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Advanced Electrical Engineering Advanced Electrical Engineering Michael E. Auer Basic Concepts Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Chapter Content Michael E.Auer • Introduction • Complex Numbers and Phasors • Circuit Theory Review • Methods of Network Analysis • Locus Diagrams • Circuit Element Variations 02.05.2012 AEE01 Advanced Electrical Engineering Chapter Content Michael E.Auer • Introduction • Complex Numbers and Phasors • Circuit Theory Review • Methods of Network Analysis • Locus Diagrams • Circuit Element Variations 02.05.2012 AEE01 Advanced Electrical Engineering This course: Fundamentals of Electro-Magnetism !!! Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Fundamental Forces of Nature Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering The Electro-magnetic Spectrum microprocessor Michael E.Auer 02.05.2012 ac power network AEE01 Advanced Electrical Engineering Scientific progress, but same fundamental laws Example Lighting Michael E.Auer Light bulb Fluorescent lamp Incandescence is the emission of light from a hot object due to its temperature. Fluoresce means to emit radiation in consequence to incident radiation of a shorter wavelength 02.05.2012 LED diode lamp When a voltage is applied in a forward-biased direction across an LED diode, current flows through the junction and some of the streaming electrons are captured by positive charges (holes). Associated with each electron-hole recombining act is the release of energy in the form of a photon. AEE01 Advanced Electrical Engineering Chapter Content Michael E.Auer • Introduction • Complex Numbers and Phasors • Circuit Theory Review • Methods of Network Analysis • Locus Diagrams • Circuit Element Variations 02.05.2012 AEE01 Advanced Electrical Engineering Complex Numbers Is useful to represent sinusoids as complex numbers. z = x + jy Rectangular coordinates z = z ∠θ = z e jθ Polar coordinates j = −1 Re(z ) = x Im( z ) = y Relations based on Euler’s Identity e ± jθ = cos θ ± j sin θ Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Relations for Complex Numbers Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Phasor Domain 1. The phasor-analysis technique transforms equations from the time domain to the phasor domain. 2. Integro-differential equations get converted into linear equations with no sinusoidal functions. 3. After solving for the desired variable -such as a particular voltage or current - in the phasor domain, conversion back to the time domain provides the same solution that would have been obtained had the original integro-differential equations been solved entirely in the time domain. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Time Domain vs. Phasor Domain (1) Rotating phasor Stationary phasor Phasor counterpart of V0 ∠ φ V0 ∠ Michael E.Auer 02.05.2012 π 2 AEE01 Advanced Electrical Engineering Time Domain vs. Phasor Domain (2) It is much easier to deal with exponentials in the phasor domain than sinusoidal relations in the time domain. Just need to track magnitude/phase, knowing that everything is at frequency ω. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Phasor Relation for Resistors Current through resistor Time Domain Frequency Domain Time domain i = I m cos (ω t + φ ) υ = iR = RI m cos (ω t + φ ) Phasor Domain V = RI m ∠φ Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Phasor Relation for Inductors Time domain Phasor Domain Time Domain Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Phasor Relation for Capacitors Time domain Phasor Domain Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Circuit Analysis in the Phasor Domain Time Domain Phasor Domain Differential Equations Algebraic Equations complex ! Transformation Solution of the Differential Equations Solution of the Algebraic Equations Solution Solution Retransformation Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Basic Approach Transform the circuit to the phasor or frequency domain. Solve the problem using circuit techniques (nodal analysis, mesh analysis, superposition, etc.). Transform the resulting phasor to the time domain. 1. 2. 3. Time to Freq Michael E.Auer Solve Equations in Freq Domain 02.05.2012 Freq to Time AEE01 Advanced Electrical Engineering Phasor Analysis: General Procedure Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Example: RL Circuit (1) Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Example: RL Circuit (2) Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Transformation Table x means a travelling wave! Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Traveling Waves in the Phasor Domain Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Chapter Content Michael E.Auer • Introduction • Complex Numbers and Phasors • Circuit Theory Review • Methods of Network Analysis • Locus Diagrams • Circuit Element Variations 02.05.2012 AEE01 Advanced Electrical Engineering Voltage Division v1 = ii R1 and v 2 = ii R2 Applying KVL to the loop, v i = v1 + v 2 = ii (R1 + R2 ) and vi ii = R1 + R2 Combining these yields the basic voltage division formula: R1 v1 = v i R1 + R2 Michael E.Auer 02.05.2012 R2 v2 = vi R1 + R2 AEE01 Advanced Electrical Engineering Current Division ii = i1 + i2 where vi i1 = R1 vi and i2 = R2 Combining and solving for vi, R1 R2 v i = ii = ii = ii (R1 || R2 ) 1 1 R1 + R2 + R1 R2 1 Combining these yields the basic current division formula: R2 i1 = ii R1 + R2 Michael E.Auer 02.05.2012 and R1 i2 = ii R1 + R2 AEE01 Advanced Electrical Engineering Thevenin and Norton Equivalent Circuits Thévenin Norton Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Thevenin Equivalent Voltage (1) Applying KCL at the output node, vo − vi vo = G1 (v o − vi )+ GS v o + βi1 = RS R1 i1 = G1 (v o − v i ) Current i1 can be written as: open circuit Combining the previous equations G1 (β +1)v i = [G1 (β +1) + GS ]v o G1 (β +1) R1RS (β +1)RS vi × = vi vo = G1 (β +1) + GS R1RS (β +1)RS + R1 Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Thevenin Equivalent Voltage (2) Using the given component values: (β +1)RS (50 +1)1 kΩ vi = v i = 0.718v i vo = (β +1)RS + R1 (50 +1)1 kΩ +1 kΩ and v th = 0.718v i Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Thevenin Equivalent Resistance Applying KCL, i x = −i1 − βi1 + G S v x = G1v x + βG1v x + G S v x = [G1 (β + 1) + G S ]v x R1 1 vx = RS = Rth = β +1 G1 (β + 1) + G S ix Rth = RS Michael E.Auer R1 20 kΩ = 1 kΩ = 1 kΩ 392 Ω = 282 Ω 50 + 1 β +1 02.05.2012 AEE01 Advanced Electrical Engineering Norton Equivalent Circuit Applying KCL, in = i1 + βi1 = G1v i + βG1vi = G1 (β +1)v i = v i (β +1) R1 short circuit Short circuit at the output causes zero current to flow through RS. Rth is equal to Rth found earlier. 50 +1 vi vi = = (2.55 mS)v i in = 20 kΩ 392 Ω Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Final Thevenin and Norton Equivalent Circuits Check of Results: Note that vth = inRth and this can be used to check the calculations: inRth=(2.55 mS)vi(282 Ω) = 0.719vi, accurate within round-off error. While the two circuits are identical in terms of voltages and currents at the output terminals, there is one difference between the two circuits. With no load connected, the Norton circuit still dissipates power! Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Find Thevenin and Norton Equivalent Circuits! Rth , vth , in Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Source Transformation (1) • An equivalent circuit is one whose v-i characteristics are identical with the original circuit. • It is the process of replacing a voltage source vS in series with a resistor R by a current source iS in parallel with a resistor R, or vice versa. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Source Transformation (2) + + - - (a) Independent source transform + + - - (b) Dependent source transform Michael E.Auer 02.05.2012 • The arrow of the current source is directed toward the positive terminal of the voltage source. • The source transformation is not possible when R = 0 for voltage source and R = ∞ for current source. AEE01 Advanced Electrical Engineering Source Transformation Example Find io in the circuit shown below using source transformation. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Superposition Theorem (1) It states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltage across (or currents through) that element due to EACH independent source acting alone. The principle of superposition helps us to analyze a linear circuit with more than one independent source by calculating the contribution of each independent source separately. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Superposition Theorem (2) We consider the effects of 8A and 20V one by one, then add the two effects together for final vo. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Superposition Theorem (3) Steps to apply superposition principle 1. Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis. 2. Repeat step 1 for each of the other independent sources. 3. Find the total contribution by adding algebraically all the contributions due to the independent sources. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Superposition Theorem (4) Two things have to be keep in mind: 1. When we say turn off all other independent sources: Independent voltage sources are replaced by 0 V (short circuit) and Independent current sources are replaced by 0 A (open circuit). 2. Dependent sources are left intact because they are controlled by circuit variables. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Superposition Theorem Example 1 Example 2 Use the superposition theorem to find v in the circuit shown below. 3A is discarded by open-circuit 6V is discarded by short-circuit Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Superposition Theorem Example 2 Use superposition to find vx in the circuit.. 2A is discarded by open-circuit 20 Ω 10 V 10V is discarded by open-circuit 20 Ω v1 + − 4Ω 0.1v1 v2 2A 4Ω 0.1v2 (b) (a) Michael E.Auer Dependant source keep unchanged !!! 02.05.2012 AEE01 Advanced Electrical Engineering Chapter Content Michael E.Auer • Introduction • Complex Numbers and Phasors • Circuit Theory Review • Methods of Network Analysis • Locus Diagrams • Circuit Element Variations 02.05.2012 AEE01 Advanced Electrical Engineering Introduction Things we need to know in solving any resistive circuit with current and voltage sources only: Number of equations • Ohm’s Law b • Kirchhoff’s Current Laws (KCL) n-1 • Kirchhoff’s Voltage Laws (KVL) b – (n-1) mesh = independend loop Number of branch currents and branch voltages = 2b (variables) Problem: Number of equations! Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Mesh Analysis (1) 1. Mesh analysis provides a general procedure for analyzing circuits using mesh currents as the circuit variables. 2. Mesh analysis applies KVL to find unknown currents. 3. A mesh is a loop which does not contain any other loops within it (independent loop). Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Mesh Analysis (2) Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Mesh Analysis (3) Example – circuit with independent voltage sources Equations: R1∙i1 + (i1 – i2) ∙ R3 = V1 R2 ∙ i2 + R3 ∙(i2 – i1) = -V2 reordered: (R1+ R3) ∙ i1 - i2 ∙ R3 = V1 - R3 ∙ i1 + (R2 + R3)∙i2 = -V2 Note: i1 and i2 are mesh current (imaginative, not measurable directly) I1, I2 and I3 are branch current (real, measurable directly) I1 = i1; I2 = i2; I3 = i1 - i2 Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Mesh Analysis (4) Formalization: Network equations by inspection. − R3 i1 V1 ( R1 + R3 ) ⋅ = ( R2 + R3 ) i2 − V2 − R3 Impedance matrix Excitation Mesh currents General rules: 1. Main diagonal: ring resistance of mesh n 2. Other elements: connection resistance between meshes n and m • Sign depends on direction of mesh currents! Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Mesh Analysis (5) Example: By inspection, write the mesh-current equations in matrix form for the circuit below. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Mesh Analysis Special Cases dependent source ideal voltage source Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Mesh Analysis Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Nodal Analysis (1) It provides a general procedure for analyzing circuits using node voltages as the circuit variables. Example 3 Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Nodal Analysis (2) Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Nodal Analysis (3) Example Apply KCL at node 1 and 2 v1 G3 G1 Michael E.Auer v2 G1∙v1 + (v1 – v2) ∙ G3 = 1A G2 ∙ v2 + G3 ∙(v2 – v1) = - 4A reordered: (G1+ G3) ∙ v1 - v2 ∙ G3 = 1A - G3 ∙ v1 + (G2 + G3)∙v2 = - 4A G2 02.05.2012 AEE01 Advanced Electrical Engineering Nodal Analysis (4) Formalization: Network equations by inspection. − G3 v1 1A (G1 + G3 ) ⋅ = (G2 + G3 ) v2 − 2A − G3 Admittance matrix Excitation Node voltages General rules: 1. Main diagonal: sum of connected admittances at node n 2. Other elements: connection admittances between nodes n and m • Sign: negative! Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Nodal Analysis (5) Example: By inspection, write the node-voltage equations in matrix form for the circuit below. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Nodal Analysis Special Case dependent source Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering AC Network Nodal Analysis Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Chapter Content Michael E.Auer • Introduction • Complex Numbers and Phasors • Circuit Theory Review • Methods of Network Analysis • Locus Diagrams • Circuit Element Variations 02.05.2012 AEE01 Advanced Electrical Engineering Phasor Inversion Example Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering R-C-Circuit Locus Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Locus Inversion Theorems 1. The inversion of a straight line through the origin is again a line through the origin. 2. The inversion of a straight line not through the origin is a circle through the origin and vice versa. 3. The inversion of a circle that does not pass through the origin is again a circle that does not pass through the zero point Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Inversion of a Straight Line through the origin Michael E.Auer not through the origin 02.05.2012 AEE01 Advanced Electrical Engineering Inversion of a Circle not Through the Origin Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Locus of a Low-Pass Transfer Function Ua 1 = T ( jω ) = U e 1 + jΩ Michael E.Auer 02.05.2012 mit ω Ω= ωg AEE01 Advanced Electrical Engineering Locus of a Parallel-Series Circuit (1) Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Locus of a Parallel-Series Circuit (2) Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Equivalent Networks Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Dual Networks (1) In other words, both circuits are described by the same pair of equations: Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Dual Networks (2) Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Dual Networks (3) Rules for obtaining the dual of a planar circuit, regardless of wether or not it is a series-parallel network: Rule1: Insight of each mesh, including the infinite region surrounding the circuit, place a node. Rule2: Suppose two of this nodes, for example a and b, are in adjacent meshes. Then there is at least one element in the boundary common to these two meshes. Place the dual of each common element between nodes a and b. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering I-V-Relations in Dual Nezworks ∑ iµ = 0 ∑ vν = 0 v = vq i = iq µ v = R ⋅i v=L ν di dt i=C i ′ = iq′ Given Network i-v-Duality Dual Network ∑µ uµ′ = 0 ∑ν iν′ = 0 dv dt i ′ = G ′ ⋅ u′ i′ = C′ du ′ dt u′ = L′ di ′ dt v' = Z 0 ⋅ i und i′ = v Z0 For example Z0 = 1Ω u ′ = uq′ Duality relations for the basic network elements: G′ = Michael E.Auer R Z 02 C′ = L Z 02 L′ = C ⋅ Z 02 iq′ = 02.05.2012 vq Z0 vq' = Z 0 ⋅ iq AEE01 Advanced Electrical Engineering Chapter Content Michael E.Auer • Introduction • Complex Numbers and Phasors • Circuit Theory Review • Methods of Network Analysis • Locus Diagrams • Circuit Element Variations 02.05.2012 AEE01 Advanced Electrical Engineering Circuit Element Variations • All electronic components have manufacturing tolerances. o Resistors can be purchased with ± 10%, ± 5%, and ± 1% tolerance. (IC resistors are often ± 10% and more.) o Capacitors can have asymmetrical tolerances such as +20%/-50%. o Power supply voltages typically vary from 1% to 10%. • • • Device parameters will also vary with temperature and age. Circuits must be designed to accommodate these variations. Mainly Worst-case Analysis and Monte Carlo Analysis (statistical) are used to examine the effects of component parameter variations. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Tolerance Modeling • For symmetrical parameter variations Pnom(1 - ε) ≤ P ≤ Pnom(1 + ε) • For example, a 10 kΩ resistor with ±5% percent tolerance could take on the following range of values: 10kΩ(1 - 0.05) ≤ R ≤ 10kΩ(1 + 0.05) 9500 Ω ≤ R ≤ 10500 Ω Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Numeric Precision • Michael E.Auer Most circuit parameters vary from less than ± 1 % to greater than ± 50%. • As a consequence, more than three significant digits is meaningless. • Results should be represented with three significant digits: 2.03 mA, 5.72 V, 0.0436 µA, and so on. 02.05.2012 AEE01 Advanced Electrical Engineering Circuit Analysis with Tolerances • Worst-case analysis – Parameters are manipulated to produce the worst-case min and max values of desired quantities. – This can lead to over design since the worst-case combination of parameters is rare. – It may be less expensive to discard a rare failure than to design for 100% yield. • Monte-Carlo analysis – Parameters are randomly varied to generate a set of statistics for desired outputs. – The design can be optimized so that failures due to parameter variation are less frequent than failures due to other mechanisms. – In this way, the design difficulty is better managed than a worst-case approach. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Worst Case Analysis Example Problem: Find the nominal and worstcase values for output voltage and source current. Solution: • Known Information and Given Data: Circuit topology and values in figure. • Nominal voltage solution: Unknowns: nom O V Michael E.Auer min O , V max O , V , I nom I , I min I , I max I 02.05.2012 nom O V R1nom =V R1nom + R2nom 18kΩ = 15V = 5V 18kΩ + 36kΩ nom I AEE01 Advanced Electrical Engineering Monte Carlo Analysis • • • Parameters are varied randomly and output statistics are gathered. We use programs like MATLAB, Mathcad, SPICE, or a spreadsheet to complete a statistically significant set of calculations. For example, with Excel, a resistor with a nominal value Rnom and tolerance ε can be expressed as: R = Rnom (1+ 2ε ( RAND() − 0.5)) The RAND() function returns random numbers uniformly distributed between 0 and 1. Michael E.Auer 02.05.2012 AEE01 Advanced Electrical Engineering Monte Carlo Analysis Results VO (V) Average 4.96 Nominal 5.00 Standard Deviation 0.30 Maximum 5.70 W/C Maximum 5.87 Minimum 4.37 W/C Minimum 4.20 Histogram of output voltage from 1000 case Monte Carlo simulation. Michael E.Auer 02.05.2012 AEE01