Node Analysis

Nodal and Loop Analysis cont’d

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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. AIHT

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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.

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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 AIHT

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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. AIHT

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Example Transistor Circuit +10V

Vin

+ –

1kΩ + Vo –

2kΩ

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Common Collector (Emitter Follower) Amplifier

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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 180kΩ resistor. • To a circuit connected to the output, the EF amplifier looks like a voltage source connected to a 10Ω resistor. AIHT

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A Linear Large Signal Equivalent 0.7V Ib 5V

+ –

1kΩ

+ – 50Ω 100Ib

2kΩ

+ Vo –

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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. AIHT

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A Linear Large Signal Equivalent 0.7V 1 5V

Ib V 2

V1 + –

1kΩ

2

+ –

V3

V4

3 50Ω 100Ib

4

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+ Vo

2kΩ

–

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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. AIHT

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KCL @ Node 4 0.7V 1 5V

Ib V2

V1 + –

1kΩ

2

+ –

V3

V4

3 50Ω 100Ib

4

+ Vo

2kΩ

–

V3 − V4 V4 + 100 I b = 50Ω 2 kΩ AIHT

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The Dependent Source • We must express Ib in terms of the node voltages: V1 − V2 Ib = 1 kΩ

• Equation from Node 4 becomes

V3 − V4 V1 − V2 V4 + 100 − =0 50Ω 1 kΩ 2kΩ AIHT

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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

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0.7V 1

Ib V2

V1 + –

1kΩ

+ –

V3

V4 50Ω 100Ib

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+ Vo

2kΩ

–

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KCL @ the Supernode V2 − V1 V3 − V4 + =0 1kΩ 50Ω

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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. AIHT

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IF Amplifier 100pF 4kΩ

100pF

80kΩ

– 1V ∠ 0°

+ –

160Ω

Vx +

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+ 100Vx

+ –

Vout –

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Nodal AC Analysis • Use AC steady-state analysis. • Start with a frequency of ω=2π 455,000.

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Impedances -j3.5kΩ 4kΩ

-j3.5kΩ

80kΩ

– 1V ∠ 0°

+ –

160Ω

Vx +

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+ 100Vx

+ –

Vout –

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Nodal Analysis -j3.5kΩ 4kΩ 1 1V ∠ 0°

+ –

160Ω

-j3.5kΩ 2 80kΩ – Vx +

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100Vx

+ + –

Vout –

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KCL @ Node 1 V1 − 100Vx V1 − V2 V1 − 1V V1 + + + =0 4kΩ 160Ω - j 3.5kΩ - j 3.5kΩ Vx = −V2  1  1 1 1  V1  + + +  4kΩ 160Ω - j 3.5kΩ - j 3.5kΩ   100 1  1V  = + V2  +  - j 3.5kΩ j 3.5kΩ  4kΩ AIHT

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KCL @ Node 2 V2 − V1 V2 − 100Vx + =0 - j 3.5kΩ 80kΩ Vx = −V2  1   1 101   + V2   = 0 V1  +  j 3.5kΩ   - j 3.5kΩ 80kΩ  AIHT

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Matrix Formulation 1 2  1  4kΩ + 160Ω − j 3.5kΩ  1   j 3.5kΩ

− 100 1  +  1V   V   j 3.5kΩ j 3.5kΩ 1    =  4kΩ  −1 101  V2    0 +   j 3.5kΩ 80kΩ 

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Solve Equations V1 = 0.0259V-j0.1228V = 0.1255V∠-78° V2 = 0.0277V-j4.15×10-4V=0.0277V ∠ -0.86° Vout = -100V2 = 2.77V ∠ 179.1°

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