Tài liệu Bài giải phần giải mạch P5 pptx
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Chapter 5, Solution 1.
(a) R
in
= 1.5 MΩ
(b) R
out
= 60 Ω
(c) A = 8x10
4
Therefore A
dB
= 20 log 8x10
4
= 98.0 dB
Chapter 5, Solution 2.
v
0
= Av
d
= A(v
2
- v
1
)
= 10
5
(20-10) x 10
-6
= 0.1V
Chapter 5, Solution 3.
v
0
= Av
d
= A(v
2
- v
1
)
= 2 x 10
5
(30 + 20) x 10
-6
= 10V
Chapter 5, Solution 4.
v
0
= Av
d
= A(v
2
- v
1
)
v
2
- v
1
= V20
10x2
4
A
v
5
0
µ−=
−
=
If v
1
and v
2
are in mV, then
v
2
- v
1
= -20 mV = 0.02
1 - v
1
= -0.02
v
1
= 1.02 mV
Chapter 5, Solution 5.
+
v
0
-
-
v
d
+
R
0
R
in
I
v
i
-
+
Av
d
+
-
-v
i
+ Av
d
+ (R
i
- R
0
) I = 0 (1)
But v
d
= R
i
I,
-v
i
+ (R
i
+ R
0
+ R
i
A) I = 0
v
d
=
i0
ii
R)A1(R
Rv
++
(2)
-Av
d
- R
0
I + v
0
= 0
v
0
= Av
d
+ R
0
I = (R
0
+ R
i
A)I =
i0
ii0
R)A1(R
v)ARR(
++
+
4
5
54
i0
i0
i
0
10
)101(100
10x10100
R)A1(R
ARR
v
v
⋅
++
+
=
++
+
=
≅
()
=⋅
+
4
5
9
10
101
10
=
001,100
000,100
0.9999990
Chapter 5, Solution 6.
-
v
d
+
+
v
o
-
R
0
R
in
I
v
i
+
-
Av
d
+
-
(R
0
+ R
i
)R + v
i
+ Av
d
= 0
But v
d
= R
i
I,
v
i
+ (R
0
+ R
i
+ R
i
A)I = 0
I =
i0
i
R)A1(R
v
++
−
(1)
-Av
d
- R
0
I + v
o
= 0
v
o
= Av
d
+ R
0
I = (R
0
+ R
i
A)I
Substituting for I in (1),
v
0
=
++
+
i0
i0
R)A1(R
ARR
−
v
i
=
()
()
65
356
10x2x10x2150
1010x2x10x250
++
⋅+
−
−
≅
mV
10x2x001,200
10x2x000,200
6
6
−
v
0
= -0.999995 mV
Chapter 5, Solution 7.
100 k
Ω
1
2
10 kΩ
-
+
+
V
d
-
+
V
out
-
R
out
= 100 Ω
R
in
AV
d
+
-
V
S
At node 1, (V
S
– V
1
)/10 k = [V
1
/100 k] + [(V
1
– V
0
)/100 k]
10
V
S
– 10 V
1
= V
1
+ V
1
– V
0
which leads to V
1
= (10V
S
+ V
0
)/12
At node 2, (V
1
– V
0
)/100 k = (V
0
– AV
d
)/100
But V
d
= V
1
and A = 100,000,
V
1
– V
0
= 1000 (V
0
– 100,000V
1
)
0= 1001V
0
– 100,000,001[(10V
S
+ V
0
)/12]
0 = -83,333,334.17 V
S
- 8,332,333.42 V
0
which gives us (V
0
/ V
S
) = -10 (for all practical purposes)
If V
S
= 1 mV, then V
0
= -10 mV
Since V
0
= A V
d
= 100,000 V
d
, then V
d
= (V
0
/10
5
) V = -100 nV
Chapter 5, Solution 8.
(a) If v
a
and v
b
are the voltages at the inverting and noninverting terminals of the op
amp.
v
a
= v
b
= 0
1mA =
k2
v0
0
−
v
0
= -2V
(b)
10 kΩ
2V
+ -
+
v
a
-
10 kΩ
i
a
+
v
o
-
+
v
o
-
v
a
v
b
i
a
2 k
Ω
2V
-
+
1V
-
+
-
+
(b)
(a)
Since v
a
= v
b
= 1V and i
a
= 0, no current flows through the 10 kΩ resistor. From Fig. (b),
-v
a
+ 2 + v
0
= 0 v
a
= v
a
- 2 = 1 - 2 = -1V
Chapter 5, Solution 9.
(a) Let v
a
and v
b
be respectively the voltages at the inverting and noninverting
terminals of the op amp
v
a
= v
b
= 4V
At the inverting terminal,
1mA =
k2
0
v4
−
v
0
= 2V
Since v
a
= v
b
= 3V,
-v
b
+ 1 + v
o
= 0 v
o
= v
b
- 1 = 2V
+
v
b
-
+
v
o
-
+ -
(b)
1V
Chapter 5, Solution 10.
Since no current enters the op amp, the voltage at the input of the op amp is v
s
.
Hence
v
s
= v
o
2
v
1010
10
o
=
+
s
o
v
v
= 2
Chapter 5, Solution 11.
8
k
Ω
v
b
= V2)3(
510
10
=
+
i
o
b
a
+
−
5
k
Ω
2
k
Ω
4
k
Ω
+
v
o
−
10
k
Ω
−
+
3 V
At node a,
8
vv
2
v3
oaa
−
=
−
12 = 5v
a
– v
o
But v
a
= v
b
= 2V,
12 = 10 – v
o
v
o
= -2V
–i
o
=
mA1
4
2
8
22
4
v0
8
vv
ooa
=+
+
=
−
+
−
i
o
= -1mA
Chapter 5, Solution 12.
4
k
Ω
b
a
+
−
2
k
Ω
1
k
Ω
+
v
o
−
4
k
Ω
−
+
1.2V
At node b, v
b
=
ooo
v
3
2
v
3
2
v
24
4
==
+
At node a,
4
vv
1
v2.1
oaa
−
=
−
, but v
a
= v
b
=
o
v
3
2
4.8 - 4 x
ooo
vv
3
2
v
3
2
−= v
o
= V0570.2
7
8.4x3
=
v
a
= v
b
=
7
6.9
v
3
2
o
=
i
s
=
7
2.1
1
v2.
a
−
=
1
−
p = v
s
i
s
= 1.2 =
−
7
2.1
-205.7 mW
Chapter 5, Solution 13.
By voltage division,
i
1
i
2
90
k
Ω
10
k
Ω
b
a
+
−
100 k
Ω
4
k
Ω
50
k
Ω
+
−
i
o
+
v
o
−
1 V
v
a
=
V9.0)1(
100
=
90
v
b
=
3
v
v
150
o
o
=
50
But v
a
= v
b
9.0
3
v
0
= v
o
= 2.7V
i
o
= i
1
+ i
2
= =+
k150
v
k10
v
oo
0.27mA + 0.018mA = 288 µA
Chapter 5, Solution 14.
Transform the current source as shown below. At node 1,
10
vv
20
vv
5
v10
o1
211
−
+
−
=
−
But v
2
= 0. Hence 40 - 4v
1
= v
1
+ 2v
1
- 2v
o
40 = 7v
1
- 2v
o
(1)
20
k
Ω
v
o
10
k
Ω
+
−
v
1
−
+
v
2
5
k
Ω
10
k
Ω
+
v
o
−
10V
At node 2,
0v,
10
vv
20
vv
2
o2
21
=
−
=
−
or v
1
= -2v
o
(2)
From (1) and (2), 40 = -14v
o
- 2v
o
v
o
= -2.5V
Chapter 5, Solution 15
(a) Let v
1
be the voltage at the node where the three resistors meet. Applying
KCL at this node gives
332
1
3
1
2
1
11
R
v
RR
v
R
vv
R
v
i
oo
s
−
+=
−
+=
(1)
At the inverting terminal,
11
1
1
0
Riv
R
v
i
ss
−=→
−
=
(2)
Combining (1) and (2) leads to
++−=→−=
++
2
31
31
33
1
2
1
1
R
RR
RR
i
v
R
v
R
R
R
R
i
s
oo
s
(b)
For this case,
Ω=Ω
++−= k 92- k
25
4020
4020
x
i
v
s
o
Chapter 5, Solution 16
10k
Ω
i
x
5k
Ω
v
a
i
y
-
v
b
+ v
o
+ 2k
Ω
0.5V
- 8k
Ω
Let currents be in mA and resistances be in k
Ω
. At node a,
oa
oaa
vv
vvv
−=→
−
=
−
31
105
5.0
(1)
But
aooba
vvvvv
8
10
28
8
=→
+
== (2)
Substituting (2) into (1) gives
14
8
8
10
31
=→−=
aaa
vvv
Thus,
A 28.14mA 70/1
5
5.0
µ
−=−=
−
=
a
x
v
i
A 85.71mA
14
8
4
6.0
)
8
10
(6.0)(6.0
102
µ
==−=−=
−
+
−
= xvvvv
vvvv
i
aaao
aobo
y
Chapter 5, Solution 17.
(a) G = =−=−=
5
12
R
R
v
v
1
2
i
o
-2.4
(b)
5
80
v
v
i
o
−= = -16
(c)
=−=
5
2000
v
v
i
o
-400
Chapter 5, Solution 18.
Converting the voltage source to current source and back to a voltage source, we have the
circuit shown below:
3
20
2010
= kΩ
1 M
Ω
3
v2
3
20
50
1000
v
i
o
⋅
+
−= =−=
17
200
v
1
o
v
-11.764
Chapter 5, Solution 19.
We convert the current source and back to a voltage source.
3
4
42
=
5
k
Ω
v
o
0V
(
4/3
)
k
Ω 10
k
Ω
−
+
4
k
Ω
+
−
(2/3)V
(
20/3
)
k
Ω
+
v
o
−
−
+
50
k
Ω
+
−
2v
i
/3
=
−=
3
2
k
3
4
x4
k10
v
o
-1.25V
=
−
+=
k
10
0v
k
5
v
i
oo
o
-0.375mA
Chapter 5, Solution 20.
8
k
Ω
+
−
v
s
+
−
4
k
Ω
+
v
o
−
−
+
2
k
Ω
4
k
Ω
a b
9 V
At node a,
4
vv
8
vv
4
v9
baoaa
−
+
−
=
−
18 = 5v
a
– v
o
- 2v
b
(1)
At node b,
2
vv
4
vv
obba
−
=
−
v
a
= 3v
b
- 2v
o
(2)
But v
b
= v
s
= 0; (2) becomes v
a
= –2v
o
and (1) becomes
-18 = -10v
o
– v
o
v
o
= -18/(11) = -1.6364V
Chapter 5, Solution 21.
Eqs. (1) and (2) remain the same. When v
b
= v
s
= 3V, eq. (2) becomes
v
a
= 3 x 3 - 2v
0
= 9 - 2v
o
Substituting this into (1), 18 = 5 (9-2v
o
) – v
o
– 6 leads to
v
o
= 21/(11) = 1.909V
Chapter 5, Solution 22.
A
v
= -R
f
/R
i
= -15.
If R
i
= 10kΩ, then R
f
= 150 kΩ.
Chapter 5, Solution 23
At the inverting terminal, v=0 so that KCL gives
121
0
0
0
R
R
v
v
R
v
RR
v
f
s
o
f
os
−=→
−
+=
−
Chapter 5, Solution 24
v
1
R
f
R
1
R
2
- v
s
+ -
+
+
R
4
R
3
v
o
v
2
-
We notice that v
1
= v
2
. Applying KCL at node 1 gives
f
os
ff
os
R
v
R
v
v
RRRR
vv
R
vv
R
v
=−
++→=
−
+
−
+
2
1
21
1
2
1
1
1
111
0
)(
(1)
Applying KCL at node 2 gives
s
s
v
RR
R
v
R
vv
R
v
43
3
1
4
1
3
1
0
+
=→=
−
+ (2)
Substituting (2) into (1) yields
s
f
fo
v
RRR
R
R
R
R
R
R
R
Rv
−
+
−+=
243
3
2
4
3
1
3
1
i.e.
−
+
−+=
243
3
2
4
3
1
3
1
RRR
R
R
R
R
R
R
R
Rk
f
f
Chapter 5, Solution 25.
v
o
= 2 V
+ −
+
v
a
+
v
o
-v
a
+ 3 + v
o
= 0 which leads to v
a
= v
o
+ 3 = 5 V.
Chapter 5, Solution 26
+
v
b
- i
o
+ +
0.4V 5k
Ω
- 2k
Ω
v
o
8k
Ω
-
V 5.08.0/4.08.0
28
8
4.0 ==→=
+
==
ooob
vvvv
Hence,
mA 1.0
5
5.0
5
===
kk
v
o
o
i
Chapter 5, Solution 27.
(a) Let v
a
be the voltage at the noninverting terminal.
v
a
= 2/(8+2) v
i
= 0.2v
i
ia0
v2.10v
20
1 =
+
1000
v
=
G = v
0
/(v
i
) = 10.2
(b) v
i
= v
0
/(G) = 15/(10.2) cos 120πt = 1.471 cos 120πt V
Chapter 5, Solution 28.
−
+
+
−
At node 1,
k
50
vv
k
10
v0
o1
1
−
=
−
But v
1
= 0.4V,
-5v
1
= v
1
– v
o
, leads to v
o
= 6v
1
= 2.4V
Alternatively, viewed as a noninverting amplifier,
v
o
= (1 + (50/10)) (0.4V) = 2.4V
i
o
= v
o
/(20k) = 2.4/(20k) = 120 µA
Chapter 5, Solution 29
R
1
v
a
+
v
b
- +
+
v
i
R
2
R
2
v
o
- R
1
-
obia
v
RR
R
vv
RR
R
v
21
1
21
2
,
+
=
+
=
But
oiba
v
RR
R
v
RR
R
v
21
1
21
2
+
=
+
→=
v
Or
1
2
R
R
v
v
i
o
=
Chapter 5, Solution 30.
The output of the voltage becomes
v
o
= v
i
= 12
Ω= k122030
By voltage division,
V2.0)2.1(
6012
12
v
x
=
+
=
===
k20
2.0
k20
v
i
x
x
10µA
===
k
20
04.0
R
v
p
2
x
2µW
Chapter 5, Solution 31.
After converting the current source to a voltage source, the circuit is as shown below:
12
k
Ω
2
v
o
6
k
Ω
+
−
+
−
6
k
Ω
3
k
Ω
1
v
1
v
o
12 V
At node 1,
12
vv
6
vv
3
v
o1o1
1
12
−
+
−
=
−
48 = 7v
1
- 3v
o
(1)
At node 2,
x
oo1
i
6
0v
6
vv
=
−
=
−
v
1
= 2v
o
(2)
From (1) and (2),
11
48
v
o
=
==
k
6
v
i
o
x
0.7272mA
Chapter 5, Solution 32.
Let v
x
= the voltage at the output of the op amp. The given circuit is a non-inverting
amplifier.
=
x
v
+
10
50
1 (4 mV) = 24 mV
Ω= k203060
By voltage division,
v
o
= mV12
2
v
v
2020
o
o
==
+
20
i
x
=
()
==
+ k40
mV24
k2020
v
x
600nA
p =
=
−
3
6
2
o
10x60
10x
=
144
R
v
204nW
Chapter 5, Solution 33.
After transforming the current source, the current is as shown below:
1
k
Ω
This is a noninverting amplifier.
3
k
Ω
v
i
v
a
+
−
+
−
2
k
Ω
4
k
Ω
v
o
4 V
iio
v
2
3
v
2
1
1v =
+=
Since the current entering the op amp is 0, the source resistor has a OV potential drop.
Hence v
i
= 4V.
V6)4(
2
3
v
o
==
Power dissipated by the 3kΩ resistor is
==
k
3
36
R
v
2
o
12mW
=
−
=
−
=
k
1
64
R
vv
i
oa
x
-2mA
Chapter 5, Solution 34
0
R
vv
R
vv
2
in1
1
in1
=
−
+
−
(1)
but
o
43
3
a
v
RR
R
v
+
= (2)
Combining (1) and (2),
0v
R
R
v
R
R
vv
a
2
1
2
2
1
a1
=−+−
2
2
1
1
2
1
a
v
R
R
v
R
R
1v +=
+
2
2
1
1
2
1
43
o3
v
R
R
v
R
R
1
RR
vR
+=
+
+
+
+
+
=
2
2
1
1
2
1
3
43
o
v
R
R
v
R
R
1R
RR
v
v
O
= )vRv(
)RR(R
RR
221
213
43
+
+
+
Chapter 5, Solution 35.
10
R
R
1
v
v
A
i
f
i
o
v
=+== R
f
= 9R
i
If R
i
= 10kΩ, R
f
= 90kΩ
Chapter 5, Solution 36
V
abTh
V=
But
abs
V
RR
R
21
1
+
=
v
. Thus,
ssabTh
v
R
R
v
R
RR
VV )1(
1
2
1
21
+=
+
==
To get R
Th
, apply a current source I
o
at terminals a-b as shown below.
v
1
+
v
2
- a
+
R
2
v
o
i
o
R
1
-
b
Since the noninverting terminal is connected to ground, v
1
= v
2
=0, i.e. no current passes
through R
1
and consequently R
2
. Thus, v
o
=0 and
0==
o
o
Th
i
v
R
Chapter 5, Solution 37.
++−=
3
3
f
2
2
f
1
1
f
o
v
R
R
v
R
R
v
R
R
v
−++−= )3(
30
30
)2(
20
30
)1(
10
30
v
o
= -3V
Chapter 5, Solution 38.
+++−=
4
4
f
3
3
f
2
2
f
1
1
f
o
v
R
R
v
R
R
v
R
R
v
R
R
v
−++−+−= )100(
50
50
)50(
10
50
)20(
20
50
)10(
25
50
=
-120mV
Chapter 5, Solution 39
This is a summing amplifier.
223
3
2
2
1
1
5.29)1(
50
50
20
50
)2(
10
50
vvv
R
R
v
R
R
v
R
R
v
fff
o
−−=
−++−=
++−=
Thus,
V 35.295.16
22
=→−−=−= vvv
o
Chapter 5, Solution 40
R
1
R
2
v
a
+
R
3
v
b -
+ +
v
1
+
- v
2
R
f
v
o
- +
v
3
R -
-
Applying KCL at node a,
)
111
(0
3213
3
2
2
1
1
3
3
2
2
1
1
RRR
v
R
v
R
v
R
v
R
vv
R
vv
R
vv
a
aaa
++=++→=
−
+
−
+
−
(1)
But
o
f
ba
v
RR
R
v
+
==v
(2)
Substituting (2) into (1)gives
)
111
(
3213
3
2
2
1
1
RRRRR
Rv
R
v
R
v
R
v
f
o
++
+
=++
or
)
111
/()(
3213
3
2
2
1
1
RRRR
v
R
v
R
v
R
RR
v
f
o
++++
+
=
Chapter 5, Solution 41.
R
f
/R
i
= 1/(4) R
i
= 4R
f
= 40kΩ
The averaging amplifier is as shown below:
Chapter 5, Solution 42
v
1
R
2
= 40
k
Ω
v
2
R
3
= 40
k
Ω
v
3
R
4
= 40
k
Ω
v
4
10
k
Ω
−
+
R
1
= 40
k
Ω
v
o
Ω== k 10R
3
1
R
1f
Chapter 5, Solution 43.
In order for
+++=
4
4
f
3
3
f
2
2
f
1
1
f
o
v
R
R
v
R
R
v
R
R
v
R
R
v
to become
()
4321o
vvvv
4
1
v +++−=
4
1
R
R
i
f
= ===
4
12
4
R
R
i
f
3kΩ
Chapter 5, Solution 44.
R
4
At node b,
0
R
vv
R
vv
2
2b
1
1b
=
−
+
−
21
2
2
1
1
b
R
1
R
1
R
v
R
v
v
+
+
= (1)
b
a
R
1
v
1
R
2
v
2
R
3
v
o
−
+
At node a,
4
oa
3
a
R
vv
R
v0 −
=
−
34
o
a
R/R1
v
v
+
= (2)
But v
a
= v
b
. We set (1) and (2) equal.
21
1112
34
o
RR
vRvR
R/R1
v
+
+
=
+
or
v
o
=
()
()
()
1112
213
43
vRvR
RRR
RR
+
+
+
Chapter 5, Solution 45.
This can be achieved as follows:
()
+−−=
21o
v
2/R
R
v
3/R
R
v
()
+−−=
2
2
f
1
1
f
v
R
R
v
R
R
i.e. R
f
= R, R
1
= R/3, and R
2
= R/2
Thus we need an inverter to invert v
1
, and a summer, as shown below (R<100kΩ).
R/3
R/2
v
2
R
−
+
R
v
1
-v
1
R
−
+
v
o
Chapter 5, Solution 46.
3
3
f
2
2
x
1
1
f
32
1
o
v
R
R
)v(
R
R
v
R
R
v
2
1
)v(
3
1
3
v
v +−+=+−+=−
i.e. R
3
= 2R
f
, R
1
= R
2
= 3R
f
. To get -v
2
, we need an inverter with R
f
= R
i
. If R
f
= 10kΩ,
a solution is given below.
v
1
30
k
Ω
20
k
Ω
v
3
10
k
Ω
−
+
10
k
Ω
v
2
-v
2
10
k
Ω
−
+
30
k
Ω
v
o
Chapter 5, Solution 47.
If a is the inverting terminal at the op amp and b is the noninverting terminal,
then,
V6vv,V6)8(
13
3
v
bab
===
+
= and at node a,
4
vv
2
v10
oaa
−
=
−
which leads to v
o
= –2 V and i
o
=
k4
)vv(
k5
v
oao
−
− = –0.4 – 2 mA = –2.4 mA
Chapter 5, Solution 48.
Since the op amp draws no current from the bridge, the bridge may be treated separately
as follows:
v
1
v
2
i
2
i
1
+
−
For loop 1, (10 + 30) i
1
= 5 i
1
= 5/(40) = 0.125µA
For loop 2, (40 + 60) i
2
= -5 i
2
= -0.05µA
But, 10i + v
1
- 5 = 0 v
1
= 5 - 10i = 3.75mV
60i + v
2
+ 5 = 0 v
2
= -5 - 60i = -2mV
As a difference amplifier,
()
[]
mV)2(75.3
20
80
vv
R
R
v
12
1
2
o
−−=−=
=
23mV
Chapter 5, Solution 49.
R
1
= R
3
= 10kΩ, R
2
/(R
1
) = 2
i.e. R
2
= 2R
1
= 20kΩ = R
4
Verify:
1
1
2
2
43
21
1
2
o
v
R
R
v
R/R1
R/R1
R
R
v −
+
+
=
()
1212
vv2v2v
5.01
)5.01(
2 −=−
+
+
=
Thus, R
1
= R
3
= 10kΩ, R
2
= R
4
= 20kΩ
Chapter 5, Solution 50.
(a) We use a difference amplifier, as shown below:
()(
,vv2vv
R
R
v
1212
1
2
o
−=−=
)
i.e. R
2
/R
1
= 2
R
1
v
2
R
2
v
1
R
2
−
+
R
1
v
o
If R
1
= 10 kΩ then R
2
= 20kΩ
(b)
We may apply the idea in Prob. 5.35.
210
v2v2v −=
()
+−−=
21
v
2/R
R
v
2/R
R
()
+−−=
2
2
f
1
1
f
v
R
R
v
R
R
i.e. R
f
= R, R
1
= R/2 = R
2
