107_26 - 1 Electrical Engineering Technology EET 107...

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Unformatted text preview: 1 Electrical Engineering Technology EET 107 Introduction to Circuit Analysis # 26 Professor Robert Herrick Purdue University © EET 107 - 26 Introduction to Circuit Analysis 2 Electrical Engineering Technology Bridge Circuit & Thévenin’s Theorem Purdue University © EET 107 - 26 Introduction to Circuit Analysis 3 Overview Ø Thevenin Model - reviewed Ø Thevenin Analysis - bridge circuit Ø What’s in the Box Modeling Ø Balanced Bridge Purdue University © EET 107 - 26 Introduction to Circuit Analysis 4 Thevenin Model - loaded IL Electronic Circuit + VL - Load Same IL and V L IL Thevenin Model Purdue University © + VL EET 107 - 26 Load Introduction to Circuit Analysis 5 Thevenin Model RTH Electronic Circuit ETH Original circuit + - Thevenin Model ETH = original circuit’s open circuit voltage VOC RTH = original circuit’s output resistance with sources 0’d Purdue University © EET 107 - 26 Introduction to Circuit Analysis 6 Thevenin Model of Circuit Electronic Circuit Load Analysis 1. Remove the load Purdue University © EET 107 - 26 Introduction to Circuit Analysis 7 Thevenin Model of Circuit + ETH - Electronic Circuit Analysis 1. Remove the load 2. Find ETH = VOC Purdue University © EET 107 - 26 Introduction to Circuit Analysis 8 Thevenin Model of Circuit Electronic Circuit ISC Analysis 1. Remove the load 2. Find E TH = VOC 3. Find ISC = IN Purdue University © EET 107 - 26 Introduction to Circuit Analysis 9 Thevenin Model of Circuit RTH ETH Analysis + - 1. Remove the load 2. Find E TH = VOC 3. Find ISC = IN 4. RTH = ETH / IN Purdue University © EET 107 - 26 Introduction to Circuit Analysis 10 Thevenin Model of Circuit Electronic Circuit Without load. RTH ETH Attach load. + - Purdue University © Find VL & IL EET 107 - 26 Introduction to Circuit Analysis 11 Use Thevenin Model to Find IL or VL RTH ETH IL + VL - + - Load 5. Draw the Thevenin Model & attach load 6. Find the load voltage or load current Purdue University © EET 107 - 26 Introduction to Circuit Analysis 12 Overview Ø Thevenin Model - reviewed Ø Thevenin Analysis - Bridge Circuit Ø What’s in the Box Modeling Ø Balanced Bridge Purdue University © EET 107 - 26 Introduction to Circuit Analysis 13 Thevenin Analysis - bridge circuit R1 + VL IL 3k E 36V X R2 6k R3 6k Z RL 2k R4 3k R1 – R2 – R3 – R4 form a BRIDGE circuit Find IL and V L using the Thevenin model approach. Our other techniques do not work! Purdue University © EET 107 - 26 Introduction to Circuit Analysis 14 Thevenin Analysis - bridge circuit E 36V R1 3k X OPEN Z R3 6k R4 3k R2 6k Remove the LOAD Purdue University © EET 107 - 26 Introduction to Circuit Analysis ETH Analysis - bridge circuit E 36V R1 3k X R2 6k + VOC V Z 15 R3 6k R4 3k VX = VR2 = (6 kΩ / 9 kΩ) × 36V = 24 V VZ = VR4 = (3 kΩ / 9kΩ) × 36 V = 12V Purdue University © EET 107 - 26 Introduction to Circuit Analysis ETH Analysis - bridge circuit E 36V R1 3k X 24 V 12 V Z R2 + E TH 6k 16 R3 6k R4 3k Node voltage difference: ETH = V OC = V XZ = 24 V − 12 V = 12 V Purdue University © EET 107 - 26 Introduction to Circuit Analysis ETH Analysis - bridge circuit E 36V R1 3k + 24 V - X + ETH - + 12V - R2 6k Z 17 R3 6k R4 3k Or walkabout starting at “− ” of ETH and walking to the “ +” ETH = −12V + 24V = 12V Purdue University © EET 107 - 26 Introduction to Circuit Analysis ETH Analysis - bridge circuit E 36V R1 3k X + 12 V − Z 18 R3 6k R4 3k R2 6k ETH = 12 V Purdue University © EET 107 - 26 Introduction to Circuit Analysis ISC Analysis - bridge circuit E 36V R1 3k X R2 6k SHORT Z 19 R3 6k R4 3k ISC Remove the LOAD & replace it with a short. Find the short circuit current ISC. Purdue University © EET 107 - 26 Introduction to Circuit Analysis ISC Analysis - bridge circuit R1 3k 2k X R2 6k Z 2k 20 R3 6k R4 3k R1 // R3 = R13 = 3kΩ // 6kΩ = 2 kΩ R2 // R4 = R24 = 6kΩ // 3kΩ = 2kΩ Rtotal = R1234 = R13 + R24 = 2kΩ + 2kΩ = 4kΩ Purdue University © EET 107 - 26 Introduction to Circuit Analysis ISC Analysis - bridge circuit 9mA R13 2k R24 2k 21 + 18V + 18V - Isupply = E / RT = 36V / 4kΩ = 9mA VR13 = 9mA × 2kΩ = 18V VR24 = 9mA × 2kΩ = 18V Purdue University © EET 107 - 26 Introduction to Circuit Analysis 22 ISC Analysis - bridge circuit IR1 = VR1 / R1 = 18V / 3kΩ = 6mA IR3 = VR3 / R3 = 18V / 6kΩ = 3mA ISHORT = 6mA – 3mA = 3mA 6mA x 3mA 3mA Purdue University © EET 107 - 26 Introduction to Circuit Analysis 23 Thevenin Analysis - bridge circuit ETH = 12V ISC = 3mA RTH = RN = ETH / IN = 12V / 3mA = 4kΩ Thevenin Model Purdue University © 12V ETH EET 107 - 26 RTH + - 4kΩ Introduction to Circuit Analysis 24 Thevenin Model - with load attached RTH ETH IL VL + - IL + RL VL 2kΩ - 4kΩ 12V = 12V / 6kΩ = 2mA = 2mA × 2kΩ = Purdue University © Load results ! 4V EET 107 - 26 Introduction to Circuit Analysis 25 Thevenin Analysis - bridge circuit R1 3k IL + VL - E 36V X Z RL 2k R2 6k IL = 2mA Purdue University © R3 6k R4 3k VL = 4V EET 107 - 26 Introduction to Circuit Analysis 26 Thevenin Analysis - bridge circuit R1 3k E 36V R3 6k X Z R2 6k ILED = IL =? IL + VL - R4 3k VL = VLED = 2V Use Thevenin model to solve. Purdue University © EET 107 - 26 Introduction to Circuit Analysis 27 Thevenin Model - with load attached 4kΩ RTH IL 12V ETH VRTH IL + 2V - + - Load results ! = 12V – 2V = 10V = IRTH = 10V / 4kΩ = Purdue University © 2.5mA EET 107 - 26 Introduction to Circuit Analysis 28 Thevenin Analysis - bridge circuit R1 3k E 36V R3 6k X Z R2 6k IL + VL - IL = 2.5mA Purdue University © EET 107 - 26 R4 3k VL = 2V Introduction to Circuit Analysis 29 Overview Ø Thevenin Model - reviewed Ø Thevenin Analysis - bridge circuit Ø What’s in the Box Modeling Ø Balanced Bridge Purdue University © EET 107 - 26 Introduction to Circuit Analysis 30 What’s in the Box Modeling Electronic Circuit + 10V - ISC - short not usually reasonable. Reasonable load needed. + Electronic Circuit Purdue University © 900Ω 9V - EET 107 - 26 Introduction to Circuit Analysis 31 What’s in the Box Modeling RTH ETH + - OPEN 10V + 10V - VOC = 10V ETH = 10V Purdue University © EET 107 - 26 Introduction to Circuit Analysis 32 What’s in the Box Modeling RTH 10mA ETH + - + 1V 10V 900Ω + 9V - Reasonable load needed. VRTH = 10V - 9V = 1V I = 9V / 900Ω = 10mA RTH = 1V / 10mA = 100Ω Purdue University © EET 107 - 26 Introduction to Circuit Analysis 33 What’s in the Circuit Modeling Electronic Circuit RTH ? 10V ETH Original circuit + - 100Ω Thevenin Model ETH = 10V RTH = 100Ω Purdue University © EET 107 - 26 Introduction to Circuit Analysis 34 Overview Ø Thevenin Model - reviewed Ø Thevenin Analysis - bridge circuit Ø What’s in the Box Modeling Ø Balanced Bridge Purdue University © EET 107 - 26 Introduction to Circuit Analysis 35 Thevenin Analysis – Balanced Bridge R1 E X R2 VXZ = 0V R2 R4 •E= •E R1 + R2 R3 + R4 R3 Z R4 R2 R4 = R1 + R2 R3 + R4 R2 • R3 = R1 • R4 Bridge VX = V Z Purdue University © R1 R3 = R2 R4 EET 107 - 26 Introduction to Circuit Analysis Balanced 36 Thevenin Analysis – Balanced Bridge E 36V R1 3k R3 6k X R2 6k Z RL 2k R2 to balance bridge ? R4 3k R1 R3 Balanced ? = R2 R4 3kΩ 6kΩ ≠ 6kΩ 3kΩ Not balanced ! Purdue University © EET 107 - 26 3kΩ 6kΩ = R 2 3kΩ R 2 = 1.5kΩ Introduction to Circuit Analysis 37 Overview Ø Thevenin Model - reviewed Ø Thevenin Analysis - bridge circuit Ø What’s in the Box Modeling Ø Balanced Bridge Purdue University © EET 107 - 26 Introduction to Circuit Analysis ...
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