Get started today!

Good to have you back!
If you've signed in to StudyBlue with Facebook in the past, please do that again.

- StudyBlue
- Michigan
- University of Michigan - Ann Arbor
- Electrical Engineering
- Electrical Engineering 314
- Ganago
- 2010+W+314+HW+08+all_solutions.pdf

Anonymous

Advertisement

Advertisement

EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 1 of 3 Problem 1 (30 points) Summing Amplifier, or Adder; Application for sensor circuits In all parts of this problem, use the “Golden Rules” Part 1 (10 points) Use the node voltage equation for node A to express the output voltage as the sum of two amplified input voltages in the form: VOUT = VS1 Gain1 + VS2 Gain2 [equation 1] Express the gain values in terms of the resistances RF , R1 and R2; pay attention to the signs. Show your work on a separate page. Your answers: Gain1 =−𝑅𝑓𝑅 1 Gain2 =−𝑅𝑓𝑅 2 The node voltage equation for the above circuit is 𝑉− − 𝑉𝑆1 𝑅1 + 𝑉− − 𝑉𝑆2 𝑅2 − 𝑉− − 𝑉𝑜𝑢𝑡 𝑅𝐹 = 0 Using the golden rule we have 𝑉𝑜𝑢𝑡 = 𝑉𝑆1 × −𝑅𝑓𝑅 1 + 𝑉𝑆2 × −𝑅𝑓𝑅 2 𝐺𝑎𝑖𝑛1 = −𝑅𝑓𝑅 1 ; 𝐺𝑎𝑖𝑛2 = −𝑅𝑓𝑅 2 EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 2 of 3 Part 2 (20 points) Practical Perspective The pressure sensor for your project has the zero offset equal to 0.25 V and the sensitivity equal to 0.15 V/psi. (See the file “Pressure sensor specs” for more detail.) Thus, at 5 psi the sensor’s output equals: V S 1 , 5 p s i 0 . 2 5 V 5 p s i 0 . 1 5 Vp s i 1 . 0 0 V You need the VOUT readings: exactly 0 V at 5 psi and exactly 10 V at 10 psi. To achieve this goal, use an adder circuit with two sources – the pressure sensor as VS1 and a 1.5 V battery as VS2. Use RF = 100 k. Calculate the input resistances R1 and R2 in k. Show your work on a separate page. Your answers: R1 = ____7.5 _____ k R2 = 11.25 k Hints: 1. Algebraically, you have to write [equation 1] twice: for the output at 5 psi and for the output at 10 psi, thus obtain two equations with two unknown resistances R1 and R2 ; then solve for R1 and R2. 2. To match the desired polarity of the output voltage, you can connect the sensor and/or the battery in any of the two ways; for example, at 5 psi, you can obtain either VS1 = +1 V or VS1 = – 1 V. Keep the chosen connection at all pressures. Clearly explain which connection you choose. Using the equation derived in the part 1 of the problem 𝑉𝑜𝑢𝑡 = 𝑉𝑆1 × −𝑅𝑓𝑅 1 + 𝑉𝑆2 × −𝑅𝑓𝑅 2 Here Rf = 100 kΩ and Vs2 = 1.5 V and R1 and R2 are in kΩ The value of Vs1 at 5 psi is 0.25 + 0.15 x 5 = 1.0 V Vs1 at 10 psi is 0.25 + 0.15 x 10 = 1.75 V Here the objective is to achieve a positive output voltage. In the inverting configuration we have to connect both the voltages sources in the reverse way as shown in the diagram. EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 3 of 3 From the node voltage equations we obtain 1 × 100𝑅 1 + 1.5 × 100𝑅 2 = 0; (𝑒𝑞𝑛. 1) 1.75 × 100𝑅 1 + 1.5 × 100𝑅 2 = 10; (𝑒𝑞𝑛. 2) Upon solving these equations we get R1 = 7.5 kΩ and R2 = 11.25 kΩ. EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 1 of 6 Problem 2 (30 points) Integrator and Differentiator based on Op Amps; Prototype of PID control circuit In both parts, apply the “Golden Rules” Part 1 (10 points) Which of the sketches 1-4 most likely represents the input voltage? Your answers: For the Integrator: ____________1_______________ For the Differentiator: _______4________________ An alternative is: None of the above. On a separate page, show your work: use equations to justify your answers. EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 2 of 6 The output signal can be represented as a sinusoidal wave of the form Vout= A Sin (ωt-π/2) = -A1 cos (ωt) The inverting integrator integrates a input function and the resulting wave will be 𝑉𝑜𝑢𝑡 = − 1𝑅𝐶 𝑉𝑖𝑛 𝑡 𝑑𝑡 Therefore Vin= -A2 Sin (ωt) This waveform is similar to #1 For an inverting differentiator we have 𝑉𝑜𝑢𝑡 = −𝑅𝐶 𝑑𝑉𝑑𝑡 Vin= A3 Sin (ωt) This waveform is similar to #4 EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 3 of 6 Problem 2 Part 2 (20 points) A control system measures the error, or difference between the set point and the actual output of the Plant (the device to be controlled). In PID (Proportional, Integral, and Differential) control system, the error is amplified (triangular symbol on the diagram), integrated, and differentiated; the sum of all three parts is further amplified and fed into an actuator (such as motor, heater, cooler, lighting, etc.) that takes action on the Plant. The output of a sensor installed in the Plant is fed back to the input of the control system. All components of such system can be implemented with Op Amps. The necessary blocks include: Integrator and Differentiator, Amplifiers (Inverting, Non-Inverting, and Difference), and the Buffer, or Voltage Follower: see the diagrams below. EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 4 of 6 Problem 2 Part 2, continued The output of the Difference Amplifier shown on this diagram equals: V O U T V IN , 1 V IN , 2 R FR A Consider the prototype of a PID control circuit shown below. Assume that the Actuator signal should be: A c t u a t o r S i g n a l 0 . 5 E r r o r dd t E r r o r E r r o r d t where Error is the output of the Difference Amplifier. For simplicity, assume RC =1. EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 5 of 6 Answer the following questions: EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 6 of 6 Problem 2 Part 2, continued 1. Given R2 = 100 k, determine R1 in k. Your answer: R1 = _____200__________ k. From the actuator signal relation we see that the proportional part of the PID controller has a factor of 0.5. This can be achieved only when the gain of the inverting amplifier that is part of the P section has a gain of 0.5. Gain = -R2/R1 = 0.5; therefore R1 = 200 kΩ. 2. Note that the diagram includes an inverting amplifier for the error signal. Is an additional inverting amplifier needed before the adder? Your answer: Yes No (circle one) If you answered Yes, calculate R1 in k for the additional inverting amplifier (assume R2 = 100 k, as above). Your answer: R1 = ______NA__________ k. Inverter is not necessary here. The polarity of the signal output from the voltage follower stages is negative. This results in the adder output voltage being positive. The input or the outcome does not in any way affect the functionality of the circuit. 3. Is an additional inverting amplifier needed after the adder? Your answer: Yes No (circle one) If you answered Yes, calculate R1 in k for the additional inverting amplifier (assume R2 = 100 k, as above). Your answer: R1 = _______N/A________ k. Justify your answers. Show your work below and/or on additional pages. The signal input from the PID section has to be subtracted from the setpoint value in order to get the error. Thus an inverter is not required as the difference amplifier stage requires positive voltage inputs in order to function as a subtractor. Comment. An inverter with R1 = R2 multiplies the signal by (– 1); thus putting two inverters in series produces multiplication by (– 1)(– 1) = +1, which is unnecessary. Your design should avoid unnecessary inverters. EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 1 of 3 Problem 3 (30 points) Active filters based on Op Amps In both parts, apply the “Golden Rules” Part 1 (10 points) Determine the type of filter (LP, HP, BP, or BR) for the circuit shown here. Your answer: High-Pass Show your work on a separate page: (A) Redraw the circuit at a very low frequency 0 and at a very high frequency (B) On each redrawn circuit, show the capacitor and inductor as an open/closed switch With small ω, inductors act as short circuits, and capacitors act as open circuits. With large ω, inductors act as open circuits, and capacitors act as short circuits. (C) Write the algebraic expression for the transfer function magnitude at 0 and at Large ω Small ω 𝑅3𝑅4 𝑅3+𝑅4 ∗ 1 𝑅2 = 𝑅3𝑅4 𝑅2(𝑅3+𝑅4) Small ω This is an inverting op-amp with Rf = R3||R4 and Ri = R2. Thus, the transfer function is 𝑅4 ∗𝑅1 + 𝑅2𝑅 1𝑅2 = 𝑅4(𝑅1 + 𝑅2)𝑅 1𝑅2 Large ω This is an inverting op-amp with Rf = R4 and Ri = R2||R1. Thus, the transfer function is We see that a R4/R2 is present in each term. Because R3/(R3+R4) < 1 and (R1+R2)/R1 > 1, the transfer function is larger for large ω. As a result, we are working with a high-pass filter. EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 2 of 3 Problem 3 Part 2 (20 points) For the circuit shown here, do the following: (A) Determine the type of filter. Your answer: Low-Pass (B) Derive the algebraic expression for the transfer function H() VOUT ()V S () . (C) Derive the algebraic expression for the cutoff frequency C max( 𝐻(𝑗𝜔) ) 2 = 𝑅2 + 𝑅𝑋 𝑅1 2 = 𝐻(𝑗𝜔𝐶) = 𝑅2 + 𝑅𝑋 𝑅12 + 𝜔𝐶2𝐶2𝑅12 𝑅2 + 𝑅𝑋 2 1 2 = 1 1 + 𝜔𝐶2𝐶2 𝑅2 + 𝑅𝑋 2 1 + 𝜔𝐶2𝐶2 𝑅2 + 𝑅𝑥 2 = 2 𝝎𝑪 = 𝟏𝑪(𝑹 𝟐 + 𝑹𝒙) (C) Derive the algebraic expression for zero-frequency gain for any value of RX H ( 0 ) V O U T ( 0 )V S ( 0 ) When ω=0, the imaginary term of the denominator drops out. Your answer: 𝑹𝟐+𝑹𝑿𝑹 𝟏 (D) Derive the algebraic expression for the product of zero-frequency gain and the cutoff frequency C (called the “Gain Bandwidth” product) for any value of RX Your answer: 1 𝐶(𝑅2 + 𝑅𝑥) ∗ 𝑅2 + 𝑅𝑋 𝑅1 = 𝟏 𝑪𝑹𝟏 Does it depend on the value of RX ? Yes No (circle one) (E) Use the following circuit parameters: R1 = R2 = 1 k; RP = 100 k; C = 100 nF. Given that VS is a sine wave at 100 mV peak amplitude, calculate the peak 𝐻 𝑗𝜔 = −𝑅𝑓𝑅 𝑖 = −(𝑅2 + 𝑅𝑋)||𝐶𝑅 1 = − (𝑅2 + 𝑅𝑋) 1𝑗𝜔𝐶 𝑅1(𝑅2 + 𝑅𝑋 + 1𝑗𝜔𝐶) = − 𝑹𝟐 + 𝑹𝑿𝑹 𝟏 + 𝒋𝝎𝑪𝑹𝟏(𝑹𝟐 + 𝑹𝑿) EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 3 of 3 amplitude of the output sine wave at the maximal and minimal values of RX for two frequencies: 200 Hz and 5 kHz. Compare with the circuit, from which the capacitor is disconnected. Write the results in the table below: Input signal frequency Peak amplitude (in volts) of the output in the given circuit Peak amplitude (in volts) of the output in the circuit without C RX = 0 RX = RP RX = 0 RX = RP 200 Hz 99.2 mV 793 mV 100 mV 10.1 V 5 kHz 30.3 mV 31.8 mV 100 mV 10.1 V On a separate page, show your work for all parts of this problem. - Row 1, Column 1: |𝐻 𝑗2𝜋 ∗ 200 | = | 1000 + 01000 + 𝑗 2𝜋 ∗ 200 100 ∗ 10−9 1000 1000 + 0 | = 0.992 Peak output = |H(ω)|*VIN = 0.992 * 100 mV = 99.2 mV - Row 1, Column 2: |𝐻 𝑗2𝜋 ∗ 200 | = | 1000 + 100 ∗ 10 3 1000 + 𝑗 2𝜋 ∗ 200 100 ∗ 10−9 1000 1000 + 100 ∗ 103 | = 7.93 Peak output = |H(ω)|*VIN = 7.93 * 100 mV = 793 mV - Row 1, Column 3: |𝐻 𝑗2𝜋 ∗ 200 | = | 10001000 | = 1.00 Peak output = |H(ω)|*VIN = 1.00 * 100 mV = 100 mV - Row 1, Column 4: |𝐻 𝑗2𝜋 ∗ 200 | = | 1000 + 100 ∗ 10 3 1000 | = 101 Peak output = |H(ω)|*VIN = 101 * 100 mV = 10.1 V - Row 2, Column 1: |𝐻 𝑗2𝜋 ∗ 5000 | = | 1000 + 01000 + 𝑗 2𝜋 ∗ 5000 100 ∗ 10−9 1000 1000 + 0 | = 0.303 Peak output = |H(ω)|*VIN = 0.303 * 100 mV = 30.3 mV - Repeat this pattern for the last three entries. EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 1 of 2 Problem 4 (30 points) Control circuits with Comparator, MOSFET, and the Wheatstone bridge: Position sensor Part 1 (10 points) Potentiometer as a sensor for linear displacement The linear potentiometer used is used to measure the amount of liquid in a cylindrical tank. This is similar to how your fuel gage measures how much gas you have in the tank of your car. Assume the following parameters: the cross-section area of the tank equals S = 800 cm2; the length of the linear potentiometer (from A to B) equals L = 12 cm; the middle position of the potentiometer’s tap corresponds to the liquid level equal to H = 7 cm; the total resistance of the potentiometer is RP = 240 k; the source voltage is VS = 14 V. Evidently, the circuit’s readings are accurate only if the volume of liquid in the tank P remains within certain limits: if the float moves too low, the output voltage equals zero regardless of the amount of liquid; if the float moves too high, the output voltage saturates at VS regardless of the amount of liquid. Obtain and record the algebraic expressions for PMIN and PMAX in terms of S, L, and H. Neglect the volume of the float. Your answers: PMIN =𝑺(𝑯− 𝑳𝟐) PMAX = 𝑺(𝑯 + 𝑳𝟐) Calculate and record below the minimal and maximal volumes of liquid in the tank PMIN and PMAX in liters (1 liter = 1,000 cm3) that can be accurately measured with this circuit. Your answers: PMIN =0.8 L PMAX = 10.4 L Show your work below and/or on additional pages. Volume = Surface Area · Depth, so we just need to find the minimum and maximum depths. We know that H = 7cm corresponds to the center point of the potentiometer, 6cm. Thus, the minimum potentiometer position will occur at H – L/2 = 1cm, and the maximum will occur at H + L/2 = 13cm. PMIN = 800cm2·1cm = 800cm3=0.8L. PMAX = 800cm2·13cm = 1040cm3=10.4L. EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 2010 Alexander Ganago Page 2 of 2 Problem 4 Part 2 (10 points) Operation of the control circuit Assume that the potentiometer as the liquid level sensor, which you studied in Part 1 of this problem, is connected to the circuit shown here: RX is the resistance between A and B; voltage VOUT of the diagram in Part 1 is labeled V1 on the diagram in Part 2. The circuit must keep the level of liquid in the tank below the safe limit. Determine what kind of Actuator should be used: a) Pump that adds liquid to the tank, or b) Valve that blocks the flow of liquid into the tank. Your answer: a b (circle one) On a separate page, explain how the circuit operates and justify your answer. In this circuit, the op-amp acts as a comparator. When the water is low, RX is also low. As a result, V1 > V2, meaning that VOUT = VOUT, MAX and the actuator is turned on. We want to fill the tank, so the “on” state of the actuator should correspond to when a pump is pumping liquid into the tank. When the water reaches the safe limit, RX will be larger and we have V1 < V2. As a result, the pump will turn off, and the water in the tank will remain safe. Part 3 (10 points) Use the circuit parameters and tank sizes given in Part 1 and design the control circuit that limits the amount of liquid in the tank at exactly 10 liters. Assume that R3 = 20 k, calculate R4 in k. Your answer: R4 = 460 k 𝑥 = 10 𝐿0.8 𝑐𝑚 = 12.5 𝑐𝑚 𝑥 − 1𝐿 = 𝑅𝑋𝑅 𝑃 = 11.512 = 0.958 𝑊𝑖𝑡ℎ 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛,𝑅𝑋𝑅 𝑃 = 𝑅3𝑅 3 + 𝑅4 = 0.958. 𝐾𝑛𝑜𝑤𝑖𝑛𝑔 𝑅3 = 20 𝑘𝛺,𝑅4 = 𝟒𝟔𝟎 𝒌𝜴 EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 5 (30 points) Resistive sensors: Temperature controller circuit; Real sensors’ parameters In all parts, use “Golden Rule” #1. Part 1 (10 points) The four circuits shown above are built to maintain the temperature within the desired limits, that is turn on a heater when it gets too cold. We will call the circuit that operates this way a “good” one. Two types of sensors can be used as R T : a) Thermistor with Negative Temperature Coefficient (NTC), and b) Temperature-Dependent Resistor (TDR) with Positive Temperature Coefficient (PTC). For each of type of sensor, briefly explain (on a separate page) which of the circuits on diagrams 1 – 4 is “good”. Explain why each of the “bad” circuits is “bad.” © 2010 Alexander Ganago Page 1 of 5 EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Notes • R T can be PTC or NTC - For PTC, as temperature increases resistance increases (T↑ R↑) - For NTC, as temperature increases resistance decreases (T↑ R↓) • Output of comparator goes to V S1 when V + > V - or V 1 > V 2 • Want heater to be turned on when temperature gets too low • Heater is turned on when V GS > V T of MOSFET, this occurs when output of comparator which equals V GS goes to V S1 • Heater is on when V + > V - or V 1 > V 2 • Heater is off when V + < V - or V 1 < V 2 Circuit 1 V 2 is constant and V 1 ↑ as R T ↑ Assume heater off, V 1 < V 2 So as T↓ want V 1 ↑ or R T ↑ NTC R T would be good choice for this PTC R T would have V 1 ↓ as T↓ and heater won’t turn on as T↓, bad choice Circuit 2 V 2 is constant and V 1 ↑ as R T ↓ Assume heater off, V 1 < V 2 So as T↓ want V 1 ↑ or R T ↓ PTC R T would be good choice for this NTC R T would have V 1 ↓ as T↓ and heater won’t turn on as T↓, bad choice © 2010 Alexander Ganago Page 2 of 5 EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Circuit 3 V 1 is constant and V 2 ↓ as R T ↑ Assume heater off, V 2 > V 1 So as T↓ want V 2 ↓ or R T ↑ NTC R T would be good choice for this PTC R T would have V 2 ↑ as T↓ and heater won’t turn on as T↓, bad choice Circuit 4 V 1 is constant and V 2 ↓ as R T ↓ Assume heater off, V 2 > V 1 So as T↓ want V 2 ↓ or R T ↓ PTC R T would be good choice for this NTC R T would have V 2 ↑ as T↓ and heater won’t turn on as T↓, bad choice Summary Circuit 1 Circuit 2 Circuit 3 Circuit 4 NTC R T Good Bad Good Bad PTC R T Bad Good Bad Good © 2010 Alexander Ganago Page 3 of 5 EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 5 In Parts 2 and 3, consider the circuits listed below regardless of whether you found them “good” in Part 1. Part 2 (10 points) Consider circuit diagram 3 where R T is a thermistor with 1000 Ω resistance at 25 ºC (see US Sensor catalog, page 40, attached), R A = 1 kΩ, and R SET = 2 kΩ. Calculate the resistance R 2 to ensure that the actuator is turned on/off at – 20 ˚C. Show your work. Actuator is turned on/off when inputs of comparator are equal or V 1 = V 2 From voltage division, 2 21 2 S T R VV R R = + 11 A S ASET R VV RR = + Setting V 1 = V 2 yields: 2 2 A TASET RR R RRR = ++ 2 11 1 1 TSET A RR R R = + + 2 SETT A RR R R = 2 A TSET R R RR = From the given resistances R A = 1 kΩ, R SET = 2 kΩ, 2 11 22 A TSET RR k RR k === 2 1 2 T R R= Looking up thermistor’s resistance from table for -20ºC, R T = 9708 Ω. So, 2 1 (9708) 48 4 5 2 R ==Ω © 2010 Alexander Ganago Page 4 of 5 EECS 314 Winter 2010 Homework set 8 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Part 3 (10 points) Consider circuit diagram 4 where R T is a TDR with 1000 Ω resistance at 0 ºC (see Vishay Beyschlag catalog, attached), R 2 = 2.2 kΩ, and R SET = 3.3 kΩ. Calculate the resistance R A to ensure that the actuator is turned on/off at – 20 ˚C. Show your work. Actuator is turned on/off when inputs of comparator are equal or V 1 = V 2 From voltage division, 21 2 T S T R VV R R = + 11 A S ASET R VV RR = + Setting V 1 = V 2 yields: 2 TA SET RR R RRR = ++ 2 11 1 1 SET T A RR R R = + + 2 SET TA RR R R = 2 SETA T RR R R = From the given resistances R 2 = 2.2 kΩ, R SET = 3.3 kΩ, 2 3.3 3 2.2 2 SETA T RRk RR k === 3 2 AT R R= Looking up TDR’s resistance from data sheet for -20ºC, R T = 920 Ω. So, 3 (920) 138 0 2 A R ==Ω © 2010 Alexander Ganago Page 5 of 5 Alexander O. Ganago

"StudyBlue is great for studying. I love the study guides, flashcards and quizzes. So extremely helpful for all of my classes!"

Alice , Arizona State University"I'm a student using StudyBlue, and I can 100% say that it helps me so much. Study materials for almost every subject in school are available in StudyBlue. It is so helpful for my education!"

Tim , University of Florida"StudyBlue provides way more features than other studying apps, and thus allows me to learn very quickly!??I actually feel much more comfortable taking my exams after I study with this app. It's amazing!"

Jennifer , Rutgers University"I love flashcards but carrying around physical flashcards is cumbersome and simply outdated. StudyBlue is exactly what I was looking for!"

Justin , LSU