Solucionário Sadiku - Problemas Práticos Cap 5

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Sunday, June 26, 2011 
CHAPTER 5 
 
 
P.P.5.1 The equivalent circuit is shown below: 
 
 
 
 
 
 
 
 
 
 
 
At node 1, 3
01
3
1
6
1s
10x40
vv
10x5
v
10x2
vv  v1 = 
451
v50v 0S  (1) 
 
At node 2, 3
0
3
010d
13x20
v
10x40
vv
50
vAv  
 
But vd = v1 - vS. 
 
 [2 x 105 (v1 - vS) - v0] 4000/(5) + v1 - v0 = 2v0 
 
 1600 x 105 (vS - v1) + 803v0  0 (2) 
 
Substituting v1 in (1) into (2) gives 
 
 1.5914523 x 108 vS - 17737556v0 = 0 
 
 
17737556
10x5964523.1
v
v 8
S
0 9.00041 
 
If vS = 1 V, v0 = 9.00041 V, v1 = 1.0000455 
 
 vd = vS - v1 = - 4.545 x 10-5 
 Avd = - 9.0909, i0 = 
50
vAv 0d 657 A 
 
- 
vd 
+ 
+ 
v0 
-
+ 
v1 
-
-
+vs 
+ 
Avd 
-
2 M 
40 k 
20 k 5 k 
50  i0 2 1 
P.P.5.2 
 
 
 
 
 
 
 
 
 
 
 
 
At node 1, At node 1, 
20
vv
10
vv 011S  
 
But v1 = v2 = 0, 
 
 
20
v
10
v 0S  
S
0
v
v
 –2 
 
 i0 = 3
0
3
0
10x20
v
10x20
v0  
 
When vs = 2V, v0 = -4, i0 = 
20
10x4 3 = 200 A 
 
P.P.5.3 v0 =   mV454
280v
R
R
i
1
2 
 
 –3.15 V 
 
 i = 
k120
v0 0 26.25 A 
 
 
P.P.5.4 (a) iS = 
R
v0 0 R
i
v
S
0 
 
-
+
-
+VS 
i
V1 
V2 
+ 
V0 
-
20 k 
10 k 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
(b) At node 2, iS = 
1
1
R
v0 
 v1 = -iSR1 (1) 
 
 At node 1, 
3
01
2
1
1
1
R
vv
R
0v
R
v0  
 
 -v1 
3
0
321 R
v
R
1
R
1
R
1 


  
 
 v0 = –iSR1R3 


 
321 R
1
R
1
R
1 
 
 


 
2
3
1
3
1
0 1
R
R
R
R
R
i
v
S
 
 
 
P.P.5.5 By voltage division 
 
 V2)3(
84
8v1  
 
where v1 is the voltage at the top end of the 8k resistor. Using the formula for 
noninverting amplifier, 
 
 v0 = 

  )2(
2
51 7 V 
 
 
-
+
iS V1 
V2 
+ 
V0 
-
R1 
20 k 
R3 
R2 
0 V 
 
P.P.5.6 This is a summer. 
 
 

  )2.1(
6
8)2(
10
8)5.1(
20
8v0 = –3.8 V 
 
 
4
8.3
8
8.3
4
v
8
v
i 000 –1.425 mA 
 
P.P.5.7 If the gain is 7.5, then 
 
 5.7
R
R
1
2  R2 = 7.5R1 
 
But 
3
4
1
2
R
R
R
R  R4 = 7.5R3 
 
If we select R1 = R3 = 20k, then R2 = R4 = 150 k. 
 
P.P.5.8 v0 = 


 
4
3
1
2
R
R2
1
R
R (v2 – v1) 
 
R3 = 0, R4 = , R2 = 40k, R1 = 20k 
 
 v0 = .04.0)798.6(20
40 volts 
 i0 =  50
04.0
50
v0 –800 A. 
 
P.P.5.9 Due to the voltage follower 
 
 va = 1.2 V 
 
For the noninverting amplifier, 
 
 vo = av50
2001 

 
 
vo = (1 + 4) (1.2) = 6 V. 
 
 i0 = mA50
vb 
 
 
 
 
 
 
 -
 
 
 
 
 
 
 
 
But vb = va = 4 
i0 = 
 
310x50
2.1 24 µA. 
 
.P.5.10 As a voltage follower, 
va = v1 = 7 V 
here va is the voltage at the left end of the 20 k resistor. 
b = 
P
 
 
 
w
 
As an inverter, v V5.15v
10
50
2  
 
Where vb is the voltage at the right end of the 50 k resistor. As a summer 
0 = 
 
 v 

  ba v30
60v
20
60 
 
 = [–21 + 31] = 10 V. 
.P.5.11 The schematic is shown below. When it is saved and run, the results are 
v0 = 9.0027V and i0 = 650.2 µA 
 
 
 
P
displayed on 1PROBE and VIEWPOINT as shown. By making vs = 1V, we obtain 
 
 
 
+ +
-
-
+vS 
+ 
200 k
v0 
– 50 k 
a 
b 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
6.502E-04
9.0027 
 
P.P.5.12 -V0 = 3
3
f
2
2
f
1
1
f V
R
R
V
R
R
V
R
R  
 
or 3210 V25.0V5.0VV  
 
(a) If [V1V2V3] = [010], 0V = 0.5V 
(b) If [V1V2V3] = [110], 0V = 1 + 0.5 = 1.5V 
(c) If 0V = 1.25, then V1 = 1, V2 = 0, V3 = 1, i.e. 
 [V1V2V3] = [101] 
(d) 0V = 1.75, then V1 = 1, V2 = 1, V3 = 1, i.e. 
 [V1V2V3] = [111] 
 
 
P.P.5.13 Av = 1 + 
GR
R2 RG = 
1A
R2
v 
 
 
RG = 1142
10x25x2 3 354.6  
 
	P.P.5.2