Assimen Tronic 2

7/26/2019 Assimen Tronic 2

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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

--------------------------------------------

112111742

--------------------------------------------LE

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!"#r""

Ma$%r

&EE '

LE '

CE '

TE '

ME '

O(E '

MCE '

NANO)

Q1

50%

Q2

50%

Total

100%

!"*lara+i%, %f %ri#i,ali+-.

e !ecl"re t#"t "ll $entence$% re$&lt$ "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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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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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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) $&k*

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) "'k*

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!igure 1(b): onostable 555 Timer ;ircuit

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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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Gut+ut voltage ave%orms 4++ 20

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ii"

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Ri

Rf

+

_

V+10

-10V

Vin

FigureQ1 (c)


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%


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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%@; 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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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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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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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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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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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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$ 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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