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plugin-HW2_TOTAL_SOLUTIONS - EECS 314 Winter 2011 Homework...

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Unformatted text preview: EECS 314 Winter 2011 Homework set 2 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 1 (40 points) MOSFET as a Switch In your analysis of this circuit with a MOSFET assume that the supply voltage Vdd is maintained constant, while the control voltage VGS is varied, and neglect the current through the gate (G) terminal. From your calculations, learn that the MOSFET here acts as a switch, delivering to the load either low power (open switch) or high power (closed switch), controlled with the voltage VGS . The plot below shows typical lab data obtained with Vdd = 6 V. From this plot, all circuit parameters can be calculated by using KCL, KVL, and Ohm’s law, at each value of the control voltage VGS. Indeed, the current through the V − VDS load resistor and through the MOSFET equals: I LOAD = dd RLOAD The apparent conductance of the MOSFET is reciprocal to its apparent resistance 1 I thus equal to: GMOSFET = = LOAD (the units are siemens (S), reciprocal of Ω). RMOSFET VDS The power absorbed by the MOSFET equals: PMOSFET = VDS ⋅ I LOAD The power absorbed by the load resistor equals: PLOAD (Vdd − VDS )2 . = RLOAD © 2011 Alexander Ganago Last printed 1/16/2011 3:05:00 AM Page 1 of 8 File: 2011 W 314 HW 02 p1-1.doc EECS 314 Winter 2011 Homework set 2 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 1 Part 1 (10 points) Use the data from the plot above From the plot of lab data on page 1, calculate the power in mW absorbed by the load and by the MOSFET at three values of the control voltage V GS: 3.16 V, 3.48 V, and 3.92 V. Assume Vdd = 6 V and RLOAD = 150 Ω. Write your results in the table below. On an additional page, show your work step‐by‐step for VGS = 3.48 V. Voltage VGS 3.16 V 3.48 V 3.92 V 0.00265 21.6 231.7 Power absorbed by the load resistor, mW 0.797 50.4 3.9 Power absorbed by the MOSFET, mW Explain step‐by‐step how the equations for GMOSFET and PMOSFET are derived: at each step of derivation indicate whether we use KCL, KVL, and Ohm’s law. NOTE: The numbers in the solutions may slightly differ from students results since problem is based on the rough graph reading. SOLUTION: From the graph Vgs=3.16 V Vds= 5.98 V Id=(6‐Vds)/150 Id=(6‐5.98)/150=1.33e‐4 A Power absorbed by load resistor is Pr= Id^2*(R)=((1.33e‐4)^2)*150=2.65e‐6 W = 2.65uW (very small, almost zero) Power absorbed by mosfet Pm=Vds*Id=5.98*1.33e‐4=7.9733e‐004 W= 0.797 mW (which is also small, can be ignorable) © 2011 Alexander Ganago Last printed 1/16/2011 3:05:00 AM Page 2 of 8 File: 2011 W 314 HW 02 p1-1.doc EECS 314 Winter 2011 Homework set 2 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 graph Vgs=3.48 V Vds= 4.2 V Id=(6‐4.2)/150=0.012 A Power absorbed by load resistor is Pr= Id^2*(R)=((0.012)^2)*150=0.0216 W = 21.6 mW Power absorbed by mosfet Pm=Vds*Id=4.2*0.012=0.0504 W= 50.4 mW From the graph Vgs=3.92 V Vds= 0.1 V Id=(6‐0.1)/150=0.0393 A Power absorbed by load resistor is Pr= Id^2*(R)=((0.0393)^2)*150=0.2317 W = 231.7 mW Power absorbed by mosfet Pm=Vds*Id=0.1*0.0393=0.0039 W= 3.9 mW © 2011 Alexander Ganago Last printed 1/16/2011 3:05:00 AM Page 3 of 8 File: 2011 W 314 HW 02 p1-1.doc EECS 314 Winter 2011 Homework set 2 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 2 (30 points) Make computer‐generated plots of lab data Download the file MOSFET 20 ohm.csv from the CTools web site (HW 02 folder). Assume Vdd = 6 V and RLOAD = 20 Ω. Use any software, with which you are comfortable (EXCEL, Matlab, etc.) to make computer‐generated plots of the following parameters as functions of the control voltage VGS: a. The apparent conductance (reciprocal of resistance) of the MOSFET, in S b. Power absorbed by the load resistor, in W c. Power absorbed by the MOSFET, W Make sure that each curve is at least ¾ of the height of your plot. Use line types distinct on the printout, and clearly explain what each curve shows. Write your name, discussion section #, and the date of work on each page of your plot. Briefly explain why the power absorbed by the load is low at low VGS and why it is low at high VGS . Calculate the ratio of these powers at VGS = 6 V and at VGS = 0 V. Voltage VGS 6 V 0 V Power absorbed by the load resistor, mW 1.65 W 6.03e‐6 W 2.7363e+005 P (V = 6 V ) Ratio of the powers LOAD GS PLOAD (VGS = 0 V ) Pr(6V)=1.65 W Pr(0V)=6.03e‐6 W Ratio= (1.65)/(6.03e‐6)= 2.7363e+005 Calculate the power absorbed by the MOSFET as percentage of the power absorbed by the load resistor at VGS = 4 V. Repeat for VGS = 5 V. Show your work. Voltage VGS 4 V 5 V Power absorbed by the MOSFET, mW Power absorbed by the load resistor, mW P Ratio of the powers MOSFET PLOAD © 2011 Alexander Ganago Last printed 1/16/2011 3:05:00 AM Page 4 of 8 File: 2011 W 314 HW 02 p1-1.doc 0.449033 0.49268457 0.082113 1.631642524 0.9114 0.0503 Just take the power values from the calculated data set. EECS 314 Winter 2011 Homework set 2 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: a) Gds= 1/Rds= Ids/Vds; Ids= (6‐Vds)/20 So , by plugging the given data in to the formulas given above, one can find the graph as Apparent Conductance 1.4 1.2 1 0.8 Gd Series1 0.6 0.4 0.2 0 0 0.3 0.61 0.91 1.21 1.52 1.82 2.12 2.42 2.73 3.03 3.33 3.64 3.94 4.24 4.55 4.85 5.15 5.45 5.76 Vgs Unit is S © 2011 Alexander Ganago Last printed 1/16/2011 3:05:00 AM Page 5 of 8 File: 2011 W 314 HW 02 p1-1.doc EECS 314 Winter 2011 Homework set 2 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 b) Pr=(Ids^2)*20 So Pr vs Vgs 1.8 1.6 1.4 1.2 Pr 1 0.8 0.6 0.4 0.2 0 0.61 0.91 1.21 1.52 1.82 2.12 2.42 2.73 3.03 3.33 3.64 3.94 4.24 4.55 4.85 5.15 5.45 5.76 0.3 0 Series1 Vgs Unit is W © 2011 Alexander Ganago Last printed 1/16/2011 3:05:00 AM Page 6 of 8 File: 2011 W 314 HW 02 p1-1.doc EECS 314 Winter 2011 Homework set 2 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 c) Pmosfet= Vds*Ids Pmosfet vs Vgs 0.5 0.45 0.4 0.35 Pmosfet 0.3 0.25 0.2 0.15 0.1 0.05 0 3.7 4 0.06 0.36 0.67 0.97 1.27 1.58 1.88 2.18 2.48 2.79 3.09 3.39 4.3 4.61 4.91 5.21 5.52 5.82 Series1 Vgs Unit is W Power absorbed by the load is low because there is very small current passing through the load at low Vgs voltages, © 2011 Alexander Ganago Last printed 1/16/2011 3:05:00 AM Page 7 of 8 File: 2011 W 314 HW 02 p1-1.doc EECS 314 Winter 2011 Homework set 2 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 Last printed 1/16/2011 3:05:00 AM Page 8 of 8 File: 2011 W 314 HW 02 p1-1.doc EECS 314 Winter 2011 Homework set 2 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 (40 points) Semiconductor diode does not obey Ohm’s law: Real lab data In the lab experiment with a semiconductor diode (its circuit symbol looks like an arrow), the source voltage Vdd is varied and recorded; the load resistance R is known; the voltage Va across the diode is measured and recorded; all other circuit parameters can be calculated by using KCL, KVL, and Ohm’s law, at each value of Vdd. Indeed, the current through the load V − Va resistor and through the diode equals: I d = dd ; R the apparent conductance Gd of the diode is reciprocal to its 1 Id 1A Ω]. apparent resistance: Gd = = [the units are siemens (S); 1 S = Rd Va 1V The power absorbed by the diode equals: Pd = Va ⋅ I d The power absorbed by the load resistor equals: PLOAD The power supplied by the source equals PS = Vdd ⋅ I d . Vdd 0.499 V 1.499 V 1.996 V 2.496 V 2.995 V 3.994 V 4.995 V 5.995 V V a 0.499 V 1.497 V 1.694 V 1.754 V 1.796 V 1.864 V 1.924 V 1.981 0 1.01011e‐05 0.0015 0.00374 0.00605 0.0107 0.0155 0.0202 Id (A) 0 6.7475e‐06 0.00090038 0.002136530 0.0033716901 0.00577123 0.00806138 0.010233582 Gd(S) 0 1.5121e‐05 0.002583 0.006573 0.010875 0.020052 0.029841 0.0401 Pd(W) (Vdd − Va )2 = R PLOAD (W) 0 2.0e‐08 0.00046 0.0027 0.00726 0.02291 0.0476 0.08137 PS(W) 0 1.51e‐ 05 Pd/PS NaN 0.9986 0.00304 0.8486 0.00935 0.7027 0.01813 0.5996 0.04296 0.4667 0.07747 0.3851 0.12153 0.3304 Part 1 (20 points) Based on lab data for the circuit above Use R = 198 Ω and fill in the table above. Clearly write the units of measure. Show your work step‐by‐step for Vdd = 3.994 V on an additional page. © 2011 Alexander Ganago Last printed 1/16/2011 3:06:00 AM Page 1 of 4 File: 2011 W 314 HW 02 p2.doc EECS 314 Winter 2011 Homework set 2 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: All the formulas are given in the text of the question as: Id = Vdd − Va R Gd = 1 Id = Rd Va 1S= 1A 1V PS = Vdd ⋅ I d So just plug the numbers in the table in to these formulas Part 1, continued Explain step‐by‐step how the equations for Id and PLOAD are derived: at each step of derivation indicate whether we use KCL, KVL, and Ohm’s law. To get Id, KVL is applied in the loop as given in the picture with the chosen direction. ‐Vdd+Id*Rload+Va=0 So Id=(Vdd‐Va)/Rload Voltage drop on the load resistance VLoad=Vdd‐Va Pload=(VLoad^2)/RLoad= ((Vdd‐Va)^2)/RLoad © 2011 Alexander Ganago Last printed 1/16/2011 3:06:00 AM Page 2 of 4 File: 2011 W 314 HW 02 p2.doc EECS 314 Winter 2011 Homework set 2 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 Part 2 (10 points) Theoretical calculations For comparison, calculate the same parameters for the circuit built of two resistors in series: Use R1 = 200 Ω and R2 = 100 Ω, and fill in the table below. VTotal 0.5 V 1.5 V 2.0 V 2.5 V 3.0 V 4.0 V 5.0 V 6.0 V V2 (V) 0.16 0.5 0.666 0.833 1 1.33 1.66 2 I (A) G2(S) P2(S) 0.00027 0.00250 0.004 0.0069 0.01 0.0177 0.027 0.04 P1(W) 0.000555 0.00500 0.00888 0.01388 0.0200 0.03555 0.0555 0.0800 PS(W) 0.00083 0.0075 0.0133 0.02083 0.030 0.05333 0.083333 0.12 P2/PS 0.333333333333333 0.333333333333333 0.333333333333333 0.333333333333333 0.333333333333333 0.333333333333333 0.333333333333333 0.333333333333333 0.0016 0.01 0.005 0.006 0.010 0.01 0.0083 0.01 0.01 0.01 0.0133 0.01 0.0166 0.01 0.02 0.01 Clearly write the units of measure (preferably, use the same units as in Part 1). Note that V2 can be calculated from Ohm’s law. Show your work step‐by‐step for Vdd = 4.0 V on an additional page. Based on your results, explain which of the circuit parameters (as functions of Vdd) most clearly help you to distinguish between (1) the circuit built of a resistor and a non‐ohmic element (in this case, the diode), and (2) the circuit built of ohmic elements (two resistors). In your explanation, compare either the numbers from similar columns of the two tables – or sketch simple plots such as Pd/PS and P2/PS (as functions of Vdd). © 2011 Alexander Ganago Last printed 1/16/2011 3:06:00 AM Page 3 of 4 File: 2011 W 314 HW 02 p2.doc EECS 314 Winter 2011 Homework set 2 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: V2=(Vtot*R2)/(R1+R2) I=Vtot/(R1+R2) G2=I/V2 P2=V2*I P1=(I^2)*R1 PS=Vtot*I By plugging the data given in the table in to the formulas above, one can fill the table quite easily. Those formulas can be derived basic circuit laws like KCL and KVL. Between two circuits, the difference between conductance values of diode and resistor helpmed me most. Basically conductance or resistance of diode depends on the source, that means it is nonlinear. Other values differs correspondingly between two circuits also verify the nonlinear property of the diode. © 2011 Alexander Ganago Last printed 1/16/2011 3:06:00 AM Page 4 of 4 File: 2011 W 314 HW 02 p2.doc EECS 314 Winter 2010 Homework set 1 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 (30 points) Power Absorbed by Resistors In all circuits shown below: a. Voltage VTotal is positive but may differ from one circuit to the next b. R1 = 10 Ω, R2 = 30 Ω, R3 = 90 Ω c. R1 absorbs 10 W (this is a very good hint: from Ohm’s law, you can easily find the current through this resistor and the voltage across it). For each circuit, determine: 1. the current in amps through each resistor, and the total current 2. the voltage in volts across each resistor, and the total voltage 3. the power in watts absorbed by each resistor and the total absorbed by all three. Justify your answers: if calculations needed, show your work; if you use KCL, say so. On each circuit diagram, circle the resistor that absorbs the most power, and show with an arrow the resistor, which absorbs the least power. Show your work on these pages and/or additional pages as needed. R1, R2 and R3 are connected in series. In a series circuit, current through each component is the same. We can find the current using the information given: R1 = 10 Ω and P1 = 10W. Using P = I2R, I can be found as sqrt(P1/R1) = sqrt(10W/10 Ω) = 1A © 2011 Alexander Ganago Page 1 of 1 Last printed 1/15/2011 11:49:00 AM File: 2011 W 314 HW 02 p3.doc EECS 314 Winter 2010 Homework set 1 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 Using Ohm’s Law, we can find the Voltage across each resistor: V = IR. Vtotal is found Using Ohm’s Law, we can find the Voltage across each resistor: V = IR. V by using KVL: Vtotal = V1+V2+V3 by using KVL: V Power absorbed can be found using P=VI for P2 and P3. The battery supplies power equal to the sum of powers absorbed by R1, R2 and R3. Therefore, the power absorbed by the battery is a negative value and is ‐130W. Resistor Voltage across (V) Current through (A) Power absorbed (W) R1 10 1 10 R2 30 1 30 R3 R3 90 1 90 Total 130 1 ‐130 Problem 3, continued R1, R2 and R3 are connected in prarallel. In a parallel circuit, voltage across each component is the same. We can find the voltage using the information given: R1 = 10 Ω and P1 = 10W. Using P = V2/R, V can be found as sqrt(P1R1) = sqrt(10W*10 Ω) = 10V. Current through each resistor can be found using Ohm’s Law: V = IR I = V/R. The current through the battery can be found using KCL: Itotal = I1+I2+I3 © 2011 Alexander Ganago Page 2 of 2 Last printed 1/15/2011 11:49:00 AM File: 2011 W 314 HW 02 p3.doc 11 314 EECS 314 Winter 2010 Homework set 1 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 Power absorbed by each resistor is P=VI. The battery supplies power equal to the sum of powers absorbed by the resistors. Therefore the battery absorbs a negative power. Resistor Voltage across (V) Current through (A) Power absorbed (W) R1 10 1 10 R2 10 1/3 10/3 R3 10 1/9 10/9 Total 10 13/9 = 1.444 ‐130/9 = ‐14.44 Using P1 and R1, we can find I1, which is same as Itotal. P1 = I12R1 I1 = sqrt(P1/R1) = sqrt(10W/10Ω) = 1A. Using Ohm’s Law, we can find V1 = I1R1 = 1A*10Ω = 10V. R2||R3 are in parallel. Therefore, we can find the equivalent resistance using: R2||R3 = 1 / (1/R2 + 1/R3) = 1/(1/30 + 1/90) = 22.5 Ω Using Ohm’s Law, we can find V23 = V2 = V3 = Itotal* (R2||R3) = 1A*22.5Ω = 22.5V Vtotal is the sum of V1 and V23. Vtotal = 10V+22.5V = 32.5V. © 2011 Alexander Ganago Page 3 of 3 Last printed 1/15/2011 11:49:00 AM File: 2011 W 314 HW 02 p3.doc EECS 314 Winter 2010 Homework set 1 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 R2 and R3 are current dividers. Therefore I2 = Itotal*(R3/R2+R3) and I3 = Itotal*(R2/R2+R3). I2 = 1A*(90/120) = 3/4A =0.75A I3 = 1A*(30/120) = 1/4A = 0.25A Power absorbed by each resistor can be calculated using P = VI. Battery supplies power each to the sum of powers absorbed by the resistors. Therefore, it has a negative value and equal to –(P1+P2+P3). Resistor Voltage across (V) Current through (A) Power absorbed (W) R1 10 1 10 R2 22.5 0.75 16.875 R3 22.5 0.25 5.625 Total 32.5 1 ‐32.5 © 2011 Alexander Ganago Page 4 of 4 Last printed 1/15/2011 11:49:00 AM File: 2011 W 314 HW 02 p3.doc EECS 314 Winter 2011 Homework set 2 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 (30 points) Electrostatic Discharge (ESD) and its prevention: 1 The Big Picture This problem requires basic knowledge of Electrostatic Discharge (ESD). To answer the questions below, read 3 documents on the web: http://www.esda.org/documents/esdfunds1print.pdf http://www.esda.org/documents/esdfunds2print.pdf http://www.esda.org/documents/esdfunds3print.pdf Grading policy: Maximal score is 30 points per problem. Deduct 1 point for each wrong answer, and deduct 1.5 points for each skipped question. Questions: 1. Static electricity was discovered by Ben Franklin in 18th century True / False (circle one) 2. Static electricity did not present any practical problems until the electronics industry was developed in 20th century True / False (circle one) 3. The sensitivity of electronic devices to static electricity reduces as the size of devices is reduced and the speed of operation is increased True / False (circle one) 4. ESD presents a great problem for product reliability, manufacturing costs, and profitability of electronics industry True / False (circle one) 5. Typical cost of devices damaged by ESD does not exceed a few cents True / False (circle one) 6. Static electricity is defined as the transfer of charges between bodies at different electric potentials True / False (circle one) 7. Electrostatic discharge is defined as electric charge caused by an imbalance of electrons on the surface of a material True / False (circle one) © 2011 Alexander Ganago Page 1 of 4 Last printed 1/15/2011 11:47:00 AM File: 2011 W 314 HW 02 p4.doc EECS 314 Winter 2011 Homework set 2 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 (continued) 8. Electrostatic discharge can change the electrical characteristics of a semiconductor device, degrading or destroying it True / False (circle one) 9. ESD can cause malfunction or failure of electronic equipment True / False (circle one) 10. Electrostatic charge is only created on the surface of certain materials after a special treatment in high‐vacuum chambers True / False (circle one) 11. Electrostatic charge is commonly created on many surfaces when two materials come in contact and then get separated True / False (circle one) 12. Everyday actions such as walking across the floor cannot create electrostatic charge True / False (circle one) 13. Triboelectric charge occurs only in certain tribes of exotic materials True / False (circle one) 14. Triboelectric charging is rare because it includes creation of electric charges on the surface of a material, which is not among commonly observed phenomena True / False (circle one) 15. Triboelectric charging occurs by contact and separation of materials, and it is the most common way of creating electrostatic charges on materials True / False (circle one) 16. Triboelectric series table shows how electrostatic charges are generated on various materials, namely, the higher in the table the material is listed, the more negative charge it accumulates True / False (circle one) 17. When wool is rubbed against polyester, electrons from wool are transferred to polyester True / False (circle one) 18. When rubbed against Teflon, human hair typically loses more electrons than silk True / False (circle one) 19. Dirt particles in the air cannot be triboelecrically charged, because they are too small to be rubbed against True / False (circle one) Page 2 of 4 © 2011 Alexander Ganago Last printed 1/15/2011 11:47:00 AM File: 2011 W 314 HW 02 p4.doc EECS 314 Winter 2011 Homework set 2 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 (continued) 20. A piece of cotton can carry negative charges at one location, and carry positive charges at another location at the same time True / False (circle one) 21. A piece of copper can carry negative charges at one location, and carry positive charges at another location at the same time True / False (circle one) 22. Electrostatic charge cannot be triboelectrically generated on the surface of a static dissipative material True / False (circle one) 23. The following statements about insulative, conductive, and static dissipative materials are correct: a.b.c.h a. Insulative materials generally keep their electrostatic charge for a very long time b. Even when grounded, a piece of insulative material can keep its electrostatic charge for a very long time c. Static dissipative materials transfer electrostatic charges faster than insulative materials d. Static dissipative materials transfer electrostatic charges faster than conductive materials e. Electric resistance of static dissipative materials is between that of conductive materials (highest) and that of insulative materials (lowest) f. Electric resistance of conductive materials is between that of static dissipative materials (highest) and that of insulative materials (lowest) g. Electric resistance of conductive materials is highest, that of insulative materials is lowest, and the resistance of static dissipative materials is between those listed above h. Electric resistance of conductive materials is lowest, that of insulative materials is highest, and the resistance of static dissipative materials is between those listed above 24. A latent defect is easier to identify than catastrophic failure, because the electronic device continues to perform its intended function True / False (circle one) Page 3 of 4 © 2011 Alexander Ganago Last printed 1/15/2011 11:47:00 AM File: 2011 W 314 HW 02 p4.doc EECS 314 Winter 2011 Homework set 2 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 (continued) 25. Installation of electronic devices with latent effects into products of systems is a good, cost‐effective practice True / False (circle one) 26. Electrostatic discharge sensitive (ESDS) device can transfer electric charges to other objects, but such ESD events are not destructive True / False (circle one) 27. When one walks across a floor, en electrostatic charge accumulates on the body. Simple contact of a finger to the leads of an ESDS device may cause damage to the device True / False (circle one) 28. ESDS device may accumulate damaging charges if it simply vibrates in a package or slides down the feeder in an automated assembly line True / False (circle one) 29. None of the electronic devices listed in Table 4 (Part One, page 11) can be damaged by electrostatic charges accumulated on the human body as a result of walking across vinyl tile at 90% relative humidity True / False (circle one) 30. None of the electronic devices listed in Table 5 (Part One, page 12) can be damaged by electrostatic charges accumulated on the human body as a result of walking across carpet at 90% relative humidity True / False (circle one) © 2011 Alexander Ganago Page 4 of 4 Last printed 1/15/2011 11:47:00 AM File: 2011 W 314 HW 02 p4.doc EECS 314 Winter 2011 Homework set 2 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 Electrostatic Discharge (ESD) and its prevention: 2 The Big Picture This problem requires basic knowledge of Electrostatic Discharge (ESD). To answer the questions below, read 3 documents on the web: http://www.esda.org/documents/esdfunds1print.pdf http://www.esda.org/documents/esdfunds2print.pdf http://www.esda.org/documents/esdfunds3print.pdf Grading policy: Maximal score is 30 points per problem. Deduct 1 point for each wrong answer, and deduct 1.5 points for each skipped question. Questions: 1. In general, electrostatic discharge (ESD) is very easy to control, especially in the electronics environment True / False (circle one) 2. The first principle of ESD control – design products and assemblies as immune as possible to the effects of ESD – becomes more and more challenging to follow because modern electronic components involve smaller sizes and more complex geometries that often are more susceptible to ESD True / False (circle one) 3. In electrostatic protected areas (EPA), you have to bond or electrically connect to the ground all conductive and dissipative materials but human bodies do not have to be grounded True / False (circle one) 4. The Fourth Principle of ESD control boils down to the basic truth: no charge – no discharge True / False (circle one) 5. The Fourth Principle of ESD control requires, in particular, that we begin by reducing as many static generating processes or materials, such as the contact and separation of dissimilar materials and common plastics, as possible from the work environment True / False (circle one) 6. The Fourth Principle of ESD control also requires that we keep other processes and materials at the same electrostatic potential True / False (circle one) © 2011 Alexander Ganago Page 1 of 4 Last printed 1/16/2011 3:07:00 AM File: 2011 W 314 HW 02 p5.doc EECS 314 Winter 2011 Homework set 2 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 (continued) 7. In addition, the Fourth Principle of ESD control also requires that we provide ground paths, such as wrist straps, flooring and work surfaces, to reduce charge generation and accumulation True / False (circle one) 8. The Fifth Principle of ESD control requires that we eliminate all generation of static in the environment True / False (circle one) 9. Workers who “carry” a charge into the work environment can rid themselves of that charge when they a. take a second cup of coffee b. lick a grounded water pipe c. attach a wrist strap connected to the ground d. change their shoes e. step on an ESD floor mat while wearing ESD control footwear (circle all recommended procedures) 10. Grounding is an excellent way to remove all electrostatic charges from common plastic bags True / False (circle one) 11. To get rid of electrostatic charges on good insulators, one has to use exotic procedures such as ionization True / False (circle one) 12. One way to prevent discharges that do occur from reaching ESDS parts is to connect everything to the common, well-known ground True / False (circle one) 13. Another way to prevent discharges that do occur from reaching ESDS parts is to package and transport susceptible devices in proper packaging that effectively shield them from charge True / False (circle one) 14. In many facilities, people are one of the prime generators of static electricity. True / False (circle one) 15. The simple act of walking around or repairing a board can generate several thousand volts on the human body True / False (circle one) © 2011 Alexander Ganago Page 2 of 4 Last printed 1/16/2011 3:07:00 AM File: 2011 W 314 HW 02 p5.doc EECS 314 Winter 2011 Homework set 2 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 (continued) 16. If not properly controlled, this static charge can easily discharge into an ESDS device and damage it True / False (circle one) 17. In highly automated assembly and test processes, people do not handle static sensitive devices True / False (circle one) 18. The movement of carts and other wheeled equipment through the facility usually does not generate static charges True / False (circle one) 19. Typically, wrist straps are the primary means of controlling static charge on personnel True / False (circle one) 20. If the person and other grounded objects in the work area are at or near the same potential, there can be hazardous discharge between them True / False (circle one) 21. Wrist straps have two major components, the cuff that goes around the person’s waist and the ground cord that connects the cuff to the uncommon point ground True / False (circle one) 22. Most wrist straps have a current limiting resistor molded into the ground cord head on the end that connects to the cuff True / False (circle one) 23. The resistor most commonly used is a one ohm, 1/4 watt with a working voltage rating of 2500 volts True / False (circle one) 24. Wrist straps should be tested on a regular basis. Yearly testing is recommended True / False (circle one) 25. A second method of controlling electrostatic charge on personnel is with the use of ESD protective floors, not necessarily in conjunction with ESD control footwear or foot straps True / False (circle one) 26. Any combination of floor materials and footwear provides a ground path for the dissipation of electrostatic charge, thus reducing the charge accumulation on personnel and other objects to safe levels True / False (circle one) © 2011 Alexander Ganago Page 3 of 4 Last printed 1/16/2011 3:07:00 AM File: 2011 W 314 HW 02 p5.doc EECS 314 Winter 2011 Homework set 2 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 (continued) 27. In addition to dissipating charge, some floor materials (and floor finishes) also reduce triboelectric charging True / False (circle one) 28. The use of any special floor materials is not required in those areas where increased personnel mobility is necessary True / False (circle one) 29. In addition, floor materials can minimize charge accumulation on chairs, carts, lift trucks and other objects that move across the floor True / False (circle one) 30. To reduce the risk of ESD, chairs, carts, lift trucks and other objects that move across the floor do not even require dissipative or conductive casters or wheels to make electrical contact with the floor True / False (circle one) © 2011 Alexander Ganago Page 4 of 4 Last printed 1/16/2011 3:07:00 AM File: 2011 W 314 HW 02 p5.doc EECS 314 Winter 2011 Homework set 2 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 6 (30 points) Electrical Safety; Ground-Fault Circuit Interrupter (GFCI) For The Big Picture, read the files Safety 1 and From Electrical Safety Handbook posted on CTools. Consider a 150-lb adult working with an electric drill plugged into a 120 V AC receptacle. The drill handle is made of aluminum; the hot (H) and neutral (N) wires are connected to the motor. The hot wire insulation inside the drill failed and the handle got energized thus the current can flow through the human body. Your goal is to determine the consequences for the human being: use Tables 1.2, 1.3, and 1.4 on pages 7/9 and 8/9 (and refer to a more realistic picture on page 6/9) of the file From Electrical Safety Handbook). For each part of the problem, do the following (on page 2 and/or additional pages): A) Determine the path of current: if the current enters the human body, describe its path (the shock circuit) and explain whether it passes through the vital organs such as heart and lungs; B) Calculate the maximal magnitude of the current through the human body (use the worst case scenario: for example, if the table lists resistances from 1 to 10 kΩ, use 1 kΩ in your calculations) and neglect the resistance of the ground; C) Describe the physiological effects (Table 1.4) of the current and consequences for the human being; © 2011 Alexander Ganago Page 1 of 5 Last printed 1/16/2011 2:56:00 AM File: 2011 W 314 HW 02 p6.doc EECS 314 Winter 2011 Homework set 2 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 D) Briefly explain whether the path and/or magnitude of the current can change with time (for example, due to the change of human body resistance, or due to interruption of the circuit by GFCI), and outline the consequences Problem 6 Part 1 (10 points) Assume dry hands and feet, dry leather sole of the footwear. A) Bulk of the current takes a low resistance path from the drill handle, through the dry skin of the hand, through the arms and legs, and then again through the dry skin of the foot, through the dry leather sole of the boot, and then to the floor (ground). Some of the current may flow through the heart and lungs too. B) Equivalent circuit: Fig. 6 -1 Condition Hand around drill handle (dry) Human body, internal, excluding skin Leather sole, including foot (dry) Resistance (kΩ) 1-3 0.2 - 1 100 - 500 To find out the total worst case current (maximum current) through the body use minimum values of the resistances for computation: The total equivalent resistance is calculated to be: Req = (1 + 0.2 + 100) kΩ = 101.2 kΩ © 2011 Alexander Ganago Page 2 of 5 Last printed 1/16/2011 2:56:00 AM File: 2011 W 314 HW 02 p6.doc EECS 314 Winter 2011 Homework set 2 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 Since the magnitude of the voltage is 120 V, the magnitude of the current will be: Ibody = V / Req = 1.186 mA C) The physiological effect of this magnitude of the current (1.186 mA) is a painful sensation (Refer: table 1.4 from the Electrical Safety Handbook) D) Instantaneous Power dissipated (Pinst) = V2 / Req Assume voltage is an AC sinusoidal signal then V(t) = V0 sin(2*Π*f*t) Average Power dissipated = ∫Pinst dt / T where, T = time period of AC signal and integration is done over one clock period Thus, on solving this integral, Average Power dissipated = 71.14mW Brief Explanation: The path and magnitude of the current would not change very much over time since only about 71 mW of average power is being delivered through the equivalent resistor, which should be easily dissipated by the body. The current is also not large enough to trip the GFCI. There are no time dependent consequences. Part 2 (10 points) Assume wet, sweaty hands, and feet immersed in water (during house repair after a flood). A) The majority of the current takes the path of least resistance from the drill handle, through the wet skin of the hand, through the arms and legs, and then again through the wet skin of the foot, through the wet leather sole of the boot, and then to the floor (ground). Some of the current may flow through the heart and lungs. B) The equivalent circuit is similar to Fig. 6-1 shown above except that the resistance ‘R_sole_including_foot’ is replaced by ‘R_foot_immersed’. The below table gives the resistance values for the given condition: © 2011 Alexander Ganago Page 3 of 5 Last printed 1/16/2011 2:56:00 AM File: 2011 W 314 HW 02 p6.doc EECS 314 Winter 2011 Homework set 2 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 Condition Hand around drill handle (wet) Human body, internal, excluding skin Foot immersed The total equivalent resistance is calculated to be: Req = (0.5 + 0.2 + 0.1) kΩ 0.8 kΩ = Since the magnitude of the voltage is 120 V, the magnitude of the current will be: Ibody = V / Req = 150 mA C) The physiological effect of this magnitude of the current (150 mA) is discoordinated heart action (probably fatal) (Refer: table 1.4 from the Electrical Safety Handbook) D) If that amount of current were to travel through the human body, the path of the current would probably change as the person falls into the water upon unconsciousness or death. As the person is still standing, the body would dramatically heat up, reducing the resistance of the body, and increasing the current flow. Resistance (kΩ) 0.5 – 1.5 0.2 - 1 0.2 – 0.3 Part 3 (10 points) Same as in Part 2 but the electric drill is plugged into the receptacle with a GFCI. A) Initially, the same as part 2, but as the GFCI kicks in it detects that the electric current is not balanced between the hot conductor and the return neutral conductor. It is able to sense a mismatch and trips at 6 - 8 mA. Thus the circuit is interrupted and majority of the current does not enter the human body. B) Same as part 2 © 2011 Alexander Ganago Page 4 of 5 Last printed 1/16/2011 2:56:00 AM File: 2011 W 314 HW 02 p6.doc EECS 314 Winter 2011 Homework set 2 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 C) The physiological effect of the magnitude of the current is just a painful sensation according to table 1.4 from the Electrical Safety Handbook (since when the current goes beyond 6 - 8 mA, the main current path would be changed such that the current going through the body reduces to about 0 mA). D) The current is large enough such that the GFCI would kick in preventing the current from going through the person’s body. © 2011 Alexander Ganago Page 5 of 5 Last printed 1/16/2011 2:56:00 AM File: 2011 W 314 HW 02 p6.doc ...
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This note was uploaded on 03/20/2011 for the course EECS 314 taught by Professor Ganago during the Spring '07 term at University of Michigan.

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