INC 111 Basic Circuit Analysis

Week 8

RL circuits

Non-periodic Signal

There are infinite number of non-periodic signal.

This course will cover only the most basic one, a step.

A step is a result from on/off switches, which is common in our daily life.

0V

9V

On switch

I

1V1Ω

I

2V1Ω

I = 1A I = 2A

Voltage source change from 1V to 2V immediatelyDoes the current change immediately too?

Voltage

Current

time

time

1V

2V

1A

2A

AC voltage

I

L1V1Ω

I

L2V1Ω

I = 1A I = 2A

Voltage source change from 1V to 2V immediatelyDoes the current change immediately too?

Voltage

Current

time

time

1V

2V

1A

2A

Forced Response

Transient Response + Forced Response

AC voltage

I am holding a ball with a rope attached, what is the movement of the ball ifI move my hand to another point?

Movements

1. Oscillation

2. Forced position change

Pendulum Example

• Transient Response or Natural Response (e.g. oscillation, position change temporarily)

Fade over timeResist changes

• Forced Response (e.g. position change permanently)

Follows inputIndependent of time passed

Forced response Natural responseat different time

Mechanical systems are similar to electrical system

Transient Response

• RL Circuit

• RC Circuit

• RLC Circuit

First-order differential equation

Second-order differential equation

RL Circuit

AC

+

-

u(t)

i(t)

L

R

KVL )()(

)( tudt

tdiLtRi

First-order Differential equation

Objective: Want to solve for i(t) (in term of function of t)

)()(

)( tudt

tdiLtRi

Assume that i(t) = g(t) make this equation true.

consider

However, g(t) alone may be incomplete. The complete answer isi(t) = f(t) + g(t)

where f(t) is the answer of the equation

0)(

)( dt

tdiLtRi

Voltage source go to zero

Proof: Answer has two parts

0)(

)( dt

tdiLtRi

f(t) is the answer ofthis equation

therefore 0)(

)( dt

tdfLtRf ------------------(1)

)()(

)( tudt

tdiLtRi

g(t) is the answer ofthis equation

therefore )()(

)( tudt

tdgLtRg ------------------(2)

from (1) = 0

)()(

)( tudt

tdiLtRi

f(t)+g(t) is alsothe answer ofthis equation

therefore )(

)()()()( tu

dt

tgtfdLtgtfR

)()()(

)()( tudt

tdgL

dt

tdfLtRgtRf

)()(

)()(

)( tudt

tdgLtRg

dt

tdfLtRf

)()(

)( tudt

tdgLtRg which is true

from (2)

If

must be true

)()()( tgtfti

Transient ResponseForced Response

i(t) consists of two parts

Therefore, we will study source-free RL circuit first

Source-free RL Circuit

LR+

-

-

+

i(t)

Inductor L has energy stored so thatthe initial current is I0

Compare this with a pendulumwith some height (potential energy) left.

height

0)()(

0)(

)(

tiL

R

dt

tdidt

tdiLtRi

There are 2 ways to solve first-order differential equations

LR+

-

-

+

i(t)

Method 1: Assume solutionwhere A and s is the parameters that we want to solve for

stAeti )(

Substitute in the equationstAeti )( 0)()(

tiL

R

dt

tdi

0)(

0

st

stst

AeL

Rs

AeL

RAse

The term that can be 0 is (s+R/L) , therefore

The answer is in format

L

Rs

tL

R

Aeti

)(

Initial condition 0)0( Ii

tL

R

Aeti

)(from Substitute t=0, i(t=0)=0

AI

AeI

0

00

We gottL

R

eIti

0)(

Method 2: Direct integration

tti

I

dtL

R

ti

tdi

dtL

R

ti

tdi

tiL

R

dt

tdi

tiL

R

dt

tdi

0

)(

0)(

)(

)(

)(

)()(

0)()(

tL

R

tti

I

eIti

tL

RIti

tL

Rti

0

0

0

)(

)(

)0(ln)(ln

)(ln0

t

i(t)

I0

Approach zero

tL

R

eIti

0)(

Natural Response only

Natural Responseof RL circuit

Time Constant

Ratio L/R is called “time constant”, symbol τ

R

L

Time constant is defined as the amount of time usedfor changing from the maximum value (100%) to 36.8%.

t

tL

R

eIeIti

00)(

Unit: second

368.01 e

t

i(t)

2AtL

R

eti

1)(

Natural Response + Forced Response

1A

Natural Response

Forced Response

Forced response = 1A comes from voltage source 1V

Approach 1A

Switch

Close at t =0 Open at t =0

t=0 t=0

t=0t=0

t=0

3-way switch

t < 0

Switch

Close at t =0 Open at t =0

t=0 t=0

t=0t=0

t=0

3-way switch

t > 0

t=0 t=0

R R1V 1V

t

v(t)

1V

t

v(t)

1V

0V 0V

Step function (unit)

t=0

R=1Ω

2VL

1V

Will divide the analysis into two parts: t<0 and t>0

When t<0, the current is stable at 2A. The inductor acts likea conductor, which has some energy stored.

When t>0, the current start changing. The inductor discharges energy.Using KVL, we can write an equation of current with constantpower supply = 1V with initial condition (current) = 2A

For t>0

1))(ln(

)(

)(

)(

)(

)()(

)()(

cttRiVR

L

dttRiV

tLdi

dttRiV

tLdi

tRiVdt

tdiL

Vdt

tdiLtRi

tL

R

tL

R

tL

R

cL

RtL

R

eR

c

R

Vti

ecVtRi

cetRiV

eetRiV

cL

Rt

L

RtRiV

cttRiVR

L

2

2

2

1

1

)(

)(

)(

)(

))(ln(

))(ln(

1

tL

R

eR

c

R

Vti

2)( We can find c2 from initial condition

i(0) = 2 A

Substitute t = 0, i(0) = 2

RVcR

c

R

V

2

12

2

2

Therefore, we havetL

R

eR

RV

R

Vti

2)(

tL

R

eti

1)( Natural Response

Forced Response

Substitute V=1, R=1

RL Circuit Conclusion

• Force Response of a step input is a step

• Natural Response is in the form where k1 is a constant, whose value depends on the initial condition.

tL

R

ek

1

Response

time

Period 1

How to Solve Problems?

Period 2 Period 3

• Divide in to several periods (3 periods as shown below)• Period 1, 3 have constant V, I -> Use DC circuit analysis• Period 2 is transient.

Calculate Transient (period 2)

• Start by finding the current of the inductor L first

• Assume the response that we want to find is in form of

t

ekk

21

• Find the time constant τ (may use Thevenin’s)

• Solve for k1, k2 using initial conditions and status at the stable point

• From the current of L, find other values that the problem ask

Examplet=0

R=1Ω

2VL=1H

1V

i(t)

The switch is at this position for a long timebefore t=0 , Find i(t)

t

ekkti

21)(

Time constant τ = 1 sec

t

ekkti

21)(

At t=0, i(0) = 2 A212 kk

At t = ∞, i(∞) = 1 A 01 1 k

Therefore, k1 = 1, k2 = 1

The answer is teti 1)(

2A

1A

teti 1)(2)( ti

)(ti

Example

R1

L

R2

R3

R4i(0)=5

+ -

i2(t)

L has an initial current of 5A at t=0Find i2(t)

The current L is in form of t

L ekkti

21)(

Time constant = R/L, find Req 421

213 R

RR

RRRReq

Time constanteqR

L

(Thevenin’s)

Find k1, k2 using i(0) = 5, i(∞) = 0

At t=0, i(0) = 5 A 215 kk

At t = ∞, i(∞) = 0 A 00 1 k

t

L ekkti

21)(

Therefore, k1=0, k2 = 5 t

L eti

5)(

i2(t) comes from current divider of the inductor current

21

12 5)(

RR

Reti

t

Graph?

)(2 ti

21

12 5)(

RR

Reti

t

Example

t=0

1Ω1V

1H2Ω

2Ω+

v1(t)-

L stores no energy at t=0Find v1(t)

5.02

1

21)2||2(

eq

eq

R

L

R

Find iL(t) first

Find k1, k2 using i(0) = 0, i(∞) = 0.25

At t=0, i(0) = 0 A 210 kk

At t = ∞, i(∞) = 0.25 A 025.0 1 k

t

L ekkti

21)(

Therefore, k1=0.25, k2 = -0.25t

L eti 225.025.0)(

v1(t) = iL(t) R

tetv 21 25.025.0)(

Graph?

tetv 21 25.025.0)(

)(1 tv

0)(1 tv