RLC transients - Iowa State...

EE 201 RLC transient – 1

RLC transients

When there is a step change (or switching) in a circuit with capacitors and inductors together, a transient also occurs. With some differences:

• Energy stored in capacitors (electric fields) and inductors (magnetic fields) can trade back and forth during the transient, leading to possible “ringing” effects.

• The transient waveform can be quite different, depending on the exact relationship of the values of C, L, and R.

• The math is more involved.


+–

Vf

t = 0Vi

+

–

vR

+vL

+

–

–

vC

R

L

Ci

VS

series RLC

Voltage source makes an abrupt change from Vi to Vf at t = 0.

96 = Y5 + Y& + Y/

t < 0: i = 0.

vR = 0.

vL = 0.

vC = Vi.

t >> τ: i = 0.

vR = 0.

vL = 0.

vC = Vf.

What is τ? What happens in between?


+–

Vf

t = 0Vi

+

–

vR

+vL

+

–

–

vC

R

L

Ci

VS

L = &GY&GW

Second-order differential equation.

G�Y&GW�

+5/GY&GW +

Y&/& =

9I/&

9I = 5&GY&GW + Y& + /&G�Y&GW�

9I = L5 + YF + /GLGWt > 0:


Diff. Eq. – a bit of a review (or preview)

G�I ([)G[� + DGI ([)G[ + EI ([) = J ([) g(x) is a “forcing function”

ftr(x) → transient solution (homogenous)

fss(x) → steady-state solution (particular)

I ([) = IWU ([) + IVV ([)

fss(x) is any function that you can find that is a solution to the full differential equation. Usually find it with trial-and-error. The form of g(x) suggests what to try.

ftr(x) is the solution to the homogenous differential equation.

G�IWU ([)G[� + DGIWU ([)

G[ + EIWU ([) = �

The solution will have two unknown constants that will be specified using initial conditions, f(0) and df(0)/dx.


G�Y&GW�

+5/GY&GW +

Y&/& =

9I/&

Y& (W) = YWU (W) + YVV (W)

Since the forcing function is a constant, try using vss(t) = constant = Vss.

G�9VV

GW�+

5/G9VVGW +

9VV/& =

9I/&

Vss = Vf . (Easy!)

(We could have determined this from the physics of the circuit, also.)

Y& (W) = YWU (W) + 9I

Now we must find the transient part.


homogenous solution

G�YWUGW�

+5/GYWUGW +

YWU/& = �

YWU = $ exp (VW)Guess:

V��$HVW

�+

5/ V

�$HVW

�+�/&

�$HVW

�= �

�V� +

5/ V +

�/&

� �$HVW

�= �

Will need to determine s.

$HVW �= � so

�V� +

5/ V+

�/&

�= �

V = � 5�/ ±

��5�/

�� /&

V� = � 5�/ +

��5�/

�� /&

V� = � 5�/ �

��5�/

�� /&

YWU (W) = $HV�W + %HV�W


Initial conditions

Need to determine A and B. Use initial conditions, vc(0) and dvc(0)/dt.

Y& (W) = $HV�W + %HV�W + 9I

GY& (W)GW =

L& (W)&

GY& (W)GW =

L/ (W)&

GY& (W)GW

��W=�

=L/ (�)& = �

vC(0) = Vi

The details will depend on s1 and s2, and, as we will see, there are three possible cases.

In any case:

But don’t necessarily know iC(0) since capacitor current can change abruptly. However, in the series circuit, iC(t) = iL(t). Inductor current cannot change abruptly. So

For this particular case. May not be true in every case.


roots of characteristic equation

Transient behavior depends on the values of s1 and s2.

V� = � 5�/ +

��5�/

�� /& V� = � 5

�/ �

��5�/

�� /&

Rename things slightly: R/2L = α and 1/LC = ωo.

V� = �� +�

�� R V� = ��

�� R

α is the damping factor or decay constant [s–1]$ωo is the resonant frequency or undamped natural frequency [radian/s].


Overdamped response5�/ >

��/&If (α > ωo) then s1 and s2 are both real and negative.

The capacitor voltage consists of two decaying exponentials.

A fast, one-way trip from Vi to Vf.

Y& (W) = $HV�W + %HV�W + 9I

Finally, use initial conditions to find A and B

YF (�) = 9L = $ + % + 9I

GY&GW

��W=�

= � = V�$ + V�%solve to give:

$ =9L � 9I�� V�

V�

% =9L � 9I�� V�

V�

Y& (W) =�9L � 9I

��

HV�W�� V�

V�+

HV�W�� V�

V�

�+ 9I


0.0!

5.0!

10.0!

15.0!

20.0!

25.0!

-1.0! 0.0! 1.0! 2.0! 3.0! 4.0! 5.0!

v c (V

)!

t (ms)!

Over-damped response: Vi = 5 V, Vf = 20 V, R = 1 k!, L = 15 mH, and C = 0.5 µF.


Underdamped response

If (α < ωo) then s1 and s2 are complex!5�/ <

��/&

V� = � 5�/ +

��5�/

�� /&

= � 5�/ + M

��/& �

�5�/

��

V� = � 5�/ �

��5�/

�� /&

= � 5�/ � M

��/& �

�5�/

��

Y& (W) = $H��WHM�GW + %H��WH�M�GW + 9I

= �� + M�

��R � �� = �� M�

��R � ��

= �� M�G= �� + M�G

Write as complex numbers. (For EE/CprEs only.)M =�

��

�G =�

��R � ��

s1 s2

(ωd is the damped frequency)


(Euler’s identity.)

= H��W [($ + %) cos �GW + M ($ � %) sin �GW] + 9I

Y& (W) = H��W [$ (cos �GW + M sin �GW) + % (cos �GW � M sin �GW)] + 9I

HM[ = cos [ + M sin [

Y& (W) = H��W �$HM�GW + %H�M�GW�+ 9I

Y& (W) = H��W �$� cos �GW + %� sin �GW�+ 9I

There is an oscillation in the response. The voltage changes from Vi to Vf, but wiggles back and forth a few times in the process. The oscillation dies out according to the damping factor over about 5 time constants, where the time constant τ = 1/α.


Use initial conditions to find A’ and B’.

YF (�) = 9L = $� + 9I

GY&GW

��W=�

= � = ��$� + �G%�solve to give:

$� =�9L � 9I

�

%� =�

�G

�9L � 9I

�

Y& (W) =�9L � 9I

�H��W

�cos �GW +

�

�Gsin �GW

�+ 9I


0.0!

5.0!

10.0!

15.0!

20.0!

25.0!

30.0!

-0.3! 0.0! 0.3! 0.5! 0.8! 1.0!

v c (V

)!

t (ms)!

Under-damped response: Vi = 5 V, Vf = 20 V, R = 300 !, L = 25 mH, and C = 60 nF.

Changing R, L, and C changes the oscillation frequency and the damping rate.


Critically damped response

If

This causes a bit of a problem, because we are left with only one term in the general solution, and hence only one coefficient – not enough to satisfy the initial conditions.

This suggests that there must be another solution lurking around in the math. In the special circumstances for the critically damped case, the homogeneous equation can be written:

5�/ =

��/&

(α = ωo) then s1 = s2 (double root).

G�YWU

GW�+ ��GYWU

GW + ��YWU = �

Y& (W) = $HV�W + %HV�W + 9I = &HV�W + 9I

= & exp

�� W�//5

�+ 9I

Note: This will almost never happen. It will be the wildest fluke if the components have exactly the correct ratios to meet the above requirement. For the most part, critical damping is only of academic interest.


G�YWU

GW�+ ��GYWU

GW + ��YWU = �

GGW

�GYWUGW + �YWU

�+ �

�GYWUGW + �YWU

�= �

G\GW + �\ = � where \ =

GYWUGW + �YWU

$H��W =GYWUGW + �YWU\ = $H��W

$ = H�W GYWUGW + �H�WYWU

$ =GGW

�H�WYWU

�

$W + % = H�WYWU

YWU (W) = ($W + %) H��W Now there are two constants.


Y& (W) = ($W + %) H��W + 9I

Also a fast trip from Vi to Vf.

Use initial conditions to find A and B.

YF (�) = 9L = % + 9I

GY&GW

��W=�

= � = $ � �%

$ = ��9L � 9I

�

% =�9L � 9I

�

Y& (W) =�9L � 9I

�(�+ �W) H��W + 9I

solve to give:


Critically-damped response: Vi = 5 V, Vf = 20 V, R = 1 k!, L = 15 mH, and C = 60 nF. (Same as overdamped plot of slide 10, except C is smaller.)

Note the significantly faster rise time compared to the overdamped response.

0.0!

5.0!

10.0!

15.0!

20.0!

25.0!

-50.0! 0.0! 50.0! 100.0! 150.0! 200.0! 250.0! 300.0!

v c (V

)!

t (µs)!


parallel RLC

Current source makes an abrupt change from Ii to If at t = 0.

If

t = 0Ii

IS R C LiR ic iL –

v+

,6 = L5 + L& + L/

=Y5 + &GY

GW + L/

t < 0:

v = 0.

iR = 0.

iL = Ii.

iC = 0.

t >> τ:v = 0.

iR = 0.

iC = 0.

iL = If.

Y = /GL/GW

,I =/5

GL/GW + /&G�L/

GW�+ L/

t > 0: initial conditions (t = 0)

L/ (�) = ,L

GL/ (W)GW

��W=�

=Y/ (�)/ =

Y& (�)/ = �G�L/

GW�+

�5&

GL/GW +

L//& =

,I/&


G�L/GW�

+�5&

GL/GW +

L//& =

,I/&

This has exactly the same form as the series case, with inductor current replacing capacitor voltage. The steady-state function will be iss = If, and the transient function will have the general form:

LWU (W) = $HV�W + %HV�W

V� = � ��5& �

��5&

�� /&

V� = � ��5& +

��5&

�� /&

= �� +�

�� R

= ��

�� R

� =��5& damping factor - note the

difference from series case

�R =��/& resonant frequency


We will obtain the exact same set of results for the parallel case:

��5& >

��/&

(� > �R) overdamped – both roots are real and negative.

L/ (W) =�,L � ,I

��

HV�W�� V�

V�+

HV�W�� V�

V�

�+ ,I

��5& =

��/&

(� = �R) critically damped – repeated root.

L/ (W) =�,L � ,I

�(�+ �W) H��W + ,I

L/ (W) =�,L � ,I

�H��W

�cos �GW +

�

�Gsin �GW

�+ ,I

��5& <

��/&

(� < �R) underdamped – roots are complex conjugates.

�G =�

��R � ��

(Virtually impossible to have critical damping.)

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