Ch3 Basic RL and RC Circuits

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;

Circuits and Analog ElectronicsCircuits and Analog Electronics


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

dtdq

i t dtdvc

2

21 cvcvdvpdtwEnergy stored

A capacitor is an energy storage device memory device.

tCtC vv



• 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

RvEi c

c

KVL around the loop: EvRi Cc

EvRdtdvC c

c

EAev RCt

C

Initial condition 000 CC vv

)1()1( t

RCt

C eEeEv

dtdvCi c

c t

eRE

Switch is thrown to 1

RCCalled time constant

Transient Response of RC Circuits

EA



)1( t

C eEv

/tc eE

dtdv

0

0

t

ct

c

dtdv

EEdtdv

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

RCt

c Aev

Initial condition Evv CC 00

0 Riv cc

0dtdvRCv c

c

// tRCtc EeEev

/tc e

REi

Transient Response of RC Circuits

cc

dvi Cdt



RCTime Constant /t

c eREi

/tc e

RE

dtdi

0

0/

t

ct

c

dtdi

RERE

dtdi

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>>



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

tdiLdt

dtv )()(

t

dxxvL

ti )(1)(

tLtL iidtdiLiivP

)(21)( 2 tLitwL 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

dtdiLvL

KVL around the loop: EviR L

iRdtdiLE

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

Called time constant RL /

Transient Response of RL Circuits

/

/

/

1

)1(

)1()1(

ttLRt

LR

L

tR

ttLR

EeeRELe

RE

dtdL

dtdiLv

eEiRv

eREe

REi



Time constant

• Indicate new fast i (t) will drop to zero precisely.• It is the amount of time for i (t) to drop zero if it is dropping at

the initial rate.

t

i (t)

0

.



Switch to 2

tLR

Aei

dtLR

idi

iRdtdiL

0

Initial condition REit 0,0

/ttLR

eREe

REi


R

1

L

2

K

E

0

0

0

0

0

: 0:

1

ln

i t t

I

i t tI

t ti I i t

Rdi dti L

Ri tL



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 and it is 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 iREiL

REi 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

0

28V 4V2 2Cv

0

8V 2A2 2

i

042A 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 state4) 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 )()0( ffA

t

t effff

][ )()0()()(

Initial Steady

t

t Aeff

)()(In general,



3.3 ExamplesP3.1 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 3V6K 3K

t

c c c c

c c

v t v v v e

v v

Apply Thevenin theoren :

6

1

6

2 10

1 1 2K6K 3K

2K 100pF 2 10

3 3 V

Th

Th

t

c

R

R C

v t e


3.3 Examples

vC

R2 3k

+

_ v

U

I

t=0

6V

C=1000PF

R1=10k

R1=20k

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

Apply Thevenin theoren :

6

1

6

2.31 10

1 1 30 K10K 3K 13

30 K 100pF 2.31 1013

4.615 4.615 V

Th

Th

t

c

R

R C

v t e

0 0

10K6V 4.615V10K 3Kc

v t

v

Ch3 Basic RL and RC Circuits

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