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### Midterm2AF07-02

Course: EE 307, Spring 2008
School: Cal Poly
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Word Count: 780

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306 EE F07-02A MIDTERM EXAM #2 CLOSED BOOK + 1 Cheat Sheet Show all work, state any assumptions, and test hypotheses. (1 point) Print Your Name: Braun Part 1 (14 Points) Part 2 (8 Points) Part 3 (6 Points) Part 4-5 (6 Points) Total (36 Points) SOLUTIONS BRAUN No unauthorized help given or received. (1 point) Signature: 1. Diode Circuits (14 points) I D = I S [exp(VD VT ) - 1] Use CVD for hand analysis For the...

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306 EE F07-02A MIDTERM EXAM #2 CLOSED BOOK + 1 Cheat Sheet Show all work, state any assumptions, and test hypotheses. (1 point) Print Your Name: Braun Part 1 (14 Points) Part 2 (8 Points) Part 3 (6 Points) Part 4-5 (6 Points) Total (36 Points) SOLUTIONS BRAUN No unauthorized help given or received. (1 point) Signature: 1. Diode Circuits (14 points) I D = I S [exp(VD VT ) - 1] Use CVD for hand analysis For the circuits shown below, assume diodes obey the ideal diode law and have VZ = 10 V. Each battery supplies 5 V. The output voltage, VOUT, appears between the upper right node of the circuit and ground. The input signal VS (dashed line) for each case is a 5 V triangle wave (10 V peak-to-peak) with a frequency of 1 kHz and no DC offset. For each circuit, plot the output voltage trace on the upper graph and the capacitor voltage trace on the lower graph. For each signal you sketch, clearly indicate maximum and minimum values 0.1 V. Repeated for emphasis: Pick VOUT as the POSITIVE terminal of the capacitor. Pick VOUT as the POSITIVE terminal of the capacitor. VC 1 F + VOUT + VOUT VS 1 F 5V VC VS 5V KVL: VOUT = VC + 5 V 15 10 VOUT [V] 4.3 V KVL: VOUT = VS + VC 15 10 VOUT [V] 14.3 V 5 0 -5 0V Diode Operation ON OFF 5 0 -5 OFF ONOFF 4.3 V -10 -15 15 10 VC [V] -10 Diode Operation 1.5 2 2.5 3 [ms] -15 15 10 -0.7 V VC [V] 0 0.5 1 0 0.5 1 1.5 2 2.5 9.3 V 3 [ms] 5 0 -5 -5.0 V 5 0 -5 5.0 V -10 -15 0 -10 1 1.5 2 2.5 3 [ms] . 0.5 -15 0 0.5 1 1.5 2 2.5 3 [ms] . List the circuit's function: Peak Detector List the circuit's function: Clamp 2. MOS Biasing F07-02A This problem asks you to analyze the circuit shown to the right and determine the Q-point of the transistor. Assume VTN = - VTP = 1.25 V, KN' = 50 A/V2, KP' = 200 A/V2, = 0 V1/2, = 0 V-1. W/L = 10. (4 points) A) Hypothesize the transistor turns on and operates in the saturation mode of operation. Determine values for VGS, VDS and ID, based on the transistor operating in saturation mode. VDD = 5 V Please enter the values in the table below. VGS = VG = VDD R1 120 k =5 V = 3.0 V > VTN R1 + R2 120 k + 80 k R2 80 k RD 6 k K N = K N (W / L) = (50 A/V 2 )(10) = 500 A/V 2 KN (VGS - VTN ) 2 2 500 A/V 2 = (3.0 V -1.25 V) 2 = 0.766 mA 2 VDS = VDD - I D RD ID = = 5.0 V - 0.766 mA (6.0 k) 0.406 = V Test: VDS = 0.406 V < VGS - VTN = 3.0 V 1.25 V = 1.75 V R1 120 k X Doesn't agree with Saturation VGS [V] VDS [V] ID [mA] (1 point) B) Is hypothesis A correct? YES (3 points) NO 3.0 V 0.406 V 0.766 mA C) Explain whether hypothesis A is correct or not. Saturation requires VDS VGS - VTN , so the hypothesis doesn't work. The problem didn't REQUIRE finding the actual Q-point based on the FET operating in LINEAR mode. Doing so yields (3.0 V, 1.08 V, 0.653 mA) 3. JFET Biasing F07-02A (6 points) This problem asks you to analyze the JFET circuit shown to the right and determine its Q-point. Please enter the values in the table below. Assume VP = -2 V, IDSS = 2 mA. VDD = 1 V VDS < VDD = 1 V < VGS - VP = 0 V (2 V) = 2 V So the JFET operates in the Linear region 2I V V - VDS I D = DSS (VGS - VP - DS )VDS = I R = DD 2 VP 2 R 2 VDS - [2(VGS - VP ) + R 1 k VP2 I DSS R ]VDS + VP2 I DSS R VDD = 0 - [2(0 V - (-2 V )) + (2 V) 2 ] (2 mA)(1 k) (2 V) 2 1V (2 mA)(1 k) 2V2 -6V 2 VDS - [2(VGS - VP ) + VP2 I DSS R ]VDS + VP2 I DSS R VDD = 0 (6 V) 2 - 4(2 V 2 ) 6V = 0.354 V , 5.65 V VDS = Agrees with linear operation 2 2 Only the lower value provides a physical solution. VDS = 0.354 V < (VGS VP) = 2 V is consistent with the transistor operating in Linear mode. V - VDS 1 V - 0.354 V = = 0.646 mA I D = I R = DD 1 k R VGS [V] VDS [V] ID [mA] 0V 0.354 V 0.646 mA Please turn page. 4. BJT Operation F07-02A Using the Gummel-Poon or Transport model, determine the operating points for the following transistors. Assume IS = 10 f A, F = 50, R = 1, and room temperature with VT = 25 mV. 0.1 V 0.70 V (3 points) IC [mA] IB [mA] VCE [V] 13.9 mA 0.554 mA 0.1 V v v I v I C = I S exp BE - exp BC - S exp BC - 1 V V V T T R T 0.7 V 0.6 V 10 fA 0.6 V =10 fA exp 25 mV - exp 25 mV - 1 exp 25 mV - 1 =13.9 mA IB = I S v BE I S v BC - 1 - 1 + exp exp F VT R VT 10 fA 0.7 V 10 fA 0.6 V - 1 + - 1 = exp exp 50 25 mV 1 25 mV = 0.554 mA (3 points) 5. LED Operation A GaP LED emits light at a wavelength of 565 nm. Please determine the energy of the light in electron volts and its color. E PH = 1240 [eV nm] 1240 [eV nm] = 565 nm = 2.19 eV, which corresponds to green light.
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Cal Poly - EE - 307
EE 306 F07-01B MIDTERM EXAM #2 CLOSED BOOK + 1 Cheat Sheet Show all work, state any assumptions, and test hypotheses. (1 point) Print Your Name:BraunPart 1 (14 Points) Part 2 (8 Points) Part 3 (6 Points) Part 4-5 (6 Points) Total (36 Points)SOLU
Cal Poly - EE - 307
Cal Poly - EE - 307
EE 306 W05 MIDTERM EXAM #2 CLOSED BOOK + 1 Cheat Sheet State any assumptions and show all work. (1 point) Print Your Name:BraunPart 1 (12 Points) Part 2 (15 Points) Part 3 (12 Points) Part 4 (3 Points) Total (44 Points)SOLUTIONSBRAUNNo unauth
Cal Poly - EE - 307
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Cal Poly - EE - 307
CHAPTER 22.1 Based upon Table 2.1, a resistivity of 2.6 -cm &lt; 1 m-cm, and aluminum is a conductor. 2.2 Based upon Table 2.1, a resistivity of 1015 -cm &gt; 105 -cm, and silicon dioxide is an insulator. 2.3 I max 2.4 10-8 cm2 7 A = 10 1 = 500 mA (
Cal Poly - EE - 307
CHAPTER 44.1 (a) VG &gt; VTN corresponds to the inversion region (b) VG &lt; VTN corresponds to the accumulation region (c) VG &lt; VTN corresponds to the depletion region 4.2 (a)&quot; ox -14 3.9o 3.9 8.854x10 F / cm F nF C = = = = 6.91x10-8 2 = 69.1 2 -9 Tox T
Cal Poly - EE - 307
CHAPTER 55.1 Base Contact = B n-type Emitter = D 5.2v BC iB + B + E iE C iCCollector Contact = A n-type Collector = FEmitter Contact = C Active Region = EFor VBE &gt; 0 and VBC = 0, IC = F I B or F =IC 275A = = 68.8 4A IBR =0.5 R = =1 1-
Cal Poly - EE - 307
CHAPTER 66.1(a ) Pavg = 1W 10-5W / gate = 10 W / gate (b) I = = 4 A / gate 105 gates 2.5V6.2(a) Pavg = 100 5x10-6W / gate = 5 W / gate (b) I = = 2 A/ gate 2.5V 2x10 7 gates(c) I total = 2(2x10 gates)= 40 A gate7A6.3 2.5 - 0 5 (a ) VH
Cal Poly - EE - 307
CHAPTER 33.1(1019 cm-3 )(1018 cm-3 ) = 0.979V NA ND j = VT ln 2 = (0.025V )ln ni 10 20 cm -62( 11.7 8.854 x10-14 F cm-1 ) 2s 1 1 1 1 w do = + 19 -3 + 18 -3 (0.979V) j = -19 10 cm q NA ND 1.602x10 C 10 cm w do = 3.73 x 10-6 cm = 0
Cal Poly - EE - 307
Cal Poly - EE - 307
Cal Poly - EE - 307
Cal Poly - EE - 307
Policy Paper#1 Chapter 12Anthony Milne PSCI 3143When considering whether the United States should pursue a strong or weak dollar, it is important to consider that either of these positions has both benefits and negative consequences. Using the se
Cal Poly - EE - 307
Anthony Milne-800628757-Urban Politics Final-Sprawl and SegregationThe two most important issues that face a major US city are those of urban sprawl and segregation. The first of these issues with sprawl, can be explained as the horizontal movement
Cal Poly - EE - 307
EE 306 S04 MIDTERM EXAM #2 CLOSED BOOK + 1 Cheat Sheet State any assumptions and show all work. (1 point) Print Your Name:BraunParts 1,2 (10 Points) Part 2 (22 Points) Part 3 (9 Points) Part 4 (4 Points) Total (47 Points)SOLUTIONS BraunNo unau
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Cal Poly - EE - 307
EE 306 F07-02B MIDTERM EXAM #2 CLOSED BOOK + 1 Cheat Sheet Show all work, state any assumptions, and test hypotheses. (1 point) Print Your Name:BraunPart 1 (14 Points) Part 2 (8 Points) Part 3 (6 Points) Part 4-5 (6 Points) Total (36 Points)SOLU
UCLA - LS - LS3
1. Living organisms on earth can be generally classified into three groups, Eukaryota, Eubacteria, and Archaea. What of the following best describes these groups? A). Eubacteria are prokaryotic whereas Archaea are eukaryotic organisms B). Archaea gen
Cal Poly - EE - 307
UCLA - PHYS - 6 B
Important math knowledge1. Solution to typical differential equations - P ( x ) dx P ( x ) dx dx + c (1) y + P ( x) y = Q ( x ), y = e Q ( x )e . 2 (2) x' '- x = 0, x = Ae-t+ Bet . ( &gt; 0)(3) x' '+ x = 0, 2. Basic expansions2x =
UCLA - LS - LS3
LS3-2 Midterm 1-KEY1. D 2. C 3. E 4. C 5. D 6. C 7. B 8. E 9. D 10. A 11. D 12. A 13. D 14. C 15. C 16. A 17. B 18. B 19. D 20. A 21. E 22. D 23. D 24. C 25. A 26. C 27. B 28. E 29. D 30. E 31. A 32. C 33. A 34. C 35. C
Cal Poly - EE - 307
CHAPTER 77.1' n -14 cm 2 (3.9) 8.854x10 F / cm 3.9o K = nC = n = n = 500 Tox Tox V - sec 10x10-9 m( 100cm / m) &quot; oxox()F A A = 173 x 10-6 2 = 173 2 V - sec V V p ' 200 A A &quot; K 'p = pCox = Kn = 173 2 = 69.1 2 n V V 500 ' Kn =
Cal Poly - EE - 307
CHAPTER 88.1(a) 256Mb = 28 210 210 = 268,435,456 bits (b) 1Gb = 210 = 1,073,741,824 bits8 10 10 28( )( ) (c) 256Mb = 2 (2 )(2 )= 2I pA 1mA = 3.73 28 bit 2 bits( )3| 128kb = 2 7 210 = 217 |( )228 = 211 = 2048 blocks 17 28.28.3(a
Cal Poly - EE - 307
CHAPTER 99.1 Since VREF = -1.25V , and v I = -1.6V , Q1 is off and Q2 is conducting.vC1 = 0 V and vC 2 = - F I EE RC -I EE RC = -(2mA)(350) = -0.700 V9.2 V IC 2 0.995 F I EE = exp BE VBE = 0.025ln = 0.132V IC1 0.005 F I EE VT (a) v I = VR
Cal Poly - EE - 307
CHAPTER 1010.1 A/C temperature Automobile coolant temperature gasoline level oil pressure sound intensity inside temperature Battery charge level Battery voltage Fluid level Computer display hue contrast brightness Electrical variables voltage ampli
Cal Poly - EE - 307
CHAPTER 1111.1v O = vS iS = v 1M 1k (1000)1k + 0.5 | Av = vO = 990 or 59.9 dB 1M + 5k S | Ai = iO 990 6 = 10 = 9.9x105 or 120 dB iS 1000 vO 5V = = 5.05 mV AV 990 vS 990vS and iO = 1M + 5k 1kAP = Av Ai = 990 9.9x105 = 9.8x108 or 89.9 dB | v S =(
Cal Poly - EE - 307
CHAPTER 1212.1(a) A = 10 20 = 2.00x104 | Av-ideal = 1+A Av = = 1+ A FGE =86150k = 13.5 12k2.00x10 4 = 13.49 4 12k 1+ 2.00x10 162k 1 13.5 -13.49 = 6.75x10-4 or 0.0675% | Note : FGE = 6.75x10-4 A 13.5 2.00x10 4 = 125 1.2k 4 1+ 2.00x1
Cal Poly - EE - 307
CHAPTER 1313.1 Assuming linear operation : vBE = 0.700 + 0.005sin 2000t V 5mV vce = (-1.65V ) sin 2000t = -1.03sin 2000t V 8mV vCE = 5.00 -1.03sin 2000t V ; 10 - 3300IC 0.700 IC 2.82 mA 13.2 Assuming linear region operation : vGS = 3.50 +
Cal Poly - EE - 307
CHAPTER 1414.1 (a) Common-collector Amplifier (npn) (emitter-follower)RIQ1viR1R2+RER3vo-(b) Not a useful circuit because the signal is injected into the drain of the transistor.RI viRDM1+R3vo-R1(c) Common-em
Cal Poly - EE - 307
CHAPTER 1515.1(a) IC= F IE =VCE = VC - (-0.7V ) = 5.87V | Q - Point = (20.7A, 5.87V )1 F 12 - VBE 1 100 12 - 0.7 = = 20.7 A | VC = 12 - 3.3x105 IC = 5.17V 5 2 F + 1 REE 2 101 2.7x10 (b) Add= -g m RC = -40(20.7A)(330k)= -273
Cal Poly - EE - 307
CHAPTER 1616.1 Av (s) = 50 s2 s2 | Amid = 50 | FL (s)= | Poles : - 2,-30 | Zeros : 0,0 (s + 2)(s + 30) (s + 2)(s + 30) s rad | L 30 s (s + 30) | fL = Yes, s = -30 | Av (s) 50 fL = L 30 = 4.77 Hz 2 22 2 1 302 + 22 - 2(0) - 2(0) = 4.79 Hz 2 50
Cal Poly - EE - 307
CHAPTER 1717.1(a) T = A = (b) A = 10Av =80 20|Av =1=5|FGE = 0= 10000 | T = 10000(0.2)= 2000A 10000 100% 100% = = 5.00 | FGE = = = 0.05% 1+ A 1+ 2000 1+ A 2001 A 10 100% (c) T = 10(0.2)= 2 | Av = 1+ A = 1+ 2 = 3.33 | FGE = 1+ 2
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
UCLA - BIOSTAT - 100B
Midterm OneFriday the 2nd of February, 2007Name:General Comments: This exam is closed book. However, you may use two pages, front and back, of notes and formulas. Write your answers on the exam sheets. If you need more space, continue your answe
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
UCLA - BIOSTAT - 100B
Midterm One SolutionsGeneral Comments: Overall I was quite happy with the performance on this exam. The mean was 84.5, the median was 87, the high score was 100.5 and the low score was 45.5 I have given approximate grade ranges below. However, reme
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
UCLA - BIOSTAT - 100B
Final Exam SolutionsGeneral Comments: The final was a little harder than the midtermsthe median was 76, the low score was 42, and the high score was 100.5 (wow!). I was a little disturbed at how much trouble people had with the last question as th
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228