Unformatted text preview: EECS 314 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 Winter 2011 Homework set 3 Problem 1 (20 points) Resistance and resistivity / Solution A friend of yours bought new 4‐Ω speakers for his audio system and asked you to determine how far from the output amplifier he could put them using No. 20 AWG aluminum wire. As a knowledgeable EECS 314 student, you solve the problem assuming that the wire resistance should be kept below 5% of the speaker resistance so that at least 95% of the output power reaches the speaker. Part 1 (5 points) Draw the equivalent circuit diagram of the setup (in each channel): show the output amplifier as a voltage source VS; the speaker as a load resistor RLOAD; and the connecting wires as additional resistances RWIRE. Your circuit diagram: Part 2 (15 points) Calculate the maximal distance between each speaker and the amplifier, in meters. L Provide two answers: (1) based on your calculations using the equation R = ! " A and (2) based on the resistance per km data listed in the Table (see the Table on next page); and discuss whether these answers agree. Note that the table shows resistance per km for copper, not aluminum. Show your work on the other side. Write your answers below: Maximal distance, meters Based on calculations Based on the Table date © 2011 Alexander Ganago Last printed 1/20/11 6:00 PM Page 1 of 2 File: 2011 W 314 HW 3 p1s.doc EECS 314 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 Winter 2011 Homework set 3 Table AWG gauge 0000 0 20 24 30 36 Conductor diameter, inches 0.46 0.3249 0.032 0.0201 0.01 0.005 Conductor Ohms per diameter, mm km (copper) 11.684 0.16072 8.25246 0.322424 0.8128 33.292 0.51054 84.1976 0.254 338.496 0.127 1360 Maximum amps (copper) for power transmission (conservative rule) 302 150 1.5 0.577 0.142 0.035 Solution Assuming the wire resistance equal 5% of the speaker resistance (4 Ω), we find the maximal wire resistance of 200 mΩ. From equation [2‐6], substituting the wire diameter from table 2‐1, we determine the maximal length of wire: 2 $ ! 8.128 ! 10 %4 ! m A L = R ! = ( 0.2 ! # ) ! 4 = 3.68 ! m " 2.82 ! 10 8 # !!m ( ( ) ) Thus the maximal distance between the speaker and the amplifier is 1.84 m (~6 ft). Look up Table 2‐1 and see that No. 20 AWG copper wire has resistance of 33.292 Ω per kilometer, or 33.292 mΩ per meter. Since we have already found that the maximal wire resistance equals 200 mΩ, we calculate that the maximal wire length is: 200 ! m! 1.72 " 10 #8 ! " m " = 3.66 ! m 33.292 m! /m 2.82 " 10 #8 ! " m This result is in excellent agreement with the maximal wire length that we determined above from equation [2‐6]. Answer: The distance equals 1.84 m (~6 ft) max. © 2011 Alexander Ganago Last printed 1/20/11 6:00 PM Page 2 of 2 File: 2011 W 314 HW 3 p1s.doc EECS 314 Winter 2011 Homework set 3 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 2 (30 points) Potentiometer used as a part of a liquid level sensor
The diagram below shows a 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 crosssection area of the tank equals S = 720 cm2; the length of the linear potentiometer (from A to B) equals L = 24 cm; the middle position of the potentiometer’s tap corresponds to the liquid level equal to H = 15 cm; the total resistance of the potentiometer is RP = 120 kΩ; the source voltage is VS = 14 V. Part 1 (10 points)
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. PMIN = [H – (L/2)] * S Solution: The float is constrained in movement by the wiper of the potentiometer. As the wiper can only move by a distance of L/2 to +L/2 beyond the middle position of potentiometer’s tap so minimum and maximum liquid levels are: © 2011 Alexander Ganago Last printed 1/24/2011 11:02:00 PM Page 1 of 5 File: 2011 W 314 HW 3 p2.doc PMAX = [H + (L/2)] * S EECS 314 Winter 2011 Homework set 3 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 HMIN = (H – L/2) and HMAX = (H + L/2) Thus, PMIN = HMIN * S = (H – L/2) * S and PMAX = HMAX * S = (H + L/2) * S 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. PMIN = 2.16 liters PMAX = 19.44 liters Show your work on additional pages. 1 liter = 1000cm3 Solution: PMIN = (H – L/2) * S PMIN = [(15 – 12) * 720] cm3 PMIN = 2.16 liters PMAX = (H + L/2) * S PMAX = [(15 + 12) * 720] cm3 PMAX = 19.44 liters Problem 2, continued Part 2 (10 points)
The output voltage can be expressed as VOUT = K ⋅ VS. When the volume of liquid in the tank varies between PMIN and PMAX , the coefficient K varies between 0 and 1. Derive and record the algebraic expression for K in terms of S, L, H, RP, and VS. K= Solution: Let ‘h’ be the height of liquid at any given instance of time. Also ‘h’ is constrained such that (PMIN/S <= h <= PMAX/S) VOUT = K*VS © 2011 Alexander Ganago Last printed 1/24/2011 11:02:00 PM Page 2 of 5 File: 2011 W 314 HW 3 p2.doc EECS 314 Winter 2011 Homework set 3 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 From the voltage divider concept we know that system can be viewed as combination of two series resistances (R1 and R2) where the potentiometer wiper sets the relative values of R1 and R2. Thus, K = R2 / RP  eq. 1 Also let ‘x’ be displacement of the wiper measured from the point ‘A’. The resistance R2 can be expressed in terms of ‘x’ and RP using a linear relation: R2 = (x/L) * RP h (height of liquid) = [H – (L/2)] + x From eq.1, eq.2 and eq.3 we have: h = [H – (L/2)] + K*L so, K = (1/L) * [h – H + (L/2)] where (PMIN/S <= h <= PMAX/S)  eq. 4
 eq. 2 At any time the height of liquid can be given by the relation:  eq. 3 © 2011 Alexander Ganago Last printed 1/24/2011 11:02:00 PM Page 3 of 5 File: 2011 W 314 HW 3 p2.doc EECS 314 Winter 2011 Homework set 3 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 Use the given numerical values to calculate the volume of liquid in the tank (in liters) P¾ that corresponds to VOUT = ¾ VS. P¾ = _________________________ Solution: Using eq. 4 we can substitute the value of K = ¾ to find ‘h’ Now, K = (1/L) * [h – H + (L/2)] or h = (K * L) + [H – (L/2)] Therefore, h = 21 cm Volume of liquid in tank: P¾ = S*h = 720*21 = 15120 cm3 P¾ = 15.12 liters  eq. 5 Part 3 (10 points)
Assume that the error of the voltage readout is ∆VOUT = 10 mV. Calculate the error of the readout in liters ∆P. ∆P = __________________________ Solution: P = S*h Also from eq. 5 above, where, so, then, h = (K * L) + [H – (L/2)] K = VOUT / VS P = (VOUT / VS)*L*S + [H – (L/2)]*S ∆P = (∆VOUT / VS)*L*S = (10mV / 14V) * 24 *720 cm3 ∆P = 12.342 * 103 liters © 2011 Alexander Ganago Last printed 1/24/2011 11:02:00 PM Page 4 of 5 File: 2011 W 314 HW 3 p2.doc EECS 314 Winter 2011 Homework set 3 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 Consider the list of measures for reducing the error ∆P (each measure assumes that only one parameter is altered, while all the rest remain as listed above). Circle the good ones. Increase RP Increase L Increase VS Increase H Increase S Decrease RP Decrease L Decrease VS Decrease H Decrease S © 2011 Alexander Ganago Last printed 1/24/2011 11:02:00 PM Page 5 of 5 File: 2011 W 314 HW 3 p2.doc EECS 314 Winter 2011 Homework set 3 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 3 (40 points) Resistive temperature sensors in voltage dividers In this and the following problem, refer to the files “US sensor catalog p.40” and “Vishay Beyschlag catalog” on CTools. Part 1 (15 points) Consider circuit 1 with VS = 12 V, R1 = 1,500 Ω, and RSENSOR = 1,000 Ω at 0˚C – a temperature‐dependent resistor made by Vishay Beyschlag. Make a computer‐ generated plot of the output voltage VOUT, 1 as function of temperature from –40 ˚C to +140 ˚C. For simplicity assume that RSENSOR (T ) = RSENSOR (T = 0) ⋅ (1 + 0.004 ⋅ T ) , where T is temperature in ˚C, and RSENSOR (T = 0)= 1,000 Ω. Given that the error of voltage measurements equals 10 mV, estimate the error of temperature measurement in ˚C at –40 ˚C and at +140 ˚C. Your answers: Error at –40 ˚C equals ______0.75___________ ˚C; Error at +140 ˚C equals ____1.302____ ˚C. Show your work below and/or on additional page. File: 2011 W 314 HW 3 p3.doc © 2011 Alexander Ganago Page 1 of 9 Last printed 1/25/2011 12:12:00 PM EECS 314 Winter 2011 Homework set 3 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 SOLUTION: Circuit 1 is a voltage divider, and one can fin Vout by using KVL as ; Vout= (Vs*Rsensor)/(R1+Rsensor) Since Rsensor RSENSOR (T ) = RSENSOR (T = 0) ⋅ (1 + 0.004 ⋅ T ) Then Vout = f(t). Just plug Rsensor in to the equation for Vout. Vout= (12* RSENSOR (T ) = RSENSOR (T = 0) ⋅ (1 + 0.004 ⋅ T ) )/(1500+ RSENSOR (T ) = RSENSOR (T = 0) ⋅ (1 + 0.004 ⋅ T ) ) Plot Vout vs T © 2011 Alexander Ganago Page 2 of 9 Last printed 1/25/2011 12:12:00 PM File: 2011 W 314 HW 3 p3.doc EECS 314 Winter 2011 Homework set 3 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 6.2 6 5.8 5.6 5.4 V ou t (V) 5.2 5 4.8 4.6 4.4 4.2 40 20 0 20 40 Temp 60 80 100 120 140 ERROR CALCULATION; To find the error, Vout = Vout= (12* RSENSOR (T ) = RSENSOR (T = 0) ⋅ (1 + 0.004 ⋅ T ) )/(1500+ RSENSOR (T ) = RSENSOR (T = 0) ⋅ (1 + 0.004 ⋅ T ) ) Vout=(12000+48*T)/(2500+4*T) And T= (2500*Vout‐12000)/(48‐4Vout) © 2011 Alexander Ganago Page 3 of 9 Last printed 1/25/2011 12:12:00 PM File: 2011 W 314 HW 3 p3.doc EECS 314 Winter 2011 Homework set 3 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 At T= ‐40 C By using the following ratio calculation, one can roughly find the amount of error in temperature which corresponds to 10 mV error in voltage Vout=(12000+48*T)/(2500+4*T) so T1=‐40 C Vout1= 4.307 V T2= ? Vout2 =4.307 V +10mV= 4.317 V By using the following equation T= (2500*Vout‐12000)/(48‐4Vout) Here Vout= Vout2=4.317 V then T2= ‐39.2455 Error = DeltaT=T1‐T2= 40‐39.2455= 0.75 C At T= 140 C By using the same method above, T1=140 C Vout1= 6.117 V T2= ? Vout2 =6.117V +10mV= 6.127 V By using the following equation T= (2500*Vout‐12000)/(48‐4Vout) Here Vout= Vout2=6.127 then T2= 141.302 C DeltaT=T1‐T2=1.302 C © 2011 Alexander Ganago Page 4 of 9 Last printed 1/25/2011 12:12:00 PM File: 2011 W 314 HW 3 p3.doc EECS 314 Winter 2011 Homework set 3 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 3 Part 2 (15 points) Consider circuit 2 (see page 1) with VS = 12 V, R1 = 1,500 Ω, and RSENSOR = 1,000 Ω at 25˚C – a thermistor made by US Sensor. Make a computer‐generated plot of the output voltage VOUT, 1 as function of temperature from –40 ˚C to +140 ˚C. Use the data from the table (a point every 10 ˚C is enough). Given that the error of voltage measurements equals 10 mV, estimate the error of temperature measurement in ˚C at –40 ˚C and at +140 ˚C. Your answers: Error at –40 ˚C equals _________0.23________ ˚C; Error at +140 ˚C equals ____2____ ˚C. Show your work below and/or on additional page. SOLUTION: Circuit is copied here for your convenience; This time Rsensor ‘s position is different (circuit 2). Also, for the Rsensor, we have data set instead of an analytical formula of it. Vout= (12*1500)/(1500+Rsensor) © 2011 Alexander Ganago Page 5 of 9 Last printed 1/25/2011 12:12:00 PM File: 2011 W 314 HW 3 p3.doc EECS 314 Winter 2011 Homework set 3 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 Resistance values are taken from ‐40 C to 140 C at each 10 C interval. Resistor values at these points are give as following. 33649 17697 9708 5533 3265 1990 1249 805.7 532.6 360.2 248.8 175.2 125.6 91.62 67.86 51.02 38.94 30.10 23.54 By using Vout= (12*1500)/(1500+Rsensor) One can plot Vot vs temperature as following; © 2011 Alexander Ganago Page 6 of 9 Last printed 1/25/2011 12:12:00 PM File: 2011 W 314 HW 3 p3.doc EECS 314 Winter 2011 Homework set 3 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 12 10 8 Vout (V) 6 4 2 0 40 20 0 20 40 T 60 80 100 120 140 Since this time there is not an analytical expression for Rsensor, but a set of data at different temperatures, I do the error calculation in the following way. ERROR at ‐40 C Take two points around ‐40 C. ‐40 C and ‐30 C . At these points, corresponding voltages are 0.5121 V and 0.9376 V © 2011 Alexander Ganago Page 7 of 9 Last printed 1/25/2011 12:12:00 PM File: 2011 W 314 HW 3 p3.doc EECS 314 Winter 2011 Homework set 3 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 DeltaT1=‐40+30=10 C DeltaV1= 0.9376‐0.5121= 0.4255 V DeltaT2=? Delta V2= 10 mV =0.01 V DeltaT2= (0.01*10)/(0.4255)= 0.23 C ERROR at 140 C At 140 C and 130 C. The corresponding voltages are 11.7639 V and 11.8146 V DeltaT1=140‐130=10 C DeltaV1= 11.8146‐11.7639= 0.05 V DeltaT2=? Delta V2= 10 mV =0.01 V DeltaT2= (0.01*10)/(0.05)= 2 C Part 3 (10 points) Briefly compare the temperature measurements with the two sensors (use your results obtained in Parts 1 and 2). Answer the following questions: Which sensor provides smaller errors in temperature measurements? Sensor 1 (Vishay Beyschlag) provides smaller errors Does any of these sensors provide nearly the same error over the entire temperature range? No Which sensor would you prefer in your project, and why? I would prefer sensor 1 (Vishay Beyschlag), because it provides smaller errors and it is more linear to the temperature © 2011 Alexander Ganago Page 8 of 9 Last printed 1/25/2011 12:12:00 PM File: 2011 W 314 HW 3 p3.doc EECS 314 Winter 2011 Homework set 3 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 © 2011 Alexander Ganago Page 9 of 9 Last printed 1/25/2011 12:12:00 PM File: 2011 W 314 HW 3 p3.doc EECS 314 Winter 2011 Homework set 3 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 4 (40 points) Resistive temperature sensors in voltage dividers In this problem, refer to the files “US sensor catalog p.40” and “Vishay Beyschlag catalog” on CTools. Take advantage of your results in Problem 3 of this set. Consider circuit 3 with VS1 = 12 V, R2 = 1000 Ω, RSET = 1,500 Ω, RA = 1,200 Ω, and RT = 1,000 Ω at 0˚C – a temperature‐ dependent resistor made by Vishay Beyschlag. Make a computer‐generated plot of the output voltage VOUT, 1 as function of temperature from –40 ˚C to +140 ˚C. For simplicity assume that RSENSOR (T ) = RSENSOR (T = 0) ⋅ (1 + 0.004 ⋅ T ) , where T is temperature in ˚C, and RSENSOR (T = 0)= 1,000 Ω. Given that the error of voltage measurements equals 10 mV, estimate the error of temperature measurement in ˚C at –40 ˚C and at +140 ˚C. Your answers: Error at –40 ˚C equals _________0.709________ ˚C; Error at +140 ˚C equals ____1.369____ ˚C. Show your work below and/or on additional page. SOLUTION: Vout= V2‐V1= (12*1000)/(Rt+1000) ‐ (12*1200)/(1200+1500) Vout=(12000/(Rt+1000)) – 5.3333 And also RSENSOR (T ) = RSENSOR (T = 0) ⋅ (1 + 0.004 ⋅ T ) © 2011 Alexander Ganago Page 1 of 6 Last printed 1/25/2011 12:13:00 PM File: 2011 W 314 HW 3 p4.doc Part 1 (15 points) EECS 314 Winter 2011 Homework set 3 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 So, one can plot Vout vs temperature as following; 1.2 1 0.8 0.6 0.4 Vout (V) 0.2 0 0.2 0.4 0.6 0.8 40 © 2011 Alexander Ganago Page 2 of 6 Last printed 1/25/2011 12:13:00 PM File: 2011 W 314 HW 3 p4.doc 20 0 20 40 T 60 80 100 120 140 EECS 314 Winter 2011 Homework set 3 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 ERROR at ‐40 C; I can follow the previous methods here Just take two points around ‐40 C, and I can take then with 1 C difference since Rsensor has its analytical expression ‐40 C and‐39 C . Corresponding voltage values are 1.1884 V and 1.1743 V DeltaT1= 1 C DeltaV1= 1.1884‐1.1743 = 0.0141V DeltaT2=? DeltaV2= 10mV = 0.01 V DeltaT2 = (0.01*1)/(0.0141)=0.709 C. ERROR at 140 C; DeltaT1=140‐139 C = 1 C Delta V1= ‐0.6385+0.6458=0.0073 DeltaT2=? DeltaV1=0.01 V DeltaT2= 1.369 C © 2011 Alexander Ganago Page 3 of 6 Last printed 1/25/2011 12:13:00 PM File: 2011 W 314 HW 3 p4.doc EECS 314 Winter 2011 Homework set 3 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 3 Part 2 (15 points) Consider circuit 4 with VS = 12 V, R2 = 1,000 Ω, RSET = 1,500 Ω, RA = 1,200 Ω, and RT = 1,000 Ω at 25˚C – a thermistor made by US Sensor. Make a computer‐generated plot of the output voltage VOUT, 1 as function of temperature from –40 ˚C to +140 ˚C. Use the data from the table (a point every 10 ˚C is enough). Given that the error of voltage measurements equals 10 mV, estimate the error of temperature measurement in ˚C at –40 ˚C and at +140 ˚C. Your answers: Error at –40 ˚C equals _________0.338 ________ ˚C; Error at +140 ˚C equals ____1,3333____ ˚C. Show your work below and/or on additional page. SOLUTION; Vout=V2‐V1=(Rt*12)/(Rt+1000) ‐ (1200*12)/(1200+1500) Vout=(Rt*12)/(Rt+1000)‐5.3333 This time resistor values are discrete, and values are taken at each 10 C from ‐40 C to +140 C. © 2011 Alexander Ganago Page 4 of 6 Last printed 1/25/2011 12:13:00 PM File: 2011 W 314 HW 3 p4.doc EECS 314 Winter 2011 Homework set 3 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 8 6 4 Voltage (V) 2 0 2 4 6 40 20 0 20 40 T 60 80 100 120 140 ERROR at ‐40 C; At ‐40 C and ‐30 C, corresponding voltage values are 6.3203 V and So © 2011 Alexander Ganago Page 5 of 6 Last printed 1/25/2011 12:13:00 PM 6.0248 V. File: 2011 W 314 HW 3 p4.doc EECS 314 Winter 2011 Homework set 3 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 DeltaT1= 10 C DeltaV1=6.3203‐6.0248= 0.295 V DeltaT2=? DeltaV2=10mV = 0.01 V DeltaT2= (0.01*10)/(0.295)=0.338 C ERROR at 140 C; At 130 and 140 C, correspondingoutput voltages are ‐4.982 V and ‐5.057 V DeltaT1=10 C DeltaV1=‐4.982+5.057= 0.075 V DeltaT2 =? DeltaV2= 10 mV = 0.01 V Delta T2 = 0.01*10/( 0.075) = 1.33333 C Part 3 (10 points) Briefly compare the temperature measurements with the two sensors (use your results obtained in Parts 1 and 2). Answer the following questions: Which sensor circuit provides smaller errors in temperature measurements? Both sensors errors are on the same order at ‐40 C and 140 C temperatures. But, sensor 1 is more linear dependency to temperature than sensor 2 (Look at the Vout vs T graphs for both sensors). Does any of these sensor circuits provide nearly the same error over the entire temperature range? No Which sensor circuit would you prefer in your project, and why? I would prefer sensor 1 since it is more linear. © 2011 Alexander Ganago Page 6 of 6 Last printed 1/25/2011 12:13:00 PM File: 2011 W 314 HW 3 p4.doc EECS 314 Winter 2011 Homework set 3 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 (50 points) Measurements of voltage and current Part 1 (10 points) In order to measure the voltage across a resistor, we connect the voltmeter in parallel with this resistor as shown on the diagram for a circuit of 2 identical resistors, with a voltmeter connected to measure the voltage across one of them. Calculate voltage VX measured by the voltmeter in terms of the source voltage VS and resistances R and RV V Also, calculate the ratio of the measured voltage to the original voltage X V
in terms of resistances R and RV Write your result: Show your work. Use additional pages as needed. SOLUTION: = VX 1 VS 2 VX = V (2 ⋅ Rv ⋅ R) (2 ⋅ Rv ⋅ R + R 2 ) Is Vx = Is ⋅ Req and Req = R // Rv = (R ⋅ Rv) (R+Rv) Is=
Req Vs so Vx = Vs *Req/(R + Req) R + Req
Vs ⋅ (R ⋅ Rv) (Rv ⋅ R) (R+Rv) Vx = => Vx =Vs ⋅ [ ] (R ⋅ Rv) (2 ⋅ Rv ⋅ R + R 2 ) R+ (R+Rv) © 2011 Alexander Ganago Last printed 1/29/2011 1:22:00 PM Page 1 of 5 File: 2011 W 314 HW 3 p5.doc EECS 314 Winter 2011 Homework set 3 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 Vx (Vs ⋅ Rv ⋅ R) (2 ⋅ Rv ⋅ R) So = = 2 V [(2 ⋅ Rv ⋅ R + R ) ⋅ Vs/2] (2 ⋅ Rv ⋅ R + R 2 ) Part 2 (10 points) Assume that the voltage in the circuit above is measured with a good voltmeter, which has RV = 500 MΩ, and a so‐so voltmeter, which has RV = 5 MΩ. Calculate the error of measurement, or percentage difference: % Difference = R = 10 Ω, 10 kΩ, 10 MΩ. Keep the sign. Write your results in the table. Good voltmeter 10 Ω 10 kΩ 10 MΩ VX − V ⋅100% for 3 values of the resistances in the circuit: V
So‐so voltmeter 10 Ω 10 kΩ R Error of voltage measurement Problem 5 10 MΩ %1e6 %1e3 %1 %1e4 %0.1 %50 Part 3 (10 points) In order to measure the current through a resistor, we connect the ammeter in series with this resistor as shown on the diagram for a circuit of 2 identical resistors. Calculate current IX measured by the ammeter in terms of the source current IS and resistances R and RA Also, calculate the ratio of the measured current to the I original current X I = IX in terms of resistances R and RA 1 IS 2 Write your result: IX (2 ⋅ R) = (2 ⋅ R+R A ) I
Page 2 of 5 File: 2011 W 314 HW 3 p5.doc Show your work. Use additional pages as needed. © 2011 Alexander Ganago Last printed 1/29/2011 1:22:00 PM EECS 314 Winter 2011 Homework set 3 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 SOLUTION: Let’s find the equivalent resistance of Rtot; Vs (voltage across the current source) =Vx (voltage across each parallel branch) Req= RA+R and Rtot= R//Req= VX R ⋅ Re q = I S R + Re q Vs Req The voltage across the parallel branches are same and equal to : R ⋅ ( R + RA ) R ⋅ Re q Vx = Is ⋅ Rtot = Is ⋅ = Is ⋅ and equal to the R + Re q R + ( R + RA ) voltage across one of the branches like Vx= Ix ⋅ ( R + RA ) . So Is ⋅ R ⋅ ( R + RA ) = Ix ⋅ ( R + RA ) R + ( R + RA ) Ix 2R = I 2 R + RA and Ix R = so Is 2 R + RA Part 4 (10 points) Assume that the current in the circuit above is measured with a good ammeter, which has RA = 0.1 Ω, and a so‐so ammeter, which has RA = 0.5 Ω. Calculate the error of measurement, or percentage difference: % Difference = = 1 Ω, 1 kΩ, 1 MΩ. Keep the sign. IX − I ⋅100% for 3 values of the resistances in the circuit: R I © 2011 Alexander Ganago Last printed 1/29/2011 1:22:00 PM Page 3 of 5 File: 2011 W 314 HW 3 p5.doc EECS 314 Winter 2011 Homework set 3 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 Write your results in the table. Good ammeter 1 Ω 1 kΩ 1 MΩ 1 Ω So‐so ammeter 1 kΩ 1 MΩ R Error of current measurement %4.76 %0.005 %5e6 %20 %0.025 %2.5e5 © 2011 Alexander Ganago Last printed 1/29/2011 1:22:00 PM Page 4 of 5 File: 2011 W 314 HW 3 p5.doc EECS 314 Winter 2011 Homework set 3 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 Part 5 (10 points) In the following text circle the correct adjectives: With the given instruments, measurements of currents are more accurate in circuits with small/large resistances R, while measurements of voltages are more accurate in circuits with small/large R. With the given resistances in the circuit, measurements of voltages are more accurate with voltmeters that have small/large internal resistances RV, while measurements of currents are more accurate in circuits with ammeters that have small/large internal resistances RA. Justify your answers using your results in Parts 1 – 4 of this problem. SOLUTION: From Table in Part 4, we can see that measurements of current yield a much less percentage difference in circuits with large resistances (‐5e‐6 or ‐2.5e‐5) than those with small resistances (‐4.76 or 20). From Table in Part 2, we can see that measurements of voltage yield a much less percentage difference in circuits with small resistances (‐1e‐6 or ‐1e‐4) than those with large resistances (‐1 or ‐50). From Vx (Vs ⋅ Rv ⋅ R) (2 ⋅ Rv ⋅ R) = = we can see that, given the same 2 V [(2 ⋅ Rv ⋅ R + R ) ⋅ Vs/2] (2 ⋅ Rv ⋅ R + R 2 ) R, the Vx/V ratio is higher with higher Rv. For an accurate voltage measurement, we want a high Vx/V ratio, and therefore a high Rv. Ix 2R = we can see that, given the same R, Ix/I has an inverse relationship I 2 R + RA with RA. For an accurate measurement, we want a high Ix/I, therefore, we want a small RA. Page 5 of 5 File: 2011 W 314 HW 3 p5.doc From © 2011 Alexander Ganago Last printed 1/29/2011 1:22:00 PM EECS 314 Winter 2011 Homework set 3 Student’sname ___________________________ 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 6 (40 points) Digital Multimeters (DMMs): Functionality and Accuracy, or Why Would Anyone Pay Double Price? In this problem, you will compare two models of hand‐held DMMs by a well‐known manufacturer. The information provided in the manufacturer’s documents and the conclusions you draw upon comparison serve purely educational purposes: they do not constitute any advertisement, promotion, or recommendation for purchase and specific applications. Refer to the files “Fluke 113 manual” and “Fluke 175 manual” [note that 3 models are covered by this manufacturer’s file] on CTools for comparison between model 113 (current price of this DMM ~ $120) and model 179 (current price ~ $260). Part 1 (20 points) Compare the functionality and specifications of the two models: fill the table below: Model 113 Y ~3K Ohms 6V 1mV 600V Y ~3K Ohms 6V Model 179 Y > 10M Ohms 600mV 0.1mV 1000V Y > 10M Ohms 600mV Allows DC voltage (DCV) measurements: Yes/No Impedance (input resistance) for DCV measurements Minimal range of DCV measurements (low voltages) Resolution of DCV measurements (at low voltages) The highest DCV that can be measured with this DMM Allows AC voltage (ACV) measurements: Yes/No Impedance (input resistance) for ACV measurements Minimal range of ACV measurements (low voltages) © 2011 Alexander Ganago Last printed 1/29/2011 1:22:00 PM Page 1 of 4 File: 2011 W 314 HW 3 p6.doc EECS 314 Winter 2011 Homework set 3 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 The highest ACV that can be measured with this DMM Allows frequency measurements of ACV: Yes/No Frequency range for ACV measurements 600V N N/A Model 113 N N/A N/A N/A N N/A N/A N/A N N/A Y 600 Ohms to 60K Ohms 1000V Y 2 Hz to 99.99k Hz Model 179 Y 60mA 0.01mA 10A Y 60mA 0.01mA 10A Y 2 Hz to 30k Hz Y 600 Ohms to 50M Ohms Problem 6 Part 1, continued Allows DC current measurements (Yes/No) Minimal range of DC current measurements (low currents) Resolution of DC current measurements (at low currents) The highest DC current to be measured with this DMM Allows AC current measurements (Yes/No) Minimal range of AC current measurements (low currents) Resolution of AC current measurements (at low currents) The highest AC current to be measured with this DMM Allows frequency measurements of AC current: Yes/No Frequency range for AC current measurements Allows resistance R measurements: Yes/No The range of R measurements (lowest to highest) © 2011 Alexander Ganago Last printed 1/29/2011 1:22:00 PM Page 2 of 4 File: 2011 W 314 HW 3 p6.doc EECS 314 Winter 2011 Homework set 3 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 Resolution of R measurements (lowest resistances) Resolution of R measurements (highest resistances) Accuracy of R measurements (lowest resistances) 0.1 Ohms 0.01 K Ohms 0.9 % + 2 0.1 Ohms 0.01 M Ohms 0.9 % + 2 1.5% + 3 Y Y Y Accuracy of R measurements (highest resistances) 0.9 % + 1 Allows capacitance measurements: Yes/No Allows continuity measurements: Yes/No Allows diode test measurements: Yes/No Y Y Y Problem 6 Part 2 (10 points) Consider the following applications and write your conclusions on which DMM models are adequate: fill the table below: in columns 2 and 3 write YES or NO. Measurements of DC and AC (60 Hz) voltages from 2 V to 240 V Measurements of DC and AC (60 Hz) voltages from 2 V to 980 V Measurements of DC and AC (60 Hz) voltages from 2 V to 240 V and currents from 0.2 A to 5 A Measurements of DC and AC (60 Hz) voltages from 2 V to 240 V and currents from 2 A to 25 A Measurements of resistances from 100 Ω to 20 kΩ Measurements of resistances from 100 Ω to 200 kΩ Model 113 Y N N N Y N Model 179 Y Y Y N Y Y Part 3 (10 points) © 2011 Alexander Ganago Last printed 1/29/2011 1:22:00 PM Page 3 of 4 File: 2011 W 314 HW 3 p6.doc EECS 314 Winter 2011 Homework set 3 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 In the list below, circle valid justifications for purchase of the more expensive DMM; if needed, refer to Part 1 and the manufacturer’s specs. • More types of measurements are available • Broader range of parameters (V, I, R) can be measured • Better resolution at low values of parameters (V, I, R) • Better resolution at high values of parameters (V, I, R) • Fancier body style of the instrument • Higher accuracy of measurements • Broader range of frequencies • Easier operation (fewer terminals for connections to the circuit) • The more advanced model is significantly lighter than the simple one • Only the more expensive model allows temperature measurements • Only the more expensive model allows diode tests. © 2011 Alexander Ganago Last printed 1/29/2011 1:22:00 PM Page 4 of 4 File: 2011 W 314 HW 3 p6.doc ...
View
Full Document
 Winter '07
 Ganago
 Litre, Alexander Ganago, ___________________________ Discussion section

Click to edit the document details