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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.

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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+.

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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

(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

= + = +

(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

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