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Module 14 - Circuits with Resistors, Inductors, and Capacitors (RLC) 1 One Reactive Element… Capacitor and Resistor Exponential Forms vC vC = Vo et /RC t vC vC = Vo (1 et /RC) t Inductor and Resistor Exponential Forms iL iL = Io et R/L t iL iL = Io (1 et /RC) t 2 Capacitor, Inductor, and Resistor • Behavior Very Different • Typical Response RLC NETWORK: sinusoid vC or iL decaying exponential 1 0.8 0.6 0.4 0.2 0 t -0.2 -0.4 -0.6 -0.8 0 10 20 30 40 50 60 70 80 90 100 Let’s look at a circuit example… 3 Series RLC Circuit + vR – t=0 R Vo i(t) Initial Conditions + C vC vC = 0 L – + vL iL = 0 – Objective: Find i(t) 4 + vR – t=0 Objective: Find i(t) R Governing Equation: KVL: Vo = vR + vC + vL Vo i(t) + C vC L vL – + – Inductor Eqn: Ohm’s Law: R i + vC + L di = Vo dt Q: How do we take care of vC term? d A: Take of the entire equation! dt d i dvC d2i dV R + + L 2 = o dt dt dt dt 5 Transform some more…. d i dvC d2 i dV R + +L 2 = o dt dt dt dt R + vR – Vo Capacitor equation i di R + dt C + C vC L vL – + i(t) – d2i dVo +L 2 = dt dt Rearrange + Divide by L 1 dVo R di d2 i i + + == 0 2 L dt L dt LC dt Note condition at t = d2 ( ddt and dt2 = 0 ) zero i=0 Makes sense: Capacitor is a dc open 6 1 R di d2 i i + + =0 2 L L LC dt dt Here’s our differential equation Let’s solve it! (Same approach as before:) Try* solution of the form est i = A est Note: d est st = s e dt • A can be anything… s2 Aest +s d2 est 2 st dt2 = s e 1 R st Ae + Aest = 0 LC L Will be determined by initial conditions • s must satisfy: s2 1 R +s + =0 L LC * (Suggested by mathematicians who came before us…) 7 R + vR – 1 R 2 s +s + =0 LC L Vo + C vC L vL – + i(t) Characteristic Equation (of the circuit) – Quadratic Equation: ax2 + bx + c = 0 Root #1 Root #2 x = –b – b2 – 4ac 2a x = –b + b2 – 4ac 2a a R Root #1 – + 2L 1 b R 2 1 – LC 2L R L 1 c R Root #2 – – 2L LC R 2 1 – LC 2L 8 Root #1 R – s1 = 2L + Root #2 R 2 1 – LC 2L R – – s2 = 2L + vR – t=0 R Vo R 2 1 – LC 2L i(t) + C vC L vL – + – i = A est Two solutions: Which one is correct? 9 Root #1 R – s1 = 2L + R 2 1 – LC 2L Root #2 R – – s2 = 2L R 2 1 – LC 2L i(t) = A est Two cases to consider: 1 R > LC 2L Positive Number 1 R < LC 2L Negative Number = IMAGINARY = REAL Let’s look at these cases one at a time… 10 Root #1 R – s1 = 2L + Root #2 R 2 1 – LC 2L 1 R > LC 2L R – – s2 = 2L Positive Number R 2 1 – LC 2L = REAL 10 k R Vo 5V R = 417 2L 1 LC = 83,300 i(t) s1 = –116 C 1 F L 12 H s2 = – 717 11 10 k R i(t) Vo 5V s1 = –116 + C 1 F vC L 12 H – s2 = – 717 Most General Solution: Linear Superposition i(t) = A1e s1t + A2e s2t Must satisfy the Initial Conditions: i(0) = 0 vC(0) = 0 12 R i(t) Vo + C L vC vC(0) = 0 vL iL(0) = 0 – + – Ohm’s Law: i(0) = 0 By KVL: vL(0) = Vo vR(0) = 0 (The voltage that’s left over after vC = 0 and vR = 0) Initial Conditions: i(0) = 0 This is good! Vo di vdtC(0)== 0 L t=0 (The initial condition on vC translates into an IC for di/dt ) 2nd order differential equation Need I.C. for i and di to solve. dt 13 Find A1 Vo = 5 V s1 = –116 sec–1 s2 = – 717 sec–1 s1 = –116 sec–1 s2 = – 717 sec–1 L = 12 H i(t) = A1e s1t + A2e s2t i(0) = 0 Vo di = dt L t=0 A1 + A2 = 0 s1A1 + s2A2 = Vo L Solve for A1 ( s2A1 + s2A2 = 0 ) Vo (s1–s2)A1 = L Vo A1 = (s1–s2) L = 0.7 mA 14 Find A2 s1 = –116 sec–1 s2 = – 717 sec–1 A1 + A2 = 0 A1 = 0.7 mA A2 = –0.7 mA i(t) = A1e s1t + A2e s2t This term decays much more quickly i(t) = 0.7 (e –166t –e –717t ) mA 15 i(t) = 0.7 (e –717t –e ) mA 0.40 0.30 0.20 0.10 60 t (ms) 59.4 54.0 50 48.6 43.2 40 37.8 30 32.4 27.0 20 21.6 16.2 10 10.8 0 5.4 0.00 0.0 Current (mA) 0.50 –166t Time (ms) Overdamped Response 16 0.80 0.80 i1(t) = 0.7 e –166t 0.60 0.60 –166 t i(t) = 0.7 (e 0.40 0.40 0.20 0.20 00 0.00 0 -0.20 –0.20 –717 t –e ) mA i1(t) + i2(t) 10 20 30 40 50 60 –-0.40 0.40 –-0.60 0.60 i2(t) = – 0.7 e –717 t – 0.8 -0.80 17 Root #1 R – s1 = 2L + Root #2 R 2 1 – LC 2L R – – s2 = 2L R 2 1 – LC 2L R R 1 1 < > 2L 2L LC LC Negative Number Positive Number = IMAGINARY = REAL 1 k 10 k R Vo 5V R = 41.7 417 2L 1 LC = 83,300 (same) i(t) C 1 F L 12 H Huh? s1 = – 41.7 + –81,600 s2 = – 41.7 – –81,600 18 Huh? –81,600 –81,600 How to deal with this term… –81,600 –1 –81,600 –1 ij (Mathematicians) (Engineers) (41.7) s1 = – 42 + j 81,600 s1 = – + j o s2 = – 42 – j –81,600 s2 = – – j o = 42 o = 81,600 = 286 units: sec–1 19 s1 = – + j o s2 = – – j o i(t) = A1e s1t + A2e s2t e s 1t =e (– + jo)t =e –t Huh? (again) jo et n! = n(n–1)(n–2)(n–3)..(1) e x =1+x + n=0 n=1 e x =1+x + x2 + 2! x3 3! + x4 4! + x5 5! x6 + 6! +… n=2 x2 2! j 2 = –1 x2 x4 + x3 3! + x4 4! j 4 = +1 x6 + x5 5! + x6 6! +… j 6 = –1 x3 x5 jx = 1 – + 4! – 6! + … + j x – 3! + 5! – … e 2! j2 j4 j6 20 j x = 1 – x2 + x4 – x6 + … + j x – x3 + x5 – … e 2! 4! 6! 3! 5! cos x = 1 – Basic definitions of sine and cosine sin x = x – e e jx j ot = cos x + j sin x = cos ot + j sin ot x2 2! + x3 3! x4 4! + – x5 5! x6 6! +… –… “Euler’s Equation” Back to the circuit… 21 s1 = – + j o = 42 R s2 = – – j o o = 16.9 + vR – Vo Do some linear algebra… + C vC L vL – + i(t) – s1t = A e (– + jo)t = A e –t cos t + j sin t A1e cos o ot + j sin oot 1 1 -+- (– –– j joo))tt (– t ––t s t 2 2 = A e j sin cosot ot––j sin otot cos A2e ==AA22ee 21 Soln #1: Soln #2: 2 (A1+ A2) e –t cos t = K1 e –t cos ot 2 j (A1– A2) e –t sin t = K2 e –t sin ot General Solution: i(t) = K1 e –t cos ot + K2 e –t sin ot 22 General Solution: Equals 1 at t = 0 Equal 0 at t = 0 Equals 1 at t = 0 Equals 1 at t = 0 i(t) = K1 e –t cos ot + K2 e –t sin ot Initial Condition #1: i(0) = 0 K1 = 0 Vo d i Initial Condition #2: = dt L t=0 Find derivative of i(t) Equals 0 at t = 0 K2 – e –t Vo K2o = L Equals 1 at t = 0 Equals 1 at t = 0 –t sin ot + o e cos ot Vo K2 = o L 23 Vo –t sin ot i(t) = L e o 1.5 K2e–t envelope 1.0 0.5 0 0.0 10 20 30 40 50 60 70 80 90 100 t (ms) –0.5 –1.0 Underdamped Response Period 2 o 24 Root #1 R – s1 = 2L + R 2 1 – LC 2L 1 R = LC 2L Root #2 R – – s2 = 2L R 2 1 – LC 2L 0 Two Double Roots R – s1 = s2 = 2L Critically Damped Response (Boundary between Overdamped and Underdamped Responses) 25 End of This Module Homework 26