1 Module 14 - Circuits with Resistors, Inductors, and Capacitors (RLC)

2 Capacitor and Resistor Exponential Forms t vCvC t vCvC Inductor and Resistor Exponential Forms v C = V o e  t /RC v C = V o (1  e  t /RC ) t iLiL t iLiL i L = I o e  t R/L i L = I o (1  e  t /RC ) One Reactive Element…

3 Behavior Very Different Capacitor, Inductor, and Resistor Typical Response  decaying exponential sinusoid t v C or i L Let’s look at a circuit example… RLC NETWORK:

4 Series RLC Circuit L C R +vC–+vC– +vL–+vL– + v R – VoVo t = 0 i(t)i(t) Find i (t) Objective: Initial Conditions v C = 0 i L = 0

5 Governing Equation: L C R +vC–+vC– +vL–+vL– + v R – VoVo t = 0 i(t)i(t) Find i (t) Objective: V o = v R + v C + v L KVL: Ohm’s Law: Inductor Eqn: L i d dt R i + v C + = V o Q: How do we take care of v C term? A: Take of the entire equation! d dt L i d2d2 dt = R i d dt v C d dt dV o dt

6 L i d2d2 dt = R i d dt i C dV o dt Capacitor equation Rearrange + Divide by L L i d2d2 dt = R i d dt v C d dt dV o dt Transform some more…. + + = dV o dt i LC i d dt L R i d2d2 dt 2 L 1 zero Note condition at t =  ( and = 0 ) d dt d2d2 dt 2 i = 0 Makes sense: Capacitor is a dc open VoVo L C R +vC–+vC– +vL–+vL– + v R – i(t)i(t) = 0

7 + + = i LC i d dt L R i d2d2 dt 2 L 1 = 0 Let’s solve it! Here’s our differential equation e st (Same approach as before:) Try solution of the form * * (Suggested by mathematicians who came before us…) i = A e st d dt e st = s e st d2d2 dt 2 e st = s 2 e st Note: s 2 Ae st Ae st L R s LC 1 Ae st ++= 0 s2s2 LC 1 L R s ++= 0 Will be determined by initial conditions A can be anything… s must satisfy:

8 x = –b –  b 2 – 4ac 2a2a x = –b +  b 2 – 4ac 2a2a s2s2 LC 1 L R s ++= 0 Characteristic Equation (of the circuit ) VoVo L C R +vC–+vC– +vL–+vL– + v R – i(t)i(t) Quadratic Equation: ax 2 + bx + c = 0 Root #1Root #2 a1 b L R c LC 1 Root #1 2L R –+ R LC 1 – 2  Root #2 2L R –– R LC 1 – 2 

9 Root #1 2L R LC 1  R –+ – 2 s 1 = Root #2 2L R –– R LC 1 – 2  s 2 = i = A e st L C R +vC–+vC– +vL–+vL– + v R – VoVo t = 0 i(t)i(t) Two solutions: Which one is correct?

10 Root #1 2L R LC 1 R –+  – 2 s 1 = Root #2 2L R –– R LC 1 – 2  s 2 = i (t) = A e st Two cases to consider: 2L R LC 1 > = REAL Positive Number  2L R LC 1 < = IMAGINARY Negative Number  Let’s look at these cases one at a time…

11 Root #1 2L R LC 1 R –+  – 2 s 1 = Root #2 2L R –– R LC 1 – 2  s 2 = 2L R LC 1 > = REAL Positive Number  L C R 1  F 12 H 10 k  V o 5 V i(t)i(t) 2L R = 417 LC 1 = 83,300 s 1 = –116s 2 = – 717

12 L C R 1  F 12 H 10 k  V o 5 V i(t)i(t) s 1 = –116s 2 = – 717 i(t)i(t) = A 1 e s1ts1t + A 2 e s2ts2t Most General Solution: Linear Superposition Must satisfy the Initial Conditions: i (0) = 0 +vC–+vC– v C (0) = 0

13 VoVo L C R i(t)i(t) +vC–+vC– i L (0) = 0 v C (0) = 0 v L (0) = V o +vL–+vL– By KVL: (The voltage that’s left over after v C = 0 and v R = 0) Initial Conditions: i (0) = 0 v C (0) = 0 (The initial condition on v C translates into an IC for d i /dt ) This is good! 2 nd order differential equation  Need I.C. for i and to solve. d i dt i (0) = 0 v R (0) = 0 Ohm’s Law: t =0 d i dt = VoVo L

14 s 1 = –116 sec –1 s 2 = – 717 sec –1 i(t)i(t) = A 1 e s1ts1t + A 2 e s2ts2t i (0) = 0 A 1 + A 2 = 0 t =0 d i dt = VoVo L s 1 A 1 + s 2 A 2 = VoVo L Solve for A 1 s 2 A 1 + s 2 A 2 = 0() (s 1 –s 2 )A 1 = VoVo L A 1 = VoVo (s 1 –s 2 ) L s 1 = – 116 sec – 1 s 2 = – 717 sec –1 V o = 5 V L = 12 H Find A 1 = 0.7 mA

15 s 1 = –116 sec –1 s 2 = – 717 sec –1 A 1 + A 2 = 0 Find A 2 A 1 = 0.7 mA A 2 = –0.7 mA i(t)i(t) = A 1 e s1ts1t + A 2 e s2ts2t This term decays much more quickly i(t)i(t) = 0.7 ( e –166t – e –717t ) mA

t (ms) i(t)i(t) = 0.7 ( e –166t – e –717t ) mA Overdamped Response

– 0.60 – 0.40 – – 0.8 i(t)i(t) = 0.7 ( e –166 t – e –717 t ) mA i 1 (t) + i 2 (t) i1(t)i1(t) = 0.7 e –166t i2(t)i2(t) = – 0.7 e –717 t

18 Root #1 2L R LC 1 R –+  – 2 s 1 = Root #2 2L R –– R LC 1 – 2  s 2 = 2L R LC 1 > = REAL Positive Number  L C R 1  F 12 H 10 k  V o 5 V i(t)i(t) 2L R = 417 2L R LC 1 < = IMAGINARY Negative Number  1 k  41.7 LC 1 = 83,300 (same) s 1 = –  – 81,600 s 2 = – 41.7 –  – 81,600 Huh?

19  – 81,600 Huh?  –81,600 How to deal with this term…  –81,600  –1 –1 i j (Mathematicians) (Engineers) s 1 = –  + j  o s 2 = –  – j  o  = 42 units: sec –1  –1 –1  –81,600  s 1 = – 42 + j  81,600 s 2 = – 42 – j  – 81,600 (41.7)  o =  81,600 = 286

20 i(t)i(t) = A 1 e s1ts1t + A 2 e s2ts2t e s1ts1t = e (–  jo)tjo)t + Huh? (again) = e –t–t jotjot e x e = 1 + x x2x2 2! x3x3 3! x4x4 4! x5x5 5! x6x6 6! … x e = 1 + x x2x2 2! x3x3 3! x4x4 4! x5x5 5! x6x6 6! … j 2 = –1 j 4 = +1 j 6 = –1 j x e = 1 – x x2x2 2! x3x3 3! x4x4 4! x5x5 5! x6x6 6! +–+ … + j +– …– s 1 = –  + j  o s 2 = –  – j  o n=0 n=1 n=2 n ! = n(n–1)(n–2)(n–3)..(1) j 2 j 4 j 6

21 j x e = 1 – x x2x2 2! x3x3 3! x4x4 4! x5x5 5! x6x6 6! +–+ … + j +– …– cos x = 1 – x2x2 2! x4x4 4! x6x6 6! +–+ … x x3x3 3! x5x5 5! +– …– sin x = Basic definitions of sine and cosine j x e= cos x+ j sin x “Euler’s Equation” j  o t e= cos  o t + j sin  o t Back to the circuit…

22 s 1 = –  + j  o s 2 = –  – j  o  = 42  o = 16.9 A1eA1e s1ts1t = A 1 e (–  jo)tjo)t + cos  o t + j sin  o t e –t–t = A 1 – Soln #1: cos  t e –t–t 2 (A 1 + A 2 ) e –t–t cos  o t = K 1 sin  o t e –t–t = K 2 sin  o t e –t–t + K 2 cos  o t e –t–t K 1 i (t) = General Solution: Do some linear algebra… VoVo L C R +vC–+vC– +vL–+vL– + v R – i(t)i(t) Soln #2: sin  t e –t–t 2 j (A 1 – A 2 ) A2eA2e s2ts2t = A 2 e (–  jo)tjo)t cos  o t – j sin  o t e –t–t = A 2 – A2eA2e s2ts2t = A 1 e (–  jo)tjo)t cos  o t – j sin  o t e –t–t = A A1eA1e s1ts1t = A 1 e (–  jo)tjo)t + cos  o t + j sin  o t e –t–t = A 1

23 sin  o t e –t–t + K 2 cos  o t e –t–t K 1 i (t) = General Solution: Initial Condition #1: i (0) = 0 K 1 = 0 Equal 0 at t = 0 Equals 1 at t = 0 Initial Condition #2: t =0 d i dt = VoVo L Equals 0 at t = 0 cos  o t  o e –t–t + sin  o t –  e –t–t K 2 Equals 1 at t = 0 K2K2 oo = VoVo L K2K2 = VoVo  o L Find derivative of i (t) Equals 1 at t = 0

t (ms) –0.5 –1.0 0 K2e–tK2e–t envelope Period 22 oo sin  o t e –t–t i (t) = VoVo  o L Underdamped Response

25 Root #1 2L R LC 1 R –+  – 2 s 1 = Root #2 2L R –– R LC 1 – 2  s 2 = R 1 2L LC = 0  Two Double Roots 2L R – s 1 = s 2 = Critically Damped Response (Boundary between Overdamped and Underdamped Responses)

26 End of This Module Homework