On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
1
GEN E 123 (week 9)
Prof Catherine Gebotys, Department of Electrical and Computer Engineering, DC3514
h
I. 1ST ORDER CIRCUITS
i(t)
R
+
v(t)
-
C
o Assume at initial time (t=0) C is charged to Vo volts v(0)=Vo {done by
some external circuitry}
o We have Wc(0) = ½ CVo 2 , circuit response is only due to energy
stored initially in capacitor
o We have to determine v(t), t>0
Apply KCL at top node
Cdv(t)/dt + v(t)/R = 0
dv(t)/dt + (1/RC)v(t) = 0
A 1st order differential equation, whose solution is
v(t) = Vo e –t/RC
v(t)
Vo
time
Exponentially decays to 0, “Natural Response” of circuit.
-“Natural Response” is a response in absence of independent sources.
- at any time t, the energy stored in C is ½ C v2(t) = Wc(t)
(resistor stores NO energy. Power dissipated by resistor is Vr2/R )
Wc(0) - Wc(t) = Wr(t)
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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EXAMPLE: if C = 1 µF, R = 1kΩ, and v(0) = 5V, What is the
maximum power dissipated by the resistor?
v(t) = Vo e –t/RC = 5 e –100t
v(t) is also voltage across the resistor so
power dissipated by the resistor is
v2(t)/R = 0.025 e –200t Watts
so maximum power is dissipated at t=0
o Now consider a RL circuit
i(t)
R
L
Energy stored in inductor wL(0) = ½ L I02
Ldi(t)/dt + Ri = 0
di/dt + R/Li = 0
i(t) = I0 e –R/L t
EXAMPLE: R=2Ω, L = 1H, i(to) = 3A, to = 10s
i(t)
3
10
time
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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o TIME CONSTANTS : τ
v(t) = Vo e –t/τ
VO
RC CIRCUIT RESPONSE
Vo(1/e)
TIME
τ = RC
i(t) = Io e –t/τ
IO
IO(1/e )
RL CIRCUIT RESPONSE
τ = RC
EXAMPLE:
4 Ω
2Ω
8 Ω
3Ω
at t=0
1F
+
40V
switch closed, circuit was in dc steady state [where capacitor is
equivalent to open circuit]
Re = 10Ω
+
v(t)
Re
At time t=0, the switch opens, so t > 0 there is no source in the circuit.
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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Time constant is τ = RC = ReC = (10Ω)(1F) = 10 seconds
Therefore v(t) = 40 e –t/10 V
(every 10 sec the voltage reduces by a factor of 1/e. 1/e=0.368
so after 50 sec voltage is reduced to a value of 40e-5 = 0.27 V)
Rf
EXAMPLE:
C
Ra
vg(t)
+
+
v2 (t)
Node analysis (at input to opamp):
-vg(t)/Ra -v2 (t)/Rf = Cdv2 (t)/dt
with vg(t) = 0 for t > 0,
-v2 (t)/Rf = Cdv2 (t)/dt
dv2 (t)/dt + v2 (t)/(Rf C) = 0
where v2 (t) = v2 (0)e-1/(Rf C) t = v0 e-1/(Rf C) t
So far, although we use a source at t < 0 (dc steady state circuit t < 0) ,
at t = 0 we have switched the source “out” to observe the natural
response of the circuit (one which decays exponentially to zero).
Now we consider the circuit which contains a “constant” source even
after t > 0
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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EXAMPLE:
t=0
iR
Io
R
+
v(t)
C
-
V(0-) = Vo
T > 0 , switch is closed:
(1) Cdv/dt + v/R = Io
Same as before except right-hand-side has a constant or “Forcing
Term”
vf = forced solution
Derivative of vf plus vf is equal to a constant, therefore vf (t) = A, a
constant
Substitute: CdA/dt + A/R = Io
First term is 0
therefore A =IoR
vf = IoR
but we’re not finished, because v(0+) = Vo volts where Vo ≠ IoR
Vo is initial condition established before t = 0
Io is independent source function, therefore not related.
Therefore we solve “unforced” version of differential equation:
(2)
dvn (t)/dt + 1/RC vn (t) = 0
vn (t) = “natural response“ solution
[vn(t) = Ke +st , substitute it in to get s +1/RC = 0]
vn(t) = Ke -t/RC
So v(t) = vn(t) + vf(t)
So v(t) = IoR + Ke -t/RC
On-line lecture notes/slides 2002 by Prof. C. Gebotys, E&CE
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v(t) = IoR + Ke -t/RC
now v(0+) = Vo = IoR + K(1) , so K = Vo – IoR
Therefore v(t) = IoR + (Vo – IoR) e-t/RC
Steps :
1. use a trial “forced “ solution, to get unknown constant A, substitute
in differential equation to find “forced “ solution vf(t).
2. using trial “natural” solution of form , vn(t) = Ke st , substitute into
“unforced” differential equation (RHS=0) solving for s.
3. total solution is sum of natural plus forced solutions, v(t) = vn(t) +
vf(t), evaluate sum at t=0 and set equal to initial value to find K.
EXAMPLE:
2 Ω
+
12V
4Ω
t=0
1H
16V
+