Nodal and Loop Analysis contd 8 8 Dr

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Nodal and Loop Analysis cont’d (8. 8) Dr. Holbert March 1, 2006 ECE 201

Nodal and Loop Analysis cont’d (8. 8) Dr. Holbert March 1, 2006 ECE 201 Lect-11 1

Advantages of Nodal Analysis • Solves directly for node voltages. • Current sources are

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. ECE 201 Lect-11 2

Advantages of Loop Analysis • Solves directly for some currents. • Voltage sources are

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. ECE 201 Lect-11 3

Disadvantages of Loop Analysis • Some currents must be computed from loop currents. •

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 ECE 201 Lect-11 4

Where We Are • Nodal analysis is a technique that allows us to analyze

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. ECE 201 Lect-11 5

Example Transistor Circuit +10 V Vin + – 1 k. W 2 k. W

Example Transistor Circuit +10 V Vin + – 1 k. W 2 k. W + Vo – ECE 201 Lect-11 Common Collector (Emitter Follower) Amplifier 6

Why an Emitter Follower Amplifier? • The output voltage is almost the same as

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 180 k. W resistor. • To a circuit connected to the output, the EF amplifier looks like a voltage source connected to a 10 W resistor. ECE 201 Lect-11 7

A Linear Large Signal Equivalent 0. 7 V Ib 5 V + – 1

A Linear Large Signal Equivalent 0. 7 V Ib 5 V + – 1 k. W + – 50 W 100 Ib 2 k. W + Vo – ECE 201 Lect-11 8

Steps of Nodal Analysis 1. Choose a reference node. 2. Assign node voltages to

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. ECE 201 Lect-11 9

A Linear Large Signal Equivalent 0. 7 V 1 5 V Ib V 2

A Linear Large Signal Equivalent 0. 7 V 1 5 V Ib V 2 V 1 + – 1 k. W 2 + – V 3 V 4 3 50 W 100 Ib 4 ECE 201 Lect-11 + Vo 2 k. W – 10

Steps of Nodal Analysis 1. Choose a reference node. 2. Assign node voltages to

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. ECE 201 Lect-11 11

KCL @ Node 4 0. 7 V 1 5 V Ib V 2 V

KCL @ Node 4 0. 7 V 1 5 V Ib V 2 V 1 + – 1 k. W 2 + – V 3 V 4 3 50 W 100 Ib 4 ECE 201 Lect-11 + Vo 2 k. W – 12

The Dependent Source • We must express Ib in terms of the node voltages:

The Dependent Source • We must express Ib in terms of the node voltages: • Equation from Node 4 becomes ECE 201 Lect-11 13

How to Proceed? • The 0. 7 V voltage supply makes it impossible to

How to Proceed? • The 0. 7 V 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 V 2 - V 3 = 0. 7 V ECE 201 Lect-11 14

0. 7 V 1 Ib V 2 V 1 + – 1 k. W

0. 7 V 1 Ib V 2 V 1 + – 1 k. W + – V 3 50 W 100 Ib ECE 201 Lect-11 V 4 4 + Vo 2 k. W – 15

KCL @ the Supernode ECE 201 Lect-11 16

KCL @ the Supernode ECE 201 Lect-11 16

Another Analysis Example • We will analyze a possible implementation of an AM Radio

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 455 k. Hz and bandwidth 40 k. Hz. ECE 201 Lect-11 17

IF Amplifier 100 p. F 4 k. W 100 p. F 80 k. W

IF Amplifier 100 p. F 4 k. W 100 p. F 80 k. W – 1 V 0 + – 160 W Vx + ECE 201 Lect-11 + 100 Vx + – Vout – 18

Nodal AC Analysis • Use AC steady-state analysis. • Start with a frequency of

Nodal AC Analysis • Use AC steady-state analysis. • Start with a frequency of w=2 p 455, 000. ECE 201 Lect-11 19

Impedances -j 3. 5 k. W 4 k. W -j 3. 5 k. W

Impedances -j 3. 5 k. W 4 k. W -j 3. 5 k. W 80 k. W – 1 V 0 + – 160 W Vx + ECE 201 Lect-11 + 100 Vx + – Vout – 20

Nodal Analysis -j 3. 5 k. W 4 k. W 1 -j 3. 5

Nodal Analysis -j 3. 5 k. W 4 k. W 1 -j 3. 5 k. W 2 80 k. W – 1 V 0 + – 160 W Vx + ECE 201 Lect-11 + 100 Vx + – Vout – 21

KCL @ Node 1 ECE 201 Lect-11 22

KCL @ Node 1 ECE 201 Lect-11 22

KCL @ Node 2 ECE 201 Lect-11 23

KCL @ Node 2 ECE 201 Lect-11 23

Matrix Formulation ECE 201 Lect-11 24

Matrix Formulation ECE 201 Lect-11 24

Solve Equations V 1 = 0. 0259 V-j 0. 1228 V = 0. 1255

Solve Equations V 1 = 0. 0259 V-j 0. 1228 V = 0. 1255 V -78 V 2 = 0. 0277 V-j 4. 15 10 -4 V=0. 0277 V -0. 86 Vout = -100 V 2 = 2. 77 V 179. 1 ECE 201 Lect-11 25

Class Examples • Learning Extension E 3. 6 • Learning Extension E 8. 13

Class Examples • Learning Extension E 3. 6 • Learning Extension E 8. 13 • Learning Extension E 8. 14(a) ECE 201 Lect-11 26

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