lecture 8 - shanghaitechsist.shanghaitech.edu.cn/faculty/zhoupq/teaching/fall15/lec8-1st... ·...

Electric Circuits (Fall 2015) Pingqiang Zhou

Lecture 8

- First-Order Circuits

10/29/2015

Reading: Chapter 7

1Lecture 8


Overview

• This chapter examines RC and LC circuits’ reaction to

switched sources.

The circuits are referred to as first-order circuits.

• Three special functions, the unit step, unit impulse, and

unit ramp function are also introduced.

• Both source free and switched sources are examined.

2Lecture 8


First-Order Circuits

• A circuit that contains only sources,

resistors and an inductor is called an RL

circuit.

• A circuit that contains only sources,

resistors and a capacitor is called an RC

circuit.

• RL and RC circuits are called first-order

circuits because their V-Is are described

by first-order differential equations.

• There are also two ways to excite the

circuits:

initial conditions

independent sources

–+

vs L

R

i–+

vs C

R

i

3Lecture 8


Response of a Circuit

• Transient response of an RL or RC circuit is

Behavior when voltage or current source are suddenly applied to

or removed from the circuit due to switching.

Temporary behavior

• Steady-state response (aka. forced response)

Response that persists long after transient has decayed

• Natural response of an RL or RC circuit is

Behavior (i.e., current and voltage) when stored energy in the

inductor or capacitor is released to the resistive part of the

network (containing no independent sources).

4Lecture 8


Natural Response

RL Circuit

• Inductor current cannot change instantaneously

• In steady state, an inductor behaves like a short circuit.

RC Circuit

• Capacitor voltage cannot change instantaneously

• In steady state, a capacitor behaves like an open circuit

R

i

L

+

v

–

RC


Source Free RC Circuit

• A source free RC circuit occurs when its dc source is

suddenly disconnected.

The energy stored in the capacitor is released to the resistors.

• Consider a series combination of a resistor and a initially

charged capacitor as shown:

6

00v Vat t=0

0C Ri i 0dv v

dt RC

a first-order differential equation

/t RCv t Ae /

0

t RCv t V eln lnt

v ARC

Lecture 8


Natural Response of RC

• The result shows that the voltage

response of the RC circuit is an

exponential decay of the initial

voltage.

• The speed at which the voltage

decays can be characterized by

how long it takes the voltage to

drop to 1/e of the initial voltage.

called the time constant and is

represented by .

𝜏 = 𝑅𝐶.

• The voltage thus be expressed as

7

/

0

tv t V e

Lecture 8


Time Constant 𝜏 (= 𝑅𝐶)

• After five time constants the voltage on the capacitor is

less than one percent.

After five time constants, a capacitor is considered to be either fully

discharged or charged

• A circuit with a small time constant has a fast response

and vice versa.

8Lecture 8


RC Discharge

• With the voltage known, we can

find the current:

• The power dissipated in the

resistor is:

• The energy absorbed by the

resistor is:

9

/0 t

R

Vi t e

R

2

2 /0 tVp t e

R

2 2 /

0

11

2

t

Rw t CV e

/

0

t RCv t V e

Lecture 8


Example

• In the circuit below, let 𝑣𝐶 0 = 15V. Find 𝑣𝐶 , 𝑣𝑥, and 𝑖𝑥 for

𝑡 > 0.

10

𝑅𝑒𝑞 = 4Ω 𝜏 = 𝑅𝑒𝑞𝐶 = 0.4s 𝑣𝐶 𝑡 = 𝑣𝐶 0 𝑒−𝑡/𝜏 = ⋯

Lecture 8


Source Free RL Circuit

• Now lets consider the series connection

of a resistor and inductor.

In this case, the value of interest is the

current through the inductor.

Since the current cannot change

instantaneously, we can determine its value

as a function of time.

11

00i I

0L Rv v 0di

L Ridt /

0

Rt Li t I e

The time constant in this case is:L

R

Lecture 8


Example

• The switch in the circuit below has been closed for a long

time. At 𝑡 = 0, the switch is opened. Calculate 𝑖(𝑡) for 𝑡 >0.

12

When 𝑡 < 0

𝑖 0 = 6A

When 𝑡 > 0

𝑖 𝑡 = 6𝑒−4𝑡A

Lecture 8


Natural Response Summary

RL Circuit

• Inductor current cannot change instantaneously

• time constant

RC Circuit

• Capacitor voltage cannot change instantaneously

• time constant

R

L

/)0()(

)0()0(

teiti

ii

R

i

L

+

v

–

RC

/)0()(

)0()0(

tevtv

vv

RC


Singularity Functions

• Before we consider the response of a circuit to an external

voltage, we need to cover some important mathematical

functions.

• Singularity functions (also known as switching functions)

serve as good approximations to the switching signals

that arise in circuits with switching operations.

• The three most common singularity functions are

unit step

unit impulse

unit ramp

14Lecture 8


The Unit Step u(t)

• A step function is one that maintains a constant value

before a certain time and then changes to another constant

afterwards.

• The prototypical form is 0 before 𝑡 = 0 and 1 afterwards.

See the graph for an illustration.

15

0, 0

1, 0

tu t

t

switching time may be shifted to 𝑡 = 𝑡0 by

0

0

0

0,

1,

t tu t t

t t

Lecture 8


Equivalent Circuit of Unit Step

• The unit step function has

an equivalent circuit to

represent when it is used

to switch on a source.

• The equivalent circuits for

a voltage and current

source are shown.

16Lecture 8


Example

• Express the voltage pulse below in terms of the unit step.

17Lecture 8


The Unit Impulse Function 𝜹(𝒕)

• The derivative of the unit step

function is the unit impulse function.

• This is expressed as:

• Voltages of this form can occur during

switching operations.

18

0 0

Undefined 0

0 0

t

t t

t

Lecture 8


The Unit Ramp Function 𝒓(𝒕)

• Integration of the unit step function

results in the unit ramp function:

• Much like the other functions, the

onset of the ramp may be adjusted.

19

0, 0

, 0

tr t

t t

Lecture 8


Example

• Express the sawtooth function in

terms of singularity functions

20Lecture 8


21Lecture 8


Step Response of RC Circuit

• When a DC source is suddenly

applied to a RC circuit, the source

can be modeled as a step function.

The circuit response is known as the step

response.

• Let’s consider the circuit shown here.

• We can find the voltage on the

capacitor as a function of time.

22Lecture 8



• We assume an initial voltage of V0 on the capacitor.

• Applying KCL:

• For t>0 this becomes:

Integrating both sides and introducing initial conditions finally yields:

23

sVdv vu t

dt RC RC

sVdv v

dt RC RC

0

/

0

, 0

0t

s s

V tv t

V V V e t

𝑣 0−

= 𝑣 0+

= 𝑣0

Lecture 8



• This is known as the complete response,

or total response.

• We can consider the response to be

broken into two separate responses:

The natural response of the capacitor or

inductor due to the energy stored in it.

The second part is the forced response.

24

0

/

0

, 0

0t

s s

V tv t

V V V e t

Lecture 8


Forced Response

• The complete response

can be written as:

where the nature response is:

and the forced response is:

• Note that the eventual response of the circuit is to reach

Vs after the natural response decays to zero.

25

n fv v v

/

0

t

nv V e

/1 t

f sv V e

0

/

0

, 0

0t

s s

V tv t

V V V e t

𝑣 0−

= 𝑣 0+

= 𝑣0

Lecture 8


Another Perspective

• Another way to look at the response is to break it up into

the transient response and the steady state response:

𝑣 𝑡 = 𝑣 ∞

steady 𝑣𝑠𝑠

+ 𝑣 0 − 𝑣(∞) 𝑒−𝑡/𝜏

transient 𝑣𝑡

where the transient is:

and the steady state is:

26

/

0

t

t sv V V e

ss sv V

0

/

0

, 0

0t

s s

V tv t

V V V e t

Lecture 8


Example

• The switch has been in position A for a long time. At 𝑡 = 0,

the switch moves to B. Find 𝑣(𝑡).

27

𝑣 0 = 15V 𝑣 ∞ = 30V 𝜏 = 2s

𝑣 𝑡 = 𝑣 ∞ + 𝑣 0 − 𝑣 ∞ 𝑒−𝑡/𝜏

Lecture 8


Step Response of RL Circuit

• Now we can look at the step response of

a RL circuit.

• We will use the transient and steady

state response approach.

• We know that the transient response will

be an exponential:

28

/t

ti Ae

Lecture 8



• After a sufficiently long time, the current

will reach the steady state:

• This yields an overall response of:

• To determine the value of A we need to keep in mind that

the current cannot change instantaneously.

29

sss

Vi

R

/t sVi Ae

R

00 0i i I

/t

ti Ae

Lecture 8



• Thus we can use the t=0 time to establish A

• The complete response of the circuit is thus:

• Without an initial current, the circuit response is shown

here.

30

0sV

A IR

/

0

ts sV Vi t I e

R R

/t sVi Ae

R

00 0i i I

Lecture 8


Example

• Find 𝑖(𝑡) in the circuit for 𝑡 > 0. Assume that the switch

has been closed for a long time.

31

𝑖 0 = 5A 𝑖 ∞ = 2A 𝜏 =1

15s

𝑖 𝑡 = 𝑖 ∞ + 𝑖 0 − 𝑖 ∞ 𝑒−𝑡/𝜏

Lecture 8


Procedure for Finding RC/RL Response

1. Identify the variable of interest

• For RL circuits, it is usually the inductor current iL(t).

• For RC circuits, it is usually the capacitor voltage vc(t).

2. Determine the initial value (at t = t0- and t0

+) of the

variable

• Recall that iL(t) and vc(t) are continuous variables:

iL(t0+) = iL(t0

) and vc(t0+) = vc(t0

)

• Assuming that the circuit reached steady state before t0 , use the

fact that an inductor behaves like a short circuit in steady state or

that a capacitor behaves like an open circuit in steady state.

32Lecture 8


Procedure (cont’d)

3. Calculate the final value of the variable (its value as

t ∞)

• Again, make use of the fact that an inductor behaves like a

short circuit in steady state (t ∞) or that a capacitor behaves

like an open circuit in steady state (t ∞).

4. Calculate the time constant for the circuit

t = L/R for an RL circuit, where R is the Thévenin equivalent

resistance “seen” by the inductor.

t = RC for an RC circuit where R is the Thévenin equivalent

resistance “seen” by the capacitor.

33Lecture 8


Application: Delay Circuit

• The RC circuit can be used to delay the turn on of a

connected device.

• For example, a neon lamp which only triggers when a

voltage exceeds a specific value can be delayed using

such a circuit.

34Lecture 8


Delay Circuit II

• When the switch is closed, the capacitor charges.

The voltage will rise at a rate determined by:

Once the voltage reaches 70 volts, the lamp triggers.

• Once on, the lamp has low resistance and discharges the

capacitor.

This shuts off the capacitor and starts the cycle over again.

35

1 2R R C

Lecture 8


Further Reading

• Refer to Ch. 7.9, [James W. Nillson 2014] and [Anant

Agarwal 2005] for more applications

Photoflash unit

Relay circuit

Automobile ignition circuit

Artificial pacemaker

Ramp and impulse responses

• Refer to Ch. 7.7 for first-order circuit with Op Amps.

36Lecture 8


Announcements

• Midterm exam 1

Nov. 5, 2015, Thursday, in class (1pm – 2:40pm)

• Venue

Room 306 (Discussion class 1&2)

Room 309 (Discussion class 3&4)

• Review class before exam

Nov. 3, 2015, Tuesday in class

By four discussion GSIs

37Lecture 8

lecture 8 - shanghaitechsist.shanghaitech.edu.cn/faculty/zhoupq/teaching/fall15/lec8-1st... ·...

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