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Schedule… Date Day 6 Oct Mon 7 Oct Tue 8 Oct Wed 9 Oct Thu 10 Oct Fri 11 Oct Sat 12 Oct Sun 13 Oct Mon Class No. 10 Title Energy Storage Chapters HW Due date 3.7, 4.1 Lab Due date NO LAB NO LAB 11 Dynamic Circuits Recitation 12 4.2 – 4.4 HW 4 Exam 1 Review LAB 4 14 Oct ECEN 301 Exam EXAM 1 Tue Discussion #10 – Energy Storage 1 Energy Moro. 7: 48 48 Wherefore, my beloved brethren, pray unto the Father with all the energy of heart, that ye may be filled with this love, which he hath bestowed upon all who are true followers of his Son, Jesus Christ; that ye may become the sons of God; that when he shall appear we shall be like him, for we shall see him as he is; that we may have this hope; that we may be purified even as he is pure. Amen. ECEN 301 Discussion #10 – Energy Storage 2 Lecture 10 – Energy Storage Maximum Power Transfer Capacitors Inductors ECEN 301 Discussion #10 – Energy Storage 3 Maximum Power Transfer Equivalent circuit representations are important in power transfer analysis RT i + vT + – v Load – Ideally all power from source is absorbed by the load But some power will be absorbed by internal circuits (represented by RT) ECEN 301 Discussion #10 – Energy Storage 4 Maximum Power Transfer Efficiently transferring power from source to load means that internal resistances (RT) must be minimized i.e. for a given RL we want RT as small as possible! RT vT + – iL RL For a given RT there is a specific RL that will maximize power transfer to the load ECEN 301 Discussion #10 – Energy Storage 5 Maximum Power Transfer Consider the power (PL) absorbed by the load RT vT + – ECEN 301 iL RL Discussion #10 – Energy Storage 6 Maximum Power Transfer Consider the power (PL) absorbed by the load PL iL RL 2 RT vT + – BUT iL RL vT iL RL RT Therefore 2 vT PL RL 2 RL RT ECEN 301 Discussion #10 – Energy Storage 7 Maximum Power Transfer The maximum power transfer can be found by calculating dPL/dRL = 0 dPL vT RL RT 2vT RL RL RT 0 4 dRL RL RT 2 RT 2 2 OR vT + – iL RL RL RT 2 2 RL RL RT 0 OR NB: assumes RT is fixed and RL is variable ECEN 301 RL RT Discussion #10 – Energy Storage 8 Maximum Power Transfer 0.30 Normalized Power (PL/VT2) 0.25 When RL < RT: PL is lowered since vL goes down 0.20 0.15 When RL > RT: 0.10 PL is lowered since less current (iL) flows 0.05 0.00 0.0 1.0 2.0 3.0 4.0 5.0 6.0 Normalized Resistance (RL/RT) To transfer maximum power to the load, the source and load resistors must be matched (i.e. RT = RL). Power is attenuated as RL departs from RT ECEN 301 Discussion #10 – Energy Storage 9 Maximum Power Transfer Maximum power theorem: maximum power is delivered by a source (represented by its Thévenin equivalent circuit) is attained when the load (RL) is equal to the Thévenin resistance RT 2 RT 2 vT + – iL RL vT PL RL 2 RL RT PLMax vT 4 RL RL i T 2 4 NB: Substitute (RT = RL) to get PLMax ECEN 301 Discussion #10 – Energy Storage 10 Maximum Power Transfer Efficiency of power transfer (η): the ratio of the power delivered to the load (Pout), to the power supplied by the source (Pin) pout pin NB: Pout = PL ECEN 301 Discussion #10 – Energy Storage 11 Maximum Power Transfer Efficiency of power transfer (η): the ratio of the power delivered to the load (Pout), to the power supplied by the source (Pin) RT vT + – ECEN 301 iL RL Discussion #10 – Energy Storage 12 Maximum Power Transfer Efficiency of power transfer (η): the ratio of the power delivered to the load (Pout), to the power supplied by the source (Pin) Pin iL vT RT vT + – iL RL vT vT RT RL 2 vT PinMax 2 RL RL iT 2 2 NB: Substitute (RT = RL) to get PinMax ECEN 301 Discussion #10 – Energy Storage 13 Maximum Power Transfer Efficiency of power transfer (η): the ratio of the power delivered to the load (Pout), to the power supplied by the source (Pin) max RT vT + – PoutMax PinMax 2 iL RL vT 2 RL 2 4 RL vT 1 2 ECEN 301 Discussion #10 – Energy Storage In other words, at best, only 50% efficiency can be achieved 14 Maximum Power Transfer Example1: Find the maximum power delivered to RL vs = 18V, R1 = 3Ω, R2 = 6Ω, R3 = 2Ω R1 vs + – ECEN 301 R3 R2 RL Discussion #10 – Energy Storage 15 Maximum Power Transfer Example1: Find the maximum power delivered to RL vs = 18V, R1 = 3Ω, R2 = 6Ω, R3 = 2Ω 1. Find Thévenin equivalent circuit RT vT + – ECEN 301 RL R2 vT vs RT R3 R1 || R2 R1 R2 4 12V Discussion #10 – Energy Storage 16 Maximum Power Transfer Example1: Find the maximum power delivered to RL vs = 18V, R1 = 3Ω, R2 = 6Ω, R3 = 2Ω 1. 2. 3. RT vT + – RL Find Thévenin equivalent circuit Set RL = RT Calculate PLMax PL max 0.5 Pin max 2 vT 0 .5 2 RL (12) 2 0 .5 8 9W ECEN 301 Discussion #10 – Energy Storage 17 Dynamic Energy Storage Elements Capacitor and Inductor ECEN 301 Discussion #10 – Energy Storage 18 Storage Elements To this point, circuits have consisted of either: DC Sources (active element, provide energy) Resistors (passive element, absorb energy) Some passive (non-source) elements can store energy Capacitors – store voltage Inductors – store current + C – + C – +C– + L – Capacitor symbols ECEN 301 Discussion #10 – Energy Storage + L – Inductor symbols 19 Ideal Capacitor A capacitor (C) is composed of: 2 parallel conducting plates • cross-sectional area A Separated distance d by either: + • Air • Dielectric (insulating) material A d C A Dielectric (or air) – d C – Capacitance ε – relative permittivity (dielectric constant) ECEN 301 NB: the parallel plates do not touch – thus under DC current a capacitor acts like an open circuit Discussion #10 – Energy Storage 20 Ideal Capacitor Capacitance: a measure of the ability to store energy in the form of separated charge (or an electric field) q Cv The charge q+ on one plate is equal to the charge q– on the other plate + + + + ++ + + ++ + – –– –– –– –– – – – ––– Farad (F): unit of capacitance. 1 farad = 1 coulomb/volt (C/V) ECEN 301 Discussion #10 – Energy Storage 21 Ideal Capacitor No current can flow through a capacitor if the voltage across it is constant (i.e. connected to a DC source) BUT when a DC source is applied to a capacitor, there is an initial circuit current as the capacitor plates are charged (‘charging’ the capacitor) dq(t ) i (t ) dt Charging the capacitor: a current flows as electrons leave one plate to accumulate on the other. This occurs until the voltage across the capacitor = source voltage ECEN 301 vs i + – Discussion #10 – Energy Storage ++ + + – – –– – – 22 Ideal Capacitor Capacitor with an AC source: since AC sources periodically reverse voltage directions, the capacitor plates will change charge polarity accordingly With each polarity change, the capacitor goes through a ‘charging’ state, thus current continuously flows Just as with a DC source, no electrons cross between the plates Just as with a DC source, voltage across a capacitor cannot change instantaneously i vs(t) + – ECEN 301 i ++ + + – – –– – – vs(t) – + Discussion #10 – Energy Storage –– – – + + +++ 23 Ideal Capacitor i – v characteristic for an ideal capacitor Differentiate both sides q (t ) Cv(t ) dq(t ) dCv(t ) dt dt dv(t ) i (t ) C dt ECEN 301 OR BUT dq(t ) i (t ) dt 1 t v(t ) iC ( )d C Discussion #10 – Energy Storage 24 Ideal Capacitor i – v characteristic for an ideal capacitor 1 t v(t ) iC ( )d C NB: Assumes we know the value of the capacitor from time (τ = -∞) until time (τ = t) Instead 1 t v(t ) iC ( )d v(t0 ) C t0 Initial condition Initial time (often t0 = 0) ECEN 301 Discussion #10 – Energy Storage 25 Ideal Capacitor Example2: calculate the current through the capacitor C = 0.1μF with the voltage as shown: v(t) = 5(1-e-t/10-6) 5.0 voltage (V) 4.0 3.0 2.0 1.0 0.0 0.00 2.00 4.00 6.00 time (us) ECEN 301 Discussion #10 – Energy Storage 26 Ideal Capacitor Example2: calculate the current through the capacitor C = 0.1μF with the voltage as shown: v(t) = 5(1-e-t/10-6) 5.0 dv(t ) i (t ) C dt voltage (V) 4.0 3.0 2.0 1.0 0.0 0.00 2.00 4.00 time (us) ECEN 301 5 t /106 (10 ) 6 e 10 7 6.00 0.5e Discussion #10 – Energy Storage t / 106 A 27 Ideal Capacitor 5.0 0.5 4.0 0.4 current (A) voltage (V) Example2: calculate the current through the capacitor C = 0.1μF with the voltage as shown: v(t) = 5(1-e-t/10-6) 3.0 2.0 1.0 0.0 0.00 0.3 0.2 0.1 2.00 4.00 6.00 0.0 0.00 time (us) 4.00 6.00 time (us) NB: the capacitor’s current jumps ‘instantaneously’ to 0.5A. The ability of a capacitor’s current to change instantaneously is an important property of capacitors ECEN 301 2.00 Discussion #10 – Energy Storage i(t ) 0.5e t /106 A 28 Ideal Capacitor Example3: find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0 1.2 current (A) 1.0 0.8 0.6 0.4 0.2 0.0 0.0 1.0 2.0 3.0 time (s) ECEN 301 Discussion #10 – Energy Storage 29 Ideal Capacitor Example3: find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0 1.2 current (A) 1.0 0.8 0.6 0.4 0.2 0.0 0.0 1.0 2.0 time (s) 3.0 Interval t0 0 t 1 1 t 2 2t current i(t) 0 t 1 0 1 t v(t ) id v(0) C 0 ECEN 301 Discussion #10 – Energy Storage 30 Ideal Capacitor Example3: find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0 1.2 Interval t0 current (A) 1.0 0.8 0.6 0 t 1 0.4 0.2 0.0 0.0 1.0 2.0 time (s) 3.0 1 t 2 2t voltage v(t) 0 t 2 d 0 0 t 2 (1)d v(1) 0 0 v(2) 1 t v(t ) id v(0) C 0 ECEN 301 Discussion #10 – Energy Storage 31 Ideal Capacitor Example3: find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0 1.2 current (A) 1.0 0.8 0.6 0.4 0.2 0.0 0.0 1.0 2.0 time (s) ECEN 301 3.0 Interval t0 0 t 1 1 t 2 2t Discussion #10 – Energy Storage voltage v(t) 0 t2 2t 1 3 32 Ideal Capacitor 1.2 3.5 1.0 3.0 voltage (V) current (A) Example3: find the voltage v(t) for a capacitor C = 0.5F with the current as shown and v(0) = 0 0.8 0.6 0.4 0.2 2.5 2.0 1.5 1.0 0.5 0.0 0.0 0.0 1.0 2.0 3.0 0.0 1.0 time (s) 3.0 4.0 time (s) NB: The final value of the capacitor voltage after the current source has stopped charging the capacitor depends on two things: 1. The initial capacitor voltage 2. The history of the capacitor current ECEN 301 2.0 Interval t0 0 t 1 1 t 2 2t Discussion #10 – Energy Storage voltage v(t) 0 t2 2t 1 3 33 Series Capacitors Capacitor Series Rule: two or more circuit elements are said to be in series if the current from one element exclusively flows into the next element. Capacitors in series add the same way resistors in parallel add 1 1 1 1 1 C EQ C1 C2 C3 CN C1 C2 C3 C EQ 1 1 1 1 1 C1 C2 C3 CN Cn ∙∙∙ CN ∙∙∙ CEQ ECEN 301 Discussion #10 – Energy Storage 34 Parallel Capacitors Parallel Rule: two or more circuit elements are said to be in parallel if the elements share the same terminals Capacitors in parallel add the same way resistors in series add N C EQ C n n 1 C1 ECEN 301 C2 C3 Cn CN Discussion #10 – Energy Storage CEQ 35 Parallel and Series Capacitors Example4: determine the equivalent capacitance CEQ C1 = 2mF, C2 = 2mF, C3 = 1mF, C4 = 1/3mF, C5 = 1/3mF, C6 = 1/3mF C1 C4 C3 C2 ECEN 301 C6 C5 Discussion #10 – Energy Storage 36 Parallel and Series Capacitors Example4: determine the equivalent capacitance CEQ C1 = 2mF, C2 = 2mF, C3 = 1mF, C4 = 1/3mF, C5 = 1/3mF, C6 = 1/3mF C1 C4 C3 C2 ECEN 301 C4C5C6 C EQ1 C4C5 C4C6 C5C6 C6 C5 (1 / 3)(1 / 3)(1 / 3) (1 / 3)(1 / 3) (1 / 3)(1 / 3) (1 / 3)(1 / 3) 1 / 27 1/ 3 1 mF 9 Discussion #10 – Energy Storage 37 Parallel and Series Capacitors Example4: determine the equivalent capacitance CEQ C1 = 2mF, C2 = 2mF, C3 = 1mF, C4 = 1/3mF, C5 = 1/3mF, C6 = 1/3mF C EQ 2 C3 C EQ1 C1 C3 CEQ1 C2 1 C EQ1 mF 9 ECEN 301 Discussion #10 – Energy Storage 1 (1) 9 10 mF 9 38 Parallel and Series Capacitors Example4: determine the equivalent capacitance CEQ C1 = 2mF, C2 = 2mF, C3 = 1mF, C4 = 1/3mF, C5 = 1/3mF, C6 = 1/3mF C1 C EQ CEQ2 C2 C EQ 2 ECEN 301 10 mF 9 C1C2C EQ 2 C1C2 C1C EQ 2 C2C EQ 2 (2)( 2)(10 / 9) (2)( 2) (2)(10 / 9) (2)(10 / 9) 40 / 9 4 20 / 9 20 / 9 10 mF 19 Discussion #10 – Energy Storage 39 Parallel and Series Capacitors Example4: determine the equivalent capacitance CEQ C1 = 2mF, C2 = 2mF, C3 = 1mF, C4 = 1/3mF, C5 = 1/3mF, C6 = 1/3mF CEQ ECEN 301 C EQ 10 mF 19 Discussion #10 – Energy Storage 40 Energy Storage in Capacitors Capacitor energy WC(t): can be found by taking the integral of power Instantaneous power PC = iv t WC (t ) PC ( )d t vC ( )iC ( )d t vC ( )C C v (t ) vdv v ( ) 1 Cv 2 2 ECEN 301 v (t ) dvC ( ) d d Since the capacitor is uncharged at t = -∞, v(- ∞) = 0, thus: 1 WC (t ) Cv(t ) 2 J 2 v ( ) Discussion #10 – Energy Storage 41 Ideal Inductor Inductors are made by winding a coil of wire around a core The core can be an insulator or ferromagnetic material + N A 2 L l 0.45d N turns l – L – Inductance μ – relative permeability • Cross-sectional area A • Diameter d ECEN 301 Discussion #10 – Energy Storage 42 Ideal Inductor Inductance: a measure of the ability of a device to store energy in the form of a magnetic field Ideally the resistance through an inductor is zero (i.e. no voltage drop), thus an inductor acts like a short circuit in the presence of a DC source. i BUT there is an initial voltage across the inductor as the current builds up (much like ‘charging’ with capacitors) Henry (H): unit of inductance. 1 henry = 1 volt-second/ampere (V-s/A) ECEN 301 Discussion #10 – Energy Storage 43 Ideal Inductor Inductor with an AC source: since AC sources periodically reverse current directions, the current flow through the inductor also changes With each current direction change, the current through the inductor must ‘build up’, thus there is a continual voltage drop across the inductor Just as with a DC source, current across an inductor cannot change instantaneously i i vs(t) + – ECEN 301 vs(t) – + Discussion #10 – Energy Storage 44 Ideal Inductor I – v characteristic for an ideal inductor di (t ) v(t ) L dt 1 t i (t ) vL ( )d L NB: note the duality between inductors and capacitors ECEN 301 Discussion #10 – Energy Storage 45 Ideal Inductor i – v characteristic for an ideal inductor 1 t i (t ) vL ( )d L NB: Assumes we know the value of the inductor from time (τ = -∞) until time (τ = t) Instead 1 t i (t ) vL ( )d i (t0 ) L t0 Initial condition Initial time (often t0 = 0) ECEN 301 Discussion #10 – Energy Storage 46 Ideal Inductor Example5: find the voltage across an inductor L = 0.1H when the current is: i(t) = 20 t e-2t current (A) 4.0 2.0 0.0 0.0 0.5 1.0 1.5 time (s) ECEN 301 Discussion #10 – Energy Storage 47 Ideal Inductor Example5: find the voltage across an inductor L = 0.1H when the current is: i(t) = 20 t e-2t current (A) 4.0 2.0 0.0 0.0 0.5 1.0 1.5 di (t ) v(t ) L dt d 0.1 (20te 2t ) dt 2(2te 2t e 2t ) 2e 2t (1 2t )V time (s) ECEN 301 Discussion #10 – Energy Storage 48 Ideal Inductor Example5: find the voltage across an inductor L = 0.1H when the current is: i(t) = 20 t e-2t 1.5 voltage (V) current (A) 4.0 2.0 0.0 0.0 0.0 0.5 1.0 1.5 0.5 1.5 time (s) time (s) NB: the inductor’s voltage jumps ‘instantaneously’ to 2V. The ability of an inductor’s voltage to change instantaneously is an important property of inductors ECEN 301 1.0 -0.5 Discussion #10 – Energy Storage v(t ) 2e 2t (1 2t )V 49 Series Inductors Series Rule: two or more circuit elements are said to be in series if the current from one element exclusively flows into the next element. Inductors in series add the same way resistors in series add N LEQ Ln n 1 L1 L2 L3 Ln ∙∙∙ LN ∙∙∙ LEQ ECEN 301 Discussion #10 – Energy Storage 50 Parallel Inductors Parallel Rule: two or more circuit elements are said to be in parallel if the elements share the same terminals Inductors in parallel add the same way resistors in parallel add 1 1 1 1 1 LEQ L1 L2 L3 LN L1 ECEN 301 L2 L3 LEQ Ln 1 1 1 1 1 L1 L2 L3 LN LN Discussion #10 – Energy Storage LEQ 51 Energy Storage in Capacitors Capacitor energy WL(t): can be found by taking the integral of power Instantaneous power PL = iv t WL (t ) PL ( )d t vL ( )iL ( )d t L L i (t ) diL ( ) iL ( )d d idi i ( ) 1 Li 2 2 ECEN 301 i (t ) Since the inductor is uncharged at t = -∞, i(- ∞) = 0, thus: 1 WL (t ) Li(t ) 2 J 2 i ( ) Discussion #10 – Energy Storage 52 Energy Storage in Capacitors Example6: calculate the power and energy stored in a 0.1-H inductor when: i = 20 t e-2t A (i = 0 for t < 0) V = 2 e-2t (1- 2t)V (for t >= 0) ECEN 301 Discussion #10 – Energy Storage 53 Energy Storage in Capacitors Example6: calculate the power and energy stored in a 0.1-H inductor when: i = 20 t e-2t A (i = 0 for t < 0) V = 2 e-2t (1- 2t)V (for t >= 0) PL (t ) vL (t )iL (t ) 2e 2t (1 2t ) 20te2t 40te4t (1 2t )W ECEN 301 Discussion #10 – Energy Storage 54 Energy Storage in Capacitors Example6: calculate the power and energy stored in a 0.1-H inductor when: i = 20 t e-2t A (i = 0 for t < 0) V = 2 e-2t (1- 2t)V (for t >= 0) PL (t ) vL (t )iL (t ) 2e 2t (1 2t ) 20te2t 4t 40te (1 2t )W ECEN 301 1 WL (t ) LiL (t ) 2 2 1 (0.1) 20te 2t 2 20t 2e 4t J Discussion #10 – Energy Storage 2 55 Ideal Capacitors and Inductors Inductors Passive sign convention Capacitors +C– + L – i i Voltage di (t ) v(t ) L dt 1 t v(t ) iC ( )d v(t0 ) C t0 Current 1 t i (t ) vL ( )d i (t0 ) L t0 Power di (t ) PL (t ) Li(t ) dt dv(t ) i (t ) C dt dv(t ) PC (t ) Cv(t ) dt ECEN 301 Discussion #10 – Energy Storage 56 Ideal Capacitors and Inductors Inductors Energy Capacitors 1 1 2 WL (t ) Li(t ) WC (t ) Cv(t ) 2 2 2 An instantaneous change is not permitted in: Current Voltage Will permit an instantaneous change in: Voltage Current With DC source element acts as a: Short Circuit Open Circuit ECEN 301 Discussion #10 – Energy Storage 57