engg 1203 tutorialculei/engg1203_14/engg1203...the op-amp provides a gain of 1, so v m+ = v +. 14...
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ENGG 1203 Tutorial
Op Amps
7 Mar
Learning Objectives
Analyze circuits with ideal operational amplifiers
News
HW2
Mid term
Revision tutorial (14 Mar 14:30-16:20, CBA)
Ack.: MIT OCW 6.01
Q1
Determine Vo in the following circuit. Assume that the op-amp is ideal.
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Solution
Since V- = V+, V- = 5V. So there must be 1/12A flowing left through the two 6 ohm resistors. There must be a corresponding 1/12 A flowing to the left through the 12 ohm resistor. Vo is then the sum of V- = 5V and the 1V across the 12 ohm resistor.
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Q2
Determine the current Ixwhen V1 = 1V and V2 = 2V.
Determine the voltage VA
when V1 = 1V and V2 = 2V.
Determine a general expressionfor VA in terms of V1 and V2.
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Solution
When V1 = 1V and V2 = 2V, Ix = 1A
When V1 = 1V and V2 = 2V, VA = 4V
A general expression for VA: 2
1 2
2
2 3
5
1
1
2
2
4
-1
Q3
Use a single op-amp and
resistors to make a circuit
that is equivalent to the
following circuit.
6
Vn
1
1
1
=1
Q4
Use the ideal op-amp model (V+ = V-) to determine an expression for the output current Ioin terms of the input voltage Vi and resistors R1
and R2.
7
vx
vi +vx
vi +vx
⇒
1
Q5
Determine R so that Vo = 2 (V1 − V2).
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Solution
No current in +ve or -ve inputs:
Ideal op-amp:
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Q6
A proportional controller that regulates the current through a motor by setting the motor voltage VC to VC = K(Id − Io)
K is the gain (ohms)
Id is the desired motor
current
Io is the actual current
through the
motor.
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Solution
Consider the circuit inside the
dotted rectangle. Determine
V1 as a function of Io.
V+ = 1/2 x Io = V-
V- = 100/(100+9900) x V1
V1 = 1/2 x Io x 100
Determine the gain K and desired motor current Id.
KCL at -ve input to right op-amp: 2.5
10002.5
1000⇒ 50 0.1
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Q7
The shaft angle of the output pot tracks that of the input pot
If the person turn the left potentiometer (the input pot),
then the motor will turn the right potentiometer (the
output pot)
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Solution
Pot resistances depends on shaft angle
Lower part of the pot is αR
Upper part is (1 − α)R, where R = 1000Ω.
α is from 0 (most counterclockwise position) to 1 (most clockwise
position)
If αi >αo, then the voltage to the motor (VM+ − VM−) is
positive, and the motor turns clockwise (so as to
increase αo) — i.e., positive motor voltage clockwise
rotation.
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Solution
Determine an expression for VM+ in terms of αi, R, and VS.
The output of the voltage divider is
The op-amp provides a gain of 1, so VM+ = V+.
14
The following circuit produces a
voltage Vo that depends on the
position of the input pot.
Determine an expression for
the voltage Vo in terms of αi,
R, R1, R2, and VS.
The positive input to the op-amp is connected to a
voltage divider with equal resistors so
The input pot is on the output of the op-amp, so
In an ideal op-amp, V+ = V− so
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The following circuit produces a voltage Vo that depends
on the positions of both pots. Determine an expression
for Vo in terms of αi, αo, R, and VS.
The positive input to the op-amp is connected to pot 1 so
that
The output pot is on the
output of the op-amp, so
In an ideal op-amp, V+ = V− so
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Assume that we are provided with a circuit whose output
is αi/αo volts. We want to design a motor controller of the
following form so that the motor shaft angle (which is
proportional to αo) will track the input pot angle (which is
proportional to αi).
Assume that R1 = R3 = R4 = 1000Ω and VC = 0. Is it
possible to choose R2 so that αo tracks αi? If yes, enter
an acceptable value for R2.
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Assume that R1 = R3 = R4 = 1000Ω and VC = 0
If R3 = R4 then the right motor input is 5V. If αi = αo
then the gain of the left op-amp circuit must be 5 so
that the motor voltage is 0. The gain is R1 + R2/R1,
so R2 must be 4000Ω.
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5551
0
1
Assume that R1 = R3 = R4 = 1000Ω and VC = 5V
If R3 = R4 then the right motor input is 5V. If αi = αo
then V+ = V− = 1 for the right op-amp. We need the
left motor input to be 5V. But if the left motor input is
5V and VC = 5V then V− must also be 5V, which
leads to a contradiction.
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5551
5
1
Q8
You have to design a hammer machine (i.e. using a hammer to hit a platform to see how strong the participants are). The design goal is to generate an output voltage (Vo) which is proportional to the force (F) applied on the hammer, i.e. Vo = m x F + C, with m > 0 and C > 0.
(a) You found a force-sensitive resistor (FSR) from the catalog, which can be modeled by
You then design a circuit as a potential divider.Will this circuit correctly implement?
No, because Vo is not linearlyproportional to F.
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10FSR
R k F= Ω
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(b) Find the gain of the following circuit:
At the two op-amp inputs, V- = V+ = Vi.
Since Vi is related to Vo through
the two resistors such that
VoR4 = (R3 + R4),
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3
4
1o i
RV V
R
⇒ = +
(c) Design (by using the non-inverting amplifier circuit) a
circuit such that the output voltage (Vo) is directly
proportional to the input force (F).
Replace R4 by the FSR. We then have
Vo is a linear function of F.
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3 31
10000 10000
i
o i i
R F R VV V F V
= + = +
(d) The system requires that
when the force F = 0 N, Vo = 4 V;
when F = 20 N, Vo = 12 V.
Construct the circuit designed in (c) using only one FSR,
one op-amp, one 12V power supply, and 1k ohm resis.
When F = 0N, Vo = Vi = 4 V. We can use R1 = 2k ohm
and R2 = 1k ohm. 2k ohm resistors can be made by two
1k ohm resistors in series.
When F = 20N,
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3
3
412 20 4 1
10000
RR k
×= × + ⇒ = Ω
3 31
10000 10000
i
o i i
R F R VV V F V
= + = +
(e) Using the above circuit, what is
the value of Vo when someone hits
the hammer too hard, generating a
force of 200 N?
12V
(f) Suggest modification(s) such that the max. allowable
force to the circuit is 60 N.
Change R3 to 1/3 k ohm. This can be done by parallel
composition of three 1 k ohm resistors.
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3 31
10000 10000
i
o i i
R F R VV V F V
= + = +
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(Appendix) Q9
Fill in the values of R1 and R2 required to satisfy the
equations in the left column of the following table. The
values must be non-negative (i.e., in the range [0,∞])
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R1 R2
Vo = 2V2 - 2V1
Vo = V2 - V1
Vo = 4V2 - 2V1
Solution
! "
!
!
!
!
!
!
!
3rd: Negative R
i.e. Impossible
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R1 R2
Vo=2V2-2V1 20kΩ 20kΩ
Vo=V2-V1 20kΩ 20kΩ
Vo=4V2-2V1 Impossible Impossible
(Appendix) Q10
What is Vo?
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Vo = 0 Vo = V1 – V2
V3
V3+ V1
V3+ V2
(Appendix) Q11
Students Kim, Pat, Jody, Chris, and Leon are trying to
design a controller for a display of three robotic mice in
the Rube Goldberg Machine, using a 10V power supply
and three motors.
The first is supposed to spin as fast as possible (in one direction
only), the second at half of the speed of the first, and the third at
half of the speed of the second.
Assume the motors have a resistance of approximately 5Ω and
that rotational speed is proportional to voltage.
For each design, indicate the voltage across each of the
motors.
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Solution (Jody’s Design)
P.D. of motor 1 = 10V
P.D. of motor 2 = 0.05V
P.D. of motor 3 = 0V
Wrong design
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10
0.05
0
Eq. R. (Red): 1K+~5 1K
Eq. R. (Blue): 1K//1K//5 ~5
Solution (Chris’s Design)
P.D. of motor 1 = 10V
P.D. of motor 2 = 0.45V
P.D. of motor 3 = 0V
Wrong design
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10
0.45
0
Eq. R. (Red): 100K+~5 100K
Eq. R. (Blue): 1K//100K//5 ~5
Solution (Pat’s Design)
P.D. of motor 1 = 10V P.D. of motor 2 = 4V
P.D. of motor 3 = 2V Wrong design
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10
4
2
4
2
Eq. R. : 1K // 2K = 2/3K
Solution (Kim’s Design)
P.D. of motor 1 = 10V P.D. of motor 2 = 5V
P.D. of motor 3 = 2.5V Correct design
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10
5
2.5
5
2.5
Eq. R. :
100 // 200K = ~100
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Solution (Leon’s Design)
P.D. of motor 1 = 10V P.D. of motor 2 = 5V
P.D. of motor 3 = 2.5V Correct design
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10
5 5
2.52.5