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Unformatted text preview: (1 point) (1 point) (3 points) (3 points) (3 points) * 1 (3 points) (3 points) BB 306 WOSA MIDTERM EXAM #1 Braun
CLOSED BOOK + 1/2 Cheat Sheet State any assumptions and show all work. Part 1 (15 Points) _
5‘ Part 2 (12 Points) — Print Your Name: Part 3 (21 Points —
Total (50 Points) — No unauthorized help given or received. I will not discuss this exam
with another human being until after 1:00 pm Friday. «ﬁr/£411” 1. PN Junction in a new semiconductor Signature: Consider a PN junction formed in piece of the semiconductor Calpolium. The uniform doping concentration
on the P—type side of the junction is 1016 cm‘3 and the uniform doping concentration on the Ntype side is
2x10l7 cm'3. Calpolium has an intrinsic carrier concentration of 8x106 cm"3 and an energy gap of 2.0 eV at a temperature of 291 K. Please determine the following quantities at 50°C. +223K ; 5 A) Determine the intrinsic electron concentration of Calpolium at 50°C. 2 3251’ ’1‘257550 472%) ( ﬂex/[£ ’ 41 M
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B) Determine the thermal voltage at$_50°C.
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C) Determine the builtin voltage of the junction at 50°C. ié ; It; 5
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17,2 —/(7}g/’7 I 3357*,IG/772.W~é 1 M72 l/ D) Determine the electron concentration in the quasi neutral region on the N—side of the junction at 50°C. For
this part only, do not assume the depletion approximation. . )1? 4,
who : MD : 255/9 at; 3 611a; 3 :‘ Vr in E) Determine the electron concentration in the quasi neutral region on the Pside of the junction at 50°C.
For this part only, do not assume the depletion approximation. V3.” 2ésilt/Eliyi' .m r ._ 3
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53'2/4? : .1." J My. it? c’ j : vi I: I For the rest of today’s exam you may assume for Silicon at room temperature that the intrinsic carrier
concentration, ni = 1010 cm'3, with VT: 0.025 V. 2. Silicon PN Junction Diode (3 points) A) Consider a Silicon PN junction with a uniform doping concentration on the Ptype side of the junction of
8.77x1012 cm"3 and an equal uniform doping concentration on the N—type side of 8.77x1012 cm‘3. Calculate
the total depletion layer width of the junction at room temperature at zero bias. tam/02;?? t ,
< : 336: my , an; _.
()3, : WE? [7th :ztsmtlﬁ? WW r V2 ‘ r V? t ' Z1
26.51 J, J, __ ’ H, 1% 3.59300 621,9 2 M9 = (Mo W) 9] ' (amtﬂ) 0' 53"” 0L 1‘— 0,0015m : 16’}de (3 points) B) Calculate the total depletion layer width of the junction at room temperature and a reverse bias of 1000 V. M? V2 /06C2V ‘ 4% We? 3 “We (’4' 3 My”; /’” 2;;91/ “Z 545/% (3 points) C) Describe the ideal diode law for the PN—junction diode. V9) rp 1%
(3 points) D) Describe the ideal diode model for the PNjunction diode. ID
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7, ) i, nlb f2? 4? 67’2"? x/ . r, v V Fiﬁ ’
WM”) “ g (3 points) (3 points) (3 points) (3 points) (3 points) (3 points) ('3 points) 3. Electron and hole currents in Silicon Consider a piece of Silicon at room temperature containing a uniform distribution of P atoms at a
concentration of 3x1017 cm"3 and B atoms at a concentration of 2X1016 cm'3. In addition, the region has a
constant electric ﬁeld of 25 V/cm. The piece of Silicon is 5 pm wide and has a uniform cross sectional area
of 10 um2.VT= 0.025 V, ,u,, = 500 cmZ/(V s), and #1, 2 200 cmZ/(V 5). Positive ﬁeld and positive current are left to right. A) The P atoms act as Q2999
2. cceptors 5. insufficient information ___1_
Explain Cb/wzm ,E Eﬂwmfs 5W1? 4'5 #5 (It? 45 dé’ﬁm‘ﬁ z»; b
a 60/502414 1E; e/emﬁméc/ 58M) (ram/7m 24"” 5“”[7 45 5/ B) The B atoms act as 3. Both 1 & 2
4. None of the above 1. Donors 3. Both 1 & 2 5. Insufﬁcient information
@ 4. None of the above .2 Explain V r 7. .
(D/Wl/Wh e/ﬂmf’ﬁf} ff’C/7 4’5 69% £7?! (“'5 ﬁfKWé/é A; A (lg/(11,7114 gé’mﬁkzﬁ/ ff’pzalélibﬂgzl‘w Sl/Cé 4,5 5/: C) The piece of Silicon is (LN—type? 4.1&2 7.1,2,&3
2. P—Type 5. 1 & 3 8. Insufﬁcient information
3. Intrinsic 6. 2 & 3 Z
Explain (9%?(61’ 26W 29;”? %% [.45 A12) 5p e/gz (llapp, D) Calculate the electron drift current density at the center of the piece of Silicon. Elvliﬂ I 76):; .; \_— [5) [@5144 V; ~ 7 355 i [Egg/'3 III II :— A}; will}; it a I r. ' “ m i _ 1/  ~/
J/ﬁﬁ I/Miyf :gék/O‘ﬂc)/5vp 2:565.) gagym 75’” [ZS—AW) :: géaa'”2 E) Calculate the hole drift current density at the center of the piece of Silicon. a Wig M72965” ‘ 3 , ,, ~ 3
r ’ — : ——. l . ., 0””
f V! 2'82)d%ﬂ{~; 77 ,2. ﬁhxi, 3f
f’*:,zW£ m ’éJ/zaoV—éerasmv/Z 2w) » W \l’ / I I ' ' * v F) Calculate the electron diffusion current density at the center of the piece of Silicon.
( 7. 1 r , A I /7 76%? [f0 1/77 '—~9 “lg/ﬁrm ’7 “‘9 [/27 : :0 ﬂ
” ’ // 237 1/»: G) Calculate the hole diffusion current density at the center of the piece of Silicon. 3 ,»v ‘ t , _ " I I ‘ , /» 51/ . ’r[/ A
1.7;; 4127,1241 {@4009 W) H W I?) "217 [AV/U “v? r"? ’, j :; (9 ﬂ,
,. // l/ ...
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 Spring '08
 BRAUN

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