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1 ENGG 1203 Tutorial Op Amps 8 Mar Learning Objectives Analyze circuits with ideal operational amplifiers News HW2 (18 Mar 23:55) Mid term (22 Mar 2:30pm-3:30pm) Revision tutorial (14 Mar 3:30pm-5:30pm, CBA) Ack.: MIT OCW 6.01 Analysis of a Circuit with Op Amp (I) Determine V o in the following circuit. Assume that the op-amp is ideal. 2 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. V o is then the sum of V - = 5V and the 1V across the 12 ohm resistor. 3 Analysis of a Circuit with Op Amp (II) Determine the current I x when V 1 = 1V and V 2 = 2V. Determine the voltage V A when V 1 = 1V and V 2 = 2V. Determine a general expression for V A in terms of V 1 and V 2 . 4

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ENGG 1203 Tutorial

Op Amps

8 Mar

Learning Objectives

Analyze circuits with ideal operational amplifiers

News

HW2 (18 Mar 23:55)

Mid term (22 Mar 2:30pm-3:30pm)

Revision tutorial (14 Mar 3:30pm-5:30pm, CBA)

Ack.: MIT OCW 6.01

Analysis of a Circuit with Op Amp (I)

Determine Vo in the following circuit. Assume that the op-amp is ideal.

2

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.

3

Analysis of a Circuit with Op Amp (II)

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.

4

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

Multi-stage Non-inverting Amplifier

Use a single op-amp and

resistors to make a circuit

that is equivalent to the

following circuit.

6

Vn

1

1

1

=1

Voltage-controlled Current Source

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

Op Amp Configurations (I)

Determine R so that Vo = 2 (V1 − V2).

8

No current in +ve or -ve inputs:

Ideal op-amp:

9

Op Amp Configurations (II)

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,∞])

10

R1 R2

Vo = 2V2 - 2V1

Vo = V2 - V1

Vo = 4V2 - 2V1

3rd: Negative R

i.e. Impossible

11

R1 R2

Vo=2V2-2V1 20kΩ 20kΩ

Vo=V2-V1 20kΩ 20kΩ

Vo=4V2-2V1 Impossible Impossible

Unusual Op Amp Configurations

What is Vo?

12

Vo = 0 Vo = V1 – V2

V3

V3+ V1

V3+ V2

Motor Control

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.

13

Motor Control (Jody’s Design)

P.D. of motor 1 = 10V

P.D. of motor 2 = 0.05V

P.D. of motor 3 = 0V

Wrong design

14

10

0.05

0

Eq. R. (Red): 1K+~5 1K

Eq. R. (Blue): 1K//1K//5 ~5

Motor Control (Chris’s Design)

P.D. of motor 1 = 10V

P.D. of motor 2 = 0.45V

P.D. of motor 3 = 0V

Wrong design

15

10

0.45

0

Eq. R. (Red): 100K+~5 100K

Eq. R. (Blue): 1K//100K//5 ~5

Motor Control (Pat’s Design)

P.D. of motor 1 = 10V P.D. of motor 2 = 4V

P.D. of motor 3 = 2V Wrong design

16

10

4

2

4

2

Eq. R. : 1K // 2K = 2/3K

Motor Control (Kim’s Design)

P.D. of motor 1 = 10V P.D. of motor 2 = 5V

P.D. of motor 3 = 2.5V Correct design

17

10

5

2.5

5

2.5

Eq. R. :

100 // 200K = ~100

Motor Control (Leon’s Design)

P.D. of motor 1 = 10V P.D. of motor 2 = 5V

P.D. of motor 3 = 2.5V Correct design

18

10

5 5

2.52.5

Motor Control

The following circuit is a proportional controller that

regulates the current through a motor by setting the

motor voltage VC to VC = K(Id − Io) where K is the gain

(notice that its dimensions

are ohms), Id is the desired

motor current, and 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

10000

2.5 0.5 100

1000⇒ 50 0.1

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Position Controller

The following figure shows a motor controller. A human

can turn the left potentiometer (the input pot). Then the

motor will turn the right potentiometer (the output pot) so

that the shaft angle of the output pot tracks that of the

input pot.

21

The dependence of the pot resistances on shaft angle is

given in terms of α, which varies from 0 (most

counterclockwise position) to 1 (most clockwise

position). The resistance of the lower part of the pot is

αR and that of the upper part is (1 − α)R, where R =

1000Ω.

Notice that 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.

22

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+.

23

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

24

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

25

Assume that we are provided with a circuit whose output

is αi/αo volts. We wish to determine if it is possible 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.

26

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 (a number) for R2.

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.

27

5551

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1

Assume that R1 = R3 = R4 = 1000Ω and VC = 5V. Is

it possible to choose R2 so that αo tracks αi?

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.

28

5551

5

1

Course Timeline (Tentative)

Lecture Tutorial Lab Homework

Systems L1 T1

Digital systems L1

Combinational

logic

L2 T2, T3 #1, #2 HW1

Sequential logic L3 HW1

FSM L3, L4 T3, T4 #3, #4 HW1

ADC/DAC L4 #4 (DAC), #7,8 (ADC)

Circuit L5 T4, T5 HW2

Project T5 #6 HW2

Op Amp L6 T6 #5, #8 HW2

……

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Tutorial Schedule (Tentative)

1/25 Introduction+System

2/1 Digital Logic

2/8 Digital Logic

2/15 N/A

2/22 Digital Logic+Circuit

3/1 Circuit+Project

3/8 Circuit

3/15 Revision Tutorial

3/22 ** Mid Term **

3/29 N/A

4/5 Signal

4/12 Signal

4/19 Signal

4/26 N/A

5/3 N/A

5/X Computer+Revision

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