Source-Free RC Circuit

Transcription

1 First Order Circuits Source-Free RC Circuit Initial charge on capacitor q = Cv(0) so that voltage at time 0 is v(0). What is v(t)? Prof Carruthers BU) EK307 Notes Summer / 264

2 First Order Circuits Solving the RC circuit Applying KCL, i C + i R = 0 and so Hence C dv dt + v R = 0. dv dt = 1 RC v This is a first-order differential equation, and the solution is v(t) = Ae t/rc Prof Carruthers BU) EK307 Notes Summer / 264

3 First Order Circuits Natural Response: Decaying Exponential The value τ = RC is called the time constant The plot of v(t) = V 0 e t/τ is: Prof Carruthers BU) EK307 Notes Summer / 264

4 First Order Circuits Where did the energy in the capacitor go? The power dissipated in the resistor is p(t) = v2 R = V 0 2 R e 2t/τ and so the total energy dissipated up to time t is w(t) = t 0 p dt = 1 2 CV 2 0 (1 e 2t/τ ) Eventually, all the energy stored in the capacitor is dissipated by the resistor. Prof Carruthers BU) EK307 Notes Summer / 264

5 First Order Circuits Example Problem 1 Determine v C, v x and i 0 for t 0. Assume v C (0) = 30 V. Prof Carruthers BU) EK307 Notes Summer / 264

6 First Order Circuits Example Problem 2 Determine v(t) for t 0 and the initial energy in capacitor. Prof Carruthers BU) EK307 Notes Summer / 264

7 First Order Circuits Source-Free RL Circuit What is i(t)? Assume initial current in inductor is I 0. Prof Carruthers BU) EK307 Notes Summer / 264

8 First Order Circuits Solving the RL circuit Applying KVL, v L + v R = 0 and so L di + ir = 0. dt Hence di dt = R L i This is a first-order differential equation, and the solution is i(t) = Ae Rt/L = Ae t/(l/r) Prof Carruthers BU) EK307 Notes Summer / 264

9 First Order Circuits Natural Response: Decaying Exponential The value τ = L/R is the time constant of an RL circuit. Prof Carruthers BU) EK307 Notes Summer / 264

10 First Order Circuits Where did the energy in the inductor go? The power dissipated in the resistor is p(t) = i 2 R = I 2 0Re 2t/τ and so the total energy dissipated up to time t is w(t) = t 0 p dt = 1 2 LI2 0(1 e 2t/τ ) Eventually, all the energy stored in the inductor is dissipated by the resistor. Prof Carruthers BU) EK307 Notes Summer / 264

11 First Order Circuits Example Problem 3 Find i and v x assuming that i(0) = 5 A. Prof Carruthers BU) EK307 Notes Summer / 264

12 First Order Circuits Example Problem 4 Determine i, i 0, and v 0 Prof Carruthers BU) EK307 Notes Summer / 264

13 First Order Circuits When a Switch Closes Voltages across capacitors cannot change instantly since i = C dv dt Currents through inductors cannot change instantly since Hence, for these devices at t = 0 v = L di dt x(0 ) = x(0 + ) The other quantity can change instantly. Prof Carruthers BU) EK307 Notes Summer / 264

14 First Order Circuits Step Function u(t) The step function is defined as follows: u(t) = 0 for t < 0 u(t) = 1 for t > 0 Prof Carruthers BU) EK307 Notes Summer / 264

15 First Order Circuits Modeling Switches using u(t) Prof Carruthers BU) EK307 Notes Summer / 264

16 First Order Circuits Delta δ(t) - Step u(t) - Ramp r(t) r(t) = tu(t) r(t) = u(t) = 1{t > 0} u(t) = δ(t) = lim a 0 u(t + a) u(t a) 2a t t u(x)dx δ(x)dx dr(t) = u(t) dt du(t) = δ(t) dt - - Prof Carruthers BU) EK307 Notes Summer / 264

17 Decaying Exponential The Decaying Exponential An important signal (arising in many physical systems including circuits) is the decaying exponential defined as v(t) = V 0 e t/τ u(t) Prof Carruthers BU) EK307 Notes Summer / 264

18 Decaying Exponential Shape of Decaying Exponential Down to 1/e or 37% by time τ. Near zero by 5 τ. Prof Carruthers BU) EK307 Notes Summer / 264

19 Decaying Exponential Longer τ means longer tail Prof Carruthers BU) EK307 Notes Summer / 264

20 Decaying Exponential Step Response of RC Circuits Find the voltage v(t) assuming an initial voltage V 0 on the capacitor. [Are these circuits equivalent?] Prof Carruthers BU) EK307 Notes Summer / 264

21 Decaying Exponential Step Response of RC Circuits v(t) = { V0 t < 0 V s + (V 0 V s )e t/τ t > 0 Prof Carruthers BU) EK307 Notes Summer / 264

22 Decaying Exponential Voltage/Current Responses Assuming V 0 = 0, the voltage and current responses are: Prof Carruthers BU) EK307 Notes Summer / 264

23 Decaying Exponential Step Response of RL Circuits Find the current i(t) assuming an initial current I 0 in the inductor. [Are these circuits equivalent?] Prof Carruthers BU) EK307 Notes Summer / 264

24 Decaying Exponential Step Response of RL Circuits i(t) = { I0 t < 0 V s /R + (I 0 V s /R)e t/τ t > 0 Prof Carruthers BU) EK307 Notes Summer / 264

25 Decaying Exponential Voltage/Current Responses Assuming I 0 = 0, the current and voltage responses are: Prof Carruthers BU) EK307 Notes Summer / 264

26 Decaying Exponential Summary RC Step: v(t) = v( ) + [v(0) v( )]e t/τ for t > 0 initial capacitor voltage v(0) final capacitor voltage v( ) time constant τ = RC RL Step: i(t) = i( ) + [i(0) i( )]e t/τ for t > 0 initial inductor current i(0) final inductor current i( ) time constant τ = L/R Prof Carruthers BU) EK307 Notes Summer / 264

27 Decaying Exponential Example Problem: RC Find i(t) and v(t) Prof Carruthers BU) EK307 Notes Summer / 264

28 Decaying Exponential Example Problem: RL Find i(t) for t > 0 Prof Carruthers BU) EK307 Notes Summer / 264

29 Decaying Exponential Example Problem: Op-Amp Find v(t) and v 0 (t) Prof Carruthers BU) EK307 Notes Summer / 264

30 Second Order Circuits Second Order Circuits a second-order circuit is one with (the equivalent of) two energy storage elements. it is characterized by a second-order differential equation: d 2 x dt 2 + Adx dt + Bx = 0 x is either current i (through inductor) or voltage v (across capacitor). Prof Carruthers BU) EK307 Notes Summer / 264

31 Second Order Circuits Finding boundary conditions To solve for the response of an RLC circuit, one needs to find final values of capacitor voltage v and inductor current i initial values of v and i initial values of v (0 + ) and i (0 + ) Prof Carruthers BU) EK307 Notes Summer / 264

32 Second Order Circuits Boundaries: example 1 Find the initial and final conditions: i(0 + ) = 2 A, v(0 + ) = 4 V; i (0 + ) = 0 A/s, v (0 + ) = 20 V/s i( ) = 0 A, v( ) = 12 V Prof Carruthers BU) EK307 Notes Summer / 264

33 Second Order Circuits Boundaries: example 2 Find the initial and final conditions: i(0 + ) = 2 A, v(0 + ) = 4 V; i (0 + ) = 50 A/s, v (0 + ) = 0 V/s i( ) = 12 A, v( ) = 24 V Prof Carruthers BU) EK307 Notes Summer / 264

34 Second Order Circuits Boundaries: example 3 Find the initial and final conditions: i L (0 + ) = 0 A, v C (0 + ) = 20 V i L (0+ ) = 0 A/s, v C (0+ ) = 2 V/s, v R (0+ ) = 2/3 V/s i L ( ) = 1 A, v R ( ) = 4 V, v C ( ) = 20 V Prof Carruthers BU) EK307 Notes Summer / 264

35 Second Order Circuits Boundaries: example 4 Find the initial and final conditions: i L (0 + ) = 3 A, v C (0 + ) = 0 V, v R (0 + ) = 0 V; i L (0+ ) = 0 A/s, v C (0+ ) = 10 V/s, v R (0+ ) = 0 V/s i L ( ) = 1 A, v R ( ) = 10 V, v C ( ) = 10 V Prof Carruthers BU) EK307 Notes Summer / 264

36 Second Order Circuits Types of Second Order circuits series RLC circuit parallel RLC circuit op-amp circuits with RC general circuits with two independent storage elements Prof Carruthers BU) EK307 Notes Summer / 264

37 Second Order Circuits Source Free series RLC circuit d 2 i dt 2 + R di L dt + 1 LC i = 0 Prof Carruthers BU) EK307 Notes Summer / 264

38 Second Order Circuits Source Free Parallel RLC circuit d 2 v dt dv RC dt + 1 LC v = 0 Prof Carruthers BU) EK307 Notes Summer / 264

39 Second Order Circuits Solving the Equation d 2 x dt 2 + bdx dt + cx = 0 [This is a second-order differential equation with constant coefficients.] From experience or a good guess: try x(t) = e st. Since x (t) = se st we get (s 2 + bs + c)e st = 0 When is x(t) = e st a solution? Since x(t) = e st 0, then only when s 2 + bs + c = 0 Prof Carruthers BU) EK307 Notes Summer / 264

40 Second Order Circuits Roots of s 2 + bs + c = 0 There are two solutions s 1 and s 2 : s 1, s 2 = (b/2) ± (b/2) 2 c Label α = b/2 and c = ω0 2 then s 1 = α + α 2 ω0 2 and s 2 = α α 2 ω0 2 Prof Carruthers BU) EK307 Notes Summer / 264

41 Second Order Circuits Three Solution Types Condition Roots are: Type of response: α > ω 0 real, unequal Overdamped α = ω 0 real, even Critically Damped α < ω 0 complex conjugates Underdamped Overdamped x(t) = A 1 e s1t + A 2 e s 2t Critically damped x(t) = (A 1 + ta 2 )e s 1t Underdamped x(t) = e αt (B 1 cos(ω d t) + B 2 sin(ω d t)) For underdamped case: s = α ± ω d j Prof Carruthers BU) EK307 Notes Summer / 264

42 Second Order Circuits Method for determining step response 1 Find initial conditions: x(0) and x (0) and final value x( ). 2 Find the transient response x t (t): 1 Turn off independent sources. 2 Apply KCL/KVL to find 2nd-order DE. 3 Solve for characteristic equation roots. 4 Find x t (t) with two unknowns. 3 The response is x(t) = x t (t) + x ss (t) = x t (t) + x( ) 4 Determine constants by imposing x(0) and x (0) constraints. Prof Carruthers BU) EK307 Notes Summer / 264

43 Second Order Circuits Example series RLC R = 5 then v(t) = ( 16e t + e 4t ) R = 4, then v(t) = (1 + t)e 2t R = 1, then v(t) = 24 + ( sin 1.936t 12 cos 1.936t)e 0.5t Prof Carruthers BU) EK307 Notes Summer / 264

44 Second Order Circuits Example: Parallel RLC Circuit Find v(t) for t > 0. Answer: v(t) = 66.67(e 10t e 2.5t ) Prof Carruthers BU) EK307 Notes Summer / 264

45 Second Order Circuits Example: Series RLC Circuit Find v(t) and v R (t) for t > 0. Answer: v(t) = 10 ( sin(3.464t) + 2 cos(3.464t))e 2t, v R (t) = 2.31e 2t sin(3.464t) Prof Carruthers BU) EK307 Notes Summer / 264

46 Part III Circuits in Frequency Prof Carruthers BU) EK307 Notes Summer / 264

47 Part 3: Circuits in Frequency 8 Phasors and Impedance 9 Methods for Steady-State Analysis 10 Transfer Functions and Bode Plots 11 Filters: Passive and Active Prof Carruthers BU) EK307 Notes Summer / 264