Fundamentals of Electric Circuit - Chapter 7: First-order Circuits

Ex 7.14: Find i(t) for t > 0. Assume that the switch has been
closed for a long time.
i i i A(0) ( 0) ( 0) 5    
 For t < 0: i A
R
1
10
( 0) 5  
 For t > 0: i A
R R
1 2
10
( ) 2  

eq
L L
s
R R R
1 2
1
15
   

 The time constant:
R 1
1 0 V
1/3H
L
R 2
t = 0
i
3Ω2Ω
 Since the iL cannot change instantaneously:
 Thus:   t
t t
i t i i i e
i t e e A t
/
15 15
( ) ( ) (0) ( )
( ) 2 (5 2) 2 3 , 0

 
    
     
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1
4 0 V
5 H
L
R 2
t = 0
t = 0
1 0 V
R 3
32
Chapter 7: Capacitors and Inductors
IV. Step response of an RC/RL circuit
IV.2. R-L circuit
Ex 7.15: At t = 0, switch 1 is closed, and switch 2 is closed
4s later. Find i(t) for t > 0. Calculate i for t = 2s and t = 5s
 We need to consider the 3 time intervals:
 For t < 0: Two switches are open
i i i( 0) (0) ( 0) 0    
eq
eq
L
i A R R R s
R R R
1 3
1 3
40
( ) 4 , 10 , 0,5        

i
2Ω
4Ω
  t t t
t t
i t i i i e e e A
i t e e A t
/ 2 2
15 15
( ) ( ) (0) ( ) 4 (0 4) 4(1 )
( ) 2 (5 2) 2 3 , 0
  
 
         
     
6ΩS2
S1
 For 0 ≤ t ≤ 4: The switch S1 is closed, S2 is open.
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1
4 0 V
5 H
L
R 2
t = 0
t = 0
1 0 V
R 3
33
Chapter 7: Capacitors and Inductors
IV. Step response of an RC/RL circuit
IV.2. R-L circuit
Ex 7.15: At t = 0, switch 1 is closed, and switch 2 is closed
4s later. Find i(t) for t > 0. Calculate i for t = 2s and t = 5s
 For t ≥ 4: S2 is closed, but not affect iL because it
cannot change abruptly. Thus the initial current is
i i e A
8(4) ( 4) 4(1 ) 4   
v v v v
v V i A
R R R
1 2 3
40 10 180
( ) 2,727
11 6
 
       
i
2Ω
4Ω
eq
R R R R
1 2 3
22
( / / )
3
   
6ΩS2
S1
To find i(∞), using KCL:
eq
L
s
R
15
22
  
  t ti t i i i e e t( 4)/ 1,4667( 4)( ) ( ) (4) ( ) 2,727 1,273 , 4          
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1
4 0 V
5 H
L
R 2
t = 0
t = 0
1 0 V
R 3
34
Chapter 7: Capacitors and Inductors
IV. Step response of an RC/RL circuit
IV.2. R-L circuit
Ex 7.15: At t = 0, switch 1 is closed, and switch 2 is closed
4s later. Find i(t) for t > 0. Calculate i for t = 2s and t = 5s
 Putting all this together:
t
t
 t
i t e t
e t
2
1,4667( 4)
0, 0
( ) 4(1 ), 0 4
2,727 1,273 , 4

 


   
  
i
2Ω
4Ω
i e A
4(2) 4(1 ) 3,93  
6ΩS2
S1
 At t = 2s:
i e A
1,4667(5) 2,727 1,273 3,02   At t = 5s:
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
35
Chapter 7: Capacitors and Inductors
V. First-order Op Amp circuits
 An op amp circuit containing a storage element will exhibit first-order behaviors
 Methods to analyze op amp circuits:
 Differentiators and integrators treated in Chapter 5 are examples of first-order op
amp circuits.
 For practical reasons, inductors are hardly ever used in op amp circuits,
therefore, the op amp circuits we consider here are of the RC type.
 Using nodal analysis.
 Using the Thevenin equivalent circuit to simpler the op amp circuit
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
36
Chapter 7: Capacitors and Inductors
V. First-order Op Amp circuits
Ex 7.16: Find v0 for t > 0, give that v(0) = 3V
 Solution 1:
 Applying KCL gives:
 Node 3 and 4 must be at the same potential
4
3
7
R i
C
R f
-
+V
V0
+ -
80kΩ
20kΩ
5μF
V dV
C
R dt
1
1


v1
t
dV V
V V V V
dt CR
V t V e R C
1 1
1
/
0 1
0 0
( ) , 
      
  
t
V t e
10( ) 3  
 At t > 0, applying KCL at node 3 gives:
t t
f
f
VdV dV
C V R C e e V
dt R dt
3 6 10 100
0
0
80.10 .5.10 ( 30 ) 12  

       
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
37
Chapter 7: Capacitors and Inductors
V. First-order Op Amp circuits
Ex 7.16: Find v0 for t > 0, give that v(0) = 3V
 Solution 2: Apply the short-cut method
 Voltage across C:
 Apply KCL at node 3:
4
3
7
R i
C
R f
-
+V
V0
+ -
80kΩ
20kΩ
5μF
V V V( 0) ( 0) 3   
v1
I f
V
V V
R R
0
0
0 ( 0)3
0 ( 0) 12
 
    
 The circuit is source free v(∞) = 0V
 To find τ, need calculate Req across C
 Remove, replace C by a 1-A current source.
 Applying KVL to the input loop yields
4
3
7
R i
R f
-
+
V0
20kΩ
80kΩ
V+
-
1A
v v kV
320.10 .1 0 20   
eq
v
R k20
1
    eqR C s0,1  
  t tV t V V V e e V t/ 100 0 0 0( ) ( ) (0) ( ) 12 , 0
       
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R 1 R 2
3 V
4
3
7
t = 0
R 4
R 3
C
38
Chapter 7: Capacitors and Inductors
V. First-order Op Amp circuits
Ex 7.17: Determine v(t) and v0(t)
 V(t) is the step response
V
+ -
20kΩ
50kΩ
1μF
  tV t V V V e t/( ) ( ) (0) ( ) , 0     
v1
-
+
V0
RC s
3 650.10 .10 0,05    
R
v V
R R
2
1
1 2
3 2  

V V VR
V V V V V
R
1 04
0 1
3
( ) (1 ) 3,5.2 7 ( ) 2 7 5
           
10kΩ 20kΩ
 No current enters the op amp  the elements
on the feedback loop constitute an RC circuit
 For t 0:
 At steady state: C acts like an open circuit op amp is a non-inverting amplifier
 Thus:   t tV t e e V t20 20( ) 5 0 ( 5) 5( 1) , 0        
t
V t V t V t e V t
20
0 1
( ) ( ) ( ) 7 5 , 0
    
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
R i
R f
2u(t)
R 2
C
4
3
7
R 3
39
Chapter 7: Capacitors and Inductors
V. First-order Op Amp circuits
Ex 7.18: Find the step response v0(t) for t > 0.
 Remove C, and find the Thevenin
equivalent at its terminal.
V0
+
-
50kΩ
2μF
10kΩ
20kΩ
 Open voltage at the terminal:
 Gives the Thevenin equivalent circuit:
f f
ab I Th ab I
I I
R RR R
V V V V V u t
R R R R R R
3 3
2 3 2 3
2,5 ( )       
 
R
Th ideal op amp
R R
R R R R k
R R
0 0 2 3
0 2 3
2 3
( ) / / 5    

10kΩVab
+
-
 Thevenin resistance:
Req
-2,5u(t)
C
2μF
t
Th
V t e u t R C s
/ 3 6
0
( ) 2,5(1 ) ( ), . 5.10 .2.10 0,01
       
 Thus, the step response for t > 0 is:
t
V t e u t V
100
0
( ) 2,5( 1) ( )
 
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
40
Chapter 7: Capacitors and Inductors
VI. Applications
VI.1. Delay circuits
Ex7.19: An RC circuit with capacitor
connected in parallel with a neon lamp.
The voltage source can provide enough voltage to fire the lamp.
 Switch closes: VC increases to 110V with the time constant (R1 + R2)C.
 The lamp will act as an open-circuit and not emit light until the voltage across it
exceeds a particular level (70V).
 When VC reaches, the lamp fires and the capacitor discharges through it  VC
drops and the lamp turn off.
 The lamp acts again as an open-circuit and C recharges: Adjusting R2, we can
introduce either short or long time delays.
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
41
Chapter 7: Capacitors and Inductors
VI. Applications
VI.2. Photoflash unit
 This application exploits the ability of the capacitor
to oppose any abrupt change in voltage.
 Switch is in 1: C charges slowly due to the large time constant
 VC rises gradually from zero to VS
 IC decreases from I1 to 0
R 1
R 2
V s
C
V
+
-
2
1
High 
voltage 
DC 
supply i
  R C
1
 Switch is in 2: C discharges
 Low resistance R2 of the photo-lamp permits a high
discharge current with peak I2 in a short duration.

disch e
t R C
arg 2
5
 The simple RC circuit provides a short-duration, high
current pulse.
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
42
Chapter 7: Capacitors and Inductors
VI. Applications
VI.2. Photoflash unit
Ex 7.20: An electronic flashgun has a current limiting
R1 = 6kΩ, and C = 2000μF charged to 240V. If the
R 1
R 2
V s
C
V
+
-
2
1
High 
voltage 
DC 
supply i
lamp resistance R2 is 12Ω. Find:
ch e
t R C s
3 6
arg 1
5 5.6.10 .2000.10 60  
a. Peak charging current: S
V
I mA
R
1 3
1
240
40
6.10
  
b. Time required for the C to fully charge:
c. Peak discharging current: S
V
I A
R
1
2
240
20
12
  
d. Energy stored:
S
W CV J
2 6 21 1 .2000.10 .240 57,6
2 2
  
e. Energy stored in C is dissipated across the lamp during the discharging period:
disch e
t R C s
6
arg 2
5 5.12.2000.10 0,12  
disch e
W
p W
t
arg
57,6
480
0,12
   
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
43
Chapter 7: Capacitors and Inductors
VI. Applications
VI.3. Relay circuits
 A relay is essentially an electromagnetic device used to
open or close a switch that controls another circuit.
R
S 1
V s
L
 The coil circuit is an RL circuit:
 When S1 is closed iL increases, produces a magnetic field
 The magnetic field is sufficiently strong to pull the movable
contact in the other circuit and close switch S2.
 Time interval td between the closure of switches S1 and S2 is
called the relay delay time.
 Relays were used in the earliest digital circuits and are still used for switching
high power circuits.
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011
44
Chapter 7: Capacitors and Inductors
VI. Applications
VI.3. Relay circuits
Ex 7.21: The coil of a certain relay is operated by a 12V battery. If the coil has a
resistance of 150Ω and an inductance of 30mH and the current needed to pull in is
50mA. Calculate the relay delay time.
where:
 The current through the coil is given by:   ti t i i i e /( ) ( ) (0) ( )     
S
L
V L
i i mA ms
R R
312 30.10
(0) 0, ( ) 80 , 0,2
150 150


       
 Thus: ti t e mA
/( ) 80 1    
 If i(td) = 50mA, then: d
tt
e e
// 850 80 1
3
     
d
t ms
8 8
ln 0,2. ln 0,1962
3 3
   