EE323 Homework 1 Solution, Au/2004
2.8
(25 points)
(a)
Voltage gain:
/
2
1
100
10
10
VV
o
i
V
R
VR
=−
Input impedance:
10
K
in
R
Ω
=
(b)
The 10Kohm resistor has not impact on the voltage gain and input impedance.
Voltage gain:
/
2
1
100
10
10
o
i
V
R
Input impedance:
10
K
in
R
Ω
=
(c)
Voltage at the virtual ground is 0, so there is no current flowing through the
resistor between the virtual ground and ground.
Voltage gain:
/
2
1
100
10
10
o
i
V
R
Input impedance:
10
K
in
R
Ω
=
(d)
Voltage gain:
/
2
1
100
10
10
o
i
V
R
Input impedance:
10
K
in
R
Ω
=
2.12
(25 points)
For an ideal op amp, voltage gain is
2
1
o
i
V
R
. Note that resistor should be at least
10
K
Ω
. A set of possible solutions are:
(a)
12
10
R
RK
==
Ω
(b)
10
,
20
R
KR
K
=Ω
(c)
20
,
10
R
K
(b)
10
,
1000
1
R
K
M
=
Ω
=
Ω
2.49
(25 points)
Let
V
+
and
V
−
denote the voltage at the positive input terminal and negative input
terminal, respectively. Note
3
R
and
4
R
work as a voltage divider, such that:
4
34
i
R
V
RR
+
−
=
=
+
The current flowing through resistors
1
R
and
2
R
is the same and equal to:
11
i
R
R
−
+
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 Spring '08
 HEMAMI
 Impedance, Volt, Voltage divider, Resistor, Jaguar Racing, Electrical impedance, R1 R1 R1

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