First Order Circuits. Objective of Lecture Explain the operation of a RC circuit in dc circuits As...
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Transcript of First Order Circuits. Objective of Lecture Explain the operation of a RC circuit in dc circuits As...
RC and RL CircuitsFirst Order Circuits
Objective of LectureExplain the operation of a RC circuit in dc circuits
As the capacitor releases energy when there is: a transition in a unit step function voltage or current
source or a voltage or current source is switched out of the
circuit.
Explain the operation of a RL circuit in dc circuitAs the inductor releases energy when there is:
a transition in a unit step function voltage or current source
or a voltage or current source is switched out of the circuit.
Natural ResponseThe behavior of the circuit with no external sources
of excitation. There was a transition in the source in the circuit
where the unit step function changed from 1 to 0 at t ≤ 0s.
There is stored energy in the capacitor or inductor at time t = 0 s.
For t > 0 s, the stored energy is released Current flows through the circuit and voltages exist across
components in the circuit as the stored energy is released. The stored energy will decays to zero as time approaches
infinite, at which point the currents and voltages in the circuit become zero.
RC CircuitSuppose there is some charge on a capacitor
at time t = 0s. This charge could have been stored because a voltage or current source had been in the circuit at t<0s, but was switched off at t = 0s.
We can use the equations relating voltage and current to determine how the charge on the capacitor is removed as a function of time.The charge flows from one plate of the
capacitor through the resistor R to the other plate to neutralize the charge on the opposite plate of the capacitor.
Equations for RC Circuit
R
vi
dt
dvCi
ii
vv
RR
CC
RC
RC
0
ottCC
C
C
C
C
CC
CR
RC
VRC
tV
dtRCV
dV
RCdt
dV
V
RC
V
dt
dV
VV
R
V
dt
dVC
lnln
1
011
0
0
to
CR
t
oC
stCo
eR
VtItI
steVtV
RCVV
)()(
0 when )(
and If0
to
t
s
R
to
RRR
eCV
dttptw
eR
VIVtp
22
0
22
12
)()(
)(
Since the voltages are equal and the currents have the opposite sign, the power that is dissipated by the resistor is the power that is being released by the capacitor.
RL Circuits
RVIdt
dILV
II
VV
RR
LL
RL
RL
0
stLL
L
L
L
L
LL
RL
ItL
RI
dtL
R
I
dI
L
R
dt
dI
I
L
RI
dt
dI
RIdt
dIL
0lnln
01
0
0
t
oLR
t
oL
stLo
eRItVtV
steItI
R
LII
)()(
0 when )(
and If0
to
t
s
R
t
oRRR
eLI
dttptw
eRIIVtp
22
0
22
12
)()(
)(
Since the voltages are equal and the currents have the opposite sign, the power that is dissipated by the resistor is the power that is being released by the capacitor.
Initial ConditionCan be obtained by inserting a d.c. source to the
circuit for a time much longer than (t at least t = -5 )t .Capacitor
Vo is the voltage calculated by replacing the capacitor with a resistor with infinite resistance (an open circuit) after the voltage across the capacitor has reached a constant value (steady state).
Inductor Io is the current flowing through the inductor calculated by
replacing the inductor with a resistor with zero resistance (a short circuit) after the current flowing through the inductor has reached a constant value (steady-state).
PSpice
You can set the initial condition on a capacitor or inductor by doubling clicking on the part symbol. Then, enter a value for IC in the pop-up window that opens.
Time constant, tThe time required for the voltage across the
capacitor or current in the inductor to decay by a factor of 1/e or 36.8% of its initial value.
V(t) = 6V [1 - u(t)]
Example #1IL(t)
Example #1 (con’t)IL(t)
Example #1 (con’t)t < 0s
VL = 0VVR = 6VIL = IR = 2mA
Therefore,
Io = 2mA
Find the initial condition.
Io
Example #1 (con’t)
IL(t)
Example #1 (con’t)t > 0st = L/R = 10mH/3k /W =
3.33msIL = IR =Ioe-t/t = 2mA
e-(t/3.33ms)
VR = 3kW IR = 6V e-(t/3.33ms)
VL = L dIL/dt = -6V e-(t/3.33ms)
Note VR + VL = 0 V
IL(t)
Example #2
+
VC
_
Example #2 (con’t)
+
VC
_
Example #2 (con’t)Calculate the initial condition - the voltage on
the capacitor. Replace the capacitor with an open circuit and find the voltage across the two terminals. Note that in this circuit, current will flow through R3 so
there will be a voltage across C, but it will not be equal to the magnitude of the voltage source in the circuit.
Example #2 (con’t)The voltage across the capacitor is equal to
the voltage across the 12kW resistor.
VC = Vo = [12kW /15kW] 5V = 4V
Example #2 (con’t)
+
VC
_
+
VC
_
Example #2 (con’t)Further simplification of the circuit
+
VC
_
kkkkReq 31231 where
Example #2 (con’t)
0)()(
33.1)()(
33.1)(
)6/4(2)(
)(
)()(
4)(
)(
2mswhen t
6)2(3
6/)2(
6/)2(
6/)2(
6/)2(
/)2(
2
tItI
emAtItI
emAtI
emsVFdt
tdVCtI
tVtV
VetV
etVV
msFkCR
CR
msmstCR
msmstC
msmstCC
RC
msmstC
mst
mstCC
eq
eq
eq
eq
IC(t)
IR(t)
SummaryThe initial condition for:
the capacitor voltage (Vo) is determined by replacing the capacitor with an open circuit and then calculating the voltage across the terminals.
The inductor current (Io) is determined by replacing the inductor with a short circuit and then calculating the current flowing through the short.
The time constant for:an RC circuit is = t RCan RL circuit is = t L/R
The general equations for the natural response of:the voltage across a capacitor is the current through an inductor is
/)(
/)(
)(
)(o
o
ttoL
ttoC
eItI
eVtV