Homework #4 Solution
October 27, 2011
Point Distribution of Homework #4 (55 points total)
Basic rule:
1. All calculation mistakes get 1 point off.
2. Mistakes in setting up necessary equations such as KCL get 1 point off if KCL is applied at the right
nodes.
3. Applying KCL at wrong nodes, such as the output of opamp without a condition of the output current,
gets no point.
4. If the answer for later parts of successive problems is affected by a wrong answer from the previous
part but the approach and the calculation are right, it gets the full point.
Point Distribution
1. U&M 4.20 (5 points total)
2. U&M 4.38 (10 points total)
(a) 4 points
(b) 4 points
(c) 2 points
3. Inverting Amplifier (15 points total)
(a) 5 points
(b) 10 points
4. U&M 4.43 (5 points total)
5. U&M 5.10 (10 points total)
(a) 3 points
(b) 2 points
(c) 3 points
(d) 2 points
6. U&M 5.17 (5 points total)
7. U&M 5.22 (5 points total)
(a) 2 points
(b) 2 points
(c) 1 points
1
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Problem 4.20
Determine the linear range of the source
v
s
in the circuit of
Fig. P4.20.
v
o
v
s
200
Ω
400
Ω
1.2 k
Ω
V
cc
= 12 V
2 V
+
_
+
_
Figure P4.20:
Circuit for Problem 4.20.
Solution:
The circuit is a standard summing amplifier.
v
o
=
−
R
f
R
1
v
1
−
R
f
R
2
v
2
=
−
1
.
2
×
10
3
200
v
s
−
1
.
2
×
10
3
400
×
2
=
−
6
v
s
−
6
(V)
.
For
v
o
=
V
cc
=
12 V,
12
=
−
6
v
s
−
6
,
or
v
s
=
−
3 V
.
For
v
o
=
−
V
cc
=
−
12 V,
−
12
=
−
6
v
s
−
6
,
or
v
s
=
1 V
.
Hence, the linear range of
v
s
is
−
3 V
≤
v
s
≤
1 V
.
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 '10
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 Operational Amplifier, kΩ, Comparator, Science Press, National Technology

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