Delta & Y Configurations, Principles of Superposition, Resistor Voltage Divider Designs
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1 BME/ISE 3511 Bioelectronics - Test Three Course Notes Fall 2016 Delta & Y Configurations, Principles of Superposition, esistor Voltage Divider Designs Use following techniques to solve for current through and voltage across a network resistor: Equivalent esistances Kirchoff s Voltage Law Kirchoff s Current Law Principle of Superposition Thevenin Equivalent Norton Equivalent Sketch Series esistors Voltage Divider (including voltage source) Calculate voltage across load resistor Voltage Divider (esistors in Series with Voltage Source) V 2 V ( 2 / ( ) Sketch Parallel esistors Current Divider (including current source) Calculate current through load resistor Current Divider (esistors in Parallel with Current Source) I 2 I ( 1 / ( ) Design Voltage Divider Network (10% rule for two resistors and voltage source) given output requirements Convert Delta configuration to equivalent Wye configuration Convert Wye configuration to equivalent Delta configuration Solve Voltage Source & Current Source Circuit Network Problems (using Superposition) Attachments: Delta & Y Configurations and Conversions (Figures & Equations) Five esistor Equivalent Y to Delta & Delta to Y Practice Conversion Problems Principle of Superposition Example Problem Principle of Superposition (Figure 30-3 Giorgio izzoni 2009).pdf Voltage Divider Design Notes & Practice Problem
2 Y to Delta & Delta to Y Conversions a b c 2 c 2 b a Y to Delta Conversion Delta to Y Conversion + + a b b c a c a c a b b c a c 1 2 b a a b b c a c c b + 2 +
3 Calculate the equivalent resistance for a five resistor network (See Figure 1.) Step 1. Divide into two parts. Delta( 5 ) and "V"(2 ) Figure 1. Step 2. Convert Delta( 5 ) to Y(c a b) c 5 b a Step 2. Delta( 5 ) to Y(c a b) - continued Step 3. Combine Delta( 5 ) with "V"(2 ) Note: otated and eflected Labeling c b a c b a Step 4. b in Series with therefore b b + a in Series with 2 therefore a2 a + 2 Step 5. b in Parallel with a2 (b) x ( a2) (b) + ( a2) Step 6. c in Series with Parallel b a2 Step 7. eq c + Parallel b a2 2
4 Y to Delta & Delta to Y Practice Problem Conversions 1. Write closed-form solutions for each of the following configurations Figure A Figure B Figure C 2. For Figures A, B, C, Find single equivalent resistive values for 1 5, 2 7, 3 8, 4 3, Answers: Figure A. eq 5.74 Figure B. eq 2.36 Figure C. eq 5.66
5 Solving Voltage and Current Circuit Problems Using the Theory of Superposition The overall effect for two or more stimuli to a linear system is equal to the sum of individual effects. efer to Figure 1. Objective: a. Calculate V Load b. Calculate I Load Given: I 3 A V 9 V 1 3 Ω 2 4 Ω Load 2 Ω 2 I 1 Load V Figure 1. Procedure: I. Short all Voltage Sources Calculate V Load (Generated by Current Sources) V Load 2.77 V II. Open all Current Sources Calculate V Load (Generated by Voltage Sources) V Load 2.08 V III. Calculate overall results by adding results from Step I and Step II V Load Volts Calculate I Load V load / Load 4.85 V / 2 Ω 2.43 Amps
6 Fundamental of Electrical Engineering, Giorgio izzoni, McGraw-Hill, 2009
7 Voltage Divider Design Note (10% ule) - Minimizes I 2 Heat Loss Given values for voltage source and Load voltage and current specifications (V Load & I Load ); Determine values for Bias esistor ( 1 ) and Bleed esistor ( 2 ). 1 (Bias) V V Supply 2 (Bleed) Load Step 1. Calculate Load esistor ( V Load / I Load ) Step 2. Calculate I 2 (Bleed Current) 10% I Load Step 3. Select 2 (Bleed esistor) based on I 2 such that 2 V 2 / I 2 (Note: V 2 V Load ) Step 4. Calculate I Total I Load + I 2 Step 5. Calculate 1 (Bias esistor) (V S - V Load ) / (I Total ) Example Problem: Given: V S 12 volts V Load ma Find: Load 150 Ohms Bleed esistor 1500 Ohms Bias esistor 82 Ohms Check esults: Note - by design I Total 1.1 x I Load 55 ma I Bleed 0.1 x I Load 5 ma Bleed 10 x Load 1500 Ohms V Bleed Bleed x I Bleed 1500 x 5 x V Sanity Check: V Bleed V Load 7.5 V!!!!! Sanity Check: V Bias V S - V Bleed V V Bias I Tota l x Bias 55 x 10-3 x V!!!!!!!!! Short Cut Closed Form: Bias (V S - V Load ) / (1.1 x I Load) (12-7.5) / (1.1 x 0.050) 4.5 / Ohms Bleed V Load / (0.1 x I Load ) 7.5 / (0.1 x 0.050) 7.5 / Ohms
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