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F2011_ECE132_HW1_Solns

# F2011_ECE132_HW1_Solns - Chapter 3 Solutions Prob 3.1...

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Unformatted text preview: Chapter 3 Solutions Prob. 3.1 Calculate the approximate donor binding energy for GaAs ( er=13.2, m: = 0.0677720). From Equation 3—8 and Appendix H, *i 4 .6 _ - '1-31 _1' _ -19 4 mu ‘1 = moo 7 (911 0 kg” 6-10 C) =8.34-10'2ZJ=5.21116V E = ______.___. 8 -(eﬁer)2-h2 8 {8.85 - 10"12 gl— -13.2)2- (6.63 - 10-34J'S)2 Prob. 3.3 Calculate electron, hole, and intrinsic carrier concentrations. Eg=1.leV NC=NV 1140157;- EC—Ed=0.26V EC—EF=O.256V T=300K 15 11:10 «4—3-— . en] a _ EC‘EF EC '13}? 0.25'6V 5 19 11=N -e H —> N =11-e H =101—lg-e‘3-0259W =1.56-10 —l—— C C CH1 01113 NV =1.56-1019 13 (EH EF-Ev 0.85ev p=NV-e' kT =1.56-1019-—m:1—3-e‘0-0259W =871-104 1 01113 -33.. Hi = ﬁ/n-p = 9.35-109 c1113 (note: ni = «(NC *NV -6 2“ may also be used] Prob. 3.6 Find Eg for Si ﬁom F tgare 3-] 7. 111 n:- for 11].l1 and Hi2 on graph _11i1=3-10M This result is approximate because the temperature dependences ofNC, NW and Eg are neglected. NCNV 1 2kT .' E E E 1111111 =1nnﬂ-h’1ni2= - g +h1 NCNV - - g +111 NCNV zi. in}. :ui2 21<T1 2kT2 2k T2 T]l 1n 11:1 In 3-1014 . 3 for Si (see above) —> E =2ki n12 =2-8.62-1014- =1.3eV g _1____1_ 4--103%—2-.103—]gE T2 T1 Prob. 3.7 (a) Find Nd for Si with 1 01601916 boron atoms and a certain number of donors so EF-E550.36€V EF-Ei r10 =11ie H EF—Ei 0.36eV 11D =Ntfl --NH —> Nﬁl =nﬂ +Nal =nie 1” +NaL =1.5-1010-1—3-e“*”259w +1016 13 = 2.634012% CH1 em (3)) Si with 101601716 In and a certain number of donors has E F—E V=0.26eV. How many In atoms are unionized (i. e. : neutral)? fraction 0f Ea states z fCEa) = —“*—1E"'"—_EF = W = 0 ' 1 + e kT 1 + e .ezseev cmg urliemjzedln=[1-f(Eﬂ)]-1\T111 =0.021-1016 1 —2.1-1014\$ - Prob. 3.12 (a) Show that the minimum conduCtivity ofa semiconductor occurs when n“ = HW “P m?“ 2 __ _ n1E G—q-(ﬂ-Hﬂ+p.“p)_q. n.uﬂ+ .HP 11 do n? _ . . . . . ‘ "a: = - LL11 - e—Z- 14p — 0 for m1n1mum conductnuty at electron concentratlon um.11 n 2 _ 2 “p __ Pip nmin_ i'— _> nmin—ni‘ — “'11 “'11 (c) Calculate 0mm and 03' for Si. Gm =2'C1'Ili- Mn 11p =2-l.6v10'19C-l.5-1010 61:13 41350 430%; =3_9.10—6 aim 6i = q-(ni -uﬂ+ni up) = 2-1.6404901540105541350 m2 430%) = 4440-6 1 V-s Q-cm or the reciprocal of o; in Appendix III may be taken ' Prob. 3.14 (a) Find m, and p for Si doped with 1017 cm‘3 boron. I NEl >> ni so p0 = Na 2 1017 1 may be assumed 2 15-1010 13 2 nﬂ = £1. = _(______107_____1m) 2225.103 13 p0 TIE? GIII Na =10” \$ gives up = 250%; from Figure 3.23 o=q-up -p0 = 1.6-10"19C-250 -1017wﬂ% =4.0 aim 0' 4.09:3] (b) Find nafor Ge doped with 3 '1 013 Sb atoms per cm3. 2 11 Assuming Na is zero and using Equation 3—28 gives no = —~i— + Nd or n: - Nd -nD - n? = 0. - 11 By quadratic formula, 11 Prob. 3.20 Find the hole concentration and mobility with Hall measurement on a p-type semiconductor bar. The voltage measured is the Hall voltage plus the ohmic drop. The Sign of VH changes with the magnetic ﬁeld, but the ohmic voltage does not. ohmic drop = 3.2mv—3.0mv = 0.2mV ' I 'B - 340—3A-lO-4ﬂé 17 p0 = _2i__z_ (Equatlon 3—50) = =3_125_10 #3 Q't'VAB 1.6-10 (3-2-10 cm-3-10 v cm -4 .2 -3 IX'L 3-10 A.2-10 cm - up=~—“—-—-=—-————1 =—T_—4—m—77—=600e: ‘1 Po q P P0 1.6-10 C-0.033Q-cm-3.125-10 c313 _ Prob. 3.21 Find VH with Hall probes misaligned. Displacement of the probes by an amount 5 give a small IR drop V5 in addition to VH. The Hall voltage reverses when the magnetic ﬁeld is reversed; however, V5 does not depend on the direction of the magnetic ﬁeld. for positive magnetic ﬁeld: VIE = VH + V5 for negative magnetic ﬁeld: VAR = —VH + V5 VIE - VAR = 2 -VH = Vgﬂ - Vg 2 So, the true Hall voltage may be obtained by subtracting VH _ the voltage with a negative magnetic ﬁeld from the voltage With a positive magnetic ﬁeld and dividing by 2. ...
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F2011_ECE132_HW1_Solns - Chapter 3 Solutions Prob 3.1...

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