RL and RC circuits first-order response

Electric circuitsENT 161/4

RL and RC circuit original response

A first-order circuit is characterized by a first-order differential equation. This circuit contain resistor and capacitor or inductor in one close circuit.

The natural response of a circuit refers to the behaviour ( in terms of voltages and currents) of the circuit itself, with no external sources of excitation.

RL circuit: circuit that have resistor and inductor.

RC circuit: circuit that have resistor and capacitor.

Natural response RC circuit

Consider these three condition :

1. At initially, t=0 -, switch doesn’t change for some time

2. At initial, t=0 +, switch doesn’t change for some time

3. At final condition, t→∞, switch doesn’t change for some time

Known t ≤ 0, v(t) = V0.

dtRCtv

tdvRC

tv

dt

tdvRC

tv

dt

tdvR

tv

dt

tdvC

ii Rc

1

)(

)(

)()(

0)()(

0)()(

0

RCt

eVtv

RC

t

V

tv

tRC

Vtv

dvRC

duu

dvRCu

du

tv

V

t

0

0

0

)(

0

)(

)(ln

)0(1

ln)(ln

11

1

0

voltage

Therefore t ≥ 0:

For t > 0,

RCt

eVtv 0)(RCt

eR

V

R

tvtiR

0)()(

RCt

eVCtvCtW2

20

2

2

1)(

2

1)(

Natural response RC circuit graph

0

0)(

0

0

teV

tVtv

RCt

This show that the voltage response of the RC circuit is an exponential decay of the initial voltage. constant, τ = RC

t

eVtv 0)(

Constant τ define how fast voltage reach stable condition :

Natural response RL circuit

Consider these three condition :

1. At initially, t=0 -, switch doesn’t change for some time

2. At initial, t=0 +, switch doesn’t change for some time

3. At final condition, t→∞, switch doesn’t change for some time

Known at t ≤ 0, i(t) = I0

Therefore t > 0,

dvL

R

u

du

dtL

R

ti

tdi

tiRdt

tdiL

tiRdt

tdiL

tiRtv

)(

)(

)()(

0)()(

0)()(

LRt

tti

i

eiti

L

Rt

i

ti

tL

Riti

dvL

Rduu

)0()(

)0(

)(ln

)0()0(ln)(ln

10

)(

)0(

Current

For t > 0,

LRteIti 0)(

LRteRI

Rtitv

0

)()(

LRteLI

tiLtw

220

2

2

1

)(2

1)(

EXAMPLESwitch in circuit for some time before

open at t=0. Calculate

a) IL (t) at t ≥ 0

b) I0 (t) at t ≥ 0+

c) V0 (t) at t ≥ 0+

d) Total energy percentage that stored in inductor 2H that absorb by 10Ω resistor.

Answer

a) Switch close for some time until t=0, known voltage at inductor should be zero at t = 0-. Therefore, initial current at inductor was 20A at t = 0-. Thus iL (0+) also become 20A, because immidiate changes for current didn’t exist in inductor.

Equivalent resistance from inductor and constant time

1010402eqR

saatR

L

eq

2.010

2

Therefore, current iL (t)

020

)0()(5

tAe

eitit

L

t

b) Current at resistor 40Ω could be calculate by using current divider law,

4010

100 Lii

This current was at t ≥ 0+ because i0 = 0 at t = 0-. Inductor will

become close circuit when switch open immediately and produce changes immediately at current i0.

Therefore, 04)( 5

0 tAeti t

c) V0 could be calculate by using Ohm’s Law,

0160

40)(5

00

tVe

itVt

d) Total power absorb by 10Ω resistor

02560

10)(

10

20

10

tWe

Vtp

t

Total energy absorb by 10Ω resistor

J

dtetW t

256

2560)(0

1010

Initial energy stored at 2H inductor

J

iLW

40040022

1

)0(2

1)0( 2

Therefore, energy percentage that absorb by 10Ω resistor

%64100400

256

Step response RC circuit The step response of a circuit is its

behaviour when the excitation is the step function, which may be a voltage or a current source.

Consider these three condition :

1. At initially, t=0 -, switch doesn’t change for some time

2. At initial, t=0 +, switch doesn’t change for some time

3. At final condition, t→∞, switch doesn’t change for some time

Known at t ≤ 0, v(t)=V0

For t > 0,

s

s

s

s

Vtv

tdvdt

RC

tvV

tdvdt

RC

dt

tdvRCtvV

tRitvV

)(

)(1

)(

)(1

)()(

)()(

s

s

ss

s

VV

Vtv

RC

t

VVVtvRC

t

Vu

dudv

RC

0

0

)(ln

ln)(ln

1

t

RCt

eVVV

eVVVtv

ss

ss

0

0)(

voltan

Current for step response RC circuit

t

t

t

eR

V

R

V

eVVR

eVVC

dt

dvCti

s

s

s

0

0

0

1

)(1

)(

t

eiti )0()(

Then, for t >0

t

t

eVVV

VV

VV

eVVVV

sn

sf

nf

ss

0

0

Where

Vf = Force voltage or known as steady-state response

Vn = known as transient response is the circuit’s temporary response that will die out with time.

Step response RC circuit graph

force

Natural

total

Step Response RL circuitStep Response RL circuit

Consider these three condition :

1. At initially, t=0 -, switch doesn’t change for some time

2. At initial, t=0 +, switch doesn’t change for some time

3. At final condition, t→∞, switch doesn’t change for some time

known i(t)=I0 at t ≤ 0. For t > 0,

RV

RV

s

s

s

s

s

i

didt

L

R

ti

tdidt

L

Rdt

tdi

R

Lti

R

Vdt

tdiLtiRV

tvtiRV

)(

)(

)()(

)()(

)()(

RV

RV

RV

RV

ti

IR

V

t

RV

s

s

ss

s

s

I

ti

L

Rt

ItiL

Rt

u

dudv

L

R

u

dudv

L

R

0

0

)(

0

)(ln

ln)(ln

0

LR

ss tR

VR

V eIti 0)(

Current

Finally,

0

0)(

0

0

teI

tIti

LR

ss tR

VR

V

0

0)(

)(

0

teIRV

tdt

tdiLtv

LRt

s

Question Switch in those circuit was at x

position for some time. At t=0, switch move to position y immediately. Calculate,

(a) Vc(t) at t ≥ 0(b) V0 (t) at t ≥ 0+(c) i0 (t) at t ≥ 0+(d) Total energy absorb by 60kΩ resistor.

Answer (a)

Constant for circuit

ms40)1080)(105.0( 36

VC (0)=100V

equivalent resistor = 80kΩ.

Then, VC(t) for t ≥ 0:

0100)( 25 tVetV tC

Answer (b)V0 (t) could be calculate by using

voltage divider law.

060

)(80

48)(

25

0

tVe

tVtV

t

C

Answer (c)

current i0 (t) can be calculated by using ohm’s law

01060

)()( 25

30

0 tmAetV

ti t

Answer (d)

Power absorb by 60kΩ resistor

060

1060)()(50

32060

tmWe

titpt

k

Total energy

mJ

dttiW k

2.1

1060)( 3

0

2060

Second-order RLC circuit

RLC circuit : circuit that contain resistor, inductor and capacitor

Second-order response : response from RLC circuit

Type of RLC circuit:1. RLC series circuit2. RLC parallel circuit

Original response for parallel RLC circuit

Take total current flows out from node

01

0 0 t

dt

dvCIvd

LR

V

differential of t,

01

2

2

dt

vdC

L

v

dt

dv

R

01

2

2

LC

v

dt

dv

RCdt

vd

Take steAv

02 ststst eLC

Ae

RC

AseAs

012

equationsticcharacteri

st

LCRC

sseA

Characteristic equation known as zero :

012

LCRC

ss

The root of the characteristic equation are

LCRCRCs

1

2

1

2

12

1

LCRCRCs

1

2

1

2

12

2

Response for RLC parallel circuit

tsts eAeAv 2121

The root of the characteristic equation are

RC2

1

20

21 s

20

22 s

where:

LC

10

summarize

20

22

20

21

s

s

RC2

1

0 LC

10

Parameter Terminology Value in natural response

s1, s2characteristic

equation

α frequency Neper

resonant radian frequency

Roots solution s1 and s2 depend on α and

Consider these cases saperately: 1. If < α , voltage response was

overdamped 2. If > α , voltage response was

underdamped3. If = α , voltage response was

critically damped

0

0

0

0

Overdamped voltage responseOverdamped voltage response

Solution for overdamped voltage

tsts eAeAv 2121

constant A1 and A2 can be determined from the initial conditions v(0+) and

Known, dt

dv )0(

21)0( AAv

2211

)0(AsAs

dt

dv

Here v(0+) = V0 and initial value for

dv/dt was

C

i

dt

dv C )0()0(

Solution for overdamped natural response, v(t) :

1. Calculate characteristic equation, s1 and s2, using R, L and C value.

2. Calculate v(0+) and

using circuit analysis. dt

dv )0(

3. Calulate A1 and A2 by solve those equation

4. Insert s1, s2, A1 and A2 value to calculate overdamped natural response for t ≥ 0.

21)0( AAv

2211

)0()0(AsAs

C

i

dt

dv C

Example for overdamped natural response for v(0) = 1V and

i(0) = 0

Underdamped voltage response

At > α2, root of the characteristic equation was complex number and those response called underdamped.

0

Therefore

ωd : damped radian frequency

dj

j

s

220

2201 )(

djs 2

underdamped voltage response for RLC parallel circuit was

teB

teBtv

dt

dt

sin

cos)(

2

1

constant B1 and B2 was real number.

10)0( BVv

211

)0()0(BB

C

i

dt

dvd

C

Solve those two linear equation

to calculate B1 and B2,

Example for underdamped voltage response for v(0) = 1V and i(0) = 0

Critically Damped Voltage Response

Second-order circuit was critically damped when = α . When circuit was critically damped, two characterictic root equation was real and same,

RCss

2

121

0

Solution for voltage tt eDetDtv 21)(

21

20

)0()0(

)0(

DDC

i

dt

dv

DVv

C

•Linear equation to calculate D1 and D2 value

Example for critically damped voltage response at v(0) = 1V and i(0) = 0

Step response RLC parallel Step response RLC parallel circuitscircuits

Step response RLC parallel Step response RLC parallel circuitscircuits

From Kirchhoff current law

Idt

dvC

R

vi

Iiii

L

CRL

Known

Therefore

dt

diLv

2

2

dt

idL

dt

dv L

Have,

Idt

idLC

dt

di

R

Li LLL

2

2

LC

I

LC

i

dt

di

RCdt

id LLL 1

2

2

There are two solution to solve the equation, direct approach and indirect approach.

Indirect approachIndirect approach

From Kirchhoff’s current law:

Idt

dvC

R

vvd

L

t0

1

Differential

01

2

2

dt

vdC

dt

dv

RL

v

01

2

2

LC

v

dt

dv

RCdt

vd

Depend on characteristic equation root :

tsts eAeAv 2121

teB

teBv

dt

dt

sin

cos

2

1

tt eDetDv 21

Insert in Kirchhoff’s current law eq:

tstsL eAeAIi 21

21

teB

teBIi

dt

dt

L

sin

cos

2

1

ttL eDetDIi 21

Direct approach

It’s simple to calculate constant for the equation

directly by using initial value response function.

212121 ,,,B,, DDBAA

Constant of the equation could be calculate from

and dt

diL )0()0(Li

The solution for a second-order differential equation with a constant forcing function equals the forced response plus a response funtion identical in form to the natural response.

If and Vf represent the final value of the response function. The final value may be zero,

responsenaturaltheas

formsametheoffunctionIi f

responsenaturaltheas

formsametheoffunctionVv f

Natural response for RLC Series circuit

The procedures for finding the natural or step responses of a series RLC circuit are the same as those used to find the natural or step responses of a parallel RLS circuit, because both circuits are described by differential equations that have the same form.

RLC series circuit

Summing the voltages around the closed path in the circuit,

01

00 Vdi

Cdt

diLRi

t

differential

02

2

C

i

dt

idL

dt

diR

02

2

LC

i

dt

di

L

R

dt

id

Characteristic equation for RLC series circuit

012 LC

sL

Rs

Characteristic equation root

LCL

R

L

Rs

1

22

2

2,1

@

20

22,1 s

Neper frequency (α) for RLC series circuit

sradL

R/

2

and resonant radian frequency was,

sradLC

/1

0

Current response

Overdamped

Underdamped

critically damped

220

220

220

Three kind of solution

tsts eAeAti 2121)(

teB

teBti

dt

dt

sin

cos)(

2

1

tt eDetDti 21)(

Step response for RLC series circuit

The procedures for finding the step responses of series RLC circuit are the same as those used to find the step response of a parallel RLC circuit.

RLC series circuit

Cvdt

diLiRv

Using Kirchhoff’s voltage law,

Current known as,

dt

dvCi C

Differential for current

2

2

dt

vdC

dt

di C

Insert in Voltage current law equation

LC

V

LC

v

dt

dv

L

R

dt

vd CCC 2

2

Three solution that possibly for vC

tstsfC eAeAVv 21

21

teB

teBVv

dt

dt

fC

sin

cos

2

1

ttfC eDetDVv 21

Contoh 1 Tenaga awal yang disimpan oleh litar

berikut adalah sifar. Pada t = 0, satu punca arus DC 24mA diberikan kepada litar. Nilai untuk perintang adalah 400Ω.

1. Apakah nilai awal untuk iL?2. Apakah nilai awal untuk ?

3. Apakah punca-punca persamaan ciri?4. Apakah ungkapan numerik untuk iL(t)

pada t ≥ 0?

dt

diL

Jawapan1. Tiada tenaga yang disimpan

dalam litar sebaik sahaja punca arus digunakan, maka arus awal bagi induktor adalah sifar. Induktor mencegah perubahan yang serta-merta pada arus induktor, oleh itu iL (0)=0 sebaik sahaja suis dibuka.

2. Nilai awal voltan kapasitor adalah sifar sebelum suis dibuka, oleh itu ia akan sifar sebaik sahaja suis dibuka. Didapati:

dt

diLv L maka 0

)0(

dt

diL

3. Dari elemen-elemen dalam litar, diperolehi

812

20 1016

)25)(25(

101

LC

srad

RC

/105

)25)(400)(2(

10

2

1

4

9

82 1025

Oleh kerana , maka punca-punca persamaan ciri adalah nyata

srad

s

srad

s

/00080

103105

/00020

103105

442

441

4. sambutan arus induktor adalah overdamped dan persamaan penyelesaian adalah

tstsfL eAeAIi 21

21

Dua persamaan serentak:

0)0(

0)0(

2211

21

AsAsdt

di

AAIi

L

fL

mAAmAA 832 21

Penyelesaian numerik:

0

8

3224)(

80000

20000

tuntuk

mAe

eti

t

t

L

Contoh 2

Tiada tenaga disimpan dalam inductor 100mH atau kapasitor 0.4µF apabila suis di dalam litar berikut ditutup. Dapatkan vC (t) untuk t ≥ 0.

JawapanPunca-punca persamaan ciri:

sradjs

sradj

s

/48001400

/48001400

4.01.0

10

2.0

280

2.0

280

2

62

1

Punca-punca adalah kompleks, maka sambutan voltan adalah underdamped. Oleh itu, diperolehi voltan vC :

04800sin

4800cos48

14002

14001

tteB

teBv

t

tC

Pada awalnya, tiada tenaga tersimpan dalam litar, maka:

12

1

140048000)0(

480)0(

BBdt

dv

Bv

C

C

Selesaikan untuk dan 1B

2B

VB

VB

14

48

2

1

penyelesaian untuk vC (t)

0

4800sin14

4800cos4848)(

1400

1400

tuntuk

Vte

tetv

t

t

C