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Diffusivity - Microelectronic Devices and Circuits - Solved Exam, Exams of Microelectronic Circuits

Main points of this past exam are: Diffusivity, Performance Depends, Minority Carriers, Signal Output Resistance, Junction Area, Small Signal, Common Drain Amplifier

Typology: Exams

2012/2013

Uploaded on 03/22/2013

raghav
raghav 🇮🇳

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Download Diffusivity - Microelectronic Devices and Circuits - Solved Exam and more Exams Microelectronic Circuits in PDF only on Docsity! EECSIO5 1 of8 Fall 1999 Microelectronic Devices and Circuits- EECS105 Second Midterm Exam Wendesday, November 17, 1999 Costas J. Spanos University of California at Berkeley College of Engineering Department of Electrical Engineering and Computer Sciences Your Name: OFFICIAL SOLVIT OMS (last) (first) Your Signature: A F7Q093 1. Print and sign your name on this page before you start. 2. You are allowed two, 8.5""x11” handwritten sheets with formulas. No books or notes! 3. Do everything on this exam, and make your methods as clear as possible. Problem 1 /30 Problem 2 /35 Problem 3 (35 TOTAL 594/100 AvetAce — (67 studans) 14.6 STAWDARD DEWATI OW EECSI05 20f 8 Fall 1999 Problem 1 of 3 Answer each question briefly and clearly. (30 points) Explain briefly why BJT performance depends so much on the diffusivity of minority carriers (Spts) lets che diflaston of welnanry tarsters her buat thers to rye re.ene.blosed colleen funerloa that reofes the praas/'sin woh How does the small signal output resistance of a BJT depend on its size (emitter-to-base junction area), when Vas is held constant? (Spts) vee /| VA Jee 1,2 BE Veh Qo = iL Te i proport/mad co Ss , wad Js 1S prapnrloaal to AE. bot Aph av Ligh Pit =7 ed Why is it desirable for Vas = 0V in MOS Common Gate applications? (Spts) The caneat belle, ote gehr 8-4 tn ybnewl ha Pin? go the bower Ba (dub to Yucb ) 1 des rable bur phe added copacirmne ( toe! EECSIOS sof8 Fall 1999 c) Size the biasing transistors and resistor in order to get the proper Vpjq, and proper supply current (L0pts) Z2.4V- ).330V 26v Vovgs = 1332. =? BE S0qp4 2 2e 500 | HA My must hove 133204 Vos =? Ves,2 29V =? mead \ 21 og oo [W\~ Frsfpmlor. pape (O97 Tae" Ap Ytalox Vout | Zh + Venez i322 (iY (ia) = 50 v (Z L [alow Yon) 108 “a L EECSI05 6 of 8 Fall 1999 Problem 3 of 3 (35 points) For each of the following questions, make sure that you show the expressions before you plug in the specific values. A correct expression is worth 70% of the credit, even if the numerical calcula- tion is incorrect! You are given the following p-channel common-source amplifier. 254 Ta ~2.5¥ a) Draw the small signal model of the amplifier. Make sure that you include the entire small signal model of transistor Mj, along with all the relevant capacitances, including ry, Cgy2 and Cyqy from the current sink transigotr)M). (10 points) Bs Cod, i Vs Norlie thay oll the blluday smabl siquat wobeages ge g7ouads: {o/) UY, Uy, Ye Yon Thi ' why a0 these cop» ow vere Alo: Uys, =0V Vosy 0 ther 15 why there 270 Wo ob pewtlud sev7es rou My in this mode... EECSIOS 7 of 8 Fall 1999 b) Apply the Miller approximation (ignore all capacitances when calculating the Miller gain), and derive a symbolic expression for the complete transfer function (hint: this function has two poles and no zeros) (7 points). (Ml) tn Aye ~ 9m (to |] orl) ih -\- yw a +Cm n Hon Hla = . Est Pe (Pee Wn Colby Tantavéa)’ Count t(n) Vout Qn , Do Hier {fhe woo [4+ jwes Cosntay| | Pot (ae Cb) J Cen __ pode pole = + We a PCCysr0x) (Bll foil In)* (Cale Lbs thy) c) Calculate de gain and the values of the two poles, given that Cg.)=78fF, Cyg:=25fF, Cyaz=251F, Cyy1=90EF, Cayz=308F (8 points) de gain Zz - am (Gor ff? fB.) = -32.114 - V2" «bs Dm = LF flor] 2354 a § tat Ca Ta)" = Book = Yor ' | pole | (doe aust) = —— 25,50 rad foc B, (Cont Gon) Nasser x (14 32 in): S054 F { pole, = ee 754M ted foc [or |} ft B)- Cnt ldbstGgdy)
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