First Order Circuit
14
First Order Circuit • Capacitors and inductors • RC and RL circuits
description
First Order Circuit. Capacitors and inductors. RC and RL circuits. Excitation from stored energy. ‘source-free’ circuits. DC source (voltage or current source). Natural response. Sources are modeled by step functions. Step response. Forced response. - PowerPoint PPT Presentation
Transcript of First Order Circuit
First Order Circuit
• Capacitors and inductors
• RC and RL circuits
RC and RL circuits (first order circuits)RC and RL circuits (first order circuits)
Circuits containing no independent sources
Circuits containing independent sources
Complete response = Natural response + forced response
• ‘source-free’ circuits
• Excitation from stored energy
• Natural response• DC source (voltage or current source)
• Sources are modeled by step functions
• Step response
• Forced response
RC circuit – natural responseRC circuit – natural response
t
o
)t(v
)0(v c
c
RC
dt
v
dvc
c
Assume that capacitor is initially charged at t = 0
vc(0) = Vo
RC
dt
v
dv
c
c
0R
v
dt
dvC cc
RC
v
dt
dv cc Taking KCL,
Objective of analysis: to find expression for vc(t) for t >0i.e. to get the voltage response of the circuit
tRC
1
)0(v
)t(vln
c
c
tRC
1
cc e)0(v)t(v
+
vc
ic iR
RC
tRC
1
oc eV)t(v
OR
RC circuit – natural responseRC circuit – natural response
• Can be written as /t
cc e)0(v)t(v , = RC time constant
• This response is known as the natural response
Voltage decays to zero exponentially
At t=, vc(t) decays to 37.68% of its initial value
The smaller the time constant the faster the decay
tRC
1
cc e)0(v)t(v
Vo
t= t
vC(t)
0.3768Vo
RC circuit – natural responseRC circuit – natural response
The capacitor current is given by: dt
dvCi c
C R
eVi
t
oC
And the current through the resistor is given by R
)t(vi CR
R
eVt
o
The power absorbed by the resistor can be calculated as:
R
eVivp
t2
2o
RRR
The energy loss (as heat) in the resistor from 0 to t:
t
0
t22
ot
o
t2
2o
t
0RR e
R
V
2dt
R
eVdtpE
t
22oR e1CV
2
1E
RC circuit – natural responseRC circuit – natural response
t
22oR e1CV
2
1E
As t , ER 2oCV
2
1
As t , energy initially stored in capacitor will be dissipated in the resistor in the form of heat
RC circuit – natural responsePSpice simulation
+
vc
ic iR
RC
0
1 RC circuitc1 1 0 1e-6 IC=100r1 1 0 1000.tran 7e-6 7e-3 0 7e-6 UIC.probe.end
Time
0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms 6.0ms 7.0msV(1)
0V
50V
100V
RC circuit – natural responsePSpice simulation
+
vc
ic iR
RC
0
1RC circuit.param c=1c1 1 0 {c} IC=100r1 1 0 1000.step param c list 0.5e-6 1e-6 3e-6.tran 7e-6 7e-3 0 7e-6.probe.end
Time
0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms 6.0ms 7.0msV(1)
0V
50V
100V
c1 = 3e-6 c1 = 1e-6
c1 = 0.5e-6
RL circuit – natural responseRL circuit – natural response
Assume initial magnetic energy stored in L at t = 0
iL(0) = Io
dtL
R
i
di
L
L
0Ridt
diL L
L L
Ri
dt
di LL Taking KVL,
Objective of analysis: to find expression for iL(t) for t >0i.e. to get the current response of the circuit
t
o
)t(i
)0(i L
L dtL
R
i
diL
L
tL
R
)0(i
)t(iln
L
L
iL
vL
+RL
+vR
tL
R
LL e)0(i)t(i
t
L
R
oL eI)t(i
OR
RL circuit – natural responseRL circuit – natural response
• Can be written as /t
LL e)0(i)t(i , = L/R time constant
• This response is known as the natural response
Current exponentially decays to zero
At t=, iL(t) decays to 37.68% of its initial value
The smaller the time constant the faster the decay
tL
R
LL e)0(i)t(i
Io
t= t
iL(t)
0.3768Io
RL circuit – natural responseRL circuit – natural response
The inductor voltage is given by: dt
diLv L
L
t
oL ReIv
And the voltage across the resistor is given by R)t(iv LR
t
o ReI
The power absorbed by the resistor can be calculated as:
t
22oRRR ReIivp
The energy loss (as heat) in the resistor from 0 to t:
t
0
t2
2o
t
o
t2
2o
t
0RR ReI
2dtReIdtpE
t
22oR e1LI
2
1E
RL circuit – natural responseRL circuit – natural response
t
22oR e1LI
2
1E
As t , ER 2oLI
2
1
As t , energy initially stored in inductor will be dissipated in the resistor in the form of heat
RL circuit – natural responsePSpice simulation
0
1 RL circuitL1 0 1 1 IC=10r1 1 0 1000.tran 7e-6 7e-3 0 7e-6 UIC.probe.end
vL
+RL
+vR
Time
0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms 6.0ms 7.0msI(L1)
0A
5A
10A
Time
0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms 6.0ms 7.0msI(L1)
0A
5A
10A
RL circuit – natural responsePSpice simulation
L1 = 3HL1 = 1H
L1 = 0.5H
RL circuit.param L=1HL1 0 1 {L} IC=10r1 1 0 1000.step param L list 0.3 1 3.tran 7e-6 7e-3 0 7e-6 UIC.probe.end0
1
vL
+RL
+vR
