ENGG 1203 Tutorial Op Amps 10 Oct Learning Objectives Analyze circuits with ideal operational amplifiers News Mid term Revision tutorial Ack.: MIT OCW 6.01 1
Q1 This circuit is controlled by the charge output from a digital system. This digital signal represent a logic 1 using 2V, and represents a 0 using 0V. The output of the circuit is the signal V charge. To charge the battery, V charge = 20V. When it is not charging, V charge = 0V. 2
Complete the following op-amp circuit by i) determining the values of the resistors R 1, R 2, R 3, R 4 ; ii) determining the connection to point A and B. For point A and B, indicate if it should be connected to (i) charge if (ii) GND or (iii) 20V. 3
0 2 2 2 20 Consider the case when charge is HIGH and V charge is HIGH The configuration is possible when R3 = 0 (i.e. short circuit) R4 = inf. (i.e. open circuit), B = charge Thus, A = GND R1:R2 = 1:9 4
0 0 0 0 0 When B = Low (0), V charge = 0 5
Q2 When V 1 = 1V and V 2 = 2V, determine the current I x and the voltage V A. Determine a general expression for V A in terms of V 1 and V 2. 2 4 2 1-1 1 6
Solution When V 1 = 1V and V 2 = 2V, I x = 1A When V 1 = 1V and V 2 = 2V, V A = 4V A general expression for V A : V A = V 2 + I x 2 = V 2 + V 2 V 1 2 1 = V 2 + 2 V 2 V 1 = 2V 1 + 3V 2 7
Q3 Use a single op-amp and resistors to make a circuit that is equivalent to the following circuit. V n V n V i = 1 + R 2 R 1 V o V i = V o V n V n V i = 1 + R 4 R 3 1 + R 2 R 1 =1 + R 1R 4 +R 2 R 3 +R 2 R 4 R 1 R 3 8
Q4 Use the ideal op-amp model (V + = V - ) to determine an expression for the output current I o in terms of the input voltage V i and resistors R 1 and R 2. v x = v i + v x R 2 R 1 + R 2 v x = v i R 2 R 1 I o = v x R 2 = 1 R 2 v i R 2 R 1 = v i R 1 v i +v x v i +v x v x 9
Q5 Determine R so that V o = 2 (V 1 V 2 ). 10
Solution No current in +ve or -ve inputs: Ideal op-amp: 11
Q6 A proportional controller that regulates the current through a motor by setting the motor voltage V C to V C = K(I d I o ) K is the gain (ohms) I d is the desired motor current I o is the actual current through the motor. 12
Solution Consider the circuit inside the dotted rectangle. Determine V 1 as a function of I o. V + = 1/2 x I o = V - V - = 100/(100+9900) x V 1 V 1 = 1/2 x I o x 100 Determine the gain K and desired motor current I d. KCL at -ve input to right op-amp: V c 2.5 1000 = 2.5 V V C = K(I d I o ) 1 1000 V c = 50 0.1 I o 13
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) 14
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 (least clockwise) to 1 (most clockwise) If α i >α o, then the voltage to the motor (V M+ V M ) is positive, and the motor turns clockwise (so as to increase α o ) 15
Solution Determine an expression for V M+ in terms of α i, R, and V S. The output of the voltage divider is The op-amp provides a gain of 1, so V M+ = V +. 16
The following circuit produces a voltage V o that depends on the position of the input pot. Determine an expression for the voltage V o in terms of α i, R, R 1, R 2, and V S. 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 17
The following circuit produces a voltage V o that depends on the positions of both pots. Determine an expression for V o in terms of α i, α o, R, and V S. 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 18
(d) Assume that we are provided with a circuit whose output is α i /α o volts. We want to design a controller of the following form so that the motor shaft angle (proportional to α o ) will track the input pot angle (proportional to α i ). Assume that R 1 = R 3 = R 4 = 1000Ω and V C = 0. Is it possible to choose R 2 so that α o tracks α i? If yes, enter an acceptable value for R 2. 19
Assume that R 1 = R 3 = R 4 = 1000Ω and V C = 0 If R 3 = R 4 then the right motor input is 5V. If α i = α o then the gain of the left op-amp circuit must be 5 so that the motor voltage is 0. The gain is R 1 + R 2 /R 1, so R 2 must be 4000Ω. 1 5 5 5 1 0 20
Q8(a) You have to design a hammer. The design goal is to generate an output voltage (V o ) which is proportional to the force (F) applied on the hammer, i.e. V o = m x F + C, with m > 0 and C > 0. You found a force-sensitive resistor (FSR) which can be modeled by R 10k F FSR You then design a circuit as a potential divider. Will this circuit correctly implement? No, because V o is not linearly proportional to F. 21
Q8(b) Find the gain of the following circuit: At the two op-amp inputs, V - = V + = V i. Since V i is related to V o through the two resistors such that V o R 4 = (R 3 + R 4 ), Vo 1 R R 3 4 V i 22
Q8(c) Design (by using the non-inverting amplifier circuit) a circuit such that the output voltage (V o ) is directly proportional to the input force (F). Replace R 4 by the FSR. We then have R F R V V V F V 3 3 i o 1 i i 10000 10000 V o = m x F + C 23
RV 10000 3 i Vo F Vi (d) The system requires that when the force F = 0 N, V o = 4 V; when F = 20 N, V o = 12 V. Construct the circuit designed in (c) using only one FSR, one op-amp, one 12V power supply, and 1k ohm resistances. When F = 0N, V o = V i = 4 V (because of the equation). We have to use R 1 = 2k ohm (in series) and R 2 = 1k ohm. When F = 20N, R 4 k 10000 3 12 20 4 R3 1 24
(e) Using the above circuit, what is the value of V o when someone hits the hammer too hard, generating a force of 200 N? 12V RV 10000 3 i Vo F Vi (f) Suggest modification(s) such that the max. allowable force to the circuit is 60 N. Change R 3 to 1/3 k ohm. This can be done by parallel composition of three 1 k ohm resistors. 25
(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, ]) R 1 R 2 V o = 2V 2-2V 1 V o = V 2 - V 1 V o = 4V 2-2V 1 26
Solution V + = R 2 10k+R 2 V 2 = V = R 1 10k+R 1 V 1 + V o = 10k+R 1 10k+R 2 R 2 10k V 2 R 1 10k V 1 10k 10k+R 1 V o 3 rd : Negative R i.e. Impossible R 1 R 2 V o =2V 2-2V 1 20kΩ 20kΩ V o =V 2 -V 1 20kΩ 20kΩ V o =4V 2-2V 1 Impossible Impossible 27
(Appendix) Q10 What is V o? V 3 + V 1 V 3 + V 2 V 3 V o = 0 V o = V 1 V 2 28
(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. 29
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 10 0.05 Eq. R. (Red): 1K+~5 1K Eq. R. (Blue): 1K//1K//5 ~5 0 30
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 10 0.45 Eq. R. (Red): 100K+~5 100K Eq. R. (Blue): 1K//100K//5 ~5 0 31
Solution (Pat s Design) P.D. of motor 1 = 10V P.D. of motor 3 = 2V P.D. of motor 2 = 4V Wrong design 10 Eq. R. : 1K // 2K = 2/3K 4 4 2 2 32
Solution (Kim s Design) P.D. of motor 1 = 10V P.D. of motor 3 = 2.5V P.D. of motor 2 = 5V Correct design 10 Eq. R. : 100 // 200K = ~100 5 2.5 5 2.5 33
Solution (Leon s Design) P.D. of motor 1 = 10V P.D. of motor 3 = 2.5V P.D. of motor 2 = 5V Correct design 10 5 5 2.5 2.5 34