Physics 401, Spring 2013. Eugene V. Colla

Transients and Oscillations in RLC Circuits.

Outline

transient ( physics ) a short-lived oscillation in a

system caused by a sudden change of voltage or

current or load

a transient response or natural response is the

response of a system to a change from equilibrium.

System under

study

R

L

C

V scope

Resistance R [Ohm]

Capacitance C [uF] (10-6F)

Inductance L [mH] (10-3H)

R

L

C

V(t) scope

VR VL

VC

According the Kirchhoff’s law VR+VL+VC=V(t)

-1.0 -0.5 0.0 0.5 1.00.0

0.5

1.0

V(t

)

time

V0

2

02

d d q(t) q(t)L q(t) +R q(t) + = 0 , = V

dt dt C C*See Lab write-up for details

*

(1)

The solution of this differential

equation can be found in the form ( ) est

q t A

This will convert (1) in

quadratic equation

Rs s

L LC

21

0

,

,

R Rs a b

L L LC

R Ra b

L L LC

2

1 2

2

1

2 2

1

2 2

with solutions:

b2>0 over-damped solution

b2=0 critically damped solution

b2<0 under-damped solution

b2>0

( ) e at bt bt

q t A e B e1 1

In this case the solution will be aperiodic

exponential decay function with no

oscillations:

( ) ( ) ( )at bt bt at bt btdq

i t ae A e B e be A e B e

dt1 1 1 1

b2>0

Taken in account the initial conditions: q(0)=q0 and i(0)=0

( ) cosh sinhat

aq t q e bt bt

b0

( )( ) a b tq a b

i t e

b

2 2

0

2

This is exponential decay function

-( - )

( - )

a b t

a b t

q ae

b

0

11

2

b2=0

For this case the general solution can be found as

q(t)=(A2+B2t)e-at. Applying the same initial condition

the current can be written as i=–a2q0te-at

Critical damped case shows the

fastest decay with no oscillations

b2=0

In this experiment R=300 ohms,

C=1mF, L=33.43mH.

The output resistance of Wavetek is 50 ohms and

resistance of coil was measured as 8.7 ohms, so actual

resistance of the network is Ra=300+50+8.7=358.7

Decay coefficient 𝒂 =𝑹

𝟐𝑳=

𝟑𝟓𝟖.𝟕

𝟐∗𝟑𝟑.𝟒𝟑𝑬−𝟑≈ 𝟓𝟑𝟔𝟓

b2=0

Now the experimental results:

Calculated decay

coefficient ~5385,

Obtained from fitting -

~5820.

Possible reason – it is

still slightly over damped

Calculated b2 is

b2=2.99e7-2.90e7>0

Vc ~q, fiiting function: Vc=Vco(1+at)e-at

If b2<0 we will have oscillating solution. Omitting the details (see

Lab write-up) we have the equations for charge and current as:

,

cos sin sin( )

sin

;

at at

at

a aq(t) q e bt bt q e bt

b b

a bi(t) q e bt

b

R R Ra b f

L L LC LC L

2

0 0 2

2 2

0

2 2

1

1 1 1

2 2 2 2

Log decrement can be defined as 𝜹 = 𝒍𝒏𝒒(𝒕𝒎𝒂𝒙

𝒒(𝒕𝒎𝒂𝒙+𝑻𝟏=

𝒍𝒏𝒆−𝒂𝒕𝒎𝒂𝒙

𝒆−𝒂𝒕𝒎𝒂𝒙+𝑻𝟏= 𝒂𝑻𝟏, where T1=1/f1

Quality factor can be

defined as 𝑸 = 𝟐𝝅𝑬

∆𝑬,

For RLC 𝑸 =𝝎𝟏𝑳

𝑹=

𝝅

𝜹

From this plot d≈0.67

Q≈4.7 -1 0 1 2 3 4 5 6 7 8 9 10

-6

-3

0

3

6

3.529

1.809

0.929620.47494

VC (

q/C

) (V

)

time (ms)

1. Pick peaks

2. Envelope

3. Exponential term

4. Nonlinear fitting

-1 0 1 2 3 4 5 6 7 8 9 10-6

-3

0

3

6

0.00115

0.0023

0.003460.00463

VC (

q/C

) (V

)

time (ms)

f=862Hz

-1 0 1 2 3 4 5 6 7 8 9 10-6

-3

0

3

6

VC (

q/C

) (V

)

time (ms)

f=862Hz

Time domain trace

Points found using “Find peaks”

Envelope curve

Fitting the “envelope data” to

exponential decay function

Zero crossing points

0ffset Manual evaluation of the

period of the oscillations

in( )at

q(t) Ae s t offset

Limited accuracy

Results can be

effected by DC offset

in( )at

q(t) Ae s t

Limited

Use Origin standard

function

Category: Waveform

Function: SineDamp

Fitting function ; y0,A,t0 xc, w – fitting parameters

in( )at

q(t) Ae s t

Limited

Data plot + fitting curve

Residuals - criteria

of quality of fitting

in( )at

q(t) Ae s t

Final results

Rf

T LC L

2 2

21 1 1

2 2

R2

L

C VC V(t)

100 1000 100000

2

4

6

8

10

12

14 1904.83204

UC

f (Hz)

f=1500Hz

𝑸 =𝒇

𝜟𝒇=

𝟏𝟗𝟎𝟒

𝟏𝟓𝟎𝟎= 𝟏. 𝟐𝟔

One important parameter for any electrical measuring equipment

(DMM, scope, amplifier) is its input resistance

Rin≠∞;

Im=I0

RRin

1+ RRin

;

𝑉=I0𝑅

1+R

Rin

Real situation:

For correct voltage measurements

it necessary to have Rin>>R

The same about current measurement, but for ampere meter the

requirement for Rin is much different.

𝑰 =𝑽

𝑹 + 𝑹𝒊𝒏

I≈I0 if Rin →0

Real situation:

For correct current measurements

it necessary to have Rin<<R

Why?

FET The current applied

by ohmmeter to the

input of amplifier or

DMM can destroy

the electronics

Correct idea Ignoring battery output

resistance Vb we can

measure if Rs=0. Next

we have to vary Rs until

DMM reading becomes

equal 0.5 Vb

𝑽 = 𝟎. 𝟓𝑽𝒃 =𝑹𝒊𝒏

𝑹𝒔 + 𝑹𝒊𝒏𝑽𝒃

𝑹𝒔 = 𝑹𝒊𝒏

+

-

V Rout RL VL

Voltage on the load:

VL = 𝑽 ×𝑹𝑳

𝑹𝑳+𝑹𝒐𝒖𝒕

VL≈V if Rout<<RL

Power on the load:

𝑷 =𝑽𝟐

𝑳

𝑹𝑳=

𝑽𝟐𝑹𝑳

𝑹𝑳+𝑹𝒐𝒖𝒕𝟐

P=Pmax= 𝐕𝟐

𝟒𝐑𝐨𝐮𝐭

if RL=Rout Voltage source circuit

diagram symbol

Equivalent circuit of voltage source

V Rout

RL

IL Ideal current source – current

in the load should not depend

on the load resistance.

IL = 𝐕

𝐑𝐋+𝐑𝐨𝐮𝐭=

𝐈𝟎

𝟏+𝐑𝐋

𝐑𝐨𝐮𝐭

,

where

𝐈𝟎 = 𝐕

𝐑𝐨𝐮𝐭

So, for ideal current

source Rout →∞ Current source circuit

diagram symbol

Equivalent circuit of current source

Varying the resistance of the parallel resistor Rp we

have to find the value of Rp corresponding the

reading of current equal I0/2. In this case

Rin=Rp

Current from some current source

1/14/2013 Physics 403 30

If the sensor is mounted in cryostat the

overall leads resistance could reach a

couple of ohms. This will In case of RTD100

the resistance at 0oC is 100W give an error

of a couple percent!

Most of DMM’s have four probe

option for resistance

measurements.

I1=Io*RT/(r2+r3+Rin) ~0

Rin~1010W

Open template

button

Very short and simple manual

which covers only main general

operations with Origin.

Document located on server

(\\Phyaplportal\PHYCS401\Comm

on\Origin manuals) and there is a

link from P401 WEB page

There are (\\Phyaplportal\PHYCS401\Common\Origin manuals) also manuals

from OriginLab.

Video Tutorials at the

site of the company

http://www.originlab.com/index.aspx?go=SUPPORT/VideoTutorials

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