engineering circuit analysis-ch3

Ch3 Basic RL and RC Circuits

3.1 First-Order RC Circuits

3.2 First-Order RL Circuits

3.3 Examples

ReferencesReferences: Hayt-Ch5, 6; Gao-Ch5;

Engineering Circuit AnalysisEngineering Circuit Analysis


Key WordsKey Words:

Transient Response of RC Circuits, Time constant




• Used for filtering signal by blocking certain frequencies and passing others. e.g. low-pass filter

• Any circuit with a single energy storage element, an arbitrary number of sources and an arbitrary number of resistors is a circuit of order 1.

• Any voltage or current in such a circuit is the solution to a 1st order differential equation.

Ideal Linear Capacitor

dt

dqti =)(

dt

dvc

2

2

1cvcvdvpdtwEnergy stored

A capacitor is an energy storage device memory device.

)(=)+( tvtv Cc



• One capacitor and one resistor

• The source and resistor may be equivalent to a circuit with many resistors and sources.

R+

-Cvs(t)

+

-

vc(t)

+ -vr(t)



R

1

C

2

K

E

R

vEi cc

KVL around the loop: EvRi Cc

EvRdt

dvC c

c

EAev RC

t

C

Initial condition 0)0()0( CC vv

)1()1( t

RC

t

C eEeEv

dt

dvCi c

c t

eR

E

Switch is thrown to 1

RCCalled time constant

Transient Response of RC Circuits

EA



)1( t

C eEv

/tc e

E

dt

dv

0

0

t

ct

c

dtdv

EE

dt

dv

RCTime Constant

R

1

C

2

K

E

Time

0s 1ms 2ms 3ms 4ms 5ms 6ms 7ms 8ms 9ms 10msV(2)

0V

5V

10V

SEL>>

RC

R=2k

C=0.1F



Switch to 2

R

1

C

2

K

E

RC

t

c Aev

Initial condition Evv CC )0()0(

0 Riv cc

0dt

dvRCv c

c

// tRCtc EeEev

/tc e

R

Ei

Transient Response of RC Circuits

cc

dvi C

dt



RCTime Constant

R

1

C

2

K

E

R=2k

C=0.1F

Time

0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms 6.0ms 7.0ms 8.0msV(2)

0V

5V

10V

SEL>>

t

RC

t

C EeEetv

)(

E

dt

dv

t

C 0

0

t

C

dt

dv

E



Time

0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms 5.5ms 6.0msV(2) V(1)

0V

2.0V

4.0V

6.0V



Key WordsKey Words:

Transient Response of RL Circuits, Time constant



Ideal Linear Inductor

i(t)+

-

v(t)

Therestofthe

circuit

Ldt

tdiL

dt

dtv

)()(

t

dxxvL

ti )(1

)(

)()( titi LLdt

diLiivP

)(2

1)( 2 tLiLidipdttwL Energy stored:

• One inductor and one resistor

• The source and resistor may be equivalent to a circuit with many resistors and sources.



Switch to 1

R

1

L

2

K

E

dt

diLvL

KVL around the loop: EviR L

iRdt

diLE

Initial condition 0)0()0(,0 iit

Called time constant RL /

Transient Response of RL Circuits

/

/

/

1

)1(

)1()1(

ttL

Rt

L

R

L

tR

ttL

R

EeeL

R

R

ELe

R

E

dt

dL

dt

diLv

eEiRv

eR

Ee

R

Ei



Time constant

• Indicate how fast i (t) will drop to zero.

• It is the amount of time for i (t) to drop to zero if it is dropping at the initial rate .

t

i (t)

0

.

0t

t

dt

di



Switch to 2

tL

R

Aei

dtL

R

i

di

iRdt

diL

0

Initial conditionR

EIt 0,0

/ttL

R

eR

Ee

R

Ei


R

1

L

2

K

E

0

0

0

0

0

: 0

:

1

ln

i t t

I

i t tI

t t

i I i t

Rdi dt

i LR

i tL

tL

R

I

ti

0

)(ln

tL

R

eIti

0)(



R

1

L

2

K

E


Time

0s 1ms 2ms 3ms 4ms 5ms 6ms 7ms 8ms 9ms 10msI(L1)

0A

2.0mA

4.0mA

SEL>>

Time

0s 1ms 2ms 3ms 4ms 5ms 6ms 7ms 8ms 9ms 10msI(L1)

0A

2.0mA

4.0mA

SEL>>

Input energy to L

L export its energy , dissipated by R


Initial Value （ t = 0）

Steady Value (t )

time constant

RL

Circuits Source(0 state)

Source-

free(0 input)

RC

Circuits

Source(0 state)

Source-free

(0 input)

00 iR

EiL

R

Ei 0 0i

00 v Ev

Ev 0 0v

RL /

RL /

RC

RC

Summary


Summary

The Time Constant

• For an RC circuit, = RC

• For an RL circuit, = L/R

• -1/ is the initial slope of an exponential with an initial value of 1

• Also, is the amount of time necessary for an exponential to decay to 36.7% of its initial value


Summary

• How to determine initial conditions for a transient circuit. When a sudden change occurs, only two types of quantities will remain the same as before the change. – IL(t), inductor current– Vc(t), capacitor voltage

• Find these two types of the values before the change and use them as the initial conditions of the circuit after change.


About Calculation for The Initial Value

iC iL

i

t=0

+

_

1A

+

-vL(0+)

vC(0+)=4V

i(0+)

iC(0+) iL(0+)

3.3 Examples

1 3/ / 2R R

20 8V 4V

2 2Cv

8V0 2A

2 2i

40 2A 1A

4 4Li

0 0C Cv v

0 0L Li i


3.3 Examples

Method 1

(Analyzing an RC circuit or RL circuit)

Simplify the circuit

2) Find Leq(Ceq), and = Leq/Req ( = CeqReq)

1) Thévenin Equivalent.(Draw out C or L)

Veq , Req

3) Substituting Leq(Ceq) and to the previous solution of differential equation for RC (RL) circuit .


3.3 Examples

Method 2


3) Find the particular solution.

1) KVL around the loop the differential equation

4) The total solution is the sum of the particular and homogeneous solutions.

2) Find the homogeneous solution.

3.3 Examples

Method 3 (step-by-step)


1) Draw the circuit for t = 0- and find v(0-) or i(0-)

2) Use the continuity of the capacitor voltage, or inductor current, draw the circuit for t = 0+ to find v(0+) or i(0+)

3) Find v(), or i() at steady state

4) Find the time constant – For an RC circuit, = RC

– For an RL circuit, = L/R

5) The solution is: /)]()0([)()( teffftf

Given f(0+) ， thus A = f(0+) – f(∞)

t

effftf

)]()0([)()(

Initial Steady

t

Aeftf

)()(In general,



3.3 Examples

P3.1 vC (0)= 0, Find vC (t) for t 0. i1

6k

R1

R2 3k +

_ E

C=1000PF

pf

i2 i3 t=0

9V

Method 3:

0

3K0 0, 9V 3V

6K 3K

t

c c c c

c c

v t v v v e

v v

Apply Thevenin theorem :

6

1

6

2 10

1 12K

6K 3K

2K 1000pF 2 10

3 3 V

Th

Th

t

c

R

R C

v t e

s


3.3 Examples

vC

R2 3k

+

_ v

U

I

t=0

6V

C=1000PF

R1=10k

R1=20k

+ - P3.2 vC (0)= 0, Find vC (t) for t 0.

Apply Thevenin’s theorem :

6

1

6

2.31 10

1 1 30K

10K 3K 13

30K 1000pF 2.31 10

13

4.615 4.615 V

Th

Th

t

c

R

R C

v t e

0 0

10K6V 4.615V

10K 3K

c

c

v

v

s

engineering circuit analysis-ch3

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