Electrical Engineering Technology 1 ECET 17700 - DAQ & Control Systems Lecture # 9 Loading, Thévenin Model & Norton Model Professors Robert Herrick & J. Michael Jacob Module 1 Circuit Loading
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
Circuit Loading 3 Based on load current. R load I load Source circuit must deliver more current. Increased loading effect on circuit.
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 12 12 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
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.
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.
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
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
Electrical Engineering Technology 9 ECET 17700 - DAQ & Control Systems Lecture # 9 Loading, Thévenin & Norton Professors Robert Herrick & J. Michael Jacob Module 2 Thévenin Model or What s in the Box
Lecture 9 Modules 10 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
Why Model? 11 Model Complex Devices and Circuits into Simple Circuit. Simplify Circuit Analysis
Thévenin s Theorem 12 Linear Bilateral Circuit R TH Supplies, resistors, lamps, switches, etc. E TH Not diodes or LEDs! Not bilateral! Thévenin Model
Thévenin s Theorem 13 R TH E TH Thévenin Model Look Familiar? A real voltage source!
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
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
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
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
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
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
Find Thévenin Model 20 Open circuit voltage = 12 V A 2 kω load produces 4 ma Find E TH
Find Thévenin Model 21 Open circuit voltage = 12 V A 2 kω load produces 4 ma Find R TH
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
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.
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
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
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
Electrical Engineering Technology 27 ECET 17700 - DAQ & Control Systems Lecture # 9 Loading, Thévenin & Norton Professors Robert Herrick & J. Michael Jacob Module 3 Intro only Watch Class Video Norton Model
Lecture 9 Modules 28 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
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
Norton Model 30 Electronic Circuit I N R N Original circuit Norton Model Valid for any load Same resulting load voltages and currents
I N Norton Current 31 Electronic Circuit I SC Short Circuit Current Careful
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.
R N Norton Resistance 33 Electronic Circuit R N R N = R TH
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
Norton Model Analysis 35 Electronic Circuit Norton Model I L V Ḻ I L V Ḻ Load Load Same I L and V L
Series-Parallel Circuit Analysis - example 36 24V - 6Ω 3Ω I L V Ḻ 6Ω V L R // = 3Ω // 6Ω = 2Ω
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
Series-Parallel Circuit Analysis - example 38 6Ω 1111 24V - 3Ω V Ḻ I L 6Ω V L 6666 I L = 6V / 6Ω = 1111 Could you now find the rest of the circuit values?
Norton Model Analysis - example 39 6Ω I L 24V - 3Ω V Ḻ 6Ω Find I L and V L using the Norton model approach!
I N Norton Model Current - example 40 24V - 6Ω 3Ω I SC Short circuit current I N = I SC = 24V / 6Ω = 4A
E TH Thévenin Voltage - example 41 6Ω 24V - 3Ω V OC - VDR Voltage Divider Rule E TH = V OC = 3 Ω 3Ω6Ω 24V = 8888
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
Norton Model our example values 43 4A Norton I N 2Ω R TH R N Model
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
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
Norton Model Analysis - example 46 20 6Ω I SC 24V 40V - 3Ω 5 6Ω 4 Q9.7 Find I SC (A) I SC = 40 V / 20 Ω = 2A
Norton Model Analysis - example 47 20 6Ω 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
Norton Model Analysis - example 48 20 6Ω 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
Norton Model Analysis - example 49 20 6Ω I L 24V 40V - 3Ω 5 6Ω 4 Q9.10 Use Norton model, attach load, and find I L
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
Electrical Engineering Technology 51 ECET 17700 - DAQ & Control Systems Lecture # 9 Loading, Thévenin & Norton Professors Robert Herrick & J. Michael Jacob Module 4 Watch Class Video Source Conversion
Lecture 9 Modules 52 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
Equivalent Models - supply conversion 53 R TH I N RN E TH - Thévenin model Norton model Same load results!
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
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
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
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!
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
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
Switch Back to Original Circuit 1 mα 3V- 60 3 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
Electrical Engineering Technology 61 ECET 17700 - DAQ & Control Systems Lecture # 9 Loading, Thévenin & Norton Professors Robert Herrick & J. Michael Jacob Module 5 What s in the Box Practical Measurements
Lecture 9 Modules 62 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
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
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
1. R TH Measurements - Using V OC and I SC 65 I N Electronic Circuit ammeter ammeter SHORT careful I SC not always possible!
1. R TH Measurements - Using V OC and I SC 66 R TH = V OC / I SC
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
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
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
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