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Unformatted text preview: CHAPTER 7  FIRSTORDER CIRCUITS List of topics for this chapter :
SourceFree RC Circuit
Source—Free RL Circuit
Singularity Functions
Step Response of an RC Circuit
Step Response of an RL Circuit
FirstOrder Op Amp Circuits
Transient Analysis with PSpice
Applications SOURCEFREE RC CIRCUIT M Problem 7.1 For the circuit in Figure 7 .1, ﬁnd VC (t) and iC (t) given VC (0) = 10 V .
10 Q
. ic(t) +
, vat) “5 F
Figure 7.1 > Carefully DEFINE the problem.
Each component is labeled completely. The problem is clear. > PRESENT everything you know about the problem.
This is a sourcefree RC circuit. The natural response of this sourcefree RC circuit is vC (t) = VO 6"“ , where V0 = VC (0) and t = RC We know the initial voltage across the capacitor. To ﬁnd the capacitor voltage for any time
greater than zero, we need to calculate the time constant of the circuit. > Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success.
The three solution techniques that can be used are nodal analysis, mesh analysis, and basic circuit analysis. Basic circuit analysis can be used to solve this problem. 109 > ATTEMPT a problem solution. V0 =vc(0)=10V and r =RC=(10)(1/5)=23
VC(t)=106‘t/2 V .  V (t) . dv t
1C0) = I: or 1C(t) = C (is )
In either case, iC (t) = e't/2 A > EVALUATE the solution and check for accuracy.
Using KVL, 10ic(t) +\vC(t) = (10)(_et/2) + 1064/2 ___ 0
Our check for accuracy was successful. > Has the problem been solved SATISFACTORILY? If so, present the solution; if not,
then return to “ALTERNATIVE solutions” and continue through the process again.
This problem has been solved satisfactorily. i C(t) = —e“'/2 amps for all t >0 . Problem 7.2_ [7.23] Express the signals in Figure 7.2 in terms of singmlarity
functions. VICE) V2(t) Figure 7.2 110 (a) v1 (t) = u(t +1) — u(t) + [u(t — 1) — u(t)]
V1 (t) = u(t + 1) — 2 u(t) + u(t —— l) (b) v20) = (4 —t>[u<t —2) —u(t — 4)]
v2(t) = (t—4)u(t—2)+(t—4)u(t—4)
V2(t) = 2u(t—2)—r(t—2)+r(t—4) (c) V3(t)=2[u(t—2)u(t4)]+4[u(t—4)—u(t—6)]
v3(t) = 2u(t—2)+2u(t—4)—4u(t—6) (d) v4(t)=t[u(t—1)—u(t—2)]=tu(t—1)+tu(t—2) V4(t)=(t+l—l)u(t—1)+(t—2+2)u(t—2)
v4(t)=r(t—l)—u(t—l)+r(t—2)+2u(t—2) Problem 7.3 Given i(t) = 3 6'“2 A , ﬁnd VC (t) for the circuit shown. in Figure 7 .3.
10 Q 1/5 F Figure 7.3 vC (t) = 30et/2 V Problem 7.4 Given VC (1) = 10 V , ﬁnd VC (t) for all t > 0 in Figure 7.4. 109. ' +
veg) 1/20 F Figure 7.4 10 at
Vc(t)=[ej)e V 111 OW SOURCE—FREE RL CIRCUIT Problem 7.5 For the circuit in Figure 7.5, ﬁnd i(t) and vL (t) given i(0) = 4 A. 10 Q i“) +
vL(t) E 5 H Figure 7.5 > Carefully DEFINE the problem.
Each component is labeled completely. The problem is clear. > PRESENT everything you know about the problem.
This is a sourcefree RL circuit. The natural response of this source—free RL circuit is iL(t) = 10 6"“, where 10 = iL(0) and z = L/R
We know the initial current through the inductor. To ﬁnd the current through the inductor for any value of time greater than zero, we need to calculate the time constant of the circuit. > Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success.
The three solution techniques that can be used are nodal analysis, mesh analysis, and basic circuit analysis. Basic circuit analysis can be used to solve this problem. ' > ATTEMPT a problem solution. IO=iL(0)=4A and t=L/R=5/10=O.5s
i(t)= 4e'2‘ A d.
VL(t) = ~10i(t) or VL(t) = L? In either case, vLa) = 40e'2‘ v 112 > EVALUATE the solution and check for. accuracy. . Using KVL,
10i(t) + vL (t) : (1o)(4e2t) ~ 406—2: = 0 Our check for accuracy was successful. > Has the problem been solved SATISFACTORILY? If so, present the solution; if not,
then return to “ALTERNATIVE solutions” and continue through the process again. This problem has been solved satisfactorily. i(t) = 4e'2‘u(t) A and VL (t) = 40 e'“u(t) V Problem 7.6 For the circuit in Figure 7.6, ﬁnd i(t) given VL (t) = 20 e'2t V .
20 Q i“)
+ vL(t) E 10H Figure 7.6 L/R=10/20= 0.5 5 KO = % Ivar) dc but it is also  v t
i(t) = L( ) = e‘2'u(t) A
R __.____
Problem 7.7 Given i(0) = 2 A , ﬁnd i(t) , p109 (power absorbed by the 10 ohm resistor),
and w109 (total energy dissipated by the 10 ohm resistor) for the circuit in Figure 7.7.
10 Q i“)
5 H
Figure 7.7 113 i(t) = 2e'2t A pIOQ = 40c”4t W wIOQ : 10 J
SINGULARITY FUNCTIONS
Problem 7.8 Solve for
du(t) dr(t)
(3) dt (b) dt 0 t<0
(a) HG):1 t>0 d j 0 t < 0 d
Eua) = undeﬁned t = 0 ———> an“) 2 5(t)
0 t > 0
{0 t s 0 (b) r“) — t t > 0
d t {0 t s 0 d
dtrU— 1 t>0 dtrm—ﬂ Problem 7.9 Given VC (t) = [5 u(t) + 6r(t)] V , ﬁnd ic(t) for the circuit in Figure 7.8. + Tic“) Figure 7.8 d
icm = (:5;va 114 . ‘ 1 d l d
. ic(t)=1—6'3;[5u(t)+6r(t)]=16.[52i’;u(t)+62§1;r(t):lI Using Problem 7.8, it is clear that 1
ice): 15[56(t)+ 6am] A Problem 7.10 Solve for
(a) I8(t)dt (b) {new <20 359. 0» ﬂ «2W STEP RESPONSE OF AN RC CIRCUIT Problem 7.11 Given V(t) = 20 u(t) V , ﬁnd VC (t) and ic(t) in Figure 7.9. 100 1/20 F Figure 7.9 'c =RC=(10)(§16)=';S vc(0)=ov vc(oo)=20V vC (t) = (20)(1 — e2t )u(t) V d t 1
10a) = (3 V5: ) = (56)(20)(—2e'2‘) = 2e'2‘ u(t) A 115 Problem 7.12 [7.37] Find the step responses V(t) and i(t) to vS = 5u(t) V in the circuit of Figure 7 . 10 129 79 Figure 7.10
Fort<0, Vs=5u(t)=0 —> V(O)=0V
0 5 V ( ) (5) 5 V
= 00 : : 
For t> , vS , v 4+12 4
Req=7+4l12=10£2, T=ReqC=(10)(1/2)=SS
«0 = V(oo) + [ v(0> — V(oo)] eW
V(t) = 1.25(1 — e"/5) V
' _ 9X _ _1~ 4/5
1(0‘ dt {2] 4 5 e
i(t) = 0.125 e"/5 A
Problem 7.13 Given V(t) = 10[u(t) —— u(t — 2)] V , ﬁnd VC (t) in Figure 7.11. 109 1/10 F Figure 7.11 116 Find the Thevenin equivalent of the circuit at the terminals of the capacitor. This will simplify
. the circuit, forming an RC circuit with a voltage source. Use the following circuit to ﬁnd the open circuit voltage.
10 Q V0c must be equal to V(t) , since i(t) + 2i(t) = 0 ——> i(t) = 0 A‘ To ﬁnd the short circuit current, 109 we a 0 15¢ . 1(1). 150 = i(t) + 21(1) = 31(1) where i(t = 1 0
I 3 (t)
= — v
SC 1 0 Thus, R  Voc — 19 Q “1 “ 156 ‘ 3
which leads to the following Thevenin equivalent circuit.
10/3 Q
V(t) 1/ 10 F Using the Thevenin equivalent circuit with the capacitor as the load, we can see that 101 1
. TZR‘hC‘isi 10)”? 117 Fort<0, V(t)=0V, vC(0)=0V, Vc(oo)=0V
vc(t)=OV For0<t<2, V(t)=10V, VC(O)=OV, V(oo)=10V
Vc(t)=(10)(1—e‘3‘)V For2<t, v(t)=0V, vC(2)=(10)(1—e~6)v, vc(oo)=OV
vC (t) = (10)(1— e'6)e'3("2) V Combining these cases, vC (t) = { [(10)(1 — e‘3' )[u(t) — u(t — 2)]]+ [9.975 e“W u(t — 2) ]} v STEP RESPONSE OF AN RL CIRCUIT Problem 7.14 Given V(t) = 40 u(t) V , ﬁnd iL (t) and VL (t) in Figure 7.12.
10 Q V(t) 0 Figure 7.12 5H NH iL(0)=OA iL (t) = (4)(1 — e'“) “(t) A 'diL(t)
dt vL (t) = L = (5)(4)(2)e'2‘ u(t) = 40 e2t u(t) v 118 . Problem 7.15 [7. 55] Find V0 (t) for t > O in the circuit ofFigure 7.13.
6 Q 4H NH Figure 7.13 Let i be the inductor current. Fort < 0, the inductor acts like a short circuit and the 2 Q resistor is
shortcircuited so that the equivalent circuit is shown in Fig. (a). 69 _ i 69 i0 v i
. 109 10o
(a) > (b)
i=i(0)=1—69=1.6667A
L 4
Fort>0, Rm=2+3H6=4Q, T=EE=Z=1S To ﬁnd i(oo) , consider the circuit in Fig. (b). 10—V v X v_i_0_V
2 a “6 i0) = i(oo) + [ i(m — i(oo)] et/T . 5 10 5 _t 5 4
t=—+ ————e =—1+ A
10 6 [6 6) 6( e) 119 V0 is the voltage across the 4 H inductor and the 2 Q resistor di 10 10 _t (5) 4 10 10 _t
Ldt—6+6e +(4)6(1)e —6—6e
vo(t)=1.6667(1—e")V vo(t):2i+ Problem 7.16 Find iL (t) and VL (t) in Figure 7.14 for V(t) = [20 u(t) — 40 u(t ~ 1)] V . 10S) Figure 7.14 iL (t) = (2)(1 — e“)[u(t) — u(t — 1)] + [2 + (4 — 2e'l)e'““1)]u(t— 1) A vL (t) = 20c" 11(t)—[20e't + (20)(2 ~ e'1)e'(‘"”]u(t — 1) V _ . FIRSTORDER OP AMP CIRCUITS Problem 7.17 Given V(t) = 10u(t) V , ﬁnd i0 (t) for the circuit in Figure 7.15.
1 kg 1 mF
10 Q
Figure 7.15
va — v(t) d +1ma‘;[va — Vo (t)] = 0 , 1k where Va=Vb=0V. 120 _vo(t) = _[v(t) dt = [10u(t)dt = 10t V io(t) = Viét) = —tu(t) A Problem 7.18 [7.59] Obtain V0 for t > 0 in the circuit of Figure 7.16.
t = 0 Figure 7.16 This is a very interesting problem and has both an important ideal solution as well as an important
practical solution. Let us look at the ideal solution ﬁrst. Just before the switch closes, the value
of the voltage across the capacitor is zero, which means that the voltage at both input terminals of
the op amp are zero. As soon as the switch closes, the output tries to go to a voltage such that the
inputs to the op amp both go to 4 volts. The ideal op amp puts out whatever current is necessary
to reach this condition. An inﬁnite (impulse) current is necessary if the voltage across the
capacitor is to go to 8 volts in zero time (8 volts across the capacitor will result in 4 volts appearing at the negative terminal of the op amp). So V() will be equal to 8 volts for all t > 0. What happens in a real circuit? Essentially, the output of the ampliﬁer portion of the op amp
goes to whatever its maximum value can be. Then, this maximum voltage appears across the
output resistance of the op amp and the capacitor that is in series with it. This then results in an
exponential rise in the capacitor voltage to the steadystate value of 8 volts. I For all values of vC (t) less than 8 V,
vC (t) = V opampmax (1 _. e't/(Routc)) V where Vop_m)_max is the maximum value of the op amp and Rout is the real output resistance of the practical op amp. When t is large enough so that the 8 V is reached,
VC (t) = 8 V 121 TRANSIENT ANALYSIS WITH PSPICE Problem 7.19 The switch in Figure 7.17 moves from position a to b at t = 0. may
Use PSpice to ﬁnd i(t) for t > 0.
a 6 Q
4 Q
i“)
108 V O E 2 H
Figure 7.17 (a) (b) When the switch is in position a, the schematic is shown below. We insert IPROBE to
display i. After simulation, we obtain, 1(0) = 7.714 A from the display of IPROBE. When the switch is in position b, the schematic is as shown below. For inductor L1, we let IC = 7.714 A. By clicking Analysis/Setup/Transient, we let Print Step = 25 ms and
Final Step = 2 5. After Simulation, we click Trace/Add in the probe menu and display I(L1) as shown below. Note that i(oo) = 12 A , which is correct. 122 129T ————————————————————————————————————————————————————————— — d
a
:: J__________L________ Time We now know the initial and ﬁnal values of the current through the inductor.
i(O) = 7.714 A i(oo) =12 A To ﬁnd the current through the inductor for any value of time, we need to know the time constant
of the circuit. Using the circuit from part (b), Req=3l6+4=2+4=6§2
t=L/Req =2/6=l/3s Therefore, i(t) = i(oo) + [ 1(0) — i(oo)] e"/‘
i(t) = 12 + [7.714 — 12] e31 = 12 — 4.286e'3‘ A 123 W0 APPLICATIONS Problem 7.20 [7.73] Figure 7.18 shows a circuit for setting the length of time voltage
is applied to the electrodes of a welding machine. The time is taken as how long it takes the
capacitor to charge from 0 to 8 V. What is the time range covered by the variable resistor? 100 kg to 1 M0 12V Figure 7.18 > Carefully DEFINE the problem.
Each component is labeled completely. The problem is clear. > PRESENT everything you know about the problem.
This is an RC circuit with a dc voltage source. When the welding machine is activated, the dc source supplies power to the RC circuit. To ﬁnd the time in which it takes the capacitor to
charge from 0 to 8 V, we need to ﬁnd the response of the RC circuit, written as v0) = W») + [ v(0) — v<oo>l er where v(0) is the initial voltage across the capacitor, v(oo) is the steady—state value of the
voltage across the capacitor, and T is the time constant of the RC circuit. > Establish a set of ALTERNATIVE solutions and determine the one that promises the greatest likelihood of success.
The three solution techniques that can be used are nodal analysis, mesh analysis, and basic
circuit analysis. Basic circuit analysis can be used to solve this problem. > ATTEMPT a problem solution.
v0) = v0») + [ v<0) — v(oo)] er V(0) = 0 V and V(oo) = 12 V W) = (12)(1€"/’) Let the voltage at an unknown time, t0 , be equal to 8V.
V(t0) = 8 = (12)(1 —e"°/‘) 124 _8_ “to/'r “to/1 1
. 12_1~e ———> e —3 t0 = T ln (3) For R=100 kQ,
1 = RC = (100x103)(2><10‘6)=0.2 s
o =0.21n(3)=0.2197s For R=1MQ, ,
1;:RC=(1X106)(2X10'6)=25
t0 =2ln(3)=2.197s Thus,
0.2197 s < to < 2.197 s > EVALUATE the solution and check for accuracy. For R = 100 kg),
1 = RC = (100x103)(2x10'6) = 0.2 s v(0.2197) = (12)(1— W) = (12)(1— e02197/°~2) = 8 V . ForR=1MQ,
I =Rc= (1x106)(2x10'6)=2s v(2.197) = (12)(1—6““) = (12)(1—6'2‘197/2) = s V
Our check for accuracy was successful. > Has the problem been solved SATISFACTORILY? If so, present the solution; if not,
then return to “ALTERNATIVE solutions” and continue through the process again. This problem has been solved satisfactorily. 0.2197 s < t0 < 2.197 s 125 ...
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This note was uploaded on 09/06/2010 for the course EE 50145 taught by Professor Chi during the Spring '10 term at UCSF.
 Spring '10
 Chi

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