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3 Pages

### Midterm2W05

Course: EE 307, Spring 2008
School: Cal Poly
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Word Count: 656

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306 EE W05 MIDTERM EXAM #2 CLOSED BOOK + 1 Cheat Sheet State any assumptions and show all work. (1 point) Print Your Name: Braun Part 1 (12 Points) Part 2 (15 Points) Part 3 (12 Points) Part 4 (3 Points) Total (44 Points) SOLUTIONS BRAUN No unauthorized help given or received. (1 point) Signature: FOR THE ENTIRE EXAM, UNLESS OTHERWISE SPECIFIED, USE: The CVD model for all diodes -- including breakdown diodes...

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306 EE W05 MIDTERM EXAM #2 CLOSED BOOK + 1 Cheat Sheet State any assumptions and show all work. (1 point) Print Your Name: Braun Part 1 (12 Points) Part 2 (15 Points) Part 3 (12 Points) Part 4 (3 Points) Total (44 Points) SOLUTIONS BRAUN No unauthorized help given or received. (1 point) Signature: FOR THE ENTIRE EXAM, UNLESS OTHERWISE SPECIFIED, USE: The CVD model for all diodes -- including breakdown diodes -- with VON = 0.7 V & VZ = 6.2 V. For FETs, VTN = - VTP = 0.8 V, KN' = 100 A/V2, KP' = 50 A/V2, = 0 V1/2, = 0 V-1. Show all work, state any assumptions, and test hypotheses. 1. Diode Circuit (12 points) Please analyze the diode logic circuit shown using the constant voltage drop model with VON = 0.7 V & VZ = 6.2 V. Determine the Q-point for each diode and the value of output voltage, VOUT. VDD = 5 V Hypothesize: D2 & D3 ON D1 OFF Analyze: VOUT = V3 - VON2 + VON3 = 3.0 V V - VOUT 5 V - 3 V I D 3 = I R 2 = DD = = 1 mA R2 2 k V - VON 2 3 V - 0.7 V I R1 = 2 = = 2.3 mA R1 1 k Since ID1 = 0 A, ID2 = IR1 ID3 = 2.3 mA 1 mA = 1.3 mA R2 D2 V2 = 3 V V1 = 1 V D3 2 k VOUT D1 R1 1 k VDD = 5 V VD1 =V1 (V2 VON2) = 1 V 2.3 V= 1.3 V V2 = 3 V 0.7 V 0.7 V R2 VOUT R1 Test: All diodes operate self-consistently. V1 = 1 V D1 D2 D3 VOUT VD [V] -1.3 V 0.7 V 0.7 V 3.0 V ID [mA] 0 mA 1.3 mA 1.0 mA 2. Diode Circuits (15 points) The plots below show output voltage traces in solid curves. The input signal (dashed) for each case is a 10 V sine wave (20 V peak-to-peak) with a frequency of 1 kHz and no DC offset. For each output voltage trace, select the number of one circuit from the circuits below that would produce the given output trace for the given input trace. 10 5 0 -5 -100 10 5 0 -5 -100 10 5 0 -5 -100 3 0.5 1 1.5 0V 2 [ms] 8.6 V 10 5 0 -5 -100 VOUT [V] VOUT [V] 4.3 V 7 0.5 1 1.5 2 [ms] VOUT [V] 5 0.5 1 1.5 2 [ms] 9.0 V 6 0.5 1 1.5 2 [ms] 25 20 15 5 10 0 -5 -10 VOUT [V] VOUT [V] 4.3 V 4 0 0.5 1 1.5 2 2.5 3 [ms] For the circuits shown below, assume that the diodes obey the ideal diode law. Each resistor has a resistance of 1k , and each battery is a 5V battery. For circuits 1-8, the output voltage, VOUT., appears between the upper right node of the circuit and ground. In circuit 9, the output voltage appears across the resistor. 1 VS VOUT 2 VS VOUT 3 VS VOUT 1 F VS VOUT VOUT VOUT VS 4 5 10 F VS 6 47 F 5V VOUT VOUT VS 5V VS 5V VS VOUT 7 8 9 3. MOS Biasing (12 points) Please analyze the circuit shown to the right and determine the Q-point of the transistor. VTN = - VTP = 0.8 V, KN' = 100 A/V2, KP' = 50 A/V2, = 0 V1/2, = 0 V-1. W/L = 5. VDD = 4.0 V R2 50 k RD 1.5 k M1 R1 150 k RS 500 R1 150 k =4V = 3 V > VTN R1 + R2 150 k + 50 k K N = K N (W / L) = 500 A/V 2 Hypothesize: M1 ON, SATURATED VG = VDD Analyze: K V - VGS I D = N (VGS - VTN ) 2 = I RS = G 2 RS (VGS - VTN ) 2 = 2 VGS + [ 1 1 = = 4V K N RS 500 A/V 2 0.5 k 2 (VG - VGS ) K N RS 2VG 2 2 - 2VTN ]VGS + VTN - =0 K N RS K N RS Solve for: 2VG 1 1 2 2 VGS = -[ - VTN ] [ - 2VTN ]2 - 4[VTN - ] K N RS 2 K N RS K N RS = -[4 V - 0.8 V] [4 V - 0.8 V] 2 - [(0.8 V) 2 - 2 (4 V)(3 V)] = - 3.2 V + 5.797 V = 2.597 V ID = VDS KN (VGS - VTN ) 2 2 500 A = (2.597 V - 0.8 V) 2 = 807 A 2 2V = VDD - I D ( RS + RD ) VGS [V] VDS [V] ID [mA] 2.60 V 2.39 V 0.807 mA = 4 V - 807 A (0.5 k + 1.5 k) = 2.39 V Test: VDS = 2.39 V > VGS - VTN = 2.60 V 0.8 V = 1.80 V (3 points) 4. BJT Operation. The NPN-BJT to the right operates in forward active mode with a base current of IB = 50 A and a collector current of IC = 1 mA. Label the terminals of the BJT: B for Base, C for Collector, and E for Emitter. Determine the values of emitter current (IE) and F. I E = I C + I B = 1 mA + 50 A = 1.05 mA I 1 mA F = C = = 20 I B 50 A C B E IE [mA] 1.05 mA 20 F
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Cal Poly - EE - 307
1.1 Answering machine Alarm clock Automatic door Automatic lights ATM Automobile: Engine controller Temperature control ABS Electronic dash Navigation system Automotive tune-up equipment Baggage scanner Bar code scanner Battery charger Cable/DSL Mode
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
Anthony Milne PSCI 3143 United States Nuclear Energy Policy The Occurrence of Nuclear Power in the World's Most Powerful CountryI. IntroductionIn today's society, the issues surrounding energy have become of ever increasing importance and concern
To Vote Or Not To VoteIntroductionLiterally meaning &quot;rule by the people&quot;, democracy is a word that has become synonymous with the United States of America. Based on the design that the government will satisfy the desires of the people, the United S
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
Cal Poly - EE - 228
Cal Poly - EE - 228
Cal Poly - EE - 228