 # Lecture 291 More Loop and Nodal Analysis. Lecture 292 Advantages of Nodal Analysis Solves directly for node voltages. Current sources are easy. Voltage.

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Lecture 291 More Loop and Nodal Analysis

Lecture 292 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.

Lecture 293 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.

Lecture 294 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.

Lecture 295 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 both nodal and mesh analysis. This circuit is a bandpass filter with center frequency 455kHz and bandwidth 40kHz.

Lecture 296 IF Amplifier + - 4k  + - 1V  0  + - V out 100pF 160  100pF 80k  - + VxVx 100V x

Lecture 297 Analysis Use AC steady state analysis. Start with a frequency of  =2  455,000. Nodal or mesh analysis?

Lecture 298 Impedances + - 4k  + - 1V  0  + - V out 160  80k  - + VxVx 100V x -j3.5k 

Lecture 299 Nodal Analysis + - 4k  + - 1V  0  + - V out 160  80k  - + VxVx 100V x -j3.5k  12

Lecture 2910 KCL @ Node 1

Lecture 2911 KCL @ Node 2

Lecture 2912 Matrix Formulation

Lecture 2913 Solve Equations V 1 = 0.0259V-j0.122V = 0.1247V  -78  V 2 = 0.0277V-j4.15  10 -4 V=0.0277V  -0.86  V out = -100V 2 = 2.7V  179.2 

Lecture 2914 Analysis Use AC steady state analysis. Use a frequency of  =2  100,000.

Lecture 2915 Impedances + - 4k  + - 1V  0  + - V out 160  80k  - + VxVx 100V x -j15.9k 

Lecture 2916 Mesh Analysis + - 4k  + - 1V  0  + - V out 160  80k  - + VxVx 100V x -j15.9k  I1I1 I2I2 I3I3

Lecture 2917 KVL Around Loop 1

Lecture 2918 KVL Around Loop 2

Lecture 2919 KVL Around Loop 2 (Cont)

Lecture 2920 KVL Around Loop 3

Lecture 2921 Solve Equations I 1 = 239.9  A-j0.23  A I 2 = -12.36  A-j5.98  A I 3 = -12.54  A-j3.45  A

Lecture 2922 Solve for V out

Lecture 2923 |V out | as a function of 

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