advertisement

THEVENIN’S AND NORTON’S THEOREMS These are some of the most powerful analysis results to be discussed. They permit to hide information that is not relevant and concentrate in what is important to the analysis THEVENIN’S EQUIVALENCE THEOREM LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A RTH vTH i a vO b _ i a LINEAR CIRCUIT vO _ PART A Thevenin Equivalent Circuit for PART A vTH Thevenin Equivalent Source RTH Thevenin Equivalent Resistance LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART B b PART B NORTON’S EQUIVALENCE THEOREM LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A RN i Norton Equivalent Circuit for PART A Thevenin Equivalent Source Thevenin Equivalent Resistance LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART B a LINEAR CIRCUIT vO PART A RN b _ _ iN a vO iN i b PART B ANOTHER VIEW OF THEVENIN’S AND NORTON’S THEOREMS i a RTH vOC + _ + i vO _ i SC RTH Norton Thevenin i SC vO b vOC RTH This equivalence can be viewed as a source transformation problem It shows how to convert a voltage source in series with a resistor into an equivalent current source in parallel with the resistor A General Procedure to Determine the Thevenin Equivalent Open Circuit voltage vTH voltage at a - b if Part B is removed i SC Short Circuit Current current through a - b if Part B is replaced by a short circuit RTH vTH i SC Thevenin Equivalent Resistance 1. Determine the Thevenin equivalent source Remove part B and compute the OPEN CIRCUIT voltage Vab One circuit problem i 0 a LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A vOC Vab _ b _ Second circuit problem 2. Determine the SHORT CIRCUIT current Remove part B and compute the SHORT CIRCUIT current I ab vTH vOC , RTH vOC i SC LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A i SC a v0 I ab _ b LEARNING EXAMPLE USE THEVENIN TO COMPUTE Vo You have the choice on the way to partition the circuit. Make “Part A” as simple as possible Since there are only independent sources, for the Thevenin resistance we set to zero all sources and determine the equivalent resistance RTH 2 (2 || 4) “Part B” 10 k 3 For the open circuit voltage we analyze the following circuit (“Part A”) ... Loop Analysis I 2 2mA 6V 4kI1 2k ( I1 I 2 ) 0 VOC 6 2I2 5 I1 mA mA 6 3 4k * I1 2k * I 2 20 / 3 4V 32 / 3[V ] The circuit becomes... LEARNING EXAMPLE: COMPUTE Vo USING NORTON 4k I RN IN I SC RN RTH 3k PART B 12V I SC I N 2mA 2mA 3k COMPUTE Vo USING THEVENIN 2k RN VO 2kI 2k I N RN 6k 3 4 VO 2 (2) [V ] 9 3 PART B VTH RTH + - VTH VTH 12 2mA 0 3k RTH 3k 4k VO 2k VO 2 4 (6V ) [V ] 27 3 LEARNING EXAMPLE Find the Thevenin Equivalent circuit at A - B Only dependent sources. Hence V_th = 0 To compute the equivalent resistance we must apply an external probe We choose to apply a current probe R VP TH VP IP “Conventional” circuit with dependent sources - use node analysis IP @V_1 @V_2 (IP ) Controlling variable 3(V1 2V1 ) 6V1 2(V1 V2 ) 0 2(V2 V1 ) 3V2 6[V ] RTH A B Thevenin equivalent 5V1 2V2 0 * / 2 2V1 5V2 6 * / 5 30 10 V2 21 7 (VP V2 ) ( I P 1mA) RTH V2 (10 / 7)k 1mA