Small Signal Amplifier - Microelectronic Devices and Circuits - Solved Exam, Exams of Microelectronic Circuits

Download Small Signal Amplifier - Microelectronic Devices and Circuits - Solved Exam and more Exams Microelectronic Circuits in PDF only on Docsity! UNIVERSITY OF CALIFORNIA, BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences EE 105; Microclectronic Devices and Circuits Fall 2010 MIDTERM EXAMINATION #1 Time allotted: 60 minutes NAME: Solutions STUDENT ID#: INSTRUCTIONS: 1. SHOW YOUR WORK. (Make your methods clear to the grader!) Specially, while using chart, make sure that you indicate how you have got your numbers, For example, if reading aff mobility, clearly write down what doping density that corresponds to. 2. Clearly mark (underline or box) your answers. 3. Specify the units on answers whenever appropriate. SCORE: 1 /14 2 116 3 {15 Total 145 PHYSICAL CONSTANTS Description Symbol Value Electronic charge q 16x16? C Boltzmann’s constant = & 8.62x10° eW/K Thermal voltage at Vr= 0.026 V 300K Kitg USEFUL NUMBERS Vz In(10) = 0.060 V at T=300K. exp(3) ~ 10% 2. Depletion region Width: W = Pa q mobiliry (or2/¥s) ta 19% 105 19" a) i PROPERTIES OF SILICON AT 300K Description Symbol Value Band gap energy Ko 1.12 eV Intrinsic carrier m 10" om? concentration Dielectric permittivity fi 10x10"? Fem aii — 10! 19's 19” 19% A+ N, total dopant concentration (a) Prob 2. [16 pts] {a} [10 pts] Consider that a SI sample has been doped as shown below: N,=108 em? Np=10"5 em3 {i) Briefly state how depletion region is created at the junction. {2 pt} bthind - ‘ ; , holes diffuse from Phan leauing , immobile a penind tmobile. posrve vhorg es negahyr Chores , electrons deffuse from nto p leaweng (if} Find out the built in potential at T=300K. [2 pt] Vp x KU ln Na-Wb a al {iii) !f the two sides of the sample Is now shorted, do you expect a current ta flow due to the built in potential? Justify your answer. (2 pt] : \ NO) because Without an Gpplred voltage da feractine wervimke bones wake bre dah envyort. SbEalbly, . o Ay efadion the butlt-in potential kontld wp to oppese ine AN curren? Mat wrald otniyweé fiw die bo défferenies ut bavyiey ceneentratioi, ouvms the jundliew . (iV) Draw the charge density and efectric profile for this junction. [2 pt] Nat Que. Hera profite. {V) Qualitatively explain what happens to charge density and electric field when a reverse bias is applied as opposed to equilibrium condition. Justify your answer. [2 pt] the ehovge profile will be wrdrwrd. Thrs 1 becanse @ Yeverse. blas Uneovyys additonal tmmobile chorge Mt Me FANON » but two whened ehayge profle lhe peak vlrerme Pd Wet de enhanitd ‘Note that ettctvle feta % found by wttegvading Vhy Chavge dearly. (b} A p-n junction diode will be designed such that: (i) the p side is much more heavily doped than the n side (ii) It has a built in potential of 0.72 V and (iii) at Veppnea=-O.88¥ (reverse bias}, it gives a capacitance of 50tF/(am)" (1 femto= 10“*}. Find out the doping concentration for acceptors and donors. Clearly state aif the approximations. [6 pt] Cz Bes N= 1 3 a= Cg i vig) We Gr¥ Varee ) Nar ha) LL wt Resi ic ay) Ob = ear) an a Wr Yeo) reap seg een . poem Gee be & Se ‘Nast _ € nef [Nasa 8 “(ai Wet) ets ¥ 4 gy tle 2 gtd Fem? : a Oo! io 8 em™ Fram (4) Naha a gab! x oe? Nana we rm axe Q5Xlb! s Qo" But Nanda 1 rg eee Spe ade = le Bat Tats Clearly violates Lae dna tur NadsNa [Hence sen aw Lode. tannot | be “designed voor a weode werk Trom practical considwattenis , we many st n 50FF WK ad bp site heaualy drped ond. a eapacttornre of san Atmign bar dio te. Vapp =~ 0188 Vo In thak tare We cer rane lt Wn a kU ght ditprervnay peat + Yn potential. vith Ys vob Y, ay Ud ~Q0- . plaNb = i ara oo a Vo ; ~ Be co baat Np wih um 2 and Nae 1 om