4904_B2008_Quiz4_solns

# 4904_B2008_Quiz4_solns - A 42—50[3 3 th C 253.3 M 38 a...

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Unformatted text preview: A 42—50 [3, 3+th C 253.3 M 38.! a- 6.5 MES 39 Nam SCUIHONS ECE Box # ____ Average _ Problem Score POints 1 26' 35 2 12,0 15 ECE4904 B2008 Semiconductor Devices Quiz 4 This is a closed book quiz! You are allowed one 8—1/2" X 11" sheet (both sides) of notes. Note: Potentially useful reference charts are given on pages 7 and 8 (tear off to use if you like). Show all your work. Partial credit may be given. If you think you need something that you can't remember, write down what you need and what you'd do if you remembered it. Unless otherwise indicated, you may assume Si in equilibrium at T=300K. Look for the simple, straightforward way to solve the problem for the level of accuracy required. Don't get entangled in unnecessary algebra. As in real life, some problems may give you more information than you need. Don't assume that all information must be used! It's your job to decide what's relevant to the solution. You will have 25 minutes to complete this exam. There are 2 problems on a total of 8 pages. b) d) EXPLAINI: MA.IORIT‘( pew DOPING ; MINORITY FQOM f) 9) The top figure on the opposite page shows a diode (assuming a step junction) with h%=5.0E+15 /0m3 and bk=l.0E+15 /Cm3. The extension of the depletion region from the metallurgical junction into the n—type region is :%=O.18pm (1.8E—5 cm). Also shown are four arrows indicating carrier motion. Using the space on the opposite page, draw the oriented!) circuit symbol for the diode. (correctly [3] Determine xp, the extent of the depletion region into the p— type material. Numerical value required! (accuracy i10%) CHARGE NEUTRAL! TY: 5E+5 Xn ND :XPNA \$Xp= [Earls O.Ig,uM [5] xp= ang For each carrier motion arrow, indicate the type of carrier (hole or electron) and type of motion (drift or diffusion). [8] Is the diode biased? forward biased, (Circle one): (EBRWARD \BIAS DIFFUSION CUQIQEMT” DOMINATES at equilibrium, or reverse EQUILIBRIUM REVERSE INAS [1] [4] MIN CAQQ {TONC INCREASED AT 9065 OF DE‘PL REGION The bottom figure on the opposite page shows carrier concentrations for both minority and majority carriers in the quasineutral regions. the (pnol write down and minority Numerical values required! Using the space on the opposite page, equilibrium values for majority (nm, pw) nw) carrier concentrations. [4] Determine the applied voltage VA (accuracy t0.00lV) __ LAW OF JUNCTION, V) SIDE O-5~'37V 29,3: 29.4 eVA/o.0259 [5] VA = Given that the minority carrier diffusion length on each side is Lfﬂw=4E—2cm, and the junction area is A=l.OE—2cm% determine the total diode current ID (accuracy 110%) ID: Jﬂﬁlkﬂg__ D FROM -£%:I<Tﬁl (SEE p.8) [4] Circuit symbol for (a): NA = L9 15 P gﬁceptor m3 CARRIER CONCENTRATIONS: nnn= 5:915 NDCZQ [2.0 E+13 /cm" t ___________________ / The figure below shows the carrier motion components for a bipolar junction transistor (BJT) in the active mode. The doping concentrations for three regions of the semiconductor are also shown. The total emitter current is IE = 5.015 mA, and the total base current is IB = 0.030 mA. NAB=5.OE+16 NDC=8 . 0E+14 ‘EEptorjcm3 (gonorZirm3 IB 003OInA a) Is this an NPN or PNP transistor? [1] EXPLAIN! : [31 EMITTER/ COLLECTOR ARE DONOIQ DOPE-ID: h WPE b) Determine the total collector current IC (accuracy i0.001mA) [51 IC = 4-.985mA FROM KCL: IC —= 15 \$3 2 5:0'5mA- 0,03 MA c) Determine the following figures of merit for this transistor (accuracy 1 0.1 %) [6] Emitter Injection Efficiency y Y = o. 9% iggéﬂ/j‘ 1 6 5.015 mA Base Transport Factor QT GT = 0,993 ICW w 2 i335 “A -T—-:Ey\ MA Common Emitter Current Gain [3‘k “ q c . ﬁdc = 6 £1.3— :, 13 0.03 IMA Common Base Current Gain acm adc = 0,994 is." iiéi ALSO 10¢,:(o.996)(a99z):o.99q / lﬁ “ 5,015 5 (cmZN-sec) ....... NAorND(cm—3) (£212 U ......l.... 51x10'4.... 1358 461 " 9 1357 460 459 ‘ “WWJ 298 .448 1248 ‘r' 1165 419 H I 986 378 z»;- ..... "g . . . . . . . . .. 1.301. '331_‘§§ D Z i I Q U) 4:. 1 U" 1 V 9 O N Va {9, .ur orﬂp 1014 1015 1016 1017 10’8 NA or ND (cm-3) W ' lﬁiﬂllﬁl E1 p (ohm-cm) nuazxmwnnammnwu : :mnmmnm 'r 11 nm' ' M —_.-n.———-_-.u w———w—--c— ———.‘ ﬁﬁéﬁiﬁaﬁi §§ ' gg IMII--IIIBI "Iliﬁ Iﬁﬂ m1 10—3 lllﬁiﬂ IE mm mm was iﬁﬁﬁig' Iiisnmi a; iihmiﬁi “NE E E 9 'r x x a z : 1 10-4 :3 1.; 1 ‘2 1013 1014 10'5 1o18 1019 1020 NA or 1vD (cm-3) Figure 3.8 Resistivity versus impurity concentration at 300 K in (a) Si 2 ...
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4904_B2008_Quiz4_solns - A 42—50[3 3 th C 253.3 M 38 a...

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