 36 V vsource and find the corresponding partial response from the partial subcircuit: We note that the two resistors are connected in parallel, giving an equivalent resistance of 2 Ω. Thus, we have v oc1 = 6 2 = 12 V. Next, we deactivate the 6 A isource, resulting in the other partial subcircuit: The two resistors and the vsource form a series circuit and that the voltage across the 3 Ω resistor is the partial terminal voltage. Thus, we have v oc2 = 36 3/(3 + 6) = 12 V. Adding, we obtain the total (actual) open circuit voltage: v oc = v ocl + v oc2 = = 24 V. In order to calculate the value of R eq, we must first deactivate the entire subcircuit: Hence, R eq = 2 Ω. 11
12 The short circuit current i sc can be calculated from the v oc and R eq, but let us first determine it from the circuit. We place a short circuit across the subcircuit terminals and identify the short circuit current: The test short circuit makes the voltage in the 3 Ω resistor zero; hence its current is also zero by Ohm's law. i sc is then the sum of the current source current and the current upward through the 6 Ω resistor.. Thus, we have i sc = 6 + (36/6) = 12 A. A quick check shows that, indeed, v oc /R eq = 24 V/2 Ω = 12 A, as expected. We may verify these parameters by direct analysis of the subcircuit. We attach a test voltage source v to the subcircuit: Application of KCL to the upper node gives: i + 6 = v 36 6 i = v 2 12 where we have used Ohm's law in order to determine the resistor currents. + v 3 Comparison with the Norton equivalent circuit expression: gives i sc = 12 A and R eq = 2 Ω as expected. Rearranging the equation in terms of voltage: confirms v oc = 24 V. i = v R eq i sc v = 2i + 24 v = ir + v oc The Thevenin and Norton equivalent subcircuits are as follows: 12
13 4.1 Resistance Matching a Subcircuit 4 PRACTICAL SOURCES AND MATCHING Consider the following problem. A crystal microphone can be assumed to possess a Thevenin equivalent circuit as follows: Furthermore, a loudspeaker can be modelled to a good degree of approximation by the resistor R L shown. The current in the speaker is i L = 1 V 1 MΩ +8 Ω 1 µa This is quite a problem, because for a loudspeaker to be audible, current needs to be several milliamps. The auditory power equals the electrical power absorbed: P L = v L i L = i L 2 RL = = 8 pw Consider the same circuit with more general element values: We first assume that the Thevenin equivalent parameters v oc and R eq are specified and we are asked to determine the value of R L that results in the power absorbed by R L being the maximum possible. Before solving this analytically, let's carry out a numerical study. Suppose that v oc = 12 V and R eq = 4 Ω. The power absorbed by R L is: 2 v P L = v L i L = i LRL = oc R eq + R L 2 2 v ocrl R L = ( R eq + R L ) 2 = 144R L ( ) R L If we evaluate this expression at R L = 0 Ω we get P L = 0 W; at R L = Ω, again P L = 0 W. If we evaluate the equation at several values of R L between zero and infinity, we obtain the following plot: We see that the power absorbed by the load exhibits a maximum of around 9 W for a finite value of R L. 13
14 Now let's solve the problem analytically. We can rewrite the expression for P L : 2 v P L = ocrl ( R eq + R L ) 2 = 2 v oc R eq R L + 2 = v oc 2 R L Maximising P L is obtained by minimising the denominator. 1 R eq R eq + R L 2 R L R eq R L appears only within the bracket and the bracket term is of the form x + x 1. It can easily be shown that x + x 1 has a minimum value of 2 for x = 1. Hence the term in the bracket has a minimum value of 2 for (R eq /R L ) = 1, ie: R L = R eq for which case, we have: P L max 2 ( ) = v oc 4R eq Therefore, for maximum power to be absorbed from a source with given v oc and R eq, the value of the load resistance must be numerically the same as R eq. Now consider the problem that the load resistance R L and the Thevenin equivalent voltage v oc are fixed and the problem is to pick R eq such that the power absorbed by the load is maximized. Consider the previous expression: 2 v ocrl P L = ( R eq + R L ) 2 We see that the condition for maximum power in the load is: R eq = 0 This means that for maximum power in a given load, the subcircuit represented by the Thevenin equivalent must be an ideal voltage source. The problem of a transferring power from a source to a load such that power in the load is maximised is known as the matching problem. 1.2 Practical Sources In theory, the Thevenin and Norton equivalent circuits for a subcircuit are equivalent to each other and therefore the choice of which to use is arbitrary. However, for certain practical sources one or the other of the equivalents may be preferred. Consider a device called a photodiode which converts light into an electrical signal; it is an important component in optical communication systems. The photodiode is usually given the Norton equivalent circuit shown, where the current of 1 ma is the value for a given light level: 14
15 The current source is the primary element in the equivalent circuit as its current represents the light signal applied to the photodiode and is amplified by the following circuitry. The following amplifier is designed so that as much of the 1 ma current goes into the amplifier and is amplified and as little as possible goes into the 100 MΩ resistor. The 100 MΩ resistor is not very significant and is a parasitic or unwanted element. If the 100 MΩ resistor is removed, the model will be less accurate but voltages and currents in the whole system will not change that much. We know that any Norton equivalent source can be turned into a Thevenin equivalent source using the source transformation. This leads to the following equivalent circuit for the photodiode: (Note that v oc = i sc R eq, = 1 ma 100 MΩ = 100 kv) The voltage of 100 kv does not exist in the physical device and is a mathematical artefact of the equivalent circuit. Also the 100 MΩ resistor is no longer a parasitic element and can not be removed without destroying the operation of the circuit completely. Thus, although this equivalence is valid, we can say that for the photodiode, the Norton equivalent is more natural than the Thevenin equivalent and closer to the physical mechanism of the device. Consider now the car battery. Assuming that the terminal voltage has an open circuit values of 12 V and falls to 6 V at a current of 600 A, we can determine the parameters of the Thevenin equivalent circuit: 0.01 Ω 12 V The Ω resistor is a parasitic or unwanted element. If its value is set to zero, the model is less accurate but still represents the real situation approximately. Let us derive the Norton equivalent circuit using the source transformation: 1200 A 0.01 Ω (Note that i sc = v oc /R eq, = 12 V / 0.01 Ω = 1200 A) Although this equivalent circuit is valid, there is nothing in the physical battery that corresponds to the huge current of 1200 A which flows into the resistor even when the terminal current is zero. The 0.01 Ω resistor is not a parasitic element and cannot be removed from the circuit without completely destroying the equivalence. Thus, in the case of the car battery, the Thevenin equivalent circuit is closer to the physical construction and operation of the real device. 15
16 In general, we see that although the Thevenin and Norton equivalent circuits are strictly equivalent to each other, when we represent a real device, such as a battery or a photodiode, one or other of the equivalents may be more appropriate than the other. Consider a 3terminal subcircuit: 5 SOURCE TRANSPORTATION We have used the double subscript notation for the voltages between terminals 1 and 3 and between 2 and 3. By KVL, we see that these two voltages suffice to determine any other voltage relative to the three terminals. Similarly, KCL shows that the terminal currents i 1 and i 2 suffice to determine the current at terminal 3. Thus relationships between v 13, v 23, i 1 and i 2 suffice to completely describe the subcircuit. 5.1 Voltage Source Transportation Consider now the following 3terminal subcircuit: For this circuit: v 13 = v 23 = v s These equations apply independently of the currents i 1 and i 2. Consider now the following subcircuit: Analysis leads to the same expression as the single source circuit. Thus, the two are equivalent. The equivalence we have just derived is called voltage source transportation because we are "transporting" a vsource through a node. Note that each circuit has the same 3 nodes. Connections of other elements to these 3 nodes are not changed by this equivalence. Note that for Thevenin and Norton equivalents, nodes inside the transformed subcircuit are changed (in most cases, they disappear). Example
17 Find the current i in the circuit shown using vsource transportation. Solution This circuit is neither a parallel circuit nor a series circuit. We first apply vsource transportation to the vsource: This circuit can be redrawn in a more familiar form: We now apply the Thevenin equivalent transformation to the vsources and their closely associated resistors to obtain an equivalent form: This is a series circuit that we can quickly analyze to determine: 5.2 Current Source Transportation i = v effective R effective = 6 6 = 1 A Consider the following 3terminal subcircuits containing current sources attached to two nodes: Both circuits are described by: i 1 = i s i 2 = i s i 3 = 0 17
18 These equations apply independently of the applied voltages, v 13 and v 23. Because the vi characteristics are the same, the two 3terminal current source subcircuits are equivalent as far as any external elements connected to the terminals are concerned. This equivalence is called current source transportation. In its usual application, a single current source between two nodes is replaced by two equal current sources of correct polarity between that node and any other node. Note that node 3 must be connected to elements through which current can flow; otherwise we have an invalid connection of two current sources in series. Example 3.8 Find the current i in the following circuit: Solution We apply current source transportation to the 18A source using the ground node as the 3 rd node: Combining the parallel isources, we have: We now can apply the source transformation twice: We easily find: 18
19 i = v effective = 6 28 = 22 R effective = 2 A 6.1 Voltage Source Substitution Consider a portion of a circuit as shown: 6 SOURCE SUBSTITUTION We have a 2terminal element e connected between two nodes, i and j, to which other elements (not shown) are connected, as indicated by conductors at these nodes. We assume that the voltage v across e is a known quantity. We now connect one lead of a voltage source to node j, leaving its other terminal free, and adjust its value to be exactly v; this creates a floating node, i': Because of the floating node the current through the added source is clearly zero  hence its addition does not affect any of the voltages or currents in the circuit. Now let us attach a resistor having a very large value, say R = Ω, between nodes i and i': What is the current through this resistor? Of course, one would expect it to be small because of the large value of resistance. But in fact the current is identically zero because the voltage across the resistor is identically zero! This means that we can decrease the value of this resistance to zero and the resistor current will continue to be zero. We have neither added nor subtracted any current at nodes i and j, so we see that our experimentation thus far has not affected any of the voltages or currents in the circuit in the slightest manner. When the resistance has reduced to zero, thus producing a short circuit, we have the folllowing: But a twoterminal subcircuit consisting of a voltage source and element in parallel can be replaced by the voltage source alone: 19
20 Hence we have the following: If the voltage across any twoterminal element or subcircuit is known, it can be removed and replaced by a voltage source having that known value, without affecting any of the other voltages or currents in the circuit. This statement is often called the voltage form of the substitution theorem The theorem applies to the case when the original element is a current source with a known voltage across it. 1.2 Current Source Substitution Consider the case of an element with known current i flowing in it: Suppose that we add a shorted independent isource and adjust its value to be i: This has no effect on the circuit, but note that the current in the segment of conductor between the i source connections is zero! Since this current is identically zero, the conductor segment can be clipped out and replaced by an open circuit without affecting any of the other voltages or currents in the rest of the circuit: However, any subcircuit consisting of an isource connected in series with any other element is equivalent to the current source alone: Hence, we have the current source form of the substitution theorem: If the current in any 2terminal element or subcircuit is known, that element or subcircuit can be replaced by a current source having the known value, as far as the external voltages and currents are concerned. The theorem applies to the case when the original element is a voltage source with a known current across it. Example 3.9 Find the value of v s required to adjust the current i to zero in the following circuit: 20
21 Solution We first use current source substitution. Suppose that v s has been adjusted to the value required to make i = 0. Then, because i = 0, the 2 Ω resistor can be replaced by an isource whose value is zero, which is an open circuit: We also know that the voltages v a and v b should be identical. The voltage divider rule gives v a = 10 V and v b = (2/3)v s. In order that these voltages are equal, we have v s = 15 V. Now we use voltage source substitution. Since we require i = 0 in the original circuit and i flows in a resistor, it follows that the voltage across the resistor is zero. Hence, by the theorem, the resistor may be replaced by a voltage source of zero value, which is a short circuit: But we also know that the current flowing between nodes a and b must be zero. It follows from the voltage divider rule that v a = 10 V. Also, v b = 10 V and by the voltage divider rule, v s = 15 V. 2 CONCLUSIONS In this topic we have introduced a number of equivalences, transformations and theorems: Source Transformation Thevenin and Norton Equivalent Subcircuits Voltage and Current Source Transportation Voltage and Current Source Substitution These theorems extend the complexity of the circuits which we can analyse. However, there is a limit on the extent to which we can extend such a piecemeal approach. We now have the background to be able, in the next topic, to develop a method for systematic analysis of a circuit of arbitrary complexity and any topology. 21
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