Second Midterm Exam ESE 232

October 29, 2007

Your Name: SolV\+1--DY\ &

Your 10 #:

You need to show all your work in order toget the full credit. There are 3 problems inthis examination.

lOV

1.(30 pts) Find the Quiescent point, (10, Vos), for the following circuit.

Given: kn(:)=25JlAly2 and ~ =1 Y

lOV

1.5MO

lMO

Pr s ~ C{ V\I\.L.. t-h.cvt tL<.... f.J HO S t·y~ VI. ~ iSfo v- i ~ ir. Sc\,:ttt y ('••..-n\:,-

LP ~ i Ct'\ (~) (V~S - Vt.) ~ LD ~~)QExI66X ~s _ I) 4'- -bLD =- I d... ~ (lI6JS - I) ><. I 0 ' , ,CD

~in (fl.- L& =- 0) \f G ::: ( '. _ '\ _ , r-, ='1. (.Sil) 10 - 4-\1

\f~::::. V9&T4(3/xt03)~ 4- ,\\0~ub.;-h'~\-e.- CD ; 0W @-.---------

\}'1s t-0:LoS- (V9$-I)~XI(J-6) (6q XIO:!.) = ~

VGj5, i 0, Lt~75 (V6JS _()'2... -=- Lf-

(), Lf ~ 15" (Vgs -I)'1.:: - V6) s + Lt-

( VqS - \ )7..-: _ 1. ,0 > VGJS T ~. 2.1

V9~2-- ~Vb1S f l "::. -.l.0S- V~~ 1" 8'"I:L J

V9s,L+ 0 ,o~ V,",S _ 7 \ 2.' ::;..0 ~ V(q$. == - ~,7/ \f 0 r + 2,66 Y'?/ s ,n UL t-~ -tyO\ I>') <: \' ~tc1r !' Sa. 0 (\ V6, <; :::- :2..66 V "\ ®

p0\t- Q) ir"1-D CD: II> =: /Q.S(2..,66- 1)<xJO-6-=. 34-.4-MA

V D S. =. I 0 - ED C 7 S-x 103> + 3 9" x 103 ) ::::. 10 - 34-. y x 10- r; ( II Lr X 103 )

Vps. =- b,03 1/ (/-'pcl:: ~<; =-C,O~ "/ V9S-V-/;-:=, 1.6-6lQ =- (ID) I/Ds) "" (31/, '-lIAA,2.6N) () IC, \~ nG1l\S;s-jor 1$ ; (\ &''\t.. •..'''''~.

/

2.(40 pts) Find the Quiescent point (ID, VDS) for the following circuit. Find the small signal

voltage gain Av = Vo , transconductance gill' ROllI, and Rill' Cj, C2, and C] are large blockingv',capacitors. (Hint: gm = k~ : (VGS - V;) For the small signal model, use the one in Figure

4.37(a) on page 291 of the text.) Given: kn (:) = 500,uNy2 and V; = 1 YlOY"

lOkD.

v~,'~

J

C' -­'2

1 tvID.

2 1vID.

::WkD.

+100 kD.

CXpo;~) W~- n-e€A==;:> CD. P~c. ITO,.... be ( 0 ~ c=S

'0 "

F-Or-

20K

pC. c..V'.lA!'1 <;\ 5.0. V\ 0 p-e •..... C\ r c:'tA.;1-\

Vps;-=- IO-L20X1O» (1.I>TL>() '" CD

-LD -:: -1 t:~ ( W ) (Ij(: ~ _ \j t)2.. (Ct~(U~ ty.o..n<;i!-/-bY-)~ L- '4 v I{'\ ~~+"'...c..-ii'\7-

- . -6 . '\ .2 r:::'\1-.0 =: 250 >< \0 (V&,s. - \) ", \2:;J

Vtqs = 2 VDS (vol-Cc'Ao-e- D'ui~ ~~ = D )2-r'2T \

V9s = D ,Y-VDS ",0Ix :::.\iDS ~ :::.0,)..)1-,0-6 Vos '" ®

(t 1" L 1" 2.) XIO~

PLAt- @ i-@ lAW CD

V - 1»( -(, L -6) r,:o-,D S - !0 - (1-0 X 10 2tsD X \ 0 ( V9:> - I ') T 0 I 2.. )( 10 VDS " , &PlA-t ® i"tv @

Vps = 10- (LOX 103 ) (2507< 10-b ( D. 4-\!DS - \ )>- -t- 0, 2..IQ<r 6 VPS )

~. , D.~VDS"2- ~ 3.0 Vcs -!:; ==-0 V DS ::: . .s V ov- -f ..2S"'·'

G Vos. = !;VV6-,S~ Q.q.VDS =- D,tr x !;-

c kck vos::: ~' "/ V9 s - Vt = .~ - ( =.' I ( 0 K)

ID ~ 2..5""0 x \ O-b C V6JS - !)2.. (fY1)VV\ ®)Ii:> :::. :l!:() ;>(\0-6 ( 2.. - l ) 2.. =. !2.50)< 10-6 Pr C o~ 2SV~-V( lOr)

:::;> Q ~ c rp) VD~) ==- ( 2~' A A,;' ,S \()

Om = t-~ (~ ') CVG,$ _ V.t,) = (500 )( \\1-6 ) ( ';).. - I) = ~Xltff1-(v

NOLJ SowA- c.. <:>t '(")~ \ ':J c;:, s ,\J £. it"<Jt- \.vL- mocAe.. \

; V". f;·S'.A\..L 4.5((0..)

\

..~

t>

1J:: (1-\ I ~2.HS1- \ 11-\ -0. t-r

lOoK..D-

:~jf\ '\J~

th-7 _ -

S'"'

e ~(")=- 2...K~ l) 2- tv\..!L~. \ tJ\ ~

ko4- -= IH0- \l 2.0 KJ2 ~ I 9 . 6 ".0-

tTo = - 5 rv"\ V5 S ( I ~a II 2-0 t<c[2.. II loof::12 )

"=. -~OO)( \06X~s0( 1(, . 't )< 10") ~ - 1\+,2.. ~s

~ s -=-- e \n 1f1n-==-O \ q 9 Vi' ()10 F::+ e.,,,...

v; -=-'8.' Lnn

A/J~ ~ -;. - B. t

VL.¥'

3.(30 pts) A CMOS inverter is shown below. Assume that the transistors QN and Qp are

matched such that k, (~ ) = kp (:: J ~ 400,uAN' aud v" = Iv,pl = 2 V.

With VDD = 10V, find the voltages labeled as A, B, C, D, E, F, 0, H, and I in the graphbelow. If VI =VDD, what is the maximum current that the inverter can sink ifvo :S0.4 V?

f I I •t I I •f I I •: sl0'Pe~·l :

" ." ." ." ., .

, ., .I I • I

_______ ~ ~ L L L __! ~ I •t j: I •• • I Ij j , •I t I •I fl'I I I •I I I •I I I II I I •I I I II I I II I I II I ~ ,I I ~ ~

_______ ~ L ~l~ ~*_~ ~_~I I t t II , I t II I I f I, , I • II t I • II I I t II I I I t I

: slope= tl: : :I I I • II 'I • II I I • I

V'LL- -==- B(3Uoct '2Vt:) := 8(3.10 t'(2)p)) -=: 4~2!:V

V;: ~ ~ ~ S; V

~ ~ Vr H ::: 8 ( s 1/DD - J-V-t)7:. 8( ~.10 -(2}(1)) :=:-~\IS-v

f-t =- I/DP - V-t ':: 10- 2- = e \II

L. -:::... VOD ~ 10'1/

(;! he" ;- = V DO) Qpi is; off- <I-- 6l.N~S <,. ::/ v. 006...Lo :::.. k ( W" ) [fl I V) " .l. 'J,/L J -be) is AOU-_ ~ (\ L", \.vGt£'/J I t-"1 v cs. - /~',v p~ 'Lp = 400 x 10 -6 l (VDf)- Vf:. ) Vu J ~ to frlC\'f htAppe'1 ~ WMV\

~ -6 .. ) -3 .. ALDm~'1 :: 4f)Ox: 10 [( 10- ( -6>~4-J =- 1.4-4- )(10 -=- r \ 4-4-1"'(';

oE :::..~=-

A -=- V f)'; ~ V IJ 'D --=-­

B:=.. '!DE..tVt :::.Z

c..:::: \J DO ..•. V -c ::...:'L.

- Vt ===-- 2-

AVDD I

BQp

Va

J; L-DC0

_.N

o

!o\J

ill t..:::L -=- 7 V2ill -2 =- 3 V:2

D E F H 1