Electrical Engineering Technology

Transcription

1 Electrical Engineering Technology 1 ECET DAQ & Control Systems Lecture # 9 Loading, Thévenin Model & Norton Model Professors Robert Herrick & J. Michael Jacob Module 1 Circuit Loading

2 Lecture 9 Modules 2 1. Circuit Loading 2. Thévenin Circuit Model What s in the Box 3. Norton Circuit Model 4. Model Conversion 5. What s in the Box Practical Measurements

3 Circuit Loading 3 Based on load current. R load I load Source circuit must deliver more current. Increased loading effect on circuit.

4 IDEAL Voltage Supply Fixed Output 4 E 12 V 0 ma open 12 V E 12 V 1 ma 1 A R 12 k 12 V E 12 V R V (a) (b) (c) E = 12 V with varying loads (a) I = 0 A (b) (c) I = 1 ma I = 1 A Ideal source maintains E with varying loads

5 Never SHORT a Voltage Supply 5 A E? Short Bad News 12 V 0V I = 12V / 0Ω = A E 0 V Excessive Loading. Many power supplies current limited.

6 Never SHORT a Voltage Supply 6 Watch your P/S voltmeter when hooking up circuit. Voltage suddenly drops toward 0 V. Current suddenly pegs. Driving SHORT circuit. Turn power supply off.

7 UNLOADED Real Voltage Supplies 7 R supply E supply External supply terminals E supply R supply I supply V supply V NL - No Load Voltage V OC - Open Circuit Voltage V supply = V NL = V OC = E supply = E TH

8 REAL Supply Resistance Series Circuit 20V 1Ω E supply R supply V Rsupply 0.2V I supply V supply 198 ma 100Ω I load R load V load 19.8 V 8 I supply = I load = V supply = V load = 20V / 101Ω = 198mΑ 198mΑ 100 Ω = 19.8V V Rsupply = 20V 19.8V = 0.2V LOST

9 Electrical Engineering Technology 9 ECET DAQ & Control Systems Lecture # 9 Loading, Thévenin & Norton Professors Robert Herrick & J. Michael Jacob Module 2 Thévenin Model or What s in the Box

10 Lecture 9 Modules Circuit Loading 2. Thévenin Circuit Model What s in the Box 3. Norton Circuit Model 4. Model Conversion 5. What s in the Box Practical Measurements

11 Why Model? 11 Model Complex Devices and Circuits into Simple Circuit. Simplify Circuit Analysis

12 Thévenin s Theorem 12 Linear Bilateral Circuit R TH Supplies, resistors, lamps, switches, etc. E TH Not diodes or LEDs! Not bilateral! Thévenin Model

13 Thévenin s Theorem 13 R TH E TH Thévenin Model Look Familiar? A real voltage source!

14 Measuring Thévenin Voltage 14 Open Circuit or No Load voltage E TH = V OC = V NL Linear Bilateral Circuit Supplies, resistors, lamps, switches, etc. red lead voltmeter black lead E TH R TH red lead voltmeter black lead

15 Finding Thévenin Resistance 15 Test Load Resistor: Measure two of the following Then calculate R TH I load, V load, R load Linear Bilateral Circuit I load V load R load E TH R TH V RTH I load V load R load

16 Example Find Thévenin Voltage 16 Open Circuit Voltage Linear Bilateral Circuit Supplies, resistors, lamps, switches, etc. red lead voltmeter black lead 10 V Test Circuit E TH = V OC = V NL = 10 V

17 Example Find Thévenin Resistance 17 R TH 2 ma Loaded circuit E TH 10 V V RTH 2V I load V load 8 V R load 4 k V load = 8 V R load = 4 kω Now just a simple Series Circuit I load = 8V / 4kΩ = 2 ma V RTH = 10 V 8 V = 2 V R TH = 2V / 2mA = 1 kω

18 Example Thévenin Model 18 E TH 10 V 1 k R TH V RTH Now Model works for any load! Attach any load and find I load & V load

19 Example Use Thévenin Model 19 E TH 10 V 1 k R TH V RTH 5 ma I load V load R load 1 k 5 V - Attach 1kΩ load. Find V load & I load I load = 10V / 2kΩ = 5 ma V load = 1kΩ 5mA = 5 V Same load values as if solved with original circuit

20 Find Thévenin Model 20 Open circuit voltage = 12 V A 2 kω load produces 4 ma Find E TH

21 Find Thévenin Model 21 Open circuit voltage = 12 V A 2 kω load produces 4 ma Find R TH

22 Find Thévenin Resistance 22 R TH 4 ma Loaded circuit E TH 10 V 12V V RTH 4V I load V load 8 V 2k R load 4 k 8V I load = 4 ma R load = 2 kω V load = 4mA 2kΩ = 8 V V RTH = 12 V 8 V = 4 V R TH = 4V / 4mA = 1 kω

23 Use Thévenin Model 23 Draw the Thévenin model from previous two quiz questions with E TH = 12V and R TH = 1kΩ A 5 kω load is attached. Find the circuit load current.

24 Example Use Thévenin Model 24 1 k R TH 2 ma Attach 5kΩ load. E TH 10 V 12V V RTH I load V load R load 1 k 5k Find I load I load = 12V / 6kΩ = 2 ma Same load value as if solved with original circuit

25 Use Thévenin Model 25 Q9.6 Circuit loading increases as A. Load resistance increases B. Load current increases C. None of the above D. All of the above

26 Use Thévenin Model 26 Circuit loading increases as A. Load resistance increases B. Load resistance decreases C. None of the above D. All of the above

27 Electrical Engineering Technology 27 ECET DAQ & Control Systems Lecture # 9 Loading, Thévenin & Norton Professors Robert Herrick & J. Michael Jacob Module 3 Intro only Watch Class Video Norton Model

28 Lecture 9 Modules Circuit Loading 2. Thévenin Circuit Model What s in the Box 3. Norton Circuit Model 4. Model Conversion 5. What s in the Box Practical Measurements

29 Norton Model 29 Electronic Circuit I N R N Original circuit Norton Model I N = original circuit s short circuit current R N = original circuit s output resistance with sources 0 d

30 Norton Model 30 Electronic Circuit I N R N Original circuit Norton Model Valid for any load Same resulting load voltages and currents

31 I N Norton Current 31 Electronic Circuit I SC Short Circuit Current Careful

32 I N Norton Current 32 Electronic Circuit I SC Short Circuit Current I N = I SC I N R N I SC Note: If I SC is down, current source must be up.

33 R N Norton Resistance 33 Electronic Circuit R N R N = R TH

34 34 Ohm s Law find R N Electronic Circuit V OC - Electronic Circuit I SC R N = V OC / I SC R N = E TH / I N

35 Norton Model Analysis 35 Electronic Circuit Norton Model I L V Ḻ I L V Ḻ Load Load Same I L and V L

36 Series-Parallel Circuit Analysis - example 36 24V - 6Ω 3Ω I L V Ḻ 6Ω V L R // = 3Ω // 6Ω = 2Ω

37 Series-Parallel Circuit Analysis - example 37 24V - 6Ω 3Ω I L 2 Ω V Ḻ 6Ω V L 6666 R // = 3Ω // 6Ω = 2Ω V L = 2 Ω 2Ω6Ω 24V = 6666 VVVVVV

38 Series-Parallel Circuit Analysis - example 38 6Ω V - 3Ω V Ḻ I L 6Ω V L 6666 I L = 6V / 6Ω = 1111 Could you now find the rest of the circuit values?

39 Norton Model Analysis - example 39 6Ω I L 24V - 3Ω V Ḻ 6Ω Find I L and V L using the Norton model approach!

40 I N Norton Model Current - example 40 24V - 6Ω 3Ω I SC Short circuit current I N = I SC = 24V / 6Ω = 4A

41 E TH Thévenin Voltage - example 41 6Ω 24V - 3Ω V OC - VDR Voltage Divider Rule E TH = V OC = 3 Ω 3Ω6Ω 24V = 8888

42 R N Norton Resistance - example 42 R N = R N = R TH = 8V / 4A E TH / I N = 2Ω Or simply find R N directly since E supply acts like short. 6Ω 24V - 3Ω R N = 6 Ω // 3 Ω = 2 Ω R N

43 Norton Model our example values 43 4A Norton I N 2Ω R TH R N Model

44 Norton Model - with load attached 44 I L 1 A 4A 2Ω 6V V Ḻ 6Ω model load R T V L = 2Ω // 6Ω = 1.5 Ω = 4A 1.5Ω = 6V

45 Norton Model - with load attached - CDR 45 I L 1 A 4A 2Ω 6V V Ḻ 6Ω model load 2 Branch CDR Current Divider Rule I L = 2Ω 2Ω6Ω 4Α = 1Α

46 Norton Model Analysis - example Ω I SC 24V 40V - 3Ω 5 6Ω 4 Q9.7 Find I SC (A) I SC = 40 V / 20 Ω = 2A

47 Norton Model Analysis - example Ω 24V 40V - 3Ω 5 V OC - 6Ω 4 Q9.8 Find V OC V OC = V 5Ω = ( 5 Ω / 25 Ω ) 40 V = 8 V

48 Norton Model Analysis - example Ω Q9.9 Find R N 24V 40V - 3Ω 5 R N 6Ω 4 Recall I SC = 2A V OC = 8V R N = R TH = Ε ΤΗ / Ι Ν = 8 V / 2 A = 4 Ω or simply R N = 20 Ω // 5 Ω = 4 Ω

49 Norton Model Analysis - example Ω I L 24V 40V - 3Ω 5 6Ω 4 Q9.10 Use Norton model, attach load, and find I L

50 Norton Model - with load attached 50 1A I L 2A 4Ω 4 V V Ḻ 6Ω 4Ω model load I L = 4Ω / 8Ω 2A = 1A 2-branch CDR V L = 1A 4Ω = 4V

51 Electrical Engineering Technology 51 ECET DAQ & Control Systems Lecture # 9 Loading, Thévenin & Norton Professors Robert Herrick & J. Michael Jacob Module 4 Watch Class Video Source Conversion

52 Lecture 9 Modules Circuit Loading 2. Thévenin Circuit Model What s in the Box 3. Norton Circuit Model 4. Model Conversion 5. What s in the Box Practical Measurements

53 Equivalent Models - supply conversion 53 R TH I N RN E TH - Thévenin model Norton model Same load results!

54 R TH & R N - supply conversion 54 E TH - R TH I N RN open Thevenin model R TH Norton model R N Zero the sources R TH = R N

55 I N - supply conversion 55 R TH I N RN E TH - I SC Thevenin model Norton model Short Circuit Current I N = E TH / R TH

56 E TH - supply conversion 56 R TH I N RN E TH - Thevenin model Norton model E TH Open Circuit Voltage E TH = I N R N

57 Thevenin Resistance 57 Electronic Circuit V OC - Electronic Circuit I SC R TH = R N = V OC / I SC Warning - SHORT can cause smoke or loading!

58 Equivalent Circuit - source conversion 58 R1 3 ma 6 kω 18 V Convert to Thevenin R TH = R N = 6 kω R2 3 k 9 V E TH = 3mA 6 kω = 18V

59 Example model - source conversion 59 1 mα 3V- 18 V 6 kω 12V R2 3 k 9 V Thevenin Model E NET = 18V 9V = 9V R T = 6 kω 3 kω = 9 kω I R2 = 9V / 9kΩ = 1 ma

60 Switch Back to Original Circuit 1 mα 3V ma 2 mα R1 6 kω 12V - R2 3 k 9 V I R1 = 3 ma - 1mA = 2 ma V R1 = V Isupply = 2mA 6kΩ = 12V

61 Electrical Engineering Technology 61 ECET DAQ & Control Systems Lecture # 9 Loading, Thévenin & Norton Professors Robert Herrick & J. Michael Jacob Module 5 What s in the Box Practical Measurements

62 Lecture 9 Modules Circuit Loading 2. Thévenin Circuit Model What s in the Box 3. Norton Circuit Model 4. Model Conversion 5. What s in the Box Practical Measurements

63 Thévenin Resistance Measurements 63 Four basic techniques 1. Measure E TH & I N, then R TH = E TH / I N 2. Live circuit, matched load 3. Live circuit, partial loading 4. Dead circuit, Ohmmeter with supplies zeroed Not all methods work for all circuits

64 1. R TH Measurements - Using V OC and I SC 64 I L Electronic Circuit V Ḻ Load Remove load Electronic Circuit E TH- voltmeter

65 1. R TH Measurements - Using V OC and I SC 65 I N Electronic Circuit ammeter ammeter SHORT careful I SC not always possible!

66 1. R TH Measurements - Using V OC and I SC 66 R TH = V OC / I SC

67 2. R TH Measurements - matched load 67 Electronic Circuit E TH - Attach & adjust pot. V POT = ½ E TH Electronic Circuit ½ E TH- voltmeter

68 2. R TH Measurements - matched load 68 E TH - R TH ½ E TH - R POT ½ E TH - Measure Potentiomenter with Ohmmeter R POT = R TH

69 3. R TH Measurements - unmatched load 69 Electronic Circuit V Ḻ R L voltmeter Series Circuit Analysis E TH - R TH V Rth V L I L R L I L = I R th = V L / R L V R th = E TH V L R TH = V R th / I L

70 4. R TH Measurements circuit not powered 70 Electronic Circuit ohmmeter Replace sources with equivalent resistances Current source Replace with Open Voltage source Replace with Short Measure at output terminals with ohmmeter