Midterm1S04 - E13 306 804 MIDTERM EXAM#1 Braun CLOSED BOOK...

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Unformatted text preview: E13 306 804 MIDTERM EXAM #1 Braun CLOSED BOOK + 1/2 Cheat Sheet Part 1 (18 Points) State any assumptions and show all work. Part 2 (29 Pomts) (1 point) Print Your Name: 5&4 “7/ Total (49 Points) No unauthorized help given or received. I will not discuss this exam with another human being until after 11:00 am Friday. (1 point) Signature: For today’s exam you may assume for Silicon at room temperature that the intrinsic carrier concentration, n,- = 1010 cm'3, with B=l.08 x 103' cm'6 K'3. 1. Semiconductor Fundamentals A piece of Silicon has a uniform donor concentration of 1018 cm‘3 and a uniform acceptor concentration of 2x1017 cm'3. An applied electric field of +10 kV/cm exists in the piece of Silicon. Please determine the following quantities at room temperature with VT = 0.025 V. You may use the mobility approximations from p. 61 of Jaeger & Blalock. figfizfi} {2%} CmZ/y.5 m1 . . . 3'1” "' 3 71 3; am (3 pomts) A) Determine the electron concentration)? 3 4/ W)”: E + ‘ T i -3 ' _ ‘ 0 ‘ I _/ M Mg : Wig”, )% ,Qx/a (m .5 L3 X m3? 011—3 ._ ’7 x3 /7 -g n z A? ~44 : m'gcm 3 “2"” M : 3M9 6’” 6% da‘kmfl/s >72»; 5a yaw/7 #27” r0 ‘(55 flMfl/WV 15/ “if mégfiwm $0LI/77 . ~ ‘ (3 points) B) Determine the hole concentration. v g 4. 3 m2 [fife _5 .63 x10“? 0; ’72:“: [75:12ch I") gx/p [M (3 points) C) Determine the electron mobilitng 3 M: Wye/£6 :/,.2X/t7 (M ;2 70 i y a J [1% 5’” 2 » ‘22 + $0" =- 1 7* : “WSW; ‘r’ : . fl” — A]? ’+//.31/a"7(»” V} V 5 * “5 (3 points) D) Determine the electron diffusivity cm a W2 . Dy, awn ; > » Vr :(2w ‘5 X29110: (.9 S V. (3 points) E) Determine the electron drift current density Jr”,sz : flu” mg : flgw‘lzjkéfflégijmlim'y/MJ : 3077 f1; (3 points) E) Determine the electron diffusion current density d] I? . [(17 - r dyFF ‘— n . C 0 fl ‘5 , -.z a 2. Silicon PN Junction Diode The solid line plots the electric field as a function of position in a PN—junction at VD = VR = 0 V. The P-side is x<0 pm, and the N-side is x>0 pm. Please analyze this problem at room temperature with VT = 0.025 V. 50 (3 points) A) The solid line plots the electric field as a function of position in a PN—junction at VD = VR = 0 V (zero bias), and the dashed line plots the electric field in the same PN— a O junction at: o 1.VD<0V(VR>0V), S 2.VD=VR=OV, fl 3.VD>OV(VR<OV),or U " 4. Insufficient information '5 1» five/F55 6%; “r —100 Explain m ,sz Y/A/é army 49/5 1 O W/pws 7M5 MgT/Mt/ 4055/5” ' 5 _O 1 O 0 1 zM/fl //1/C'/17./:WS£S 7/505 £255 toe/é X [Hm] ' / é/fl/C . . . . . £752” flW 7 " Fig. 2A — E—field vs. Posrtion 1n PN—Junction (4 points) B) On the lower plot, graph the charge density normalized to q, p/q, as a function of position for the junction at zero bias. cpl—i Hint: complete subsequent sections of a _ [7, V3 problem 2 first. Label y-axis. fll ’9 0"” \Z +- s'x/o'flm'i CD 30 0 CG I c 6 C 3300 [rim ; ‘D - g __ I. 0}?” 3 Q m Fig. 2B — Space Charge vs. Position in PN-Junction (3 points) C) Calculate the total depletion layer width of the junction at zero bias. W6!” : ‘Xfm + X”; I 0,0521)» ~l— flay/w I fl/fj/Iw (3 points) D) Calculate the built-in potential, ¢j. (1/ / - qfl» : ~ [MM 2 ;’ Wflvgw : 3/a/52MMWM — 4 WV J (3 points) E) Calculate the doping concentration on the P-type side of the junction. , , ~ , . x: 5!) w 6' . ,¥x§,3€4x/D ’5’va .m I; w is,” = 3L” X’” r? M = "m" = Wm 0" __..._¢ : mm a, 5 65“ (’23, Xpo Géx lD§qu)[0, 057"”) (3 points) F) Calculate the doping concentration on the N-type side of the junction. "" x ‘A .lg—[f - ,_,. M) ___ 0.5”” :01} 3.35;: m m){/0z>cm : alga/0 am 3 5,. Km; (Aéx/c? 'HC ,) {fix/fl”) (3 points) G) Determine the voltage applied to the diode that produces the electric field distribution indicated by the dashed line in Fig. 2A. Cl); rl/K 1 ~/z€§<)/>l< : “fa/£51m 3321/0,pw+a/5;W)//5'0kV/cw) : [.5525'1/ Vk - «My name? :A9775'V wisy: ~ny fizz/£435 3/2125 (7 points) H) For each region in the PN—junction, determine if the description in the left applies to it at zero bias. Place an “X” in the appropriate cell, if the description fits the region. Your instructor filled out the first row as an example. I II 7—111 IV R M instructor loves this region X X X X J Space Charge—Region _r X X 4 Neutral Region X4 _J X l N—Type Region X X I P-Type Regio_n X X Doped with Donors [—7— >4 X ' Doped with Acceptors X X Depletion Region I—X Kfl— ...
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