Response of First-Order RL and RC Circuits

Transcription

1 Response of First-Order RL and RC Circuits Qi Xuan Zhejiang University of Technology Nov 2015 Electric Circuits 1

2 Structure The Natural Response of an RL Circuit The Natural Response of an RC Circuit The Step Response of RL and RC Circuits A General Solu8on for Step and Natural Responses Sequen8al Switching Unbounded Response The Integra8ng Amplifier Electric Circuits 2

3 A Flashing Light Circuit Electric Circuits 3

4 The Natural Response of an RL Circuit a) The switch has been in a closed posi8on for a long 8me: all currents and voltages have reached a constant value. b) The inductor appears as a short circuit (Ldi/dt = 0) prior to the release of the stored energy. Electric Circuits 4

5 Expression for the Current Electric Circuits 5

6 Expression for the Power and Energy Electric Circuits 6

7 The Significance of the Time Constant The 8me constant is an important parameter for first- order circuits, and it is convenient to think of elapsed aaer switching in terms of integral of τ: one 8me constant ajer the inductor has begun to release its stored energy to the resistor, the current has been reduced to e - 1, or approximately 0.37 of its ini8al value. Electric Circuits 7

8 Steady- State Response transient response <1% The response that exists a long time after the switching has taken place is called the steady-state response Electric Circuits 8

9 Calcula8ng the natural response of an RL circuit can be summarized as follows: 1. Find the ini8al current, I 0, through the inductor. 2. Find the 8me constant of the circuit, τ = L/R. 3. Use I 0 e -t/τ to generate i(t) from I 0 and τ. Electric Circuits 9

10 Example #1 The switch in the circuit shown in Figure has been closed for a long 8me before it is opened at t = 0. Find a) i L (t) for t 0, b) i o (t) for t 0 +, c) v o (t) for t 0 +, d) the percentage of the total energy stored in the 2 H inductor that is dissipated in the 10 Ω resistor. Electric Circuits 10

11 for Example #1 a) i L (0 + )=20A, τ = L/R eq = 0.2 s b) c) Electric Circuits 11

12 d) Electric Circuits 12

13 The Natural Response of an RC Circuit Electric Circuits 13

14 The Step Response of an RL Circuit Kirchhoff s voltage law Electric Circuits 14

15 I 0 = 0 Electric Circuits 15

16 Step Response Curve τ Electric Circuits 16

17 The voltage across an inductor is Ldi/dt, so for t > 0 +, we have: I 0 = 0 Electric Circuits 17

18 Another Way to Get the Voltage Expression (From Eq. 7.29) (Compare with Eq. 7.1) Electric Circuits 18

19 Example #2 The switch in the circuit has been in posi8on a for a long 8me. At t = 0, the switch moves from posi8on a to posi8on b. The switch is a make- before- break type; that is, the connec@on at posi@on b is established before the connec@on at posi@on a is broken, so there is no interrup8on of current through the inductor. a) Find the expression for i(t) for t 0. b) What is the ini8al voltage across the inductor just ajer the switch has been moved to posi8on b? c) Does this ini8al voltage make sense in terms of circuit behavior? d) How many milliseconds ajer the switch has been moved does the inductor voltage equal 24 V? e) Plot both i(t) and v(t) versus t. Electric Circuits 19

20 for Example #2 I 0 = -8 A τ = L/R = 0.2 H / 2 Ω = 0.1 s a) b) c) Yes; in the instant ajer the switch has been moved to posi8on b, the inductor sustains a current of 8 A counterclockwise around the newly formed closed path. This current causes a 16 V drop across the 2 Ω resistor. This voltage drop adds to the drop across the source, producing a 40 V drop across the inductor. Electric Circuits 20

21 d) We find the 8me at which the inductor voltage equals 24 V by solving the expression e) Electric Circuits 21

22 The Step of an RC Circuit By comparing with Eq Electric Circuits 22

23 A General Solu8on for Step and Natural Response τ = L/R for RL circuit τ = RC for RC circuit Electric Circuits 23

24 Electric Circuits 24

25 Helpful Notes 1. Iden8fy the variable of interest for the circuit. For RC circuits, it is most convenient to choose the voltage; for RL circuits, it is best to choose the current. 2. Determine the ini8al value of the variable, which is its value at t 0. Note that if you choose capaci8ve voltage or induc8ve current as your variable of interest, it is not necessary to dis8nguish between t = t 0 - and t = t 0+. This is because they both are con@nuous variables. If you choose another variable, you need to remember that its ini8al value is defined at t = t Calculate the final value of the variable, which is its value as t >. 4. Calculate constant for the circuit. Electric Circuits 25

26 Example #3 The switch in the circuit has been in posi8on a for a long 8me. At t = 0 the switch is moved to posi8on b. a) What is the ini8al value of v C? b) What is the final value of v C? c) What is the 8me constant of the circuit when the switch is in posi8on b? d) What is the expression for v C (t) when t 0? e) What is the expression for i(t) when t 0 +? f) How long ajer the switch is in posi8on b does the capacitor voltage equal zero? g) Plot v C (t) and i(t) versus t. Electric Circuits 26

27 for Example #3 a) b) c) d) e) v C (t) = 0 v C (0) = 40 V 60 / ( ) = 30 V v C ( ) = 90 V τ = RC = ( )( ) = 0.2 s v C (t) = v C ( ) + [v C (0) v C ( )] e t/τ = e 5t V, t 0 i(t) = i( ) + [i(0) i( )] e t/τ = 300e 5t µa, t 0 i( ) = [90 v C (0)] V / 400 kω = 300 µa f) t = 0.2 ln(120/90) = ms g) Electric Circuits 27

28 Sequen8al Switching Defini8on: switching occurs more than once in a circuit. Method: derive the expressions for v(t) and i(t) for a given posi8on of the switch or switches and then use these solu8ons to determine the ini8al condi8ons for the next posi8on of the switch or switches. Note: anything but induc8ve currents and capaci8ve voltages can change instantaneously at the 8me of switching. Electric Circuits 28

29 Example #4 The uncharged capacitor in the circuit is ini8ally switched to terminal a of the three- posi8on switch. At t = 0, the switch is moved to posi8on b, where it remains for 15 ms. AJer the 15 ms delay, the switch is moved to posi8on c, where it remains indefinitely. a) Derive the numerical expression for the voltage across the capacitor. b) Plot the capacitor voltage versus 8me. c) When will the voltage on the capacitor equal 200 V? Electric Circuits 29

30 a) for Example #4 Touch b: v(0) = 0 V; v( ) = 400 V; τ = RC = 10 ms v(t) = v( ) + [v(0) v( )]e t/τ = e 100t V, 0 t 15 ms Touch c: v(0.015) = 400(1 e 1.5 ) = V; v( ) = 0 V; τ = RC = 5 ms v(t) = v( ) + [v(0) v( )]e (t 0.015)/τ = e 200(t 0.015) V, t 15 ms b) c) t 1 = 6.93 ms; t 2 = ms Electric Circuits 30

31 The Integra8ng Amplifier v n = v p = 0 i f + i s = 0 = 0 Electric Circuits 31

32 Example #5 Assume that the numerical values for the signal voltage are V m = 50 mv and t 1 = 1 s. This signal voltage is applied to the integra8ng- amplifier circuit shown in Fig The circuit parameters of the amplifier are R s = 100 kω, C f = 0.1 µf, and V CC = 6 V. The ini8al voltage on the capacitor is zero. a) Calculate v o (t). b) Plot v o (t) versus t. Electric Circuits 32

33 a) for Example #5 b) Electric Circuits 33

34 The Differen8a8ng Amplifier Electric Circuits 34

35 A Flashing Light Circuit The lamp in this circuit starts to conduct whenever the lamp voltage reaches a value V max. During the 8me the lamp conducts, it can be modeled as a resistor whose resistance is R L. The lamp will con8nue to conduct un8l the lamp voltage drops to the value V min. When the lamp is not conduc8ng, it behaves as an open circuit. Electric Circuits 35

36 Electric Circuits 36

37 Electric Circuits 37

38 Summary Natural response for RL and RC circuits Time constants RL and RC circuits (L/R, RC) Step response for RL and RC circuits Sequen8al switching Unbounded response Integra8ng and differen8a8ng amplifier Electric Circuits 38