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

Download Junction - Microelectronic Devices and Circuits - Solved Exam and more Exams Microelectronic Circuits in PDF only on Docsity! University of California at Berkeley College of Engineering Dept. of Electrical Engineering and Computer Sciences EE 105 Midterm If Spring 2002 Prof, Roger T. Howe April 17, 2002 (SObprots, Your Name (Last, First) Guidelines Closed book and notes; one 8.5” x 11” page (both sides) of your own notes is allowed You may use a calculator, Do not unstaple the exam. Show all your work and reasoning on the exam in order lo receive full or partial credit. Score Paints Problem | Possible | Score | 16 Totat 50 . Junction Meld-Hifect ‘Transistor (JPET) Modet. [16 ports]. b Device parameters: ' fogs = 125 WA Vp=-15V G vps hy = 0.05 V" A simplified large-signal model for an n-channel JF ET 1s: 2L ass Vos ss 7 . | in= — (veg —Vp = Woy (L+4,¥ig) for Yay SYe¢ —Pp and F,, < v4 SOV (triode) | £ Faye j ‘i ; i ings _ Gig Kn) 4 Avy HOF ¥ yy 2% yy —K) and V, SVpy $0 ¥ (saturation) where Mp is the pinch-off voltage and ,, is the “fudge factor.” (a) [4 pts.) Skewwh the drain characteristics for this JPET on the graph beluw for Vos= 0 V,-0.5 V,-1 V, and —1.5 V. You can set 4,= 0 for this part. Your current values in saturation should be accurate. Vasa 150 ov ney -O5¥ sr iv 125 My j4# 05 -1-5V oO e 100 in (WAl wi  (c) [3 pts.] Find the numerical value of the output resistance R,., of this amplifier in kQ. If you couldn’t solve part (a), you can assume for this part that the channel width W’= 100 um (not the correct answer to (a), of course.) ) Pont = x Rag (Ves = 0 mm sgemre Ine. geoerat é fl t i > Guy 4+) ~ fen gpaN Vina O2s 1 Ruf = TORN SAD [a= a] = $000, (d) [3 pts.] Find the numerical value of the two-pori parameter 4,, the open-circuit vollage gain, for this amplifier. Again, if you couldn't solve part (a), you can assume for this part that the channel width = 100 jm (not the correet answer to (a), of course.) [ae = i] ance Gut cm de gna (e) [3 pts.] Find the overall voltage gain vy / v, with Ry and R, present (values of which are given next to the schematic on the previous page). [If you couldn't solve (c) or (d), you can assume for this part that Ri = 2.5 kQ, and 4, = 0.85. Needless to say, these are not correct answers to either (c) or (d). Re j — 4 V5 9 Ye sy RB (02 Yndt Aye me Be Si Ss Kat + Ke $S¢L +/ 00008 | * fo = 0.89  (£) [3 pts.] We now remove the small-signal source and its resistance and replace iL with a large-signal source ¥jy; we also remove the load resistor. Assuming the MOSFET remains in the saturation (constant-current) region, find an equation for yyy in terms Of vpy7. What is the numerical value of v,, for the case when Your = 2 ¥? Ifyou couldn't solve part (a), you can assume that W’ = 100 jum for this part. 2-5V . = 2.5V—€ by) Por : Sn” om ss an Reve = pela MAL) (Me + \) (i+ a ied Vue Veg = Urine ~ Vins his We, = Mp sa oe Ww, 2 én, = Ap lox C Pat.) (vy Hy +E V4, Vee) Your = 25M — Aplin Roop (Wa) Canin Von) (14 in Yon) (4% Con Pvp (4 WRU 4 Yn Your) ivr = tas + View y* - 2 tec Vor Your ~ Ving + Vee =} pare Your a Moca I ——_ | i = | = We * “ — if 25voy — a fee —— WwW) Veur = 2Y =  3. npn bipolar transistors [16 pts.] ., + 4 pA Pour Va=2¥ e Given: Base width = Wy = 100 nm = 0.1 um Emitter-base junction area = Ay =4.5 am? Emitter width = Wy = 75 nm = 0.075 pm Base-collector junction area = Ae =)5 jum Electron diffusion constant in base: Dy = 10 em*/s [lole diffusion constant in emitter: Py = 5 em"/s Electron charge: ¢ =-1.6x 101.C Intrinsic concentration: + = 10" em™ Fin = 26 mV (a) [4 pts.] Find the numerical value of the electron diffusion current density J,g in the base [units pAYpor]. Neglect the base current / for this part. L. = Tia Ae La = = Te A Ing = Ae oat (b) [4 pts.] What is the numerical value of n,g(x¢ = 0), the minority electron concentration in the base at the edge of the emitter-base depletion region? Again, vou can neglect the base current J, for this part. # Th Mpg Cx cee Tipp (x0) = ~ tao i W, a ee - fie Bt a a Mae (ie x19 FC) (Zoe fe) a | Thalt)= 1-56 x 10" =