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Fixed Thickness - Microelectronic Devices and Circuits - Solved Exam, Exams of Microelectronic Circuits

Main points of this past exam are: Fixed Thickness, Diffused Resistor, Doping Level Doubles, Happens, Mobility, Doubles, Reduced

Typology: Exams

2012/2013

Uploaded on 03/22/2013

raghav
raghav šŸ‡®šŸ‡³

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Download Fixed Thickness - Microelectronic Devices and Circuits - Solved Exam and more Exams Microelectronic Circuits in PDF only on Docsity! 20f7 Problem 1 of 4: Answer each question briefly and clearly. (5 pts each, 40 total) 1.1 Assume a diffused resistor with a fixed thickness. What happens to the Sheet Resistance (Rs) if the doping level doubles (assuring that there is only 1 single dopant and that mobility does not change)? . stays the same doubles is reduced by 50% Oo: apn viz Wo _ none of the above. My doub los oo oubles + ers duced hy F2% 1.2 Find the ā€œbuilt-in potentialā€ $3 of a junction that has N, |= 10*fem? and Np=10"emā€™, bo Koa Pa > Gay x T= Mom"? = Yop Wy 7 ED Ome : A= Pps = 6OAV 6 2-3E0mY Ya te lusivy the 60mV rb) 1.3 Place check marks where appropriate to indicate the comrect-region of upecation, assuming that Vin=-Vap= 1V. Transistor [Ves [Vos | Vas on Triode Saturation_| NMOS 2 2 9 Be x NMOS 2 05 a x FM0s 2 2 8 x PMOS 2 05 [8 x 1.4 Chaose the most appropriste answer. ā€˜A nceatively charged ā€œdepletionā€ region in doped Si is characterized by. an abundance of electrons and negative ions ā€œ100 many electrons, too few holes ZL Ā«severe imbalance between negative fons and holes // jou, et owrly the complete depletion of charge density have Sxed, neyarioe / lors ) oxce 3of7 15 We have a 2-terminal device that accumul i n lates charge as a function of volta according to the equation: Q = SV +3V 42 , where Q is in pico Cb and V is in volte Find the smell signal capacitance of this device in pF at 2V. : ae eo Co Cr: - fas. Pleo ovā€ C Ā° Ā„+3) v fr very IB. 23 roth _ by yo 23 v = Up 1.6 Write the expression of a transfer function that corresponds to the Bode plot drawn. eo et 1 ; -to ooo (I+ jwhoĀ®) a A - Bike ee POLE Ā„ (ir 22) (W38) ull ir to? jo? jo* Io? on 20f055 (1000) = B08! ted free) 1.7 The value of a voltage source is 5 + 0.02cos(10Ā° + 45Ā°) Volts, Please complete the list below: ve 3 Ce shynol ne 0.07005 (tt us) sel siyuae - 5 cots Got eas) total siyacā‚¬ Vee 1.8 For the n-channel MOS transisior shown below, please consider the two ends of the channel (near the source and near the drain) and indicate whether or not each musl be inverted so that this device is in saturation Ā© This end is inverted This end is NOT inverted we This end is inverted This end is NOT inverted 4of7 Problem 2 of 4: Answer each question briefly and clearly. (20 points) The C/V plot below was taken by ā€œscanningā€ the gate-to-buik voltage of an MOS structure. We know (from other measurements) that dy+ = 0.55V and Cox = 2.36F/pm?. Also, ā‚¬ = 8.85 10 Flom, Gox = 3.9 ā‚¬9 and 83) 7 11.7 Fe ā€˜Use the graph to calculate: accumulation es inversion Minima Clq" 03 Vo OMY Ont qn O8Ā„ a | F ā€œa { Figg on 2.1 The oxide thickness tox. expression _value & unils 7 BR WRI F885 1 Fern 0.0 Wyn Cox Cox DRONES pry # 2.2 The maximum depth of the depletion region (What is the Vag voltage that yields the maximum depth of the depletion region?). i ' \ i. typ to op eto DT Cam fox (Cee Gept ā€” min Cox Ć©si Cool = Ces b Xdueay expression value and units [= bi (Ze - ZL): (sn &) | 6.105 pm Van when maximum depletion depth is achieved = O6Y @ Show f figure Sof? 2.3 The doping concentration of acceptors (N,) in the channel ~O.42 ow, Voss - (Guts) =? fe* ~Vex - Po ve GOmif p= bon Ā£04. May Mae yl? to" Jem expression _ value and units ~ at VERT ent) Ne Wi Ogee | 10 Lan | 2.4 The charge density of the inversion layer (in Cb/emt") when Ve is 2V. Above Vin, sttucrute acta as a brnear lapacitet vy value = Cox. Cont 4g 2 Aa: Log AV =-Cow (2Ā„-Ven ) Ineprvlos charge vobage aboe tlneshold expression value and units 2 a S/H foe Cox C Voe- Ven) 3.27 $cb/y f | she tavenion layer (& nade ot elections
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