elec 201 – electric circuit analysis i - department of...

12/4/2017

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Solving RLC Circuits: Instructions

When you are presented with a switched or pulsed RLC circuit

and asked to solve for v(t) or i(t) somewhere in the circuit, for all time…

(1) Identify the circuit as containing two (and only two) energy storage elements

(i.e. a single Leq and a single Ceq).

(2) Determine 1 initial condition just before the switch/pulse, e.g. v(0–) .

(3) Determine 2 conditions just after the switch/pulse, e.g. v(0+) and d/dtv(0+) ,

with the knowledge that vC(0–) = vC(0+) , iL(0–) = iL(0+) ,

iC(0+) = C d/dtvC(0+), vL(0+) = L d/dtiL(0+) .

(4) Determine final conditions (a long time after the switch/pulse), e.g. v(∞) .

(5) Identify the circuit as a series or parallel RLC circuit

and find the Thevenin equivalent resistance seen by the LC pair, Rth .

(6) Calculate α and ω0 and determine the circuit damping (over, under, critical) .

(7) Assume the appropriate solution form and use α, ω0, initial conditions,

& the final condition to determine the complete (particular) solution.

All slides and content courtesy of Dr. Gregory J. Mazzaro

ELEC 201 – Electric Circuit Analysis I

Lecture 9(d)

RLC Circuit

Example

3

Solve for vR(t) for all values of t

and determine the settling time ts .

Example: RLC Circuit #1

4



( )0 48 VRv−

=

( )0 48 VRv+

=

( ) 0 VR

v ∞ =

( )0 0R

dv

dt

+=

overdamped

• Find 4 boundary conditions

& decide on the damping…

• Assume the appropriate form for t > 0 and solve for all 5 unknowns…

( ) 1 2

1 2 3

s t s tx t X e X e X= + + 2 2

1,2 0s α α ω= − ± −


( ) ( ) ( )

( ) ( )0

15

2 24 1 240

124

10 1 240

α

ω

= =

= =

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( )0 48 VRv −=

( )0 48 VRv +=

( ) 0 VRv ∞ =

( )0 0R

dv

dt

+=

( ) 1 2

1 2 3

c t c t

Rv t V e V e V= + +05.0 , 4.9α ω= =

2 2

1,2 1 25.0 5.0 4.9 4, 6c c c= − ± − ⇒ ≈ − ≈ −2 2

1,2 0c α α ω= − ± −

( ) 4 6

1 2 3

t t

Rv t V e V e V

− −= + +

Substituting 2 of the 5 unknowns

into the overdamped solution…


6



( )0 48 VRv −=

( )0 48 VRv +=

( ) 0 VRv ∞ =

( )0 0R

dv

dt

+=

( ) 4 6

1 2 3

t t

Rv t V e V e V

− −= + +

Making use of the initial & final conditions…

( ) 1 2 30 48 VR

v V V V+

= + + =

( ) 3 0 VR

v V∞ = =( ) 1 20 4 6 0R

dv V V

dt

+= − − =

Solving the 3 equations with 3 unknowns…1 2 3144 V , 96 V, 0V V V= = − =


7



( ) 4 6

48 V 0

144 96 V 0R t t

tv t

e e t− −

<=

− >


8



The circuit’s settling time ts is the time

after the switch/pulse for v/i to stay

within 1% of x(∞∞∞∞) up/down from x(0+) :

( ) ( ) ( ) ( )1

0100

sx t x x x+

≈ ∞ + ⋅ − ∞

( ) ( )1

48 V 0.48 V100

1.42 s

R s

s

v t

t

= =

≈

( ) 4 6

48 V 0

144 96 V 0R t t

tv t

e e t− −

<=

− >


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t = 0:1e-3:1.5;

v_R = 144*exp(-4*t) - 96*exp(-6*t);

plot(t,v_R)

axis([0 1.5 0 50])

grid

ylabel('v_R (volts)')

xlabel('Time (seconds)')

( ) 4 6

48 V 0

144 96 V 0R t t

tv t

e e t− −

<=

− >


All content courtesy of Dr. Gregory J. Mazzaro

ELEC 201 – Electric Circuit Analysis I

Lecture 9(e)

More RLC Circuit

Examples

11

Is the circuit (at right) underdamped,

overdamped, or critically damped?


12



th0

1,

2

R

L LCα ω= =series RLC:

R is the Thevenin equivalent resistance

seen by the L & C in series…

A

B

no independent sources

must use a test source to find RthItest

Vtest

+

–


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13



th0

1,

2

R


ItestVtest

+

–

test 1 A, 1 AI i= =

( ) ( ) ( ) ( )test

test

9 1 3 1 2 1 0

8 V

V

V

− + − + =

=

choose the test current:

solve for the test voltage:

their ratio is Rth: th test test8R V I= = Ω


14



th0

1,

2

R


( ) ( )th 8 rad

0.82 2 5 s

R

Lα = = =

α < ω0, underdamped

( ) ( )0

3

1 1 rad10

s5 2 10LCω

−= = =

⋅


15

PSpice: RLC Circuit #2



oscillates, therefore

underdamped

16

( )2 Vsv u t=

Is the circuit (below) underdamped,


235 Ω

4.7

nF

1

mHsv

Analog Discovery: RLC Circuit #3

12/4/2017

5

17

( )2 Vsv u t=


18


19

no oscillation, therefore

overdamped


20

1.5 kΩ

4.7

nF

1

mHsv

Is the circuit (below) underdamped,


( )2 Vsv u t=


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21


22


oscillates, therefore

underdamped

23


Determine the damping of this circuit

(after t = 0).

24

After the switch closes, the circuit becomes

0

1 1,

2RC LCα ω= =For a parallel RLC circuit,

( ) ( )5

9

1 rad1.3 10

s2 200 20 10α

−= = ⋅

⋅ ( )( )5

03 9

1 rad10

s5 10 20 10ω

− −= =

⋅ ⋅

α > ω0, overdamped



(after t = 0).

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7

25



(after t = 0).


overdamped

26



(after t = 0).


overdamped

27

Solve for i1(t) for all values of t

and plot i1 from t = 0 to t = ts .


28





( ) ( )1

10 3 A 1.5 A

2i

−= − = −

( )0 0Li−

=

( ) ( ) ( )

( ) ( )

1 1

1

0 2 0 0 0

0 3 0 4.5 V

C

C

v i i

v i

− − −

− −

− − =

= = −

Before the step…


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Just after the step…

( ) ( )0 0 0L L

i i− += =

( ) ( ) ( ) ( ) ( )1 10 0 2 0 0 0

C Lv i i v+ + + +

− + − − − =

KVL around right loop…

KVL around outer loop…

( ) ( ) ( ) ( ) ( )1 10 0 2 0 0 0C L

v i i v+ + + +

− − − − − =

( )1 0 0i+

=

( ) ( )0 0

4.5 V

L Cv v+ +

= −

=

vL

+

–

i1


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Also note:

( ) ( )1

10 0

2L

i i+ += − ( ) ( )

( )

( )

1

10 0

2

1 10

2

1 1 A4.5 0.225

2 10 s

L

L

d di i

dt dt

vL

+ +

+

= −

= −

= − = −

Just after the step…


31





A long time after the step…

( ) 0Li ∞ =

( ) 0Cv ∞ =

( )1 0i ∞ =

(no independent source)


32





We need Rth seen by the L, C to decide on the damping…

A

BVtest

+

–

Itest

choose Itest = 2 A…

test1 1 A

2

Ii = =

( ) ( ) ( )test

test

1 1 2 1 0

3 V

V

V

− + − − =

=

th test test1.5R V I= = Ω


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33





For a series RLC circuit,

th0

1,

2

R

L LCα ω= =

( ) ( )

1.5 rad0.075

2 10 sα = =

( ) ( )0

1 rad0.316

s10 1ω = =

0α ω<

underdamped

( ) ( ) ( )1 1 2 3cos sint

d di t e I t I t I

α ω ω−= + +

…therefore the solution is of the form


34

• Solve for I1, I2, I3…



( ) ( ) ( )1 1 2 3cos sint

d di t e I t I t I

α ω ω−= + +

( )10i ∞ =( )1

A0 0.225

s

di

dt

+= −

( )1 0 1.5 Ai−

= − ( )1 0 0 Ai+

=

( ) ( ) ( )0

1 1 2 3 1 30 cos 0 sin 0 0

d di e I I I I Iω ω+ −

= ⋅ + ⋅ + = + =

( ) ( )2 22 2

0 .316 .075

0.307 rad s

dω ω α= − = −

=

( ) ( ) ( )1 1 2 3 30 cos sin 0

d di I t I t I Iω ω∞ = ⋅ ⋅ + ⋅ + = =

( ) ( ) ( ) ( ) ( )1 1 2 1 2

1 2

0 cos sin sin cos

0.075 0.307 0.225

t t

d d d d d d

di e I t I t e I t I t

dt

I I

α αα ω ω ω ω ω ω+ − −= − + + − +

= − ⋅ + ⋅ = −

1

2

3

0 A

733 mA

0 A

I

I

I

=

= −

=


35



( ) ( ) ( ) ( )

( ) ( )

1 1 1 1

1

10

100

1733 mA

100

s

s

i t i i i

i t

+≈ ∞ + − ∞

≈( )

0.0751

100

ln 1 100 0.075

4.6 0.075

st

s

s

e

t

t

−≈

≈ − ⋅

− ≈ − ⋅

60 ss

t ≈

( )( )1 0.075

1500 mA 0

733 sin 0.307 mA 0t

ti t

e t t−

− <=

− ⋅ ⋅ >


36



t = -10:.1:100;

alpha = 0.075;

omega_d = 0.307;

I_1 = 0; I_2 = -733e-3; I_3 = 0;

i = (t>0).*(exp(-alpha*t).*(I_1*cos(omega_d*t) ...

+ I_2*sin(omega_d*t)) + I_3) + ...

(t<0).* -1.5;

plot(t,i,'Linewidth',2)

axis([-10 50 -1.6 0.4])

grid

ylabel('i_1 (A)')

xlabel('Time (s)')

( )( )1 0.075

1500 mA 0

733 sin 0.307 mA 0t

ti t

e t t−

− <=

− ⋅ ⋅ >

Matlab: RLC Circuit #6

12/4/2017

10

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( )( )1 0.075

1500 mA 0

733 sin 0.307 mA 0t

ti t

e t t−

− <=

− ⋅ ⋅ >


elec 201 – electric circuit analysis i - department of...

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