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Lecture 111 Thévenin's Theorem (4.3) Prof. Phillips February 24, 2003.

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Presentation on theme: "Lecture 111 Thévenin's Theorem (4.3) Prof. Phillips February 24, 2003."— Presentation transcript:

1 lecture 111 Thévenin's Theorem (4.3) Prof. Phillips February 24, 2003

2 lecture 112 Thevenin’s Theorem Any circuit with sources (dependent and/or independent) and resistors can be replaced by an equivalent circuit containing a single voltage source and a single resistor. Thevenin’s theorem implies that we can replace arbitrarily complicated networks with simple networks for purposes of analysis.

3 lecture 113 Implications We use Thevenin’s theorem to justify the concept of input and output resistance for amplifier circuits. We model transducers as equivalent sources and resistances. We model stereo speakers as an equivalent resistance.

4 lecture 114 Independent Sources (Thevenin) Circuit with independent sources R Th V oc Thevenin equivalent circuit +–+–

5 lecture 115 No Independent Sources Circuit without independent sources R Th Thevenin equivalent circuit

6 lecture 116 Example: CE Amplifier 1k  V in 2k  +10V + – VoVo +–+–

7 lecture 117 Small Signal Equivalent 1k  V in 100I b + – VoVo 50  IbIb 2k  +–+–

8 lecture 118 Thevenin Equivalent @ Output 1k  V in 100I b + - VoVo 50  IbIb 2k  R Th V oc + – VoVo +–+– +–+–

9 lecture 119 Computing Thevenin Equivalent Basic steps to determining Thevenin equivalent are –Find v oc –Find R Th (= v oc / i sc )

10 lecture 1110 Thevenin/Norton Analysis 1. Pick a good breaking point in the circuit (cannot split a dependent source and its control variable). 2. Thevenin: Compute the open circuit voltage, V OC. Norton: Compute the short circuit current, I SC. For case 3(b) both V OC =0 and I SC =0 [so skip step 2]

11 lecture 1111 Thevenin/Norton Analysis 3. Compute the Thevenin equivalent resistance, R Th. (a) If there are only independent sources, then short circuit all the voltage sources and open circuit the current sources (just like superposition). (b) If there are only dependent sources, then must use a test voltage or current source in order to calculate R Th = V Test /I test (c) If there are both independent and dependent sources, then compute R Th from V OC /I SC.

12 lecture 1112 Thevenin/Norton Analysis 4. Thevenin: Replace circuit with V OC in series with R Th. Norton: Replace circuit with I SC in parallel with R Th. Note: for 3(b) the equivalent network is merely R Th, that is, no voltage (or current) source. Only steps 2 & 4 differ from Thevenin & Norton!

13 lecture 1113 Class Examples

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