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Circuit Theorems

Circuit Theorems. Dr. Mustafa Kemal Uyguro ğlu. C ircuit Theorems Overview. Introduction Linearity Superpositions Source Transformation Th é v e nin and Norton Equivalents Maximum Power Transfer. INTRODUCTION. A large complex circuits. Simplify circuit analysis. Circuit Theorems.

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Circuit Theorems

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  1. Circuit Theorems Dr. Mustafa Kemal Uyguroğlu

  2. Circuit Theorems Overview • Introduction • Linearity • Superpositions • Source Transformation • Thévenin and Norton Equivalents • Maximum Power Transfer

  3. INTRODUCTION A large complex circuits Simplify circuit analysis Circuit Theorems ‧Thevenin’s theorem ‧ Norton theorem ‧Circuit linearity ‧ Superposition ‧source transformation ‧ max. power transfer

  4. Linearity Property A linear element or circuit satisfies the properties of • Additivity: requires that the response to a sum of inputs is the sum of the responses to each input applied separately. If v1 = i1R and v2 = i2R then applying (i1 + i2) v = (i1 + i2)R = i1R + i2R = v1 + v2

  5. Linearity Property • Homogeneity: If you multiply the input (i.e. current) by some constant K, then the output response (voltage) is scaled by the same constant. If v1 = i1R then K v1 =Ki1R

  6. i V0 v I0 Linearity Property • A linear circuit is one whose output is linearly related (or directly proportional) to its input. Suppose vs = 10 V gives i = 2 A. According to the linearity principle, vs = 5 V will give i = 1 A.

  7. i0 Linearity Property - Example Solve for v0 and i0 as a function of Vs

  8. Linearity Property – Example (continued)

  9. Linearity Property - Example Ladder Circuit 3 A 5 A 2 A + 6 V - + 3 V - 1 A 2 A + 5 V - + 8V - + 14 V - This shows that assuming I0 = 1 A givesIs = 5 A; the actual source current of 15 A will give I0 = 3 A as the actual value.

  10. Superposition • The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.

  11. Steps to apply superposition principle • Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis. • Turn off voltages sources = short voltage sources; make it equal to zero voltage • Turn off current sources = open current sources; make it equal to zero current • Repeat step 1 for each of the other independent sources. • Find the total contribution by adding algebraically all the contributions due to the independent sources. • Dependent sources are left intact.

  12. 12V 4mA 2kW – + 2mA 1kW 2kW I0 Superposition - Problem

  13. 2kW 2mA 1kW 2kW I’0 2mA Source Contribution I’0 = -4/3 mA

  14. 4mA 2kW 1kW 2kW I’’0 4mA Source Contribution I’’0 = 0

  15. 12V 2kW – + 1kW 2kW I’’’0 12V Source Contribution I’’’0 = -4 mA

  16. Final Result I’0 = -4/3 mA I’’0 = 0 I’’’0 = -4 mA I0 = I’0+ I’’0+ I’’’0 = -16/3 mA

  17. Example • find v using superposition

  18. one independent source at a time, dependent source remains KCL: i = i1 + i2 Ohm's law: i = v1 / 1 = v1 KVL: 5 = i (1 + 1) + i2(2) KVL: 5 = i(1 + 1) + i1(2) + 2v1 10 = i(4) + (i1+i2)(2) + 2v1 10 = v1(4) + v1(2) + 2v1 v1 = 10/8 V

  19. Consider the other independent source KCL: i = i1 + i2 KVL: i(1 + 1) + i2(2) + 5 = 0i2(2) + 5 = i1(2) + 2v2Ohm's law: i(1) = v2v2(2) + i2(2) +5 = 0 => i2 = -(5+2v2)/2i2(2) + 5 = i1(2) + 2v2-2v2 = (i - i2)(2) + 2v2-2v2 = [v2 + (5+2v2)/2](2) + 2v2-4v2 = 2v2 + 5 +2v2-8v2 = 5 => v2 = - 5/8 V from superposition: v = -5/8 + 10/8 v = 5/8 V

  20. Source Transformation • A source transformation is the process of replacing a voltage source vsin series with a resistor R by a current source is in parallel with a resistor R, or vice versa

  21. Source Transformation

  22. Source Transformation

  23. Source Transformation • Equivalent sources can be used to simplify the analysis of some circuits. • A voltage source in series with a resistor is transformed into a current source in parallel with a resistor. • A current source in parallel with a resistor is transformed into a voltage source in series with a resistor.

  24. Example 4.6 • Use source transformation to find vo in the circuit in Fig 4.17.

  25. Example 4.6 Fig 4.18

  26. Example 4.6 we use current division in Fig.4.18(c) to get and

  27. Example 4.7 • Find vxin Fig.4.20 using source transformation

  28. Example 4.7 • Applying KVL around the loop in Fig 4.21(b) gives (4.7.1) • Appling KVL to the loop containing only the 3V voltage source, the resistor, and vx yields (4.7.2)

  29. Example 4.7 Substituting this into Eq.(4.7.1), we obtain Alternatively thus

  30. 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.

  31. 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.

  32. RTh + – Voc Circuit with independent sources Thevenin equivalent circuit Independent Sources (Thevenin)

  33. RTh Circuit without independent sources Thevenin equivalent circuit No Independent Sources

  34. Introduction • Any Thevenin equivalent circuit is in turn equivalent to a current source in parallel with a resistor [source transformation]. • A current source in parallel with a resistor is called a Norton equivalent circuit. • Finding a Norton equivalent circuit requires essentially the same process as finding a Thevenin equivalent circuit.

  35. Computing Thevenin Equivalent • Basic steps to determining Thevenin equivalent are • Find voc • Find RTh

  36. 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, VOC. Norton: Compute the short circuit current, ISC. For case 3(b) both VOC=0 and ISC=0 [so skip step 2]

  37. Thevenin/Norton Analysis 3. Compute the Thevenin equivalent resistance, RTh (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 RTh = VTest/Itest (c) If there are both independent and dependent sources, then compute RTh from VOC/ISC.

  38. Thevenin/Norton Analysis 4. Thevenin: Replace circuit with VOC in series with RTh Norton: Replace circuit with ISC in parallel with RTh Note: for 3(b) the equivalent network is merely RTh , that is, no voltage (or current) source. Only steps 2 & 4 differ from Thevenin & Norton!

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