MIDTERM EXAM [[ Dec IS 110 min

Transcription

1 izmir UNVERSTY OF ECONOMCS Department of Electronics and Communication Engineering EEE 331 Analog Electronics Fall 2015/2016 SOLL(r Ol.J$ MDTERM EXAM [[ Dec S 110 min NSTRUCTONS W W.ll til W Read all of the instructions and all of th e questions before beginning the exam. There are 4 questions on this exa m, Totaling 100 points. The credit for each problem is given 10 help you allocate your {ime accordingly. Do not spend all your tim e on one problem and on one part and anempt to sol ve all of them. Cal culators are allowed, but borrowing is not allowed. Your mobile phones must be turned off during the exam. w Turn in the entire exam, including this cover sheet. You must show your wo rk for all problems to receive full credit; simply providing answers will result in only partial credit, even jf the answers are correct. Please indicate the number of page where yo ur work is 10 be continued. Put your name on any additional malerial that you submit. Be sure 10 provide units where necessary. Last Name Name Question J 2 Points Grade Section J 25 Student No TOTA L The basic equations of the output characteristics of an NMOS transistor Ves Vos 10 i) VGS < Vln - 0 ii) VGS > Vrn wherek = K' ' (W) L a) VDS < VGS - VTn Kn [2 (VGS - VTJVO~ - V~sl b) VGS - VTn 5 VDS Kn (Vcs - VTJ' and K:, = ~l n Cox

2 Q1. (35 pts) Consider the Class B output stage shown in the figure (V cc = 15 V). Assume the transis tors are ideal with Vnc! =V CB2 =0 V VCEsn tl =VECsal2 = 0 V. The input v,(t), is a sinusoidal waveform v,(t) =V p sin root volts Vee i) Determine the average power dissipation over the load as a function ofvp. ii) iii) iv) Determine the total average power drawn from the sources as a function of Vp. Determine the average power dissipation over one transistor as a function ofvp. Determine the power efficiency as a function ofvp. v) What is the power efficiency and the power dissipation P T over each transistor when Vp =10 V. vi) For which value ofvp, the power efficiency 11 is maximum? What is the maximum power efficiency 11",,, then? vii) For which value ofvp, the power dissipation over any transistor is maximum? What is the maximum power PT(,""l over any transistor then? viii) f the ambient temperature TA = 25 C. The thermal resistances of the transistor are e,c = 20 0 CjW (junctionto-case) and eea = 50 0 CjW (case-to-ambient), determine the maximum junction temperature T, of the transistor, for maximum PT. ix) Nowa heat sink is placed over the transistor with e CH = 5 CjW (case-to-heat sink) and eha = 1 OOCjW (heat sink-to-ambient), what will be the junction temperature T, then? x) Assuming the transistors are not ideal (Le. VBE = VC02 = 0.8 V and Vee,,,! = VCC,,,2 = 0 V), draw the output signal waveform Vo. i) V S (+\::. \lp S/~ ln~+. p., S Ve., =- V T;..f!.,. : 0\/ Vo -:=. Ve; :: Vf' S'~ LN, t-, ----rhq. S;S~ O"Q.r ~ L. is; S;f\v:,,,:c<ai',, n 2. p~ _ v~ 2. RL ) lo-kj. fl!>wer 0\ rw.u (\ e,...,... -\v..o J;:<:>.1ft ~ l 2

3 iii) PD<4 r \)"'~ o(\1c ~,-o,...-. tv) P-r:::. PCco - P L =.1 ( J,..\(J/cc" :l.. '2-7i fl' VP~2-a1.. -._ " V('. LVfv~X '-t VeL Till... (v) VU,-::::1)V Vp : OV, 1Z... =bjl tfl", ~, ~ ~;,)2.3;. S'"2. 3%, dpr -::: :> dlif Vp=-.2.vc..c. _ l.j.,.'\';: 0, )\ ~ V,, - 7( 1i (v; ~ i ) ejc:;:20"c/w -rc.,.j - 7",:. p it """,\ ( G-jc. +Sell) Ti- 2rc=-l'?>, ~) (1-0 c/.. ) f",-.:t'2.foc 7fT"'t,,,\..r ~~ b,.rn31 owt!.9-c~::?9t./w 7" '/,,2.;- C J &,,,:, = 10 </...,; )(), Vf' ~l'o.1' " T<- ',-:.2\'+ (3./l)( 3f) 9<1" ': lot ", f=-.;j)~ "c LT" "lr 0c. +~\y,.b~ ~\)'~+-O' s~%. 'r ~m';'f

4 Q2. (25 pts) Consider the amplifier with active load given below. (a) Derive the small signal voltage gain, A,,= Vo/Vi Determine the values ofa, for (b) RL = 10 K (c) RL =100 K (d) RL =rt) Vi ( r " Q2,, - C co t Vee, Q l ( :4t { J, R, - Cvrr~('\~ (,{\,aor Circu it Pa rameters Vee = 15 V R J = 150 K Transjstor parameters {J = 60 VBEON} = 0 V Vr = 26 mv V Aln" n} = 80 V VAlpop}= 120 V -Vee Vce - V e (?,.J) - ( - V Ce ) ~ :: O.2.",A {'0 k. fl. ~ J-c 0 = Loti = 0, 2..",,~. 1 ", 100 c. 0 4 ~9M"; \T\;, S'Tlo -.,. eo 1"0 Q.... ~ ro:l -\ ~ 0-;,:: l- j\v\ V,,)( \'0/1 R.. '- /1 r:>1) -5""

5 e '> O,l.... A '-0 V~("(>") gov O.OcQ) ""g - - ;' D. CD - b'}- ",S; ~ co O.2{'t\A r'0:z. fr (""' ) ) 0.1 O. '2.rn -irn ~(j + 10 (3) 1lJ" c) ~L ~ ioo(..l '3

6 QJ. (25 ps) Consider the differential amplifier given below. TO indicates the small signal resistance of th e current source 1 0, (a) Determine the DC values of the transistors when Vi =Vz =O. Check whether the transistors are in forward active region or not. (b) Find the differential-mode gain. (c) Find the common-mode gain. (d) Find CMRR. (el f V, = 3 sin(2n*50tj sin (2n*200t) volts, and vz = 3 sin (2n*50t) sin (2n*200t) volts, then find Vo = Vcl - vcz. Re Vee vel Ve2 10, TO Rc Circuit Parameters Vee =18 V Rc= 2 K Ra= 10 K 10 = lorna TO = 50 K Transistor Parameters fj = 100 V ac(on) = 0.7 V V T = 26 mv VA = CJ) - Vee 0,+ - ~(j. (AO'(.) = -O, ~ -(r;ot)( 10k) = -O,:}-o.('= -1,21 fl "6 - { - {, 2J = '.2v? V(.(i, (~M) =,) Forw-evtl cmiw..( b)! "V c= 0,010 5,;' (vn){lqjt- ) J. J-Ol-< ~ "1-1- (J-,... = V,t ll 2. = 35,;' (2 7i.~ 5?+-) v, 2 -'-O r u _ tj" , 6

7 c J r fe, l b l r2c. c r (~...) v. (1. jm~ r 'i, 1 "" L '~,-:, ~ -== (i R(,-+ rn d-f(311rd ~ 7 2ro u, ==- {)Clf, - t{e,::o 0 ~ Ac-. ~ " ArJ". J '= Aq.O'~(O~ - <0 ""ce) 19Q -=- \ t.., lj.,l"" + V rjc._ = ({'f, 01/.(0)) (0,ob) 5.~ (1 7i ~ 2..<>.:11-) o!'!. 4,1~ ){(O'l S. ~ ('2l\2!>:rtJ V~tt~ -----r

8 Q4. (S pis) Consider the feedback amplifier given below. + v, R, + v" Circuit Parameters AM" = 10' w, = 10 5 rad/sec R, = 100 kq Ro = 200 Q /3= 0.03 a) dentify the type of feedback. b) Derive the percentage variation in closed loop voltage gain t1g,/g" c) Determine the percentage variation in closed loop voltage gain if open loop voltage gain AMB = 10. d) Derive the closed loop voltage gain G,Ciw) and determine the bandwidth if A v (j co ) = A Ṃ " 1+.lW Ole e) Find and compare the gain-bandwidth products of the basic amplifier and the feedback amplifier. 0) S'c:ui e.,. _ ':::~, ~ -~re <) e.- ~e.cl bc.j.:.... b) A....1c.-, C-r = CA~S"M'i P' S Mr ~~~ '\ v!-t- ft~ -= dl Ac.) dpr c.l 11<. 1-+ ~~ " - ( 1-+ '1(1) - 0,~ -::. = ~ Av (-+~r-) C!+ A,~)2- ~r--; "Tk" tl G." d p,,, 6&" 6Av ::.:?)." fi." (1-'~f") (::;" /'V -"A,~ A./ f.. A;l G:- " G" ( :AVf} t:.a"" -10~- A.,~ \D~ ~O = qr,'j.q 8

9 ~) &,, (1"') := - A s;v""~ Av(iw ) p i S 1"..Nf~.0~rd ~ +A~{jwl ~ l-,.,a """ A+ 1-- A""1.3 ~"' 1+1" ""t. W",A""e, -p, U '" A 'l "'6, vel (,.t r~ W t ' Wl t.v. A "(j e) &..;.,- baru>iw't.lfh f'~ A" ~ ~r 1tn(3' lvc =- -folt,40 ) = 10 1