1. ELECTRIC CIRCUITS ECSE-2010 Spring 2003 Class 7

2. ASSIGNMENTS DUE Today (Tuesday/Wednesday): Will do Experiment #2 In Class (EP-2) Activities 7-1, 7-2, (In Class) Today is a Very Busy Day! Thursday: Will do Experiment #3 In Class (EP-3) Activities 8-1, 8-2, (In Class) Next Monday: Homework #3 Due Activity 9-1, Op-Amp ILM

3. REVIEW Node Equations: Technique to Solve Any Linear Circuit Label Unknown Node Voltages, v 1, v 2, v 3, etc. Write KCL at Each Unknown Node Sum of Currents Out of Node= 0 Relate Currents to Node Voltages using Ohm’s Law Always get same number of equations as unknowns Solve linear, algebraic equations for v 1, v 2, v 3, etc.

4. REVIEW Mesh Equations: Another Technique to Solve Any Linear Circuit Define All Mesh Currents Unknown Mesh Currents (i 1, i 2, i 3, etc.) and Current from Current Sources (Independent and Controlled) Write a KVL around Each Unknown Mesh Current Sum of Voltages around Mesh due to all Mesh Currents = 0 Relate Voltages to Mesh Currents using Ohm’s Law Always get same number of equations as unknowns Solve linear, algebraic equations for i 1, i 2, i 3, etc.

5. REVIEW Add Controlled Sources: Things get harder! Must find a Constraint Equation: Relate Controlling Voltage or Current to Unknown Node Voltages or Unknown Mesh Currents Must Do By Inspection; THINK! Find Constraint Equation directly from circuit Try Ohm’s Law, KCL, KVL, etc. It’s there; you just have to find it!

6. NODE OR MESH? # Unknown Node Voltages = # Nodes - # Voltage Sources - 1 (Ref): # Unknown Mesh Currents = # Meshes - # Current Sources: Choose the Method with the Smaller # of Unknowns: If Same: Choose Method that Gives Desired Output (Node Equations for v or Mesh Equations for i):

7. MORE CIRCUIT ANALYSIS TECHNIQUES Source Conversions: Will Use Many Times During Course Superposition: Will Practice with Activity 7-1 Thevenin and Norton Equivalent Circuits : Will Practice with Activity 7-2 today and Activity 8-1 Tomorrow Maximum Power Transfer: Tomorrow – Activity 8-2 using PSpice

8. SOURCE CONVERSIONS Any Voltage Source, v s, in Series with a Resistor, R s, May be Replaced by a Current Source, i s = v s / R s, in Parallel with Same Resistor, R s : (And Vice Versa) Sometimes a Useful Technique Often Simplifies a Problem Not Always the Best Technique to Use Be Very Careful with Controlled Sources Sometimes “lose” controlling v, i, when make source conversion

9. SOURCE CONVERSIONS Define v, i using Active Convention

10. EXAMPLE

11. SOURCE CONVERSIONS Source Conversion Reduces Ckt from 2 Meshes to 1 Mesh in Example: Source Conversion Appears to Simplify Ckt v x is No Longer the Voltage Across the 4 Ohm Resistor: v x still exists in this case Just need to be careful Sometimes controlling v,i will completely disappear

12. SUPERPOSITION Technique to use when there is more than 1 Independent Source in a circuit: Not always the best technique to use Will learn lots of techniques; Experience helps us learn which technique to choose Find Output due to each independent source with all other independent sources set = 0; then Add to find Total Output: Source of 0 is called a “dead source” “Dead” voltage source = 0 V = Short Circuit “Dead” current source = 0 A = Open Circuit

13. SUPERPOSITION Total Output = Sum of all Outputs due to each independent source with all other independent sources “dead”: Simply Add them Works only for Linear Circuits; Only kind we will consider Leave Controlled Sources Alone!!: Find Outputs due to Independent Sources only!! Do NOT set Controlled Sources = 0!! Will Practice with Activity 7-1

14. ACTIVITY 7-1

15. 2 Independent Sources: Output defined as i 1 Need to find i 1 due to each Independent Source with other Independent Source set = 0 Add i 1-1 + i 1-2 = i 1

16. ACTIVITY 7-1 Let’s First Make the Current Source Active and the Voltage Source “Dead”: Set 16 V Source = 0 => Short Circuit Find i 1-1

17. ACTIVITY 7-1

19. Current Source Active: Set 16 V source = 0 => Short Circuit 6 // 6 = 3 ohms v x-1 = 3 ohms x -10 Amps = - 30 Volts KCL at a: 10 + (-15) = i 1-1 => i 1-1 = - 5 Amps

20. ACTIVITY 7-1 Now Make the Voltage Source Active and the Current Source “Dead”: Set 10 A source = 0 => Open Circuit Find i 1-2

21. ACTIVITY 7-1

22. Voltage Source Active: Set 10 A source = 0 => Open Circuit i = 0; => i 1-2 = v x / 2 Voltage Divider Rule v x-2 = [6/(6+6)] 16 = 8 V i 1-2 = v x-2 / 2 = 4 Amps Total Output: i 1-1 + i 1-2 = i 1 = - 5 + 4 = - 1 Amp

23. THEVENIN EQUIVALENT CIRCUITS SOURCE Network: At Least 1 Independent Source R’s, Controlled Sources Define v, i at Terminals Using Active Convention Thevenin’s Theorem: Any Source Network can be replaced by a Voltage Source, v oc, in Series with a Resistor, R T v oc is called the “Open Circuit Voltage” Sometimes also called the “Thevenin Voltage”, v T v oc = v T = v when i = 0 R T is called the “Thevenin Resistance” Thevenin Equivalent Circuit

24. THEVENIN EQUIVALENT CIRCUIT Define v, i using Active Convention

25. NORTON EQUIVALENT CIRCUIT Norton’s Theorem: Any Source Network can be replaced by a Current Source, i sc, in Parallel with a Resistor, R T i sc is called the “Short Circuit Current” i sc = i when v = 0 R T is the same Thevenin Resistance Norton Equivalent Circuit R T = v oc / i sc : R T = R eq of “Dead Source Network”: Source Network with Independent Sources set = 0 Dead Source Network is a Load Network

26. NORTON EQUIVALENT CIRCUIT Define v, i using Active Convention

27. THEVENIN/NORTON CIRCUITS

28. Find v oc, i sc, R T using Ckt Analysis: Need only find 2 out of 3 R T = v oc / i sc Let’s Practice with Activity 7-2

29. ACTIVITY 7-2

30. Find v oc : Open Circuit the Output: i = 0 Defines v oc

31. ACTIVITY 7-2

32. Find v oc : Open Circuit the Output: i = 0 Defines v oc KVL: v oc = v x - 6 v x = - 5 v x v x = 2 mA x 4 kohms = 8 V v oc = - 5 x 8 = - 40 V

33. ACTIVITY 7-2 Find i sc : Short Circuit the Output: v = 0 Defines i sc

34. ACTIVITY 7-2

35. Find i sc : Short Circuit the Output: v = 0 Defines i sc KCL: i x = 2 - i sc Also: v x = 4 i x = 8 - 4 i sc KVL: v x - 6 v x - 12 i sc = 0 - 5 v x - 12 i sc = 0 - 40 + 20 i sc = 12 i sc => 8 i sc = 40 i sc = 5 mA

36. ACTIVITY 7-2 Find R T : Create Dead Source Network Independent Sources = 0 Leave Controlled Sources Alone Now have Load Network Find R eq = R T Connect Test Voltage v t Define i t using Active Convention R eq = v t / i t

37. ACTIVITY 7-2

38. v x = 4 i t KVL: v x - 6 v x + 12 i t - v t = 0 - 5 v x + 12 i t = v t - 20 i t + 12 i t = v t = - 8 i t v t / i t = R eq = R T = - 8 kohms Check: R T = v oc / i sc R T = - 40 V / 5mA = - 8 k OK

39. BRIDGE CIRCUITS Bridge Circuits are circuits used to accurately measure Circuit Elements: A Wheatstone Bridge is used to Measure Resistance Will Use a Maxwell Bridge later to Measure Inductance

40. WHEATSTONE BRIDGE

41. See Circuit: 3 Accurately Known Resistors; 1 Variable 1 Unknown Resistor; To be Measured Connect Voltmeter Across “Bridge” Adjust R 3 to Balance Bridge: i m = 0; v m = 0 (R 1 + R 3 ) i 1 = (R u + R 2 ) i 2 ; R 3 i 1 = R u i 2 and R 1 i 1 = R 2 i 2

42. WHEATSTONE BRIDGE

43. Solve for R u : R u = (i 1 / i 2 ) R 3 => R u = (R 2 / R 1 ) R 3 = R 2 R 3 / R 1 Very Accurate Method of Measuring an Unknown Resistance: Wheatstone Bridges can be very expensive Will build an “inexpensive” Wheatstone Bridge in Exp. 2b today

44. EXPERIMENT 2a

45. Determine the Value of an Unknown Resistor: R u is in Plastic Box in Drawer Statistical Treatment of Data: No measurements are “perfect” Need to know how to handle “errors” See Notes in Supplement

46. EXPERIMENT 2a Take Several Measurements of v Across R u and i Thru R u : Will Measure voltage v across R u directly Will Measure i thru R u indirectly Will Measure voltage v Pot across Pot and the resistance R Pot of the Pot using MM Calculate i = v Pot /R Pot Plot i vs v; (or v vs i) and Use Statistics to calculate “best” value for R u Use 5-6 Significant Digits in all your Measurements!

47. EXPERIMENT 2a Know that i vs v for a Resistor should be Linear and go through i = v = 0: Slope = 1 / R u : (Slope = R u if Plot v vs i) “Best” Slope When Have Random Errors => Method of Least Squares: See Notes on Statistical Treatment of Data Take Data Today - Do Analysis Later

48. EXPERIMENT 2a

49. EXPERIMENT 2b Wheatstone Bridge: “ Cheap” Version; Not Very Accurate Use 100k Pot for R 2 ; 10k Pot for R 3 Use same R u as Exp 2a

50. EXPERIMENT 2b

51. Try 4 Different Values for R 2 : 10k, 20k, 1k, 0.1k R 2 Affects “Range” of Measurement Need to be able to make R 2 R 3 / R 1 = R u If R 2 is too Small; Cannot Balance Bridge