This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ’ ECE 110
ProfeSSOrs Brunet and Loui October 15, 2007 HOUR EXAMINATION #2 1) Write your: (y
Last Name (use capital letters): ”So L“. l (,0 N S
First Name (use capital letters):
Signature: 2) Write your name and sectiOn at the back of the test DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD Make sure to Write your name AGAIN at the top of every page of your exam. A. Write or print clearly. Answer each problem on the exam itself. If you
need extra paper, there is an extra sheet at the end of this exam. Clearly
identify the problem number on any additional pages. The Boolean Algebra
identities are also attached to the exam. B. In order to receive partial or full credit, you must show all your work,
e.g., your solution process, the equation(s) that you use, the values of the
variables used in the equation(s), etc. You must also include the unit of
measurement in each answer. Students caught cheating on this exam will earn a grade of F for the
entire course. Other penalties may include suspension and/or dismissal
from the university. Problem 1 (20 points) The circuit below has an AC voltage source with a peak voltage of 4 V, and
a Zener diode with Von = 0.6 V and breakdown voltage VZ = 3 V. (a) [5 pts.] Determine the current iD through the diode when the input voltage reaches its peak,+4 V.
iD: ‘ 0' G8 MA   .. o. (a
60 — 32—— : o. (’8’ MA
5‘ .
(b) [9 pts.] Determine Vout as a function of Vin for all possible states of the Zener diode. List
them in the table below. Explain your reasoning for all parts. C210»: Vent : VI) “31 'oélndbo'ﬂ, Vm:U«D=O.éav Wax 050:9!! CA #4, iv: 0) bag, V06=\Ipz\lcu‘SZD—= Vw
wimp oliocQQ C6'uo. (ma«MW, Vw = ‘9: ~sz ”3" Brnoion
" 3V (c) [6 pts.] Sketch the output voltage Vout. ‘Label the vertical axis with the voltages at which
the behavior of the circuit changes. The dotted line is Vin(t), plotted for convenience. Explain your reasoning. '
UpéOL aha2?, “3169“ V w) 0.6:) Siam
‘  A (Var—“9’ 0, Ac atoota Ca On, «WNW9V. ‘12— g '
Wk» —3 «mama (haze “.345,
i cut! ﬁve/6(1)), Vow}: g, Vout (volts) 0 t (sec) U £7th “9?— ’3V $98308") ‘ — 1619; 340) AAJW‘L‘Q "“ éﬁﬂk “0157‘ M VM£VD=—3V' Ug<~3 Problem 2 (20 points) Consider the common emitter circuit below. C ZkQ [3: 100
VBEON = 0.7 V 6 V VCESAT = 02 V (a) [3 pts.] Using KVL, give an expression for V2 as a function of VCE. (b) [3 pts.] Using Ohm’s law, give an expression for V2 as a function of ic. (c) [8 pts. ] For all possible states of the transistor, give an expression for V2 as a function of Vin.
Explain and clearly justify all three parts. Problem 3 (10 points) Consider the CMOS circuit below. l
.
l
o .—
9‘ MiG—n 8:0 or (kso “Ni 8‘0) (a) [8 pts.] Fill in the truth table for F. 'Show your work. (b) [2 pts.] What gate does the circuit implement? N A ND Problem 4 (15 points) ' 7segment
display N06
eja r\4 Problem 5 (20 points) The Boolean function h(x, y, z) is deﬁned by the truth table below. ix: $352: +383? 4.1%?
Jr “F “‘32 ..
3 (55+ 13(2); Ali13 «82 +7032:
._: ‘. CZSQHg’i—wZ—ﬁ :%%+\8Z+1¢32: TagM8? +1763; 0r “)2; +i‘%+'7‘(%—Z
" ’  or fi+§e+7x§
“SEtrjzv 79: W 1§+35+Y§z (b) [6 ptS. ] Find a canonical SOP expression for the complement of h. ~—— h’= 7082, + R32 +11}, (c) [4 ptS.] Draw a twolevel AND—to—NOR circuit that implements the original function h.
Assume that the variables x, y, z and their complements x’, y’, z’ are available as inputs. [Hint use the result of part (b).] r\‘ Problem 6 (15 points) For each question, circle one answer. Show your work. aﬁ>~ ":DO— 6.1. [3 ptS.] The output is 6.2. [3 pts.] The output is (3) Always o (b) Always 1 (b) Always 1 (c)x Q<+03 .232 (c)x 83.43 :ﬂ': 0
6.3. [3 pts.] Which statement below is _N_(_)l always true? (M943 T 7 l
(a)x®y=yex (etao
(b)x®l=x' (d)x(y®z)=xy(sz 6.4. [3 pts.] If representing the positive integer N requires 5 hexadecimal digits (without
leading zeroes), then to represent N in binary (without leading zeroes) requires (a) between 5 and 8 bits (c) between 21 and 24 bits (b) between 17 and 20 bits (d) between 65 and 80 bits 6.5. [3 pts.] The complement of x y’ (w + z) + x’ z is equivalent to (a)x’y(w’+z’)+xz’ ' (c)(x+y’+wz)(x’+z)
(b)x'+y+w’z’x+z’
/—_‘ X? (we) \— 3E7: ”
 “K 7504+?) ‘ ﬁ
7* (figkm) .605)
. (a: +3 «ii3 (74+?) ...
View
Full Document
 Fall '08
 HAKEN

Click to edit the document details