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Physics 401, Fall 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
Measuring of the input resistance by applying the
ohmmeter to the input of DMM is bot correct. 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
9/1/2013 Physics 403 31
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
Do not forget about
Origin Help