Notes for course EE1.1 Circuit Analysis TOPIC 8 DEPENDENT SOURCES

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1 Objectives Notes for course EE1.1 Circuit Analysis To introduce dependent sources TOPIC 8 DEPENDENT SOURCES To study active sub-circuits containing dependent sources To perform nodal analysis of circuits with dependent sources 1.1 General 1 INTRODUCTION TO DEPENDENT SOURCES The elements we have introduced so far are the resistor, the capacitor, the inductor, the independent voltage source and the independent current source. These are all -terminal elements The power absorbed by a resistor is non-negative at all times, that is it is always positive or zero The inductor and capacitor can absorb power or deliver power at different time instants, but the average power over a period of an AC steady state signal must be zero; these elements are called lossless. Since the resistor, inductor and capacitor cannot deliver net power, they are passive elements. The independent voltage source and current source can deliver power into a suitable load, such as a resistor. The independent voltage and current source are active elements. In many situations, we separate the sources from the circuit and refer to them as excitations to the circuit. If we do this, our circuit elements are all passive. In this topic, we introduce four new elements which we describe as dependent (or controlled) sources. Like independent sources, dependent sources are either voltage sources or current sources. However, unlike independent sources, they receive a stimulus from somewhere else in the circuit and that stimulus may also be a voltage or a current, leading to four versions of the element Dependent sources are considered part of the circuit rather than the excitation and have the function of providing circuit elements which are active; they can be used to model transistors and operational amplifiers. Since there are two source terminals and also two terminals where the stimulus exists, dependent sources are in general 4-terminal elements. 1. An Intuitive Analogy for Dependent Sources The basic idea of the voltage-controlled voltage source can be illustrated by considering a voltage source v c connected between two nodes and a voltmeter connected between two different nodes between which there exists a voltage difference v x :

2 The voltmeter measures the voltage v x and then controls the voltage source such that v c = µv x. Since the electrical equivalent of an ideal voltmeter is an open circuit, we can represent the element as follows: The basic idea of the current-controlled current source can be illustrated by considering a current source i c connected between two nodes and an ammeter connected between two different nodes between which a current i x flows: The ammeter measures the current i x and then controls the current source such that i c = βi x. Since the electrical equivalent of an ideal ammeter is a short circuit, we can represent the element as follows: In these examples, the output variable v c or i c is called the controlled variable and the input variable v x or i x is called the controlling variable. The remaining two source types are obtained by combining controlled and controlling variables in different ways. For instance, the voltage-controlled current source combines open-circuit voltage measurement with a current source at the output; it is described by i c = g m v x : The current-controlled voltage source combines short-circuit current measurement with a voltage source at the output; it is described by v c = r m i x : Thus the four types of dependent sources are as follows:

3 VCVS VCCS CCVS CCCS Dependent (or controlled) sources can be thought of intuitively as amplifiers, where the input and output variables can both be either a voltage or a current. In many cases, there is some flexibility over which type of source can be used. For example, when the load is resistive, it may be driven by a voltage source or a current source because the voltage and current are related by Ohm's law. 1.3 The Dependent Voltage Source Although dependent sources are strictly 4-terminal elements, we can show them simply as a voltage or current source and write next to it an expression for the voltage or current as a function of another voltage or current defined elsewhere in the circuit. Using this notation, the symbol for a dependent voltage source is as follows: where v x and i x are defined elsewhere in the circuit. Its outline is a diamond shape to show that it is a dependent source and to distinguish it from the round-shaped independent voltage source. Inside the diamond shape, a pair of '+' and ' ' signs denote the positive reference for the voltage generated. The terminal voltage v of the dependent voltage source, when plotted as a function of its terminal current i, is exactly the same as that of an independent voltage source: This plot shows that v c is independent of i. However, whereas for an independent voltage source, the voltage is also constant with time, in the case of the dependent voltage source, the voltage depends upon either a voltage v x or a current i x somewhere else in the circuit and is therefore not, in general, constant with time. Voltage v c is referred to as the controlled voltage (or dependent voltage). When the controlling variable is a voltage, the dependent voltage source is called a voltagecontrolled voltage source, or VCVS; the dependency relationship has the form: v c = µv x The parameter µ is a unit-less real constant called the voltage gain. When the controlling variable is a current, the dependent voltage source is called a currentcontrolled voltage source (or CCVS); the linear dependency relationship takes the form 3

4 v c = r m i x The real constant r m has units of Ohms and is called the trans-resistance. The prefix "trans" is used because the dependent source "transfers" the effect of a current somewhere else in the circuit to the dependent voltage source; the term "resistance" is used because it multiplies that current by a constant having the unit of Ω to turn it into a voltage. The subscript "m" denotes "mutual" which is a synonym for "trans". The CCVS, however, is quite different from the resistor element because the current in a resistor element must be defined in the same element as that across which the voltage is defined. We could say that a resistor has self-resistance (or just resistance) and that a CCVS has mutual resistance or trans-resistance. 1.4 The Dependent Current Source The symbol for a dependent current source is as follows: It has the same diamond shape as that of the dependent voltage source, but instead of '+' and ' ' signs, it has an arrow denoting the positive current reference for the controlled current i c. The v-i characteristic has the following form: It has the same terminal behaviour as that of an independent current source, but the value of the controlled current i c depends upon either a voltage v x or a current i x somewhere else in the circuit and will therefore in general vary with time. When the controlling variable is a voltage, we refer to the dependent i-source as a VCCS, for voltage-controlled current source; the dependency relationship is: i c = g m v x The real constant g m has units of Siemens and is therefore called the trans-conductance. When the controlling variable is a current,, we refer to the dependent i-source as a CCCS, for current-controlled current source; the dependency relationship is: i c = βi x The real unit-less constant β is called the current gain. We develop a method for analysing a circuit containing dependent sources by considering some examples. 1.5 Dependent Source Examples Example 5.1: Find the value of current i in the circuit shown which contains a CCCS: 4

5 Solution Note that the current-controlled current source is defined by i c = i x, where i x is the current in the 6 Ω resistor. Note that for a dependent source, the polarities for the controlling variable and controlled variable are inter-related and neither should be changed separately. We treat the dependent source constraint relationship i c = i x as a label on the source. We temporarily cover this label with a small piece of tape on which we write the symbol i c ; we call this procedure taping the dependent source. This leads to the following: We now treat i c as if it was an independent current source, just like the 9 A current source. The current divider rule leads to: ( ) i = 9 + i c = 6 3 i c If i c was an independent current source, we would be through; however, for a dependent source there is a further stage: We next un-tape the dependent source and express its controlled variable i c in terms of its controlling variable: i c = i x We then express the controlling variable in terms of our analysis variables, in this case the unknown current i. Using KCL, we have: i c = i x = ( 9 + i + i c ) Note that i c appears on both sides of the equation. We can now solve the two equations: ( ) = 18 + i + i c i c = 9 + i + i c i c = 18 i i = 6 3 i c = i 3 i = = 18 A We can state the general method of analysis as follows: 5 ( ) = i 1) Tape all dependent sources, thus treating them temporarily as independent sources.

6 ) Analyze the circuit for the unknown variable or variables you have chosen using the analysis technique of your choice. 3) Un-tape the dependent sources and express their values in terms of the unknowns you have chosen to use in step. 4) Solve the resulting equations. Example 5.: Find the voltage v in the following circuit which contains a CCVS: Solution We show the circuit also with the dependent source taped. We can use voltage division to compute v: v = If v c were a known quantity, we would be through. 4 ( v c 4) = 5 v c 48 5 As it is not, we un-tape the dependent source and express its value in terms of the unknown voltage v: Then we use the previous equation to obtain v:.1 General v c = i x = 4 v c 10 = 4 5 v c 5 6v c = 4 v c = 4 V v = = 8 V ACTIVE SUB-CIRCUITS In this section, we use hand analysis techniques to look at unusual behaviour of circuits containing dependent sources and at their Thevenin and Norton equivalents. In the following section we will cover the extension of the systematic nodal analysis method to include dependent sources.. Unusual behaviour of sub-circuits containing dependent sources Active subcircuits may be defined as any subcircuit containing one or more dependent sources. Circuits consisting of passive elements and independent voltage and current sources always have a solution, provided we avoid some obviously problematical situations, such as a loop of voltage sources or a cut-set of current sources. An example of a loop of voltages sources is as follows: 6

7 Circuit De-activated circuit A cut-set of current sources describes the situation where de-activating current sources divides the circuit into two separate parts: Circuit De-activated circuit If we avoid these situations, any circuit with any element values is soluble. In the case of active circuits containing dependent sources, this is not the case and it is very easy to generate circuits which do not have any solution; in practice, they are unstable; there is no unique set of finite nodal voltages and branch currents which satisfies Kirchhoff's laws. We now give some examples of such behaviour. Consider the simple subcircuit shown: The circuit consists of a CCVS whose controlling variable i x is equal to the terminal current i. Analysis of the circuit to find its v i relationship gives: v = i v i = where we have used the normal polarity conventions for v and i. We see that this subcircuit is equivalent to a resistor; however, the resistance value is negative. The power absorbed by the subcircuit is: which is negative if the current is nonzero. P = v i = ( i) i = i Hence, the energy (the time integral of power) is also negative and the negative resistor can deliver energy to the external circuit; the subcircuit is clearly an active one. Consider a slightly more complicated subcircuit: 7

8 We have just added an ordinary resistor element in series with the same CCVS with trans-resistance r m. We know from our preceding discussion that the CCVS is equivalent to a negative resistor whose value is r m. Thus the equivalent subcircuit is as shown above. Now, if r m is set equal to R, then the equivalent resistance is zero! This means that the subcircuit is equivalent to a short circuit. So, if this sub-circuit was connected across a voltage source, there would be a conflict of voltages and the circuit would not be soluble, even though this is not obvious from the circuit topology..3 Thevenin and Norton equivalents for circuits containing dependent sources We explore this topic by considering an example. Example 5.4: For the -terminal subcircuit shown: a. Find the Thevenin and Norton equivalents for general values of β. b. Evaluate the parameters of these equivalents for β = 1 and β =. Solution Note that this circuit contains a CCCS controlled by the current in the Ω resistor. We start by taping the dependent source: We have attached an independent current source to the terminals as a test source. 8

9 We have drawn an imaginary closed surface around the dependent source and its parallel resistor to help understand the derivation of the following KCL equation: i x = i s + i Furthermore, KCL applied to the node at the junction of the applied test source and the dependent source gives a current of i i c through the resistor inside the surface. Thus, by KVL around the loop consisting of the two resistors and the applied test source, we have v = ( i + i s ) + ( i i c ) We next un-tape the dependent source: i c = βi x = β ( i + i s ) Using this in the other equation gives: v = ( i + i s ) + ( i i c ) = ( i + i s ) + i β ( i + i s ) = ( 1 β)i s + ( β)i Since v has a constant part and a part proportional to i, we can now identify the Thevenin parameters, leading to the Thevenin equivalent circuit: The Norton equivalent obtained by rearranging the equation for v in terms of i: i = 1 β β i s + 1 ( β) v We identify the constant term as the short circuit current i sc, leading to the Norton equivalent circuit: If β = 1, we see that the Thevenin parameters are: v oc = ( 1 β)i s = 0 V and R eq = ( β) = Ω This is saying that the entire subcircuit is equivalent to a pure Ω resistance, even though there is an independent source. When β =, we see that: and v oc = ( 1 )i s = i s = 8 V R eq = ( β) = 0 Ω 9

10 Thus, we have two completely different Thevenin equivalent circuits for different β values: Example 5.6: Consider the circuit shown: β = 1 β = a) If β = 0.5, find the value of R that results in maximum power absorbed by R. b) If β = 1.5, find the value of R that results in maximum power absorbed by R. Solution The first step is to tape the dependent source; we also attach a test source in the place of the resistor R: We have shown the resistor currents obtained by applying KCL to the top and middle nodes. KVL now implies that: Un-taping the dependent source, we obtain: v = 4( i + i s ) + 4( i + i s + i c ) = 8i s + 4i c + 8i ( ) i c = 0.5i x = 0.5 i + i s + i c Solving this last equation for i c and using the resulting value in the previous equation gives: v = 4 + 1i Thus, we know that R eq = 1 Ω and we must therefore make R = 1 Ω for maximum power transfer. If we let β = 1.5 when we un-tape, the resulting equation (5.-37) becomes i c = 1.5i x = 1.5( i + i s + i c ) Solving for i c and using the resulting value above now gives: v = 8 4i The equivalent resistance is now negative and the original derivation of the maximum power transfer condition, which assumed positive resistor values, is no longer valid. 10

11 For example, if we make R = 4 Ω, then an infinite current will flow and the power transferred will be infinite! 3 NODAL ANALYSIS OF CIRCUITS WITH DEPENDENT SOURCES We have stated that dependent sources have exactly the same v-i characteristics as their independent cousins; the only difference is that there is an added algebraic constraint requiring that their values be proportional to some other voltage or current in the circuit. We will now use this observation to extend our techniques of nodal and mesh analysis to active circuits with the following step-by-step algorithm. 1) Tape all the dependent sources, thus temporarily treating them as independent sources. ) Perform nodal analysis in order to determine the nodal equations or write them by inspection. 3) Un-tape the dependent sources and express the value of each in terms of the node voltages. 4) Solve the resulting modified nodal equations for the node voltages. The procedure for nodal analysis for circuits with dependent sources are similar to those for circuits without them. Therefore we will illustrating the method by means of examples. In these examples, the dependent source parameter values are such that a unique solution exists. Example 5.7: Find the value of i in the circuit in the following circuit using nodal analysis: Solution Our first step is to tape the single dependent source and to prepare the circuit for nodal analysis by assigning a reference node and labelling the non-reference nodes with node voltage symbols: KCL can be applied at the two labelled essential nodes. Alternatively, the inspection method yields the matrix form of the nodal equations: v = 15 i c i c

12 Next, we un-tape the dependent source and express its value in terms of the unknown node voltages and v : Inserting this in the nodal equation gives: i c = i a = 4 = v v 15 1 = The v1 terms on the RHS of the equation really belong on the LHS and they can be simply transferred across: v = 15 0 Notice that it is only the elements in the first column that are multiplied by that are modified by this transfer. Note that the terms proportional to on the RHS disappear leaving only constants. The solution is: = 8 V, v = 64 V Referring to the circuit diagram, we have: i = v /8 = 8 A Example 5.9: Find the value of v in the following circuit using nodal analysis. Solution The circuit prepared for nodal analysis is as follows: Both dependent sources have been taped. We have identified a supernode (shown shaded) and an essential node (also shaded). The matrix equations may be written using the by-inspection method: 1

13 Un-taping the CCVS, we have: Un-taping the VCCS, we have: The un-taped nodal equations are now: v v c = 8i x = i c = 5 + v c = v c i c = 3 4 v y = 3 4 v ( + v c ) = 3 4 v v = v 7 Moving the terms involving the node voltages on the RHS to the LHS yields: This may be tidied up: v v = 9 7 The solution is = 0 V and v = 4 V. Hence, v = v = 4 V. = 9 7 For a passive circuit, we noticed that the nodal conductance matrix was always symmetrical. We can see form the examples in this section that the effect of introducing dependent sources into a circuit is to change the admittance matrix from a symmetrical one to an unsymmetrical one. If a conductance matrix is unsymmetrical, then it must be the conductance matrix for an active circuit containing dependent sources. 4 CONCLUSIONS We have introduced the dependent sources as a new type of element that has 4 terminals and is active, ie it can deliver power to other elements. We have studied active sub-circuits containing dependent sources and shown that they can realise negative resistance and that circuits using dependent sources may not always be soluble. Finally, we found that the systematic by-inspection method of nodal analysis can be extended in a straightforward way to analyse circuits with dependent sources 13