Response of First-Order RL and RC Circuits

1 Electric Circuits

Response of First-Order RL and RC Circuits

Qi Xuan Zhejiang University of Technology

Nov 2015

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

2 Electric Circuits

A Flashing Light Circuit

Electric Circuits 3

The Natural Response of an RL Circuit

Electric Circuits 4

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.

Expression for the Current

Electric Circuits 5

Expression for the Power and Energy

Electric Circuits 6

The Significance of the Time Constant

Electric Circuits 7

The 8me constant is an important parameter for first-‐order circuits, and it is convenient to think of the @me elapsed aAer switching in terms of integral mul@ples 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.

Steady-‐State Response

Electric Circuits 8

The response that exists a long time after the switching has taken place is called the steady-state response

<1%

transient response

Calcula8ng the natural response of an RL circuit can be summarized as follows: 1.  Find the ini8al current, I0, through the inductor. 2.  Find the 8me constant of the circuit, τ = L/R. 3.  Use I0e-t/τ to generate i(t) from I0 and τ.

Electric Circuits 9

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)  iL(t) for t ≥ 0, b)  io(t) for t ≥ 0+, c)  vo(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

Solu@on for Example #1


iL(0+)=20A, τ = L/Req= 0.2 s a)

b)

c)


d)

The Natural Response of an RC Circuit


The Step Response of an RL Circuit


Kirchhoff’s voltage law


I0 = 0

Step Response Curve


τ


The voltage across an inductor is Ldi/dt, so for t > 0+, we have:

I0 = 0

Another Way to Get the Voltage Expression


(From Eq. 7.29)

(Compare with Eq. 7.1)

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.




I0 = -8 A τ = L/R = 0.2 H / 2 Ω = 0.1 s

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.

a)

b)

c)


We find the 8me at which the inductor voltage equals 24 V by solving the expression

d)

e)

The Step of an RC Circuit


By comparing with Eq. 7.48

A General Solu8on for Step and Natural Response


τ = L/R for RL circuit

τ = RC for RC circuit


Helpful Notes 1.  Iden8fy the variable of interest for the circuit. For RC circuits, it is

most convenient to choose the capaci@ve voltage; for RL circuits, it is best to choose the induc@ve current.

2.  Determine the ini8al value of the variable, which is its value at t0. Note that if you choose capaci8ve voltage or induc8ve current as your variable of interest, it is not necessary to dis8nguish between t = t0

- and t = t0+. 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 = t0

+. 3.  Calculate the final value of the variable, which is its value as t —> ∞. 4.  Calculate the @me constant for the circuit.


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 vC ? b)  What is the final value of vC ? c)  What is the 8me constant of the circuit when the switch is in posi8on b? d)  What is the expression for vC(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 vC(t) and i(t) versus t.



vC(0) = −40 V × 60 / (20 + 60) = −30 V

vC(∞) = 90 V

vC(t) = vC(∞) + [vC(0) − vC(∞)] e−t/τ = 90 − 120e−5t V, t ≥ 0 τ = RC = (400 × 103)(0.5 × 10−6) = 0.2 s

i(t) = i(∞) + [i(0) − i(∞)] e−t/τ = 300e−5t µA, t ≥ 0 i(∞) = [90 − vC(0)] V / 400 kΩ = 300 µA

t = 0.2 ln(120/90) = 57.54 ms

vC(t) = 0

a) b) c) d) e)

f) g)

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.


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?




Touch b: v(0) = 0 V; v(∞) = 400 V; τ = RC = 10 ms

v(t) = v(∞) + [v(0) − v(∞)]e−t/τ= 400 − 400e−100t V, 0 ≤ t ≤ 15 ms

Touch c: v(0.015) = 400(1−e−1.5) = 310.75 V; v(∞) = 0 V; τ = RC = 5 ms

v(t) = v(∞) + [v(0) − v(∞)]e−(t−0.015)/τ= 310.75e−200(t−0.015) V, t ≥ 15 ms

a)

b)

c) t1 = 6.93 ms; t2 = 17.20 ms

The Integra8ng Amplifier


if + is = 0

vn = vp = 0

= 0

Example #5


Assume that the numerical values for the signal voltage are Vm = 50 mV and t1 = 1 s. This signal voltage is applied to the integra8ng-‐amplifier circuit shown in Fig. 7.40. The circuit parameters of the amplifier are Rs = 100 kΩ, Cf = 0.1 µF, and VCC = 6 V. The ini8al voltage on the capacitor is zero.

a) Calculate vo(t). b) Plot vo(t) versus t.



a)

b)

The Differen8a8ng Amplifier


A Flashing Light Circuit


The lamp in this circuit starts to conduct whenever the lamp voltage reaches a value Vmax. During the 8me the lamp conducts, it can be modeled as a resistor whose resistance is RL. The lamp will con8nue to conduct un8l the lamp voltage drops to the value Vmin. When the lamp is not conduc8ng, it behaves as an open circuit.



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


Documents

Response of First-Order RL and RC Circuits