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Unformatted text preview: ECE 110 3/” 1L8” Professors Brunet and Trick October 17, 2005 HOUR EXAMINATION #2 LAST Name (use capital letters):
First Name (use capital letters):
Signature: Circle your section: AL1(3pm)Trick BL1(1pm)Brunet DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD 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 additionalpages. The Boolean Algebra
identities are also given at the end of 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) At least one of the four circuits below is the following waveform rectiﬁer: the output waveform
Vout is the negative portion of the input waveform Vin (i.e., Vow: Vin when Vin < 0, Vout= 0 when
Vin > 0). State which circuit it is and explain why, using the ideal diode model (V on = 0): (‘1‘) You must draw the linear circuit for each assumption,
completely analyze the circuit, and validate each assumption made. NOTE: If you start with the wrong circuit, start analyzing it, and show it does not work While
following directions (at), you will receive some partial credit. R (d) R
(a) I > + (b) + (c) ‘ + +
Vin an Vout Vin v Vout Vin Vout Vin A Vout
o The circuit is  [:1 (a) I] (b) (C) [El ((1) , Mﬁoﬂowﬁ)” W When the diode is on: ‘ ﬁrm: . r analyze: validate:
( § \
+ z : "Vin/[Z L 20 OIL/{S wine—n
e trout __ ~ z
_ v.0 —' t K (‘yL  O
Summag: when Vm .... Vent ...... ..
When the diode is off:
m: analyze: validate:
[2 O"E . 040 onla 40V
"r'l‘ + ‘l‘ _ 014 :0l/
E Von? % ' > U—[yl > O
 :  UEVI
Summag: when Vin .... .... .., V0“; Problem 2 (20 points) For the given BJT' inverter circuit 25 k9 Vi 9 6V ‘3 = 50, VBEON = 0.7 V, Vegan: 0.2 V (a) assume that the transistor is operating in the active region and write the KVL' equations for
the input and output loops. Find the function V0 = f(Vi) in terms of the circuit and tran51stor ' parameters.
1 v— 566 04: El“
0
'1’ ': 50.1],
522/4587?" — .
v = av— 542:2. (SOIL) vo=¢v—§Lsugs—o) [Va0.71;)
. 0 v\__0'7y _ oZS’EV. W (b) If Vi = l + 0.4 sin cot, use the result in Part (a) to ﬁnd Vo(t). V0 = /BV w (/+0.slsanwz—) No 11C: {ramse‘szé/ a c {TA/E 79% 0.2 i V i (0V 0 Problem 2 (continued) I (c) If your result in Part (b) is not a good approximation to the actual output voltage in the given BJT inverter circuit, sketch a better approximation of the output waveform over one period
below. ' V00) 10 oNAChOO" Problem 3 (I Opoints) , Complete the truth table for the given CMOS circuit. A high voltage represents a logic “1,” and
a low voltage a logic “0.” A
0
0
0
0
l
l
l
1 ’1 §0V each incorrec‘é 2.21% CS “6" a“; F inVerfed Problem 4 (I Opoints) Implement the exclusive NOR gate below using only NAND gates. Problem 5 (20 points) Consider the Boolean function F given by the truth table below: (a) (4 pts.) Write the canonical sum of product expression for F. F(X,Y,z)= {rid—X7211; x72 (1)) (16pts.) You must chooSe an optimized circuit for F. For each circuit below explain why you
would choose it or not. Wn'te neatly! ' {.2 gauges f cenxcc‘lc‘eng F4 ‘ a? 3 341‘ €$
of‘éutm/ éui 7 couxeci‘c‘l Problem 6 (20 points)
For each question below, check the one most correct answer. 1) The “sum” output of a full adder can be implemented using NAND gates only. E] always [I sometimes [:1 never 2) The “carry” output of a half adder can be implemented using NAND gates only. IX] always I] sometimes [I never 3) A half adder can be implemented using a ﬁll adder. always E] sometimes ' [I never 4) A full adder can be implemented using half adders and" few gates. always I] sometimes D never 5) With 8 bits it is possible to represent: [3 8 numbers Z] 256 numbers [I 3 numbers [:I 24 numbers 6) The largest decimal number that can be represented using 3 hexadecimal digits is: [:13x16—1 [13‘6—1 163—1 [123“6—1 7) To display the decimal value of one hexadecimal digit we need: ( [:I one E two E] four E] sixteen ) 7—segment displays 8) A 7segment display can be used to display (check all that apply): [I some decimal digits E some letters g all decimal digits I] all letters ...
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