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ENGG 1203 Tutorial_05 Example 1 : Last Q5 1 Solution 1 : use KCL I1 I2 From B to M I4 2 Solution 1 : I1 Solution 1 : I2 I4 4 Solution 1 : 5 Example 2 : You have connected the lamp, with Vcc = 12V. The datasheet of the lamp states that it only turns on when VL > 8V. The lamp has an internal resistance of 1k ohm. What is the range of R that would allow the circuit to function correctly with all input combinations. 6 Solution 2 : The range of R When the lamp is turned on, x = “1”, VL > 8 When the lamp is off, x = 0, and VL < 8. It will function regardless of the value of R. Function as an AND gate 7 Example 3(a) : A force sensitive resistor (FSR) is a resistor with its resistance changed according to the force applied to it. For simplicity sake, your partner has wired up the FSR using a simple potential divider circuit 8 Solution 3(a) : Calculate the following quantities when Rref = 10k ohm : ◦ Voltage across the FSR ; ◦ Voltage at Vout ; ◦ Current owing through the FSR. 9 Example 3(b) : The output Vout is used to detect the presence of a ball. Due to its light weight, the ball produces only 0.5N when it is located on top of the sensor. The rest of the system requires that VIL = 2V and VIH = 10V ◦ VIL : Max. voltage that the system regards as logical LOW ◦ VIH : Min. voltage that the system regards as logical HIGH Determine the range of value that Rref may take for correct functioning of the circuit. ◦ It should output a logical HIGH when a ball is presence and a logical LOW otherwise. 10 Solution 3(b) : VDD: 12V (Presence of the ball / 0.5N); Rfsr = RL = 10k ohm VIH: 10V; Minimum HIGH; 10V – 2V: (Unknown; Not in use) VIL: 2V; Maximum LOW; Rfsr = RL = 1M ohm (Absence of the ball / 0N) VGND: 0V 11 Solution 3(b) : When the ball is on the sensor, Rfsr = RH = 10k ohm. The output voltage at that time should be higher than VIH. When the ball is not on the sensor, then Rfsr = RL = 1M ohm. The output voltage should be lower than VIL. 12 Solution 3(c) : Your partner suggests that it may be possible to use 2 FSRs connected to perform a logical OR operation: When the ball rolls over either one of the 2 FSRs, the output Vout is HIGH, and is LOW otherwise. What is the output voltage Vout ? ◦ ◦ ◦ ◦ one of the FSRs is under pressure of 0.5N, both FSRs are under a pressure of 0.5N each, none of the FSRs is under pressure. assume Rref is 100k ohm 13 Solution 3(c) : If Rref = 100k, ◦ ◦ ◦ ◦ ◦ (i) one of the FSRs is under pressure of 0.5N, (ii) both FSRs are under a pressure of 0.5N each, (iii) none of the FSRs is under pressure. Reqv is the equivalent resistance of the parallel combination of the two FSRs. VIL is not MORE than 2V and VIH is not less than10V, the circuit still functioning correctly as a 2input OR 14 Solution 3(d) : If there are 3 FSRs connected in parallel, assume Rref remains at 100k THEN, will it behave as a 3-input OR? NOT, Vout > 2 V 15 Solution 3 : Yes, it works correctly as a 2-input OR gate because the output is HIGH when there is a ball on top of at least one of the input. Even if we connect 3 FSRs in parallel, the circuit cannot correctly function as a 3-input OR gate. ◦ In the case when there is no ball falling on the circuit, the equivalent resistance of the parallel combination of the 3 FSRs drops too low that Vout > 2V . ◦ As a result, the output FAILED to represent a logical LOW in this case. ◦ A new R’ref is needed 16 Operational Amplifier uA mA Output stage Input Differential Pair Ack : http://www.ti.com/lit/ds/symlink/ua741.pdf https://www.triadsemi.com/2014/06/02/asic-integration-power-savings/ 17 Ideal Op-Amp Ideal Op-Amp Ack : http://physics.tutorvista.com/waves/direct-sensing.html • Infinite Gain K, Vo = K (V+ - V-) • V+ = V• Input Current in Op-Amp is ZERO • (Input Resistance is Ideally infinite) • Vo,max = Vcc, • Vo,min = -Vcc 18 Recap : Vi to + 19 Recap : Vi to - 20 Recap : Buffer Op-Amp Buffer ◦ Non-Inverting Op-Amp ◦ R1 = Infinite, R2 = 0 R1 = Infinite, R2 = 0 GAIN = ?? 21 Example 4 : The circuit is controlled by the charge output from a digital system. ◦ The digital signal represent a logic “1” using 2V, and represents a “0” using 0V. The output of the circuit is the signal Vcharge. ◦ To charge the battery, Vcharge = 20V. ◦ When it is not charging, Vcharge = 0V. 22 Solution 4 : Complete the following op-amp circuit by i) determining the values of the resistors R1, R2, R3, R4; 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. • 23 Solution 4 : 1 : 9 • Consider the case when charge is HIGH and Vcharge is HIGH • The configuration is possible • when R3 = 0 (i.e. short circuit) • R4 = infinite (i.e. open circuit), • B = charge 2V • Thus, A = GND R1:R2 = 1:9 24 Solution 4 : • When B = Low (0), Vcharge = 0 25 Example 5 : • When V1 = 1V and V2 = 2V, determine the current Ix and the voltage VA. • Determine a general expression for VA in terms of V1 and V2. 26 Solution 5 : When V1= 1V and V2= 2V, Ix = 1A When V1= 1V and V2 = 2V, VA= 4V A general expression for VA : VA = V2 + Ix X 2 = V2 + ( V2 – V1 )/1 X 2 = V2 + 2( V2 – V1) = -2V1 + 3 V2 27 Example 6 : Use a single op-amp and resistors to make a circuit that is equivalent to the following circuit. Vn/Vi = (1+R2/R1) ……. Vo/Vi = (Vo/Vn)(Vn/Vi) = (1+R4/R3)(1+R2/R1) =1+ (R1R4+R2R3+R2R4)/R1R3 28 Example 7 Use the ideal op-amp model (V+ = V-) to determine an expression for the output current Io in terms of the input voltage Vi and resistors R1 and R2. Vx = (Vi+Vx) R2 /(R1+R2) Vx = ViR2/R1 Io = Vx/R2 =(ViR2/R1)/R2 = Vi/R1 Example 8(a) • 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) 30 Solution 8(a) • 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 αi >αo, then the voltage to the motor (VM+ − VM−) is positive, and the motor turns clockwise (so as to increase αo) 31 Solution 8(a) • 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+. V+ 32 Solution 8(b) 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 33 Solution 8(c) 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 34 Solution 8(d) Assume that we are provided with a circuit whose output is αi/αo volts. We want to design a controller of the following form so that the motor shaft angle (proportional to αo) will track the input pot angle (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. Solution 8(d) 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 (1 + R2/R1), so R2 must be 4000Ω. 36 Lab (5) Can Vp go to 6V ? & How ? Vp max = VccRpot/(Ro+R pot) 37 Lab (5) Ack : http://www.electronics-tutorials.ws/diode/diode_4.html Vdrop 38