48520 Electronics & Circuits: Web Tutor

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1 852 Electronics & Circuits: Web Tutor Topic : Resistive Circuits 2 Help for Exercise.: Nodal Analysis, circuits with I, R and controlled sources. The purpose of this exercise is to further extend Nodal Analysis, introduced in Problem 2., to circuits containing all four types of controlled sources (VCCS, VCVS, CCCS and CCVS). In this exercise, you are presented with a randomly generated circuit containing resistors (R), independent current source (I) and one of the four controlled sources (please note that the symbols used on our diagrams are different than the ones in your textbook; the symbols used here conform to the Australian Standard AS 2 and the International Electrotechnical Commission Standards IEC 667). Let us start with a circuit containing a voltage-controlled current source (VCCS). An example of such circuit is shown in Fig.... Component values: R 27Ω R 2 5Ω R 7Ω R 6Ω G m 6S I 7A In further calculations we'll need the conductances: G G R 2 G R 2 R Fig...: A sample resistive circuit with a voltage-controlled current source. G R In order to formulate the node voltage equations for a circuit we must first choose one node as a reference node (ground, datum). Potential at this node is considered to be zero and all other potentials are calculated with reference to this one. Customarily, the reference node is numbered as. It is important to note that the choice of the reference node is absolutely arbitrary; it may be sometimes more convenient to choose one particular node instead of the other, but the results of analysis - voltages across circuit components - will be identical, regardless of our choice. Once the ground node is designated, we number all remaining nodes with consecutive integers, starting from. This has already been done on the diagram in Fig.... Now, we apply the Kirchhoff Current Law (KCL) to the currents leaving each node (i.e. we assume that the current leaving a node is positive). For the circuit in Fig... we can write: WebTutor Help: Exercise. Page 852 Electronics & Circuits

2 KCL at node : KCL at node 2: v R G m V c I G m V c v R v = v R 2 v = where V c = v 2 v KCL at node : v R 2 v R v 2 v R v = Substituting for V c, sorting on variables and rearranging the KCL equations, we get: R R G m R v R v G m v 2 G R m v = v 2 G m R v = I v R 2 = R 2 R R v Rewriting in the matrix form (and using conductances, G i = /R i ), we obtain a matrix equation G*v=i, where: G G G G G m G m G G G m G G m G G 2 G G G Ω i I i 7 A Numerical solution: v G i v V WebTutor Help: Exercise. Page Electronics & Circuits

3 As the second example we'll consider a circuit with a voltage-controlled voltage source (VCVS), shown in Fig...2. Circuit components: R R 2 26 R R 7 I E m 2 G G R 2 G R 2 R G R Fig...2: A sample resistive circuit with a voltage-controlled voltage source. Here, the KCL can only be applied directly to nodes and, providing two equations with three unknowns. The third equation arises from the voltage source constraint. KCL at node : v R v R v 2 I = Voltage source constraint: v 2 = E m V c = E m v v KCL at node : v R 2 v R v 2 I = Sorting on variables and rearranging equations, we get: R R v v R 2 = I E m v v 2 E m v = v R 2 = I R 2 R v Rewriting in the matrix form (and using conductances, G i = /R i ), we obtain a matrix equation G*v=i, where: G G G E m G G E m G 2 G i I I Numerical solution: v G i v WebTutor Help: Exercise. Page 852 Electronics & Circuits

4 Next, we'll consider a circuit with a VCVS that is not grounded, as shown in Fig.... Circuit elements: R 5 R 2 R 22 R I 6 E m 8 G G R 2 G R 2 R Fig...: A resistive circuit with a floating voltage-controlled voltage source. G R In this circuit, the KCL can be directly written for nodes:, and a supernode around the floating VCVS. We may use any two of these three equations; we choose the KCL equations at node and the supernode. The third equation describes the voltage source constraint. KCL at node : Voltage source constraint: v v 2 v R v R v 2 v R v = = E m V C = E m v 2 v KCL at supernode: v R 2 v v R v v R I = 2 Sorting on variables and rearranging equations, we get: R R R v v R 2 v R = v 2 E m v E m v = R R v v R 2 = I R 2 R v Rewriting in the matrix form (and using conductances, G i = /R i ), we obtain a matrix equation G*v=i, where: G G G G E m G G G E m G G G 2 G i I Numerical solution: v G i v WebTutor Help: Exercise. Page 852 Electronics & Circuits

5 Another type of dependent source is the current-controlled current source (CCCS). A sample circuit with a CCCS is shown in Fig.... Circuit components: R 2 R 2 29 R 8 F m I 8 G G R 2 G R 2 R Fig...: A resistive circuit with a current-controlled current source. Writing the KCL equations at each node, we obtain: KCL at node : G v G v v 2 KCL at node 2: G 2 v 2 G v 2 v KCL at node : I c F m I c I = I c = F m I c = The short-circuit between nodes and makes potentials v and v identical, so we can write the fourth equation: Short-circuit constraint: v v = We have now four equations and four unknowns ( v, v 2, v and I c ). We may, of course, directly solve this set of four equations, but we choose to eliminate I c. Using the third equation we can write: I I c F m = I --> I c I F c.889 m Now, substituting the expression for I c into the first two equations, sorting on variables, rearranging and adding the short-circuit constraint, we obtain a set of three node voltage equations: KCL at node : G G v G v 2 = I F m v 2 KCL at node 2: G v G 2 G = F m I F m Short-circuit constraint: v v = WebTutor Help: Exercise. Page Electronics & Circuits

6 Solving this system of equations, we get: v G G G G G 2 G I F m F m I F m v The control variable can be also eliminated using different technique. Consider again the original KCL equations: KCL at node : G v G v v 2 KCL at node 2: G 2 v 2 G v 2 v KCL at node : I c F m I c I = I c = F m I c = Adding equations and 2 to the third equation yields: G v G 2 v 2 = I (you should notice that this is nothing else, but the KCL at node ) Now, multiplying the first equation by F m and adding it to the second equation we eliminate I c from the second equation: F m G v F m G v v 2 G 2 v 2 G v 2 v = Collecting on variables, we get: F m G F m G G v G 2 G F m G v 2 = The short-circuit constraint, v v =, completes the set of three equations. Solving the set we obtain the node voltages: v G F m G F m G G G 2 G 2 G F m G I v WebTutor Help: Exercise. Page Electronics & Circuits

7 In the final example we'll analyse a circuit with the current-controlled voltage source (CCVS). The sample circuit is shown in Fig...5. Circuit components: H m 8 I 5 R R 2 2 R 2 G G R 2 G R 2 R Fig...5: A resistive circuit with a current-controlled voltage source. KCL at node 2: G 2 v 2 G v 2 v I = KCL at node : G v G 2 v 2 I c = ==> I c = G v G 2 v 2 Voltage source constraint: v v = H m I c Short-circuit constraint: v = Substituting the expression for I c to the voltage source constraint gives: v v = H m G v G 2 v 2 or H m G v H m G 2 v 2 v = The final set of three equations is: H m G G 2 G H m G 2 G v v 2 v = I Solving this set we obtain the node equations: v H m G G 2 G H m G 2 G I v WebTutor Help: Exercise. Page Electronics & Circuits