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Nodal and Loop Analysis cont’d (8.8) Dr. Holbert March 1, 2006 ECE201 Lect-11 1 Advantages of Nodal Analysis • Solves directly for node voltages. • Current sources are easy. • Voltage sources are either very easy or somewhat difficult. • Works best for circuits with few nodes. • Works for any circuit. ECE201 Lect-11 2 Advantages of Loop Analysis • Solves directly for some currents. • Voltage sources are easy. • Current sources are either very easy or somewhat difficult. • Works best for circuits with few loops. ECE201 Lect-11 3 Disadvantages of Loop Analysis • Some currents must be computed from loop currents. • Does not work with non-planar circuits. • Choosing the supermesh may be difficult. • FYI: PSpice uses a nodal analysis approach ECE201 Lect-11 4 Where We Are • Nodal analysis is a technique that allows us to analyze more complicated circuits than those in Chapter 2. • We have developed nodal analysis for circuits with independent current sources. • We now look at circuits with dependent sources and with voltage sources. ECE201 Lect-11 5 Example Transistor Circuit +10V Vin + – 1kW + 2kW Vo – ECE201 Lect-11 Common Collector (Emitter Follower) Amplifier 6 Why an Emitter Follower Amplifier? • The output voltage is almost the same as the input voltage (for small signals, at least). • To a circuit connected to the input, the EF amplifier looks like a 180kW resistor. • To a circuit connected to the output, the EF amplifier looks like a voltage source connected to a 10W resistor. ECE201 Lect-11 7 A Linear Large Signal Equivalent 0.7V Ib 5V + – 1kW + – 50W 100Ib 2kW + Vo – ECE201 Lect-11 8 Steps of Nodal Analysis 1. Choose a reference node. 2. Assign node voltages to the other nodes. 3. Apply KCL to each node other than the reference node; express currents in terms of node voltages. 4. Solve the resulting system of linear equations. ECE201 Lect-11 9 A Linear Large Signal Equivalent 0.7V 1 5V Ib V2 V1 + – 1kW 2 + – V3 V4 3 50W 4 Vo 100Ib 2kW ECE201 Lect-11 + – 10 Steps of Nodal Analysis 1. Choose a reference node. 2. Assign node voltages to the other nodes. 3. Apply KCL to each node other than the reference node; express currents in terms of node voltages. 4. Solve the resulting system of linear equations. ECE201 Lect-11 11 KCL @ Node 4 0.7V 1 5V Ib V2 V1 + – 1kW 2 + – V3 V4 3 50W 100Ib 4 + Vo 2kW – V3 V4 V4 100 I b 50W 2 kW ECE201 Lect-11 12 The Dependent Source • We must express Ib in terms of the node voltages: V1 V2 Ib 1 kW • Equation from Node 4 becomes V3 V4 V1 V2 V4 100 0 50W 1 kW 2kW ECE201 Lect-11 13 How to Proceed? • The 0.7V voltage supply makes it impossible to apply KCL to nodes 2 and 3, since we don’t know what current is passing through the supply. • We do know that V2 - V3 = 0.7V ECE201 Lect-11 14 0.7V 1 Ib V2 V1 + – 1kW + – V3 50W 100Ib V4 4 Vo 2kW ECE201 Lect-11 + – 15 KCL @ the Supernode V2 V1 V3 V4 0 1kW 50W ECE201 Lect-11 16 Another Analysis Example • We will analyze a possible implementation of an AM Radio IF amplifier. (Actually, this would be one of four stages in the IF amplifier.) • We will solve for output voltages using nodal (and eventually) mesh analysis. • This circuit is a bandpass filter with center frequency 455kHz and bandwidth 40kHz. ECE201 Lect-11 17 IF Amplifier 100pF 4kW 100pF 80kW – 1V 0 + – 160W Vx + ECE201 Lect-11 + 100Vx + – Vout – 18 Nodal AC Analysis • Use AC steady-state analysis. • Start with a frequency of w=2p 455,000. ECE201 Lect-11 19 Impedances -j3.5kW 4kW -j3.5kW 80kW – 1V 0 + – 160W Vx + ECE201 Lect-11 + 100Vx + – Vout – 20 Nodal Analysis -j3.5kW 4kW 1 1V 0 + – 160W -j3.5kW 2 80kW – Vx + ECE201 Lect-11 100Vx + + – Vout – 21 KCL @ Node 1 V1 100Vx V1 V2 V1 1V V1 0 4kW 160W - j3.5kW - j3.5kW Vx V2 1 1 1 1 V1 4kW 160W - j 3.5kW - j3.5kW 100 1 1V V2 - j 3.5kW j 3.5kW 4kW ECE201 Lect-11 22 KCL @ Node 2 V2 V1 V2 100Vx 0 - j3.5kW 80kW Vx V2 1 1 101 V2 0 V1 j 3.5kW - j 3.5kW 80kW ECE201 Lect-11 23 Matrix Formulation 1 2 1 4kW 160W j 3.5kW 1 j 3.5kW 100 1 1V V j 3.5kW j 3.5kW 1 4kW 1 101 V2 0 j 3.5kW 80kW ECE201 Lect-11 24 Solve Equations V1 = 0.0259V-j0.1228V = 0.1255V-78 V2 = 0.0277V-j4.1510-4V=0.0277V -0.86 Vout = -100V2 = 2.77V 179.1 ECE201 Lect-11 25 Class Examples • Learning Extension E3.6 • Learning Extension E8.13 • Learning Extension E8.14(a) ECE201 Lect-11 26