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Unformatted text preview: ECE ZB MidTerm Exam Winter 2010 MidTerm EXAM (”>0 I ocl’tawg Name: You have 75 minutes to complete this exam. Two pages of notes are allowed, otherwise
there should be no books or other materials in your purview. You may use a calculator if
necessary. When you are ﬁnished, check that your name appears prominently and legibly on
the front page. The exam is worth 100 points total. IMPORTANT POINTS FOR GRADING: SHOW ALL WORK. ANSWERS GIVEN WITHOUT CLEAR
SUPPORTING EVIDENCE OF INDEPENDENT WORK WILL BE IGNORED. CIRCLE ANSWERS OR WRITE THEM IN THE DESIGNATED
SPACES. THOSE NOT CLEARLY MARKED AS YOUR FINAL
ANSWER WILL BE IGNORED. BE NEAT. ILLEGIBLE OR SLOPPY WORK WILL BE IGNORED. The Stone Age was marked by man's clever use of crude tools;
The Information Age, to date, has been marked by man's crude use of clever tools. Anonymous Page 1 of10 ECE 2B MidTerm Exam Winter 2010 Problem 1 (15 points total, 5 points for each part)
Unless otherwise speciﬁed, you can assume the diodes have a 0.7V drop under forward bias. SHOW YOUR WORK: unjustiﬁed answers will be disregarded. 270 Q a) For the zener biasing circuit at right, ﬁnd the output voltage
V and the diode current I : I 0 mA V =2g V
f
6~Eg . :51
Q70 3.3V Zener
(1N746A) [Sit/WI} b) For the circuit at right, ﬁnd the output voltage V; and the diode current I : III e) For the circuit at right, ﬁnd the output voltage V0", and the diode current I : Page 2 of 10 ECE 2B MidTerm Exam Winter 2010 Problem 2 (20 points)
A circuit has the following sdomain output voltage
4(s +1)2 V0", (S) : s(s + 2)(s2 + 2s + 2) Find the step response of the circuit, vom(t). A table of Laplace transforms is found at the
end of this exam. , I 7 .
 __, ”o x’ 07‘ 5 C a ’ 'i ’
gpr/w “C(ZLcjkf/V‘ WMMSIUM , A ‘ / A ,
dataL33 '2 j+ Page 3 of 10 ECE 2B MidTerm Exam Winter 2010 Problem 3 (15 points)
In the circuit below, the switch is closed for t<0 and the circuit is in the steady state.
Suddenly the switch is opened at t = 0. t=0 19 Page 4 of 10 ECE 2B MidTerm Exam Winter 2010 Problem 4 (20 points) In this circuit the switch has been closed
for a long time, such that the circuit is in
a steady state. Suddenly the switch is
opened at t: 0. c) Find the output volt:g:nthe S'domain’ ,Vi’”(s) 2 3/25 I}
\lMt/SV ; ”"i A W ”2:;t {97/53
7, (9413 S i :5
/ 5;? t ‘31 (Continued on next page...) Page 5 oflO ECE 2B MidTerm Exam Winter 2010 ( ............. problem 4, continued from previous page.) d) Find the output voltage van, (t) for t> 0 e) Is this an overdamped, underdamped,Q/criticallydamped response? Page 6 of 10 ECE 2B MidTerm Exam Winter 2010 Problem 5 (20 points total)
In the circuits below, each device is an enhancementmode NMOS FET with V; 21V and K" =1 mA/VZ. The constant Kn is deﬁned as in lecture such that K" (ng — V,)2 Saturation (Vgs 2 V,, Vds > VgS — V,) D Kn[2(VgS—V,)Vds—Vd§] Ohmic (0<Vds ngS—K) SHOW YOUR WORK: unjustiﬁed answers will be disregarded. +5 V
a) (5 pts) For the bias conditions shown at right, determine the region of Vg—IV
operation (saturation or ohmic) and then ﬁnd the drain current Id i Id
/. x 5 LI
// Saturation 9r Ohmic? (circle one) Id = 7 mA if
k__,/ / ., \ 3 v1 —
\ffs;(’/l\+{f$ / in I .1 5V
: L1 V :1; ;l./Z.(Vﬁ§’vé/B
<> ﬂ ”V [A +5 V
b) (5 pts) For the circuit shown at right, ﬁndV V0", . If)", = w E V _I:_
? V time :7 lé—t/Vii's’ V63 3 —
[i7 Vag :— EV 4mA Vivi; ”EV +10 V c) (10 pts) For the circuit shown at right, ﬁnd the required
source and drain resistances RS and RD that will make
Id =1mA and V =6V: 3M!) 0” ‘— .Vout
Ré— 2k Q RD— Mk Q
s _ A W
%.f Id 7 \Mzﬁ f1) 13(25 Page 7 of 10 ECE 2B MidTerm Exam Winter 2010 Problem 6 (10 points)
The circuit below uses a MOSFET whose IV characteristic are shown. é
Vds, Volts fly/LA
6/ Draw the loadline for this circuit on top of the graph, and circle the appropriate intersection.
What is the output voltage and drain current predicted by your loadline analysis? § {“l
V = V Id = mA 0!” Page 8 of 10 ECE 2B MidTerm Exam Table of Laplace Transform Pairs Winter 2010 Page 9 of 10 Signal Waveform f (t) Transform F (s)
1 Impulse 5U) 1
l
2 Step u(t) _
s
3 Ramp tu(t) :2
s
tn~l 1
4 Power oft u(t) n = 1,2,3... _
(n — 1)! Sn
5 Exponential e‘a’ u(t) 1
_ s + a
l
6 Da ed a t C" t
mp r mp e u() (“(1)2
n—l 1
7 Damped Power I e—atu(t) n 21,233.” n
oft (n—w <s+a>
8 Sine sin ﬁt u(t) ﬂ
s2 + ﬁ2
s
9 C ' cos t u t
osme ,8 < > 32 + ’32
10 Sinusoid Vaz + b2 cosEﬁt—tan‘l é]u(t) a:+ bZ’B
a s + ﬁ
11 Damped Sine [at sin ﬁt u(t) ﬂ
(s + (2)2 + ,82
12 Dam ed Cosine 6—6" cos ﬁt (t) S + a
u
p (s + (2)2 + ,82
k k‘
at + It
Damped 2ke cos(ﬁt + 4k)u(t) (s —p) (s —p )
13 Sinusoid where P = —05 + LB
(Complex Poles) 2 2 wt 4 b a(s+a)+bﬁ
V a + b e cos ﬁt — tan — u(t) —22
a (s + a) + ﬁ
k + k‘
Double Complex _a, _ 2 . 2
14 Poles 2ke tcos(ﬁt + 1k)u(t) (S p) (s — p ) where p = —a + jﬁ ECE 2B MidTerm Exam Winter 2010 Initial Conditions in the sdomain l [(3) +J— + SC C V _ V(S)
fl‘ (f) “(0) var)
— s
Capacitor has an i 1 v(0_)
impedance: SC 1(s) 2 SC V(s) — Cv(0_) V(s) = —I(s) +
SC s
l 1(5) + 1(3)
+ sL
+ éL m) i(0_) V(s)
_ _ s Li(0“)
Inductor has an L 1 . 0_ _
impedance: S [(s) : _Z V(s) + l( ) V(s) = sL[(s) — Li(0_)
S S Page 10 of 10 ...
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