source-free series
DESCRIPTION
Source-Free Series. RLC Circuits. Objective of Lecture. Derive the equations that relate the voltages across and currents flowing through a resistor, an inductor, and a capacitor in series as: the unit step function associated with voltage or current source changes from 1 to 0 or - PowerPoint PPT PresentationTRANSCRIPT
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RLC Circuits
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Objective of LectureDerive the equations that relate the voltages
across and currents flowing through a resistor, an inductor, and a capacitor in series as:the unit step function associated with voltage or
current source changes from 1 to 0 ora switch disconnects a voltage or current source
into the circuit.Describe the solution to the 2nd order equations
when the condition is:OverdampedCritically DampedUnderdamped
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Series RLC NetworkWith a step function voltage source.
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Boundary ConditionsYou must determine the initial condition of the
inductor and capacitor at t < to and then find the final conditions at t = ∞s.Since the voltage source has a magnitude of 0V at t <
to i(to
-) = iL(to-) = 0A and vC(to
-) = Vs vL(to
-) = 0V and iC(to-) = 0A
Once the steady state is reached after the voltage source has a magnitude of Vs at t > to, replace the capacitor with an open circuit and the inductor with a short circuit. i(∞s) = iL(∞s) = 0A and vC(∞s) = 0V vL(∞s) = 0V and iC(∞s) = 0A
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Selection of ParameterInitial Conditions
i(to-) = iL(to
-) = 0A and vC(to-) = Vs
vL(to-) = 0V and iC(to
-) = 0AFinal Conditions
i(∞s) = iL(∞s) = 0A and vC(∞s) = oV vL(∞s) = 0V and iC(∞s) = 0A
Since the voltage across the capacitor is the only parameter that has a non-zero boundary condition, the first set of solutions will be for vC(t).
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Kirchhoff’s Voltage Law
0)(1)()(
0)()()(
)()()(
)()(
)()(
0)()()(
0)(
2
2
2
2
tvLCdt
tdvLR
dttvd
tvdttdvRC
dttvdLC
tititidttdvCti
RivdttdiLtv
tvtvtv
tv
CCC
CCC
RCL
CC
RR
LL
RLC
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General SolutionLet vC(t) = Aest where t=t-to
01
0)1(
0
2
2
2
LCs
LRs
LCs
LRsAe
eLCAse
LAReAs
ts
tststs
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LCLR
LRs
LCLR
LRs
122
122
2
2
2
1
012 LC
sLRs
General Solution (con’t)
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LC
LR
o1
2
222
221
o
o
s
s
02 22 oss
General Solution (con’t)
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tstsCCC
tsC
tsC
eAeAtvtvtv
eAtv
eAtv
21
2
1
2121
22
11
)()()(
)(
)(
General Solution (con’t)
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Solve for Coefficients A1 and A2Use the boundary conditions at to
- and t = ∞s
to solve for A1 and A2.
Since the voltage across a capacitor must be a continuous function of time.
Also know that
SoC Vtv )(
S
ssss
SoCoCoCoC
VAAeAeA
Vtvtvtvtv
210
20
1
21
21
)()()()(
0
0)()()()(
22110
220
11
21
21
AsAseAseAs
tvtvdtd
dttdvCti
ssss
oCoCoC
oC
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Overdamped Case
implies that C > 4L/R2
s1 and s2 are negative and real numberststs
C eAeAtv 2121)(
222
221
o
o
s
s
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Critically Damped Case
implies that C = 4L/R2
s1 = s2 = - = -R/2L
ttC teAeAtv
21)(
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Underdamped Case
implies that C < 4L/R2
, i is used by the mathematicians for imaginary numbers
22
222
221
od
do
do
js
js
1j
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]sincos[)(
]sin)(cos)[()(
)]sin(cos)sin(cos[)(
sincos
sincos
)()(
21
2121
21
21
tAtAetv
tBBjtBBetv
tjtBtjtBetv
je
je
eBeBetv
ddt
C
ddt
C
ddddt
C
j
j
tjtjtC
dd
211 BBA 212 BBjA
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Angular Frequencieso is called the undamped natural frequency
The frequency at which the energy stored in the capacitor flows to the inductor and then flows back to the capacitor. If R = 0, this will occur forever.
d is called the damped natural frequencySince the resistance of R is not usually equal to
zero, some energy will be dissipated through the resistor as energy is transferred between the inductor and capacitor. determined the rate of the damping response.
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Properties of RLC networkBehavior of RLC network is described as damping,
which is a gradual loss of the initial stored energyThe resistor R causes the loss determined the rate of the damping response
If R = 0, the circuit is loss-less and energy is shifted back and forth between the inductor and capacitor forever at the natural frequency.
Oscillatory response of a lossy RLC network is possible because the energy in the inductor and capacitor can be transferred from one component to the other. Underdamped response is a damped oscillation, which is
called ringing.
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Properties of RLC networkCritically damped circuits reach the final
steady state in the shortest amount of time as compared to overdamped and underdamped circuits.However, the initial change of an overdamped
or underdamped circuit may be greater than that obtained using a critically damped circuit.
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Set of Solutions when t > toThere are three different solutions which
depend on the magnitudes of the coefficients of the and the terms. To determine which one to use, you need to
calculate the natural angular frequency of the series RLC network and the term .
LRLCo
2
1
)(tvC dttdvC )(
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Transient Solutions when t > toOverdamped response ( > o)
Critically damped response ( = o)
Underdamped response ( < o)
20
22
20
21
2121)(
s
s
eAeAtv tstsC
tC etAAtv )()( 21
22
21 )]sin()cos([)(
od
tddC etAtAtv
ottt where
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Find CoefficientsAfter you have selected the form for the
solution based upon the values of o and Solve for the coefficients in the equation by
evaluating the equation and its first derivate at t = to
- using the initial boundary conditions. vC(to
-) = Vs and dvC(to-)/dt = iC (to
-)/C = 0V/s
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Other Voltages and CurrentsOnce the voltage across the capacitor is
known, the following equations for the case where t > to can be used to find:
)()(
)()(
)()()()(
)()(
tRitvdttdiLtv
titititidttdvCti
RR
LL
RLC
CC
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Solutions when t < toThe initial conditions of all of the components are
the solutions for all times -∞s < t < to.vC(t) = VsiC(t) = 0A
vL(t) = 0ViL(t) = 0A
vR(t) = 0ViR(t) = 0A
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SummaryThe set of solutions when t > to for the voltage across
the capacitor in a RLC network in series was obtained.Selection of equations is determine by comparing the
natural frequency oto Coefficients are found by evaluating the equation and its
first derivation at t = to-.
The voltage across the capacitor is equal to the initial condition when t < to
Using the relationships between current and voltage, the current through the capacitor and the voltages and currents for the inductor and resistor can be calculated.