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Lecture slides for my course on circuit analysis (DC circuits).

Lecture slides for my course on circuit analysis (DC circuits).

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- 1. Circuit analysis: DC Circuits (3 cr) Fall 2009 / Class AS09 Vesa Linja-aho Metropolia October 8, 2010 The slides are licensed with CC By 1.0. http://creativecommons.org/licenses/by/1.0/ Slideset version: 1.1 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 1 / 125
- 2. Table of Contents Click the lecture name to jump onto the ﬁrst slide of the lecture. 1 1. lecture 7 7. lecture 2 2. lecture 8 8. lecture 3 3. lecture 9 9. lecture 4 4. lecture 10 10. lecture 5 5. lecture 11 11. lecture 6 6. lecture 12 12. lecture Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 2 / 125
- 3. 1. lecture About the Course Lecturer: M.Sc. Vesa Linja-aho Lectures on Mon 11:00-14:00 and Thu 14:00-16:30, room P113 To pass the course: Home assignments and ﬁnal exam. The exam is on Monday 12th October 2009 at 11:00-14:00. All changes to the schedule are announced in the Tuubi-portal. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 3 / 125
- 4. 1. lecture The Home Assignments There are 12 home assignments. Each assignment is graded with 0, 0,5 or 1 points. To pass the course, the student must have at least 4 points from the assignments. Each point exceeding the minimum of 4 points will give you 0,5 extra points in the exam. In the exam, there are 5 assignments, with maximum of 6 points each. To pass the exam, you need to get 15 points from the exam. All other grade limits (for grades 2-5) are ﬂexible. Example The student has 8 points from the home assignments. He gets 13 points from the exam. He will pass the exam, because he gets extra points from the home assignments and his total score is (8 − 4) · 0,5 + 13 = 15 points. However, is one gets 8 of 12 points from the home assignments, he usually gets more than 13 points from the exam :-). Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 4 / 125
- 5. 1. lecture The Course Objectives From the curriculum: Learning outcomes of the course unit Basic concepts and basic laws of electrical engineering. Analysis of direct current (DC) circuits. Course contents Basic concepts and basic laws of electrical engineering, analysis methods, controlled sources. Examples and exercises. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 5 / 125
- 6. 1. lecture The Course Schedule 1 The basic quantities and units. Voltage source and resistance. Kirchhoﬀ’s laws and Ohm’s law. 2 Conductance. Electric power. Series and parallel circuits. Node. Ground. 3 Current source. Applying the Kirchhoﬀ’s laws to solve the circuit. Node-voltage analysis. 4 Exercises on node-voltage analysis. 5 Source transformation. 6 Th´venin equivalent and Norton equivalent. e 7 Superposition principle. 8 Voltage divider and current divider. 9 Inductance and capacitance in DC circuits. 10 Controlled sources. 11 Recap. 12 Recap. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 6 / 125
- 7. 1. lecture The Course is Solid Ground for Further Studies in Electronics The basic knowledge on DC circuits is needed on the courses Circuit Analysis: Basic AC-Theory, Measuring Technology, Automotive Electronics 1, Automotive Electrical Engineering Labs, . . . Important! By studying this course well, studying the upcoming courses will be easier! The basics of DC circuits are vital for automotive electronics engineer, just like the basics of accounting are vital for an auditor, and basics of strength of materials are vital for a bridge-building engineer etc. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 7 / 125
- 8. 1. lecture What is Not Covered on This Course The basic physical characteristics of electricity is not covered on this course. Questions like ”What is electricity?” are covered on the course Rotational motion and electromagnetism. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 8 / 125
- 9. 1. lecture Studying in Our School You have an opportunity to learn on the lectures. I can not force you to learn. You have more responsibility on your learning than you had in vocational school or senior high school. 1 cr ≈ 26,7 hours of work. 3 cr = 80 hours of work. You will spend 39 hours on the lectures. Which means that you should use about 40 hours of your own time for studying! If I proceed too fast or too slow, please interject me (or tell me by email). Do not hesitate to ask. Ask also the ”stupid questions”. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 9 / 125
- 10. 1. lecture What Is Easy and What Is Hard? Diﬀerent things are hard for diﬀerent people. But my own experience shows that DC analysis is easy, because the math involved is very basic. DC analysis is hard, because the circuits are not as intuitive as, for example, mechanical systems are. Studying your math courses well is important for the upcoming courses on circuit analysis. For example, in AC circuits analysis you have to use complex arithmetics. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 10 / 125
- 11. 1. lecture Now, Let’s Get into Business Any questions on the practical arrangements of the course? Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 11 / 125
- 12. 1. lecture Electric Current Electric current is a ﬂow of electric charge. The unit for electric current is the ampere (A). The abbreviation for the quantity is I . One may compare the electric current with water ﬂowing in a pipe (so called hydraulic analogy). The current always circulates in a loop: current does not compress nor vanish. The current in a wire is denoted like this: - I = 2 mA Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 12 / 125
- 13. 1. lecture Kirchhoﬀ’s Current Law As mentioned on the previous slide, the current can not vanish anywhere. Kirchhoﬀ’s Current Law (or: Kirchhoﬀ’s First Law) At any area in an electrical circuit, the sum of currents ﬂowing into that area is equal to the sum of currents ﬂowing out of that area. I3 = 1 mA 6 - - I1 = 3 mA I2 = 2 mA If you draw a circle in any place in the circuit, you can observe that there is as the same amount of current ﬂowing into the circle and out from the circle! Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 13 / 125
- 14. 1. lecture Be Careful with Signs One can say: ”The balance of my account -50 euros” or equally ”I owe 50 euros to my bank”. One can say: ”The proﬁt of the company was -500000 euros” or equally ”The loss of the company was 500000 euros”. If you measure a current with an ammeter and it reads −15 mA, by reversing the wires of the ammeter it will show 15 mA. The sign of the current shows the direction of the current. The two circuits below are exactly identical. I3 = 1 mA I3 = 1 mA 6 6 - - I1 = 3 mA I2 = 2 mA Ia = −3 mA Ib = −2 mA Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 14 / 125
- 15. 1. lecture Voltage The potential diﬀerence between two points is called voltage. The abbreviation for the quantity is U. In circuit theory, it is insigniﬁcant how the potential diﬀerence is generated (chemically, by induction etc.). The unit of voltage is the volt (V). One may compare the voltage with a pressure diﬀerence in hydraulic system, or to a diﬀerence in altitude. Voltage is denoted with an arrow between two points. + 12 V U = 12 V − c Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 15 / 125
- 16. 1. lecture Kirchhoﬀ’s voltage law The voltage between two points is the same, regardless of the path chosen. This is easy to understand by using the analogy of diﬀerences in altitude. If you leave your home, go somewhere and return to your home, you have traveled uphill as much as you have traveled downhill. Kirchhoﬀ’s Voltage Law (or: Kirchhoﬀ’s Second Law) The directed sum of the voltages around any closed circuit is zero. r' 4,5 V r − − − + + + 1,5 V 1,5 V 1,5 V Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 16 / 125
- 17. 1. lecture Ohm’s law Resistance is a measure of the degree to which an object opposes an electric current through it. The larger the current, the larger the voltage – and vice versa. The abbreviation of the quantity is R and the unit is ( Ω) (ohm). The deﬁnition of resistance is the ratio of the voltage over the element divided with the current through the element. R = U/I U = RI U E - I R Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 17 / 125
- 18. 1. lecture Deﬁnitions Electric circuit A system consisting of compontents, in where electric current ﬂows. Direct current (DC) The electrical quantities (voltage and current) are constant (or nearly constant) over time. Direct current circuit An electric circuit, where voltages and currents are constant over time. Example In a ﬂashlight, there is a direct current circuit consisting of a battery/batteries, a switch and a bulb. In a bicycle there is an alternating current circuit (dynamo and bulb). Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 18 / 125
- 19. 1. lecture An alternate deﬁnition for direct current One may deﬁne also that direct current means a current, which does not change its direction (sign), but the magnitude of the current can vary over time. For example, a simple lead acid battery charger outputs a pulsating voltage, which varies between 0 V ... ≈ 18 V. This can be also called DC voltage. Agreement On this course, we deﬁne DC to mean constant voltage and current. The magnitude and sign are constant over time. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 19 / 125
- 20. 1. lecture A simple DC circuit A light bulb is wired to a battery. The resistance of the ﬁlament is 10 Ω. - I =? + 12 V d d − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 20 / 125
- 21. 1. lecture A simple DC circuit A light bulb is wired to a battery. The resistance of the ﬁlament is 10 Ω. - + I =? 12 V 10 Ω − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 20 / 125
- 22. 1. lecture A simple DC circuit A light bulb is wired to a battery. The resistance of the ﬁlament is 10 Ω. - + I =? 12 V 10 Ω 12 V − c Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 20 / 125
- 23. 1. lecture A simple DC circuit A light bulb is wired to a battery. The resistance of the ﬁlament is 10 Ω. - + I = 1,2 A 12 V 10 Ω 12 V − c U = RI 12 V I =U = R 10 Ω = 1,2 A Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 20 / 125
- 24. 1. lecture Homework 1 (released 31st Aug, to be returned 3rd Sep) The homework are to be returned at the beginning of the next lecture. Remember to include your name and student number. Homework 1 Find the current I . + 1,5 V − R = 20 Ω + 1,5 V ?I − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 21 / 125
- 25. 2. lecture Homework 1 - Model solution Homework 1 Find the current I . + 1,5 V − R = 20 Ω + 1,5 V ?I − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 22 / 125
- 26. 2. lecture Homework 1 - Model solution Homework 1 Find the current I . + 1,5 V U2 − c UR R = 20 Ω c + 1,5 V U1 ?I − c U1 + U2 − UR = 0 ⇔ UR = U1 + U2 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 22 / 125
- 27. 2. lecture Homework 1 - Model solution Homework 1 Find the current I . + 1,5 V U2 − c UR R = 20 Ω c + 1,5 V U1 ?I − c U1 + U2 − UR = 0 ⇔ UR = U1 + U2 UR U1 + U2 1,5 V + 1,5 V U = RI ⇒ UR = RI ⇒ I = = = = 150 mA R R 20 Ω Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 22 / 125
- 28. 2. lecture Conductance Resistance is a measure of the degree to which an object opposes an electric current through it. The inverse of resistance is conductance. The symbol for conductance is G and the unit is Siemens (S). Conductance measures how easily electricity ﬂows along certain element. For example, if resistance R = 10 Ω then conductance G = 0,1 S. 1 G= R U = RI ⇔ GU = I U E - IG = 1 R Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 23 / 125
- 29. 2. lecture Electric Power In physics, power is the rate at which work is performed. The symbol for power is P and the unit is the Watt (W). The DC power consumed by an electric element is P = UI U E - I If the formula outputs a positive power, the element is consuming power from the circuit. If the formula outputs a negative power, the element is delivering power to the circuit. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 24 / 125
- 30. 2. lecture Electric Power Energy can not be created nor destroyed The power consumed by the elements in the circuit = the power delivered by the elements in the circuit. U ? I = I I R +6 U2 E R PR = UI = U U = R R − 2 PE = U · (−I ) = U −U = − U R R The power delivered by the voltage source is consumed by the resistor. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 25 / 125
- 31. 2. lecture Series and Parallel Circuits Deﬁnition: series circuit The elements are in series, if they are connected so that the same current ﬂows through the elements. Deﬁnition: parallel circuit The elements are in parallel, if they are connected so that there is the same voltage across them. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 26 / 125
- 32. 2. lecture Series and Parallel Circuits Series circuit - - I I Parallel Circuit U E U E Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 27 / 125
- 33. 2. lecture Resistors in series and in parallel In series ⇐⇒ R1 R2 R = R1 + R2 In parallel R2 ⇐⇒ 1 R= 1 +R1 R1 2 R1 Or, by using conductances: G = G1 + G2 . Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 28 / 125
- 34. 2. lecture Resistors in series and in parallel The formulae on the previous slide can be applied to an arbitrary number of resistors. For instance, the total resistance of ﬁve resistors in series is R = R1 + R2 + R3 + R4 + R5 . Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 29 / 125
- 35. 2. lecture Voltage Sources in Series The voltages can be summed like resistances, but be careful with correct signs. Voltage sources in parallel are inadmissible in circuit theory. There can not be two diﬀerent voltages between two nodes at the same time. r r − + − + − + E1 E2 E3 ⇐⇒ r r − + E = E1 − E2 + E3 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 30 / 125
- 36. 2. lecture What Series and Parallel Circuits are NOT Just the fact that two components seem to be one after the other, does not mean that they are in series. Just the fact that two components seem to be side by side, does not mean that they are in parallel. In the ﬁgure below, which of the resistors are in parallel and which are in series with each other? + R1 R2 + E1 R3 E2 − − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 31 / 125
- 37. 2. lecture What Series and Parallel Circuits are NOT Just the fact that two components seem to be one after the other, does not mean that they are in series. Just the fact that two components seem to be side by side, does not mean that they are in parallel. In the ﬁgure below, which of the resistors are in parallel and which are in series with each other? + R1 R2 + E1 R3 E2 − − Solution None! E1 ja R1 are in series and E2 ja R2 are in series. Both of these serial circuits are in parallel with R3 . But no two resistors are in parallel nor in series. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 31 / 125
- 38. 2. lecture Terminal and Gate A point which provides a point of connection to external circuits is called a terminal (or pole). Two terminals form a gate. An easy example: a car battery with internal resistance. ˜ + RS E − ˜ Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 32 / 125
- 39. 2. lecture Node A node means an area in the circuit where there are no potential diﬀerences, or alternatively a place where two or more circuit elements meet. A ”for dummies” –way to ﬁnd nodes in the circuit: put your pen on a wire in the circuit. Start coloring the wire, and backtrack when your pen meets a circuit element. The area you colored is one node. How many nodes are there in the circuit below? - + I R1 R3 R5 E R2 R4 R6 − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 33 / 125
- 40. 2. lecture Ground One of the nodes in the circuit can be appointed the ground node. By selecting one of the nodes to be the ground node, the circuit diagram usually appear cleaner. The car battery is connected to the chassis of the car. Therefore it is convenient to handle the chassis as the ground node. When we say ”the voltage of this node is 12 volts” it means that the voltage between that node and the ground node is 12 volts. - + I R1 R3 R5 E R2 R4 R6 − r Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 34 / 125
- 41. 2. lecture Ground The ground node can be connected to the chassis of the device or it can be leave not connected to the chassis. Therefore, the existence of the ground node does not mean that the device is ”grounded”. The circuit on the previous slide can be presented also like this: - + I R1 R3 R5 E R2 R4 R6 − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 35 / 125
- 42. 2. lecture Homework 2 (released 3rd Sep, to be returned 7th Sep) Homework 2 Find the current I . - + I R1 R3 R5 E R2 R4 R6 − R1 = R2 = R3 = R4 = R5 = R6 = 1 Ω E = 9V Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 36 / 125
- 43. 3. lecture Homework 2 - Model solution Homework 2 Find the current I . - + I R1 R3 R5 E R2 R4 R6 − R1 = R2 = R3 = R4 = R5 = R6 = 1 Ω E = 9V R5 ja R6 are in series. The total resistance of the serial connection is R5 + R6 = 2 Ω. Furthermore, the serial connection is in parallel with R4 . The resistance of this parallel circuit is 1 + 1 Ω = 2 Ω. 1 3 1 2 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 37 / 125
- 44. 3. lecture Solution continues R3 is in series with the parallel circuit calculated on the previous slide. The resistance for this circuit is R3 + 3 Ω = 5 Ω. 2 3 And the serial connection is in parallel with R2 . The resistance for the 1 parallel circuit is ( 5 )−1 + 1 = 5 Ω. 8 3 1 Lastly, R1 is in series with the resistance computed in the previous step. Therefore, the total resistance seen by voltage source E is 5 13 8 Ω + R1 = 8 Ω. E 72 The current I is computed from Ohm’s law I = 13 Ω = 13 A ≈ 5,5 A. 8 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 38 / 125
- 45. 3. lecture The Current Source The current source is a circuit element which delivers a certain current throught it, just like the voltage source keeps a certain voltage between its nodes. The current can be constant or it can vary by some rule. 6 J R Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 39 / 125
- 46. 3. lecture The Current Source If there is a current source in a wire, you know the current of that wire. - I =1 A 6 J = 1A R1 R2 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 40 / 125
- 47. 3. lecture Applying Kirchhoﬀ’s Laws Systematically to the Circuit When solving a circuit, it is highly recommended to use a systematic mehtod to ﬁnd the voltages and/or currents. Otherwise it is easy to end up with writing a bunch of equations which can not be solved. One systematic method is called the nodal analysis: 1 Name each current in the circuit. 2 Select one node as the ground node. Assign a variable for each voltage between each node and ground node. 3 Write an equation based on Kirchhoﬀ’s current law for each node (except the ground node). 4 State the voltage of each resistor by using the node voltage variables in step 2. Draw the voltage arrows at the same direction you used for the current arrows (this makes it easier to avoid sign mistakes). 5 State every current by using the voltages and substitute them into the current equations in step 2. 6 Solve the set of equations to ﬁnd the voltage(s) asked. 7 If desired, solve the currents by using the voltages you solved. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 41 / 125
- 48. 3. lecture Example Find the current I . R1 R2 + + E1 R3 E2 − − ?I Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
- 49. 3. lecture Example Find the current I . I2 R1 - R2 + I1 + E1 R3 E2 − − ?I Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
- 50. 3. lecture Example Find the current I . I2 R1 - R2 + I1 + E1 R3 U3 E2 ?c − − I Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
- 51. 3. lecture Example Find the current I . I2 R1 - R2 + I1 + E1 R3 U3 E2 I = I1 + I2 ?c − − I Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
- 52. 3. lecture Example Find the current I . E1 − U3 '2 − U3 E I2 E R1 - R2 + I1 + E1 R3 U3 E2 I = I1 + I2 ?c − − I Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
- 53. 3. lecture Example Find the current I . E1 − U3 '2 − U3 E I2 E R1 - R2 + I1 + E1 R3 U3 E2 I = I1 + I2 ?c − − I U3 E1 − U3 E2 − U3 = + R3 R1 R2 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
- 54. 3. lecture Example Find the current I . E1 − U3 '2 − U3 E I2 E R1 - R2 + I1 + E1 R3 U3 E2 I = I1 + I2 ?c − − I U3 E1 − U3 E2 − U3 R2 E1 + R1 E2 = + =⇒ U3 = R3 R3 R1 R2 R1 R2 + R2 R3 + R1 R3 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
- 55. 3. lecture Example Find the current I . E1 − U3 '2 − U3 E I2 E R1 - R2 + I1 + E1 R3 U3 E2 I = I1 + I2 ?c − − I U3 E1 − U3 E2 − U3 R2 E1 + R1 E2 = + =⇒ U3 = R3 R3 R1 R2 R1 R2 + R2 R3 + R1 R3 U3 R2 E1 + R1 E2 I = = R3 R1 R2 + R2 R3 + R1 R3 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 42 / 125
- 56. 3. lecture Some Remarks There are many methods for writing the circuit equations, and there is no such thing as ”right” method. The only requirement is that you follow Kirchhoﬀ’s laws and Ohm’s law1 and you have an equal number of equations and unknowns. If there is a current source in the circuit, it will (usually) make the circuit easier to solve, as you then have one unknown less to solve. By using conductances instead of resistances, the equations look a little cleaner. 1 Ohm’s law can only be utilized for resistors. If you have other elements, you must know their current-voltage equation. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 43 / 125
- 57. 3. lecture Another Example -R1 - R2 - R5 + I1 I2 I5 + E1 R3 U 3 R4 U4 E2 I1 = I2 + I3 − c c − I2 = I4 + I5 ?3 I ?4 I E1 − U3 U3 − U4 U3 U3 − U4 U4 U4 − E2 = + ja = + R1 R2 R3 R2 R4 R5 G1 (E1 − U3 ) = G2 (U3 − U4 ) + G3 U3 ja G2 (U3 − U4 ) = G4 U4 + G5 (U4 − E2 ) Two equations, two unknowns → can be solved. Use conductances! Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 44 / 125
- 58. 3. lecture Some Remarks There are many other methods available too: mesh analysis, modiﬁed nodal analysis, branch current method . . . If there are ideal voltage sources in the circuit (=voltage sources which are connected to a node without a series resistance), you need one more unknown (the current of the voltage source) and one more equation (the voltage source will determine the voltage between the nodes it is connected to). Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 45 / 125
- 59. 3. lecture Homework 3 (released 7th Sep, to be returned 10th Sep) Homework 3a) Find the current I4 . Homework 3b) Verify your solution by writing down all the voltages and currents to the circuit diagram and checking that the solution does not contradict Ohm’s and Kirchhoﬀ’s laws. − + 6 R1 R4 R2 ER R5 J 3 ?4 I R1 = R2 = R3 = R4 = R5 = 1 Ω E = 9V J = 1A Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 46 / 125
- 60. 4. lecture Homework 3 - Model Solution Homework 3a) Find current I4 . Homework 3b) Verify your solution by writing down all the voltages and currents to the circuit diagram and checking that the solution does not contradict Ohm’s and Kirchhoﬀ’s laws. − + 6 R1 R4 R2 ER R5 J 3 ?4 I R1 = R2 = R3 = R4 = R5 = 1 Ω E = 9V J = 1A Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 47 / 125
- 61. 4. lecture Solution - − +I 6 R1 R4 J R2 U2ER3 U 3 R5 c ?4c I R1 = R2 = R3 = R4 = R5 = 1 Ω E = 9V J = 1A First we write two current equations and one voltage equation. The conductance of the series circuit formed by R4 ja R5 is denoted with G45 . J = U 2 G2 + I I = U3 G3 + U3 G45 U2 + E = U3 By substituting I from the second equation to the ﬁrst equation and then substituting U2 from the third equation, we get J = (U3 − E )G2 + U3 (G3 + G45 ) Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 48 / 125
- 62. 4. lecture By substituting the component values, we get U3 = 4 V Therefore the current I is 4 V · 1 S = 4 A. From the voltage equation U2 + E = U3 we can solve U2 = −5 V, therefore the current through R2 is 5 A upwards. The current I is therefore 1 A + 5 A = 6 A, of which 4 A goes through R3 :n and the remaining 2 A goes through R4 ja R5 . There is no contradiction with Kirchhoﬀ’s laws and therefore we can be certain that our solution is correct. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 49 / 125
- 63. 4. lecture Example 1 Find I and U. − + E3 ? + I R1 + 6 E1 E2 R2 U J − − c Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 49 / 125
- 64. 4. lecture Example 1 Find I and U. I3 − + E3 ? + I R1 + 6 E1 E2 R2 U J − − c J = UG2 + I3 I3 = I + (E1 − E2 )G1 U = E1 + E3 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 49 / 125
- 65. 4. lecture Example 2 Find U2 and I1 . R ' U2 - + + + J1 J2 E1 E2 E3 − − − I1 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 50 / 125
- 66. 4. lecture Example 2 Find U2 and I1 . R ' U2 - + + + J1 J2 E1 E2 E3 − − − I1 I1 = (E1 − E3 )G + J1 E2 + U2 = E3 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 50 / 125
- 67. 4. lecture How To Get Extra Exercise? There are plenty of problems with solutions available at http://users.tkk.fi/~ksilvone/Lisamateriaali/lisamateriaal For example, you can ﬁnd 175 DC circuit problems at http://users.tkk.fi/~ksilvone/Lisamateriaali/teht100.pdf At the end of the pdf ﬁle you can ﬁnd the model solutions, so you can check your solution. If you are enthusiastic, you can install and learn to use a circuit simulator: http://www.linear.com/designtools/software/ltspice.jsp Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 51 / 125
- 68. 4. lecture Example 3 Find U4 . + R1 R3 E R2 R4 U4 − c Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 52 / 125
- 69. 4. lecture Example 3 Find U4 . + R1 R3 E R2 U 2 R4 U4 − c c (E − U2 )G1 = U2 G2 + (U2 − U4 )G3 (U2 − U4 )G3 = G4 U4 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 52 / 125
- 70. 4. lecture Homework 4 (released 10th Sep, to be returned 14th Sep) Homework 4 Find the voltage U1 . All resistors have the value 10 Ω, E = 10 V ja J = 1 A. 6 R2 + J R1 U 1 R3 E c − R4 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 53 / 125
- 71. 5. lecture Homework 4 - Model Solution Homework 4 Find the voltage U1 . All resistors have the value 10 Ω, E = 10 V ja J = 1 A. U1 − U2 E - 6 R2 I + R1 2 U R3 U J 1 E U2 − U3− c c © ' R4 U3 J = U1 G1 + (U1 − U2 )G2 (U1 − U2 )G2 = (U2 − U3 )G3 + I G3 (U2 − U3 ) + I = U 3 G4 U2 − U3 = E Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 54 / 125
- 72. 5. lecture Solution J = U1 G1 + (U1 − U2 )G2 (U1 − U2 )G2 = EG3 + I G3 E + I = U 3 G4 U2 − U3 = E I is solved from the third equation and substituted into the second equation, then U3 is solved from the equation and substituted. J = U1 G1 + (U1 − U2 )G2 (U1 − U2 )G2 = EG3 + (U2 − E )G4 − G3 E 1 = 0,2U1 − 0,1U2 0,1U1 − 0,1U2 = 0,1U2 − 1 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 55 / 125
- 73. 5. lecture Solution 1 = 0,2U1 − 0,1U2 0,1U1 − 0,1U2 = 0,1U2 − 1 Which is solved U1 = 10 U2 = 10 Therefore the voltage U1 is 10 Volts. This is easy to verify with a circuit simulator. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 56 / 125
- 74. 5. lecture Circuit Transformation 1 An operation which transforms a part of the circuit into an internally diﬀerent, but externally equally acting circuit, is called a circuit transformation. 2 For example, combining series resistors or parallel resistors into one resistor, is a circuit transformation. Combining series voltage sources into one voltage source is a circuit transformation too. 3 On this lecture, we learn dealing with parallel current sources and the source transformation, with which we can transform a voltage source with series resistance into a current source with parallel resistance. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 57 / 125
- 75. 5. lecture An Example of a Circuit Transformation Two (or more) resistors are combined to a single resistor, which acts just like the original circuit of resistors. Resistors in series ⇐⇒ R1 R2 R = R1 + R2 Resistors in parallel R2 ⇐⇒ 1 R= 1 +R1 R1 2 R1 Or, by using conductance: G = G1 + G2 . Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 58 / 125
- 76. 5. lecture Current Sources in Parallel One or more current sources are transformed into single current source, which acts just like the original parallel circuit of current sources. Current sources in parallel ˜ ˜ 6 6 6 J1 J2 J3 ⇐⇒ J = J1 + J2 − J3 ? ˜ ˜ Just like connecting two or more voltage sources in parallel, connecting current sources in series is an undeﬁned (read: forbidden) operation in circuit theory, just like divide by zero is undeﬁned in mathematics. There can not be two currents in one wire! Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 59 / 125
- 77. 5. lecture The Source Transformation A voltage source with series resistance acts just like current source with parallel resistance. The source transformation ˜ ˜ + R 6 E ⇐⇒ J R E = RJ − ˜ ˜ Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 60 / 125
- 78. 5. lecture Important Note that an ideal voltage or current source can not be transformed like in previous slide. The voltage source to be transformed must have series resistance and the current source must have parallel resistance. The resistance remains the same, and the value for the source is found from formula E = RJ, which is based on Ohm’s law. The source transform is not just a curiosity. It can save from many lines of manual calculations, for example when analyzing a transistor ampliﬁer. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 61 / 125
- 79. 5. lecture Rationale for the Source Transformation The source transformation - - + I I R 6 E E U R U R − c c In the ﬁgure left: E −U I = U = E − RI R In the ﬁgure right: E U E −U E I = − = U=( − I )R = E − RI R R R R Both the circuits function equally. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 62 / 125
- 80. 5. lecture Example Solve U. + R1 R2 + E1 R3 U E − c − The circuit is transformed 6 6 J1 R1 R2 R3 J2 And we get the result: J1 + J2 U= G1 + G2 + G3 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 63 / 125
- 81. 5. lecture A Very Important Notice! The value of the resistance remains the same, but the resistor is not the same resistor! For instance, in the previous example the current through the original resistor is not same as current through the transformed resistor! Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 64 / 125
- 82. 5. lecture Homework 5 (released 14th Sep, to be returned 17th Sep) Homework 5 Find current I by using source transformation. J1 = 10 A, J2 = 1 A, R1 = 100 Ω, R2 = 200 Ω ja R3 = 300 Ω. - I R2 6 6 J1 R1 R3 J2 This is easy and fast assignment. If you ﬁnd yourself writing many lines of equations, you have done something wrong. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 65 / 125
- 83. 6. lecture Homework 5 - Model Solution Homework 5 Find current I by using source transformation. J1 = 10 A, J2 = 1 A, R1 = 100 Ω, R2 = 200 Ω ja R3 = 300 Ω. - I R2 6 6 J1 R1 R3 J2 - + R1 I R2 R3 + R 1 J1 R 3 J2 − − R 1 J1 − R 3 J2 1000 V − 300 V 7 I = = = A ≈ 1,17 A. R1 + R2 + R3 600 Ω 6 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 66 / 125
- 84. 6. lecture Th´venin’s Theorem and Norton’s Theorem e So far we have learned the following circuit transformations: voltage sources in series, current sources in parallel, resistances in parallel and in series and the source transformation. Th´venin’s theorem and Norton’s theorem relate to circuit e transformations too. By Th´venin’s and Norton’s theorems an arbitrary circuit constisting e of voltage sources, current sources and resistances can be transformed into a single voltage source with series resistance (Th´venin’s e equivalent) or a single current source with parallel resistance (Norton’s equivalent). Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 67 / 125
- 85. 6. lecture Th´venin’s Theorem and Norton’s Theorem e Th´venin’s Theorem e An arbitrary linear circuit with two terminals is electrically equivalent to a single voltage source and a single series resistor, called Th´venin’s e equivalent. Norton’s Theorem An arbitrary linear circuit with two terminals is electrically equivalent to a single current source and a single parallel resistor, called Norton’s equivalent. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 68 / 125
- 86. 6. lecture Calculating the Th´venin Equivalent e ˜ ˜ + R1 + RT E R2 ⇐⇒ ET − − ˜ ˜ The voltage ET in the Th´venin equivalent is solved simply by calculating e the voltage between the terminals. For solving RT , there are two ways: By turning oﬀ all independent (= non-controlled) sources in the circuit, and calculating the resistance between the terminals. By calculating the short circuit current of the port and applying Ohm’s law. Independent source is a source, whose value does not depend on any other voltage or current in the circuit. All sources we have dealt with for now, have been independent. Controlled sources covered later in this course. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 69 / 125
- 87. 6. lecture Calculating the Th´venin Equivalent e ˜ ˜ + R1 + RT E R2 ⇐⇒ ET − − ˜ ˜ The voltage at the port is found by calculating the current through the resistors and multiplying it with R2 . The voltage at the port, called also the idle voltage of the port, is equal to ET . E ET = R2 R1 + R2 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 70 / 125
- 88. 6. lecture Calculating the Th´venin Equivalent e There are two ways to solve RT . Way 1: turn oﬀ all the (independent) sources, and calculate the voltage at the port. A turned-oﬀ voltage source is a voltage source, whose voltage is zero, which is same as just a wire: ˜ ˜ R1 RT R2 ⇐⇒ ˜ ˜ Now it is easy to solve the resistance between the nodes of the port: R1 ja R2 are in parallel, and therefore the resistance is 1 R1 R2 RT = = . G1 + G2 R1 + R2 This method is usually simpler than the other way with short circuit current! Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 71 / 125
- 89. 6. lecture Solving RT by Using the Short-circuit Current There are two ways to solve RT . Way 2: short-circuit the port, and calculate the current through the short-circuit wire. This current is called the short-circuit current: ˜ ˜ + R1 + RT E R2 ? IK ⇐⇒ ET ?K I − − ˜ ˜ The value for the short-circuit current is E IK = R1 and the resistance RT is (by applying Ohm’s law to the ﬁgure on the right): E ET ET R1 +R2 R2 R1 R2 RT = = E = E = IK R R1 R1 + R2 1 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 72 / 125
- 90. 6. lecture Norton Equivalent The Norton equivalent is simply a Th´venin equivalent, which has been e source transformed into a current source and parallel resistance (or vice versa). The resistance has the same value in both equivalents. The value of the current source is the same as the short-circuit current of the port. ˜ 6 JN RN ˜ Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 73 / 125
- 91. 6. lecture Example 1 Calculate the Th´venin equivalent. e All component values = 1. ˜ − + 6 J1 R1 ER 2 ˜ Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 74 / 125
- 92. 6. lecture Example 2 Calculate the Th´venin equivalent. All component values = 1. e ˜ + R1 R3 6 E R2 J − ˜ Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 75 / 125
- 93. 6. lecture Homework 6 (released 17th Sep, to be returned 21th Sep) Homework 6 Calculate the Th´venin equivalent. All the component values are 1. (Every e resistance is 1 Ω and the current source is J1 = 1 A.) ˜ 6 R2 J1 R1 R3 ˜ Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 76 / 125
- 94. 7. lecture Homework 6 - Model Solution Homework 6 Calculate the Th´venin equivalent. All the component values are 1. (Every e resistance is 1 Ω and the current source is J1 = 1 A.) ˜ 6 R2 J1 R1 R3 ˜ First we solve the voltage ET . This can be done by using applying source transformation: ˜ + R1 R2 J1 R1 1 J1 R 1 R3 ET = R1 +R2 +R3 R3 = V 3 − c ˜ Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 77 / 125
- 95. 7. lecture Solution Next we solve the resistance RT of the Th´venin equivalent. The easiest e way to do it is to turn oﬀ all the sources and calculate the resistance between the output port. (The other way is to ﬁnd out the short-circuit current.) The resistance can be solver either from the original or the transformed circuit. Let’s use the transformed circuit and turn oﬀ the voltage source: ˜ R1 R2 1 2 R3 RT = 1 +R1 = 3 Ω R1 +R2 3 ˜ Now the resistors R1 ja R2 are in series, and the series circuit is in parallel with R3 . Now we know both ET ja RT and we can draw the Th´venin e equivalent (on the next slide). Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 78 / 125
- 96. 7. lecture The Final Circuit ˜ + RT = 2 3 Ω 1 ET = V 3 − ˜ Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 79 / 125
- 97. 7. lecture Superposition Principle A circuit consisting of resistances and constant-valued current and voltage sources is linear. If a circuit is linear, all the voltages and currents can be solved by calculating the eﬀect of each source one at the time. This principle is called the method of superposition. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 80 / 125
- 98. 7. lecture Superposition Principle The method of superposition is applied as follows The current(s) and/or voltage(s) caused by each source is calculated one at a time so that all other sources are turned oﬀ. A turned-oﬀ voltage source = short circuit (a wire), a turned-oﬀ current source = open circuit (no wire). Finally, all results are summed together. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 81 / 125
- 99. 7. lecture Superposition Principle: an Example Find current I3 by using the superposition principle. + R1 R2 + E1 R3 E2 − − ? I3 First, we turn oﬀ the rightmost voltage source: + R1 R2 E1 1 E1 R3 I31 = G R1 + G +G G2 +G3 3 1 2 3 − ? I31 Next, we turn oﬀ the leftmost voltage source: R1 R2 + E2 1 I32 = G R2 + G +G G1 +G3 3 1 R3 E2 1 3 − ? I31 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 82 / 125
- 100. 7. lecture Superposition Principle: an Example The current I3 is obtained by summing the partial currents I31 and I32 . E1 1 E2 1 I3 = I31 + I32 = 1 G3 + 1 G3 R1 + G2 +G3 G2 + G3 R2 + G1 +G3 G1 + G3 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 83 / 125
- 101. 7. lecture When Is It Handy to Use the Superposition Principle? If one doesn’t like solving equations but likes ﬁddling with the circuit. If there are many sources and few resistors, the method of superposition is usually fast. If there are sources with diﬀerent frequencies (as we learn on the AC Circuits course), the analysis of such a circuit is based on the superposition principle. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 84 / 125
- 102. 7. lecture Linearity and the Justiﬁcation for the Superposition Principle The method of superposition is based on the linearity of the circuit, which means that every source aﬀects every voltage and current with a constant factor. This means that if there are sources E1 , E2 , E3 , J1 , J2 in the circuit, then every voltage and current is of form k1 E1 + k2 E2 + k3 E3 + k4 J1 + k5 J2 , where constants kn are real numbers. If all the sources are turned oﬀ (= 0), then all currents and voltages in the circuit are zero. Therefore, by nullifying all sources except one, we can ﬁnd out the multiplier for the source in question. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 85 / 125
- 103. 7. lecture Homework 7 (released 21st Sep, to be returned 24th Sep) Homework 7 Find current I2 by using the superposition principle. J1 = 1 A R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω E1 = 5 V - I2 R + 6 2 J1 R1 R3 E1 − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 86 / 125
- 104. 8. lecture Homework 7 - Model Solution Homework 7 Find current I2 by using the superposition principle. J1 = 1 A R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω E1 = 5 V - I2 R + 6 2 J1 R1 R3 E1 − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 87 / 125
- 105. 8. lecture Ratkaisu First, we ﬁnd the eﬀect of the current source: - I21 R 6 2 J1 R1 R3 The voltage over R1 and R2 is the same (they are in parallel) and R2 is twice as large as R1 and therefore the current through R2 is half of the current of R1 . Because the total current through the resistors is J1 = 1 A, the current through R1 :n is 2/3 A and the current through R2 isI21 = 1/3 A. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 88 / 125
- 106. 8. lecture Ratkaisu Next, we ﬁnd out the eﬀect of the voltage source: - I22 R + 2 R1 R3 E1 − The resistors R1 and R2 are now in series and the total voltage over them is E = 5 V, and therefore E 5V 1 I22 = − =− = − V. R1 + R2 10 Ω + 20 Ω 6 The minus sign comes from the fact that the direction of the current I22 is upwards and the direction of the voltage E is downwards. Finally, we sum the partial results: 1 1 1 I2 = I21 + I22 = A − A = A. 3 6 6 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 89 / 125
- 107. 8. lecture Voltage Divider U1 E R1 U R2 U2 c c U1 = U R1R1 2 ja U2 = U R1R2 2 +R +R It is quite common in electronic circuit design, that we need a reference voltage formed from another voltage in the circuit. The formula is valid also for multiple resistors in series. The denominator is formed by summing all the resistances and the resistor whose voltage is to be solved is in the numerator. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 90 / 125
- 108. 8. lecture Current Divider - I ?1 I ?2 I R1 R2 I1 = I G1G1 2 ja I2 = I G1G2 2 +G +G The formula is valid also for multiple resistors in parallel. The formula for current divider is not used as frequently as the voltage divider, but it is natural to discuss it in this concept. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 91 / 125
- 109. 8. lecture Example 1 ?1 I ?2 I ?3 I 6 R4 J R1 R2 R3 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 92 / 125
- 110. 8. lecture Example 1 ?1 I ?2 I ?3 I 6 R4 J R1 R2 R3 G1 I1 = J G1 +G2 +G3 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 92 / 125
- 111. 8. lecture Example 1 ?1 I ?2 I ?3 I 6 R4 J R1 R2 R3 G1 G I1 = J G1 +G2 +G3 I2 = J G1 +G2 +G3 2 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 92 / 125
- 112. 8. lecture Example 1 ?1 I ?2 I ?3 I 6 R4 J R1 R2 R3 G1 G G I1 = J G1 +G2 +G3 I2 = J G1 +G2 +G3 2 I3 = J G1 +G3 +G3 2 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 92 / 125
- 113. 8. lecture Example 2 U1 U2 U3 ‡ ‡ ‡ + R1 R2 R3 E R4 U4 − W Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 93 / 125
- 114. 8. lecture Example 2 U1 U2 U3 ‡ ‡ ‡ + R1 R2 R3 E R4 U4 − W U1 = E R1 +R2R1 3 +R4 +R Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 93 / 125
- 115. 8. lecture Example 2 U1 U2 U3 ‡ ‡ ‡ + R1 R2 R3 E R4 U4 − W U1 = E R1 +R2R1 3 +R4 +R U2 = E R1 +R2R2 3 +R4 +R Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 93 / 125
- 116. 8. lecture Example 2 U1 U2 U3 ‡ ‡ ‡ + R1 R2 R3 E R4 U4 − W U1 = E R1 +R2R1 3 +R4 +R U2 = E R1 +R2R2 3 +R4 +R U3 = E R1 +R2R3 3 +R4 +R Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 93 / 125
- 117. 8. lecture Example 2 U1 U2 U3 ‡ ‡ ‡ + R1 R2 R3 E R4 U4 − W U1 = E R1 +R2R1 3 +R4 +R U2 = E R1 +R2R2 3 +R4 +R U3 = E R1 +R2R3 3 +R4 +R U4 = E R1 +R2R4 3 +R4 +R Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 93 / 125
- 118. 8. lecture Homework 8 (released 24th Sep, to be returned 28th Sep) Homework 8 Find the voltage U by applying the voltage divider formula. E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω R4 = 40 Ω R5 = 50 Ω E2 = 15 V U E + R1 R4 R5 + E1 R2 R3 E2 − − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 94 / 125
- 119. 9. lecture Homework 8 - Model Solution Homework 8 Find the voltage U by applying the voltage divider formula. E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω R4 = 40 Ω R5 = 50 Ω E2 = 15 V U E + R1 R4 R5 + E1 R2 U 2 R3 U3 E2 − c c − 20 Ω 2 U2 = E1 R1R2 2 = 10 V 10 Ω+20 Ω = 6 3 V +R R 30 Ω U3 = E2 R3 +R3 +R5 = 15 V 30 Ω+40 Ω+50 Ω = 3,75 V 4 U = U2 − U3 = 2 11 V ≈ 2,92 V. 12 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 95 / 125
- 120. 9. lecture Inductors and Capacitors u u u - ‡ - §¤¤¤ §§ ‡ - ‡ i R i L i C di u = Ri u = L dt i = C du dt Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 96 / 125
- 121. 9. lecture Inductors and Capacitors in DC Circuit u u u - ‡ - §§ ‡ §¤¤¤ - ‡ i R i L i C di u = Ri u = L dt i = C du dt DC voltage and current remain constant as function of time or the time derivatives of the voltage and current is zero. Therefore the voltage of an inductor and the current of a capacitor is zero in a DC circuit. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 97 / 125
- 122. 9. lecture Exception 1 The capacitor is fed with DC current so that the current has no other route. 6 J C J i = C du ⇒ J = C du ⇒ dt dt du dt = C. The voltage of the capacitor rises at constant speed. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 98 / 125
- 123. 9. lecture Exception 2 A constant voltage source is connected to the terminals of an inductor. + ¤ ¥ ¤ E L ¤ ¥ − ¥ di di di E u = L dt ⇒ E = L dt ⇒ dt = L. The current of the inductor rises at constant speed. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 99 / 125
- 124. 9. lecture Dealing with all other cases involving inductors and capacitors in DC circuits Inductors are replaced with short circuits and capacitors are replaced with open circuits. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 100 / 125
- 125. 9. lecture Homework 9 (released 28th Sep, to be returned 1st Oct) Homework 9 Solve the voltage U from this DC circuit. E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω R4 = 40 Ω L = 500 mH C = 2F E2 = 15 V §¤¤¤ §§ + R1 L R4 + C E1 R2 R3 U E2 − W − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 101 / 125
- 126. 10. lecture Homework 9 - Model Solution Homework 9 Solve the voltage U from this DC circuit. E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω R4 = 40 Ω L = 500 mH C = 2F E2 = 15 V §¤¤¤ §§ + R1 L R4 + C E1 R2 R3 U E2 − W − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 102 / 125
- 127. 10. lecture Homework 9 - Model Solution Because there are no parallel connections of inductors and voltage sources and no serial connections of capacitors and current sources and the circuit is a DC circuit (= constant voltages and currents), we can replace the inductors with short circuits and the capacitors with open circuits. + R1 R4 + E1 R2 R3 U E2 − W − in which case we obtain U easily by applying the voltage divider formula: R3 3 U = E2 = 6 V ≈ 6,4 V R3 + R4 7 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 103 / 125
- 128. 10. lecture Controlled Sources So far, all of our sources have been constant valued. If the value of a source does not depend on any of the voltages or currents in the circuit, the source is an independent source. For example, constant valued sources and sources varying as function of time (only) are independent sources. If the value of a source is a function of a voltage and/or current in the circuit, the source is a controlled source. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 104 / 125
- 129. 10. lecture Voltage Controlled Voltage Source (VCVS) r + u e = Au c − r The voltage e of VCVS is dependent of some voltage u. The multiplier A is called voltage gain. A real-world example: an audio ampliﬁer. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 105 / 125
- 130. 10. lecture Current Controlled Voltage Source (CCVS) r + ?i e = ri − r The voltage e of CCVS is dependent of some current i . The multiplier r is called transresistance. There is no good everyday example of this source available (of course we can construct this kind of source by using an operational ampliﬁer). Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 106 / 125
- 131. 10. lecture Voltage Controlled Current Source (VCCS) r 6 u j = gu c r The current j of VCCS is dependent of some voltage u. The multiplier g is called transconductance. A real-world example: a ﬁeld-eﬀect transistor (JFET or MOSFET). Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 107 / 125
- 132. 10. lecture Current Controlled Current Source (CCCS) r 6 ?i j = βi r The current j of CCCS is dependent of some current i . The multiplier β is called current gain. A real-world example: a (bipolar junction) transistor. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 108 / 125
- 133. 10. lecture Homework 10 (released 1st Oct, to be returned 5th Oct) Homework 10 Find the voltage U. E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω R4 = 40 Ω r = 2Ω - + R1 i R2 R4 + E1 R3 U e2 = ri − W − Note that the source on the right is a controlled source. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 109 / 125
- 134. 11. lecture Homework 10 - Model Solution Homework 10 Find the voltage U. E1 = 10 V R1 = 10 Ω R2 = 20 Ω R3 = 30 Ω R4 = 40 Ω r = 2Ω - + R1 i R2 R4 + E1 R3 U e2 = ri − W − Note that the source on the right is a controlled source. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 110 / 125
- 135. 11. lecture - + R1 i R2 R4 + E1 R3 U e2 = ri − W − Let’s denote the total resistance of R1 :n ja R2 with symbol R12 and write a nodal equation: UG3 = (E1 − U)G12 + (ri − U)G4 There are two unknowns in the circuit and therefore we need another equation with the same unknowns: i = (E1 − U)G12 Then we substitute i to the ﬁrst equation: E1 G12 − UG12 + rG4 G12 E1 − rG4 G12 U − UG4 = UG3 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 111 / 125
- 136. 11. lecture E1 G12 − UG12 + rG4 G12 E1 − rG4 G12 U − UG4 = UG3 from which we get G12 E1 (1 + rG4 ) = U(G3 + G12 + G4 + rG4 G12 ). Then we substitute the component values and solve U: 10 2 30 (1 + 40 ) U= 1 1 1 2 = 3,75 V 30 + 30 + 40 + 40·30 Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 112 / 125
- 137. 11. lecture Recapitulation On this lesson, we solve some refresher assignments. If you have solved all the circuits, solve the home assignment. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 113 / 125
- 138. 11. lecture Recap assignment 1 Recap assignment 1 Find U and I ﬁrst by using the method of superposition and then by some other method of your choice. R1 = 1 Ω R2 = 2 Ω J = 1A E = 3V - + I R2 E R1 J U − ?c Vastaus: I = 4 A ja U = 1 V. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 114 / 125
- 139. 11. lecture Recap assignment 2 Recap assignment 2 Form a Th´venin equivalent of the circuit on the left. Then, compute the e current IX , when the switches are closed and RX is a) 0 Ω, b) 8 Ω ja c) 12 Ω. R1 = 5 Ω R2 = 3 Ω R3 = 8 Ω R4 = 4 Ω ¨ E = 16 V ˜ ¨ ˜ R2 R4 R3 RX + R1 E ?X ¨¨˜ − I ˜ : Vastaus: RT = 8 Ω, ET = 8 V. a) 1 A b) 0,5 A c)0,4 A. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 115 / 125
- 140. 11. lecture Recap assignment 3 Recap assignment 3 Find U3 . G1 = 1 S G2 = 2 S G3 = 3 S G4 = 4 S G5 = 5 S g = 6 S J = 3A gU1 r 6 G4 G5 J U 1 G1 G2 G3 U3 c c 48 U3 = − 115 V ≈ −417 mV Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 116 / 125
- 141. 11. lecture Homework 11 (released 5th Oct, to be returned 8th Oct) Homework 11 We are given a fact that the current I3 = 0 A. Find E1 . R1 = 5 Ω R2 = 4 Ω R3 = 2 Ω R4 = 5 Ω R5 = 6 Ω E2 = 30 V - + R1 R3 I3 R2 + E1 R4 R5 E2 − − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 117 / 125
- 142. 12. lecture Homework 11 - Model Solution Homework 11 We are given a fact that the current I3 = 0 A. Find E1 . R1 = 5 Ω R2 = 4 Ω R3 = 2 Ω R4 = 5 Ω R5 = 6 Ω E2 = 30 V - + R1 R3 I3 R2 + E1 R4 R5 E2 − − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 118 / 125
- 143. 12. lecture Homework 11 - Model Solution Homework 11 We are given a fact that the current I3 = 0 A. Find E1 . R1 = 5 Ω R2 = 4 Ω R3 = 2 Ω R4 = 5 Ω R5 = 6 Ω E2 = 30 V - + R1 R3 I3 R2 + E1 R4 U 4 R5 U5 E2 − c c − Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 118 / 125
- 144. 12. lecture Because I3 = 0 A, the current through R1 equals the current through R4 and the current through R2 equals the current through R5 . Therefore, the resistors are in series2 and we may use the voltage divider formula to ﬁnd voltages over R4 and R5 . The voltage over R5 is U5 = E2 R2R5 5 = 18 V. +R Therefore the voltage over R4 is 18 V too. Now, by the voltage divider rule: R4 5Ω U4 = E1 ⇒ 18 V = E1 R1 + R4 5Ω + 5Ω from which we can solve E1 = 36 V. Note: it is completely correct to write nodal equations for the circuit and solve E1 from them, too. 2 Because and only because we know that I3 is zero. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 119 / 125
- 145. 12. lecture Recapitulation On this lesson, we solve some refresher assignments. I can also demonstrate some examples on the blackboard or to your booklets, too. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 120 / 125
- 146. 12. lecture Recap assignment 4 R2 = 5 Ω E1 = 3 V E2 = 2 V - I1 I2 + R1 E2 + − E1 − R2 a) How should we choose R1 , if we want I2 to be 0 A? b) How large is I1 then? a) 10 Ω ja b) 0,2 A. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 121 / 125
- 147. 12. lecture Recap assignment 5 R1 = 100 Ω R2 = 500 Ω R3 = 1,5 kΩ R4 = 1 kΩ E1 = 5 V J1 = 100 mA J2 = 150 mA - J2 r 6 R2 +R3 J1 R1 E1 R4 U4 − c Find U4 . U4 = 92 V Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 122 / 125
- 148. 12. lecture Recap assignment 6 R1 = 12 Ω R2 = 25 Ω J = 1 A E1 = 1 V E2 = 27 V − + R1 6 + E2 E1 J U R2 − c Find voltage U. 1 Solution: 37 V ≈ 27 mV Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 123 / 125
- 149. 12. lecture Homework 12 (released 8th Oct, to be returned 12th Oct) Write a short essay on following subjects: What did you learn on the course? Did the course suck or was it worthwhile? What could the lecturer do better? How should this course be improved? The essay will not aﬀect the grading of the exam — please give honest feedback3 . How to return this homework: Write the essay as a plain text email (no attachments) and send it to me no later than the exam day at 18:00. The subject of the email message must be ’DC Circuits course feedback 2009 Firstname Surname’. 3 I am really interested in how I could make the course better. Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 124 / 125
- 150. 12. lecture Final Notices on these Slides The slides are licensed with CC By 1.0 4 . In short: you can use and modify the slides freely as long as you mention my name (= Vesa Linja-aho) somewhere. Single examples and circuits can be of course used without any name mentioning, because they are not an object of copyright (legal term: ”Threshold of originality”). The origin of these slides is the DC Circuits course in Metropolia polytechnic in Helsinki, Finland. If you ﬁnd typos, misspellings or errors in facts, please give me feedback. 4 http://creativecommons.org/licenses/by/1.0/ Vesa Linja-aho (Metropolia) Circuit analysis: DC Circuits (3 cr) October 8, 2010 125 / 125

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