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Date & Time Submittion 26 jan 15:
EEE1026 Electronics II Assignment T2 2015-2016
FACULTY OF ENGINEERINGASSIGNMENT REPORT
COVER PAGE
EEE1026 Electronic II
(TRIMESTER 2 SESSION 2015/2016)
Muhammad Firdaus Zulkifli
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112111742
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Ma$%r
&EE '
LE '
CE '
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O(E '
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Q1
50%
Q2
50%
Total
100%
!"*lara+i%, %f %ri#i,ali+-.
e !ecl"re t#"t "ll $entence$% re$<$ "n! !"t" 'entione! in t#i$ reort "re ro' o&ro*n *or+, All *or+ !erie! ro' ot#er "&t#or$ #"e .een li$te! in t#e reerence$, e
&n!er$t"n! t#"t "il&re to !o
t#i$ i$ con$i!ere! l"i"ri$'"n! *ill .e $eerelen"li$e!,
C!
/u0mi++"d.
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M"''"
! ir!"&$
&l+ili
A'ir&l $"i3
4,Nor"#'"! $#"#
(11112510) Sin"t&re7 Sin"t&re7
Yes
No
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EEE1026 Electronics II Assignment T2 2015-2016
Question 1
Astable Multivibrator
(a) The astable multivibrator circuit is shon in !igure 1 (b)" The #ut$ c$cle an#
%re&uenc$ o% the 555 astable is 60' an# 10 * res+ectivel$"
!igure 1 (a)
i" ,+eci%$ a value %or ensure that the average current through the I #io#e #oesnot e.cee# /0mA"
I+ea /0mA 0"6 50mA
In 3T4 e assume E 0"7
E I+ea 0"7 50mA
189
ii" E.+lain (2 sentences ma.imum ) the +ur+ose o% the #ecou+ling ca+asitor
- the +ur+ose o% the #ecou+ling ca+acitor is to shunt an# absorb the noise "on the
other han# 4it maes the signal smoother "
iii" give the reasonable value %or the #ecou+ling ca+acitor
- the reasonable value %or the #ecou+ling ca+acitor is 0"1u! "
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EEE1026 Electronics II Assignment T2 2015-2016
Monostable 555 Timer
(b)
Table 1: Monostable Operation for variousR
R (k Wcal W
!! !"!#000 !"0#!$u
$& 51 52$#5$5u
"' &$'#000 &$5#&&u
) !!k*
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EEE1026 Electronics II Assignment T2 2015-2016
) $&k*
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EEE1026 Electronics II Assignment T2 2015-2016
) "'k*
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EEE1026 Electronics II Assignment T2 2015-2016
!igure 1(b): onostable 555 Timer ;ircuit
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EEE1026 Electronics II Assignment T2 2015-2016
Question 1 (+
i#
1"?
@TP -1"?
in 10
=sing the %olloing e&uation :
=TP - i % ( out - )
@TP - i % ( out >)
To calculate out 4 assume the voltage loss across circuit 1
out> in B 1
out- in >1
,ubstitube in 10C"into out (- ) an# (>)
out > >10 -1 D
out - -10 >1 -D
!in# the ratio beteen i an# %C
>1"? -i %(-D)
% 5i
F %rom the calculation 4the value % larger than i4""
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EEE1026 Electronics II Assignment T2 2015-2016
Gut+ut voltage ave%orms 4++ 20
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ii"
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Ri
Rf
+
_
V+10
-10V
Vin
FigureQ1 (c)
EEE1026 Electronics II Assignment T2 2015-2016
F ase# on calculation an# P,+ice simulation there have a slightl$ #i%%erence values o%out" It is because the voltage loss #uring carrie# out %rom the circuit "
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Cwi= 3pF
Cwo= 5pF
Cgd= 4pFCgs= 6pF
Cds= 1pF
IDSS= 6mAVp= -6V, rd=
!"#3$1%
EEE1026 Electronics II Assignment T2 2015-2016
Question 2
i# Hetermine v
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rearrange the e&uation ""
(0"2) v /"8v 7"2 0
,olve the e&uation 4ill get 2 #i%%erence values %or v i );< 1 (2 (19 > 19 )( 0"1O! )
1"5D*
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EEE1026 Electronics II Assignment T2 2015-2016
%@; 1 ( 2 (o > @ );c 1 (2 (/9 > /"D9 )( 8"7O! )
8"D*
%@, 1 ( 2
(e&);s e& s KK (1 gm ) 1 ( 2 ( 8D6"D69 )( 10O! ) 1"2 9 KK ( 1 1"1?ms )
/2"0/* 8D6"6D 9
v# %@ %@, /2"0/ *
vi# #ra the high %re&uenc$ ac e&uivalent circuit %or the netor
< H
r ! R 8 R 9V / $
/ ' V / $
,
vii#
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EEE1026 Electronics II Assignment T2 2015-2016
thi 19 > 19
0"DDD9
;mi (1 B (-2))8+!
12+! ;i /+! > 6+! > 12+!
21+!
%i 1 (2 ( 0"DDD9)(21+!) )
1/2"7/ *
tho /9 KK /"D9 ;o 5+! > 1+! > 6+!
1"6D69 12+!
;mo (1 B 1(-2))8+!
6+!
%o 1 2 ( 1"6D6 )(12+!) 7"?2 *
iii"
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&'(
,i6ure Q2 (b
Ans7er
igh B %re&uenc$ ac e&uivalent circuit %or the netor be%ore a++l$ing illers Theorem
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EEE1026 Electronics II Assignment T2 2015-2016
igh B %re&uenc$ ac e&uivalent circuit %or the netor a%ter a++l$ing illers Theorem
here ;a ;gs (1 B Av ) an# ;b ;gs (1 B 1 Av)
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EEE1026 Electronics II Assignment T2 2015-2016
There%ore4 the value o% Qti an# Qto are %ollos
Qti Qg# >Qgs (1 B Av)
Qto g# > Q#s >Qgs (1- 1 Av) > Qsn > Q@
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EEE1026 Electronics II Assignment T2 2015-2016
Av 2m+j 471.239 n
j 471.239n+2m+20u+j157.080n+6.667u+0
Av
2m+j471.239n2.027m+j785.399 n D?6"6?0m B R18D"?/O
Av D?6"6?0m B R18D"?/O
To %in# the KAvK
KAvK 986.680m2+(149.83 u)2
8Av8 ) 0#'&
To %in# Ni
Ni 1 Qti
The value o% Qti is %rom
Qti Qg# > Qgs (1 B Av)
,ince value o% Av e alrea#$ calculate# earl$ 4substitube into Qti e&uation
Qti 1u > R157"0?0n > R871"2/Dn ( 1 B D?6"6?0m-R18D"?/u)
DDD"D/n > R16/"/57n9-1
,o 4
Nti 1 Qti
1
999.930 n+j 163.357n
D78"07/ B R15D"1//9
9ti ) &$#0&!k ;15#1!!k*
KNtiK 974.037 k2+(1.59.133k)2
D?6"D519
89ti8 ) '"#51k*
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EEE1026 Electronics II Assignment T2 2015-2016
The value o% ;ti can be obtaine# b$ using "
;ti ;g# > ;a
here
;a ;gs (1 B Av)
/+(0"D?7)
2"D61+!
,ubstitube the values into ;ti
;ti 1+! > 2"D61+!
/"D61+!
To %in# No 1 Qo
here Qo Io o
$ re%erring the the %igure 4o can be %oun# b$ a++l$ing S@ circuit "
gs -o
Qto Qgs >g# > Q#s >Qsn >
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EEE1026 Electronics II Assignment T2 2015-2016
$ a++l$ing S;@ into the circuit 4the e&uation can be obtaine#
Io I$t B gmgs Qto > gm o
Qo Qt > gm
Qo Qgs > g# > Q#s Qsn >gs > gm
R871"2/Dn > 20u > R157"0?0n > R517"0?0n >6"667u >2m 2"027m > 7?5"8n9-1
,ubstitube Qo into the e&uation
No 1 Qo
1
2.027m+785.4n1 ) $!#!$0 ;0#11*
There%ore to %in% the KNoK
KNo K 493.3402+(0.191)2
8D/"/809
KNoK 8D/"/809
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