Device Data - Microelectronic Devices and Circuits - Solved Exam, Exams of Microelectronic Circuits

Download Device Data - Microelectronic Devices and Circuits - Solved Exam and more Exams Microelectronic Circuits in PDF only on Docsity! EECSIOS dof 15 Fall 1999 | | Microelectronic Devices and Circuits- EECS105 | Final Exam Wednesday, December 8, 1999 Costas J. Spanos ; University of California at Berkeley College of Engineering Department of Electrical Engineering and Computer Sciences Your Name: Ot licla Guts (last) dirs} Your Signature: - 1. Print and sign your name on this page before you start. 2. You are allowed three, 8,5"%11" handwritten sheets, No books or notes! 3. Do everything on this exam, and make your methods as clear ax possible. Problem I é24 - Problem 2 123A WV G=69 Problem 3 _/24 0 & 2 193 Problem 4 (24 TOTAL 100 MOS Device Data (vou may not have to use all of these... UnCox= SOHAIW?, thpCox= 25HAV?, Vipq = “Wry = 1V, Lin = 2m. Vas = 0. Aq ® ep = O.LV" when L = 2pm, and it is otherwise proportional to V/L. Cox = 2:38 hum”, Cjq = 0.14F um, Cjp = 0.3 fF ham?, Cisiyn = 0.50F/um, Coup = O-35EE um, Coy = 0.5EF/um, Coyy = 0.58 /um. npn Data 15= 10717A, B= 100, V4 =25V, 9 o'74, p= 50, Va=25V, ¥ pnp Data Is EBCSIOS 2of 15 Fall 1999 Problem 1 of 4: Answer each question briefly and clearly. (4 points each, total 24) ‘Mark in the table below the npn Bipolar Transistor in forward action mode the direction of flow, and the ie of flow (drift or diffusion) of electrons, in each of the bulk and depletion regions. waitin elehon o> Fe BE juneticn BE jumeciom drift . v Tv yi LL (place a mark at the diffusion Viviv appropriate box to indicale your answer. You can choose more. ~ viviviw | than one box if appropriate.) ~ | Where is the maximem [Fl field in a forward-biased junction? Please place a mark on the graph below. depletion pape og SOP nype in (Gone) ‘What happens to the drain current of an n-channel MOS transistor in saturation, when L and W increase proportionally? (ie. L and W increase, but W/L stays constant, Assume that Visg, Vpg and Vips stay constant. Do take A into account!) J i ii — v L at Tos L dalog (bs-vr)% 8 (te Aas} 2 L  EBCSIOS 3 of 15 Fall 1999 ‘Name one advantage and one disadvantage of'a MOS current source employing a CG buffer, ver- sus one that docs not employ one. 5Vv Current Mirror advange: yg ho, 7 suprly cheat Common, oa vo less. Gate Tsupply RO fe t tee Disadvantages Rerbrer! V0 rage ‘The following multistage amplifier is meant to deliver a voltage signal to a relatively small ohmic toad of IK. Mark your choice of the last stage, and write a brief justification, k A to hos very Cow S Catpor Pesistanit. {bes c - RL thang CP) Bad 5" seme [| opm [| ce [SE sheet 07) rd 0 y diye Saput Pesijronie, hr moles 0 youd Weuol Wage whoye bdler “The following voltage amplifier bas a voltage gain Av = -100, What is the capacitance “seen” by the signal source, due to the added capacitor C,=1pF as shown? c i eins Cael t~ A) Las ‘¢- 6100}, Av=-100 Leas tok A-ere0) -— Cn at EECSIOS 4 of 15 Fall 1999 Problem 2 of 4 (28 points) ‘The following p-channel common source amplifier uses a rather primitive current supply: itis a simple resistor Rp 10k2 tied between drain and ground. L = 2g, 29 = ove For each of the following questions, make sure that you show the expressions before you plug in the specific vahies, A correct expression ts worth 70% of the credit. even ifthe numerical caleuia- lon iy incorrect! a) Find WIL so that when Vpras-3-5 and Vin = OV, then Vout ~ 2.5V. Do take Ay into account! Note that L=2um. (4 points) Tos 25”. 25qWa et Mpton ¢ Vas Mipl (t+ tes. re) Han rc 5nd bos OYE MERE 25H ak sagas f OB t Adidas wy 2EHA HOSP iba) OE wi= 28 i2 >) What is the minimum and the maximum ontput voltage for this amplifier? Make sure you men tion the limiting reason for each case (i.e. transistor X falls out of saturation, or current source Y hits its minimum voltage drop, ele.), (4 points) MAX Voulm 4. gv MAX Limited by: X fe wut! of coninatiOg MIN Vout= — Q) MIN Limited by: 6«/2  EBCSIOS Dof 15 Fall 1999 by Draw the two slage amp 2-port model (1.2. draw one 2-port of each stage and connect them properly together. CS is a transconductance amp, C'C isa voltage amp). Write the expressions for the quantities shown below. in terms of the device parameters, roe, Zoe aNd R, and Ry as needed. (OK to use the simplified formulae). (6 points) TT outs eg MY t Parameter Expression Gm; dm Rint oo d Raw 4 Tor Hoe, ‘Ava Vg t fo Coall roc, WA) Rout : L EBCS105 10 of 15 Fall 1999 c) Draw the overall voltage amp 2-port for the entire amp (i.e. draw one 2-port that represents the entire 2-stage amp), and derive expressions for the Ay, Rey- Roy a8 Well 45 Vaux! ¥s in. terms of the deviee parameters, and ts), fpe2» Rg and Ry, as needed. (6 poinis) l _ Rout ae roa $ Ris Ay % ? | A | pot YT Po Ay= ~Gun,CRewt liRing) Vout ay. Re Vs Rot Rost a ‘Parameter Expression Rn ro Oi) Rout ay Re gma pe A — ~den log Ir Beal Vocal RY) outs, Re —Bralfo, Boe, LP teal (ora WRAY) Fopasy ae  EECSIOS Hof 15 Fall 1999 4) Assume that Viaiag'= 1:5¥ sand that the minimum-vokage-aoross the current sources is 0.SV. Find the maximum and minimum voltages at the drain of M1-and at the emituer df M2: Make sure you mention the limiling reason for each case (Le. transistor X falls out of saluration, or current source ¥ hits its minimum voltage drop, ete.) (6 points} re © Viena = Ves-¥y = | ev-lv4.0.8¥ | wy Memin = OSVHOFVE LIV |____—_t Peed * Vimox # %SV becouse Ve > ASV tO7V= 5.2V > Vpo Viwax = 4-5V-O.7V Zev] Min Tv Node + Voltage Max Voltage Reason for Min Voltage Reason for Max Voltage drain of Vpsat’ ®, Min vatiage actors " mer of Mi inpn 9 & OSV 3.V Ves of M2 “a Went of mi | Min oltge WAV | ge Veg of M2 ia across WSup2 | EBCSIOS 12 of 15 Fail 1999 Problem 4/4 (24 points) The following is the 2-port representation of a vollage amplifier, where the “Mill and Ry have been added as shown. ‘Ay = 100, R= SK, Ri = SKO, Root = 2, Ry 7 SKA, C= O4pF, Ry = LOOKA. Jemenis C., Miller Approximation Refresher: Note that the Miller approxiniation applies to any kind of complex impedance connected between the nodes that exhibit voltage gain Ay. In geveral, Zy4= Z,/(1-A,). (Here a bold symbol indicates a complex number). As you know, for capaci 1, = LoC,, 50 it tums out that Cay = Cy ‘A), Below you will be asked to apply the Miller approximation for expacivors, as well as for resistors Miller approximation refresher: these two circuits are almost equivalent. Zy 5 1 . ‘1 Vp “1 Ya tn a =I == — Zag = Zyl(l-A)  I of 15 Fall 1999 a) For the voltage amplifier shown in the previous page, find the overall DC gain (v,9/¥,) with no Miller resistor in place (R,, = infinity). (5 points) Rerbhe tt Aut Rint Re _ tlods 8k 2 Fhe Ms Chiat PRL + Rot) “Touatoray = w28 8) Find the overall DC gain expression snd evalua Then R,, = 100KA. (5 points) Ry =100e= Ry? rian ~ |F? Beall By = Skartke _ 9,63 KA bea Vout . AviRinlim) Re Hor oes Kayser) oy jug Cen Ren t Re) Rit Rot Ses kR1OKR ' Expression Ra LOOK Nous Ayr CRinll yg) Re Pn Fu ft-Av) | ziu3 Chiat te) (Rt Roe) | ERCSIOS 14 of 15 Fatt 1999 ©) Find the esgp of Ivguy'¥e when Ry, is infinity and when Ry, is 100kQ. Hint: apply the Miller approximation in the vesisfor and the capacitor separately. It is not nevessury to salve this consid- ering the complex impedance of the resistor and capacitor taken together. (5 points) Gy ot rf (y= Vole Coe % 40 pF Lor Bure Bs all® [Rad R-Gye 1OF sec SD Wadb= Tee” 10 Mend/roe for Rye 10KR Rn 1k# ‘- eked R= Rian ees LES = 07M _ bead Mad Rly = 2.66 x17 see > Wed 8 = rst. * * sec Express T Vatue in | xpression |. Mradsee | \ I eggptoe | 4 En =pleAs) | 10 Ry infinity Wea)" splay, " | Gat Rin Cen Lay Byli-A) | xg or te Ps rohan | Rall Ba Re) Cm | 3