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Unformatted text preview: Prob. 3.1 Chapter 3 Calculate Bohr radius for donor in Si (ma. = 0.26 mg). From Eq.(2IO) with n == 1 and using err— 11.8 for Si: r
' 2
mnq r: 2.41 x10‘9m = 24.1.11 z 41:6,.60112 z l1.8(8.85x10‘12)(6.63x10‘34)2
7c(026)(9.l1x10'3‘)(1.6x10“9)2 Note that this is more than four lattice spacings a (5.435;) ﬂab. 3,2 Plot F ermi ﬁtnction for E; = 1 eV. f(E)={1+e(EE,)lkT]—I We will choose E in eV and therefore use [CT = 0.0259 Egeﬂ gEEpykT 0.75 9.6525 0.99994
0.90 3.8610 0.97939
0.95 1 .9305 0.87330
0.98 —0.7722 0.68399
LOZ +0.7722 0.31600
1.05 +1.9305 0.12669
1.10 +3.8610 0.02061
1.25 +9.6525 0.00006 18 18 Prob. 3.3 Calculate the demity—af—statas' eﬁ'ective mass associated with the X minimum for the
given band structure. Given that near the energy minimum along [I O O], the band structure is :
E = EO  _ + which can be Taylor expanded near the minima : Mela—2r:Tl—lzzlﬁ‘fl‘zﬁgﬂ awe ~A—2B)+:§a2k3 +§Bz(k: +k3) 2
The effective mass is deﬁned as : m' = .32}.
, dkz ' Along (l 0 0) direction (longitudinal direction), the eﬂ‘ectlve mass becomes 2
. h2 7:2
m = ———— z m
’ dzE Au 2
die? Along the two tansverse directions, the effective mass becomes : Finally, the densityofstates effective mass is given by:
hz
“320‘ 2 B 4)1/ 3 t i '2
0.0.x»: =(m,m, )1”: l9 x mmwmwmw «WNW m 19 Mb. 3.4
F or the given band structure, ﬁnd the temperature at which the number of electrons in the 1‘ minimal and the X minima equal. From Eq.(3 ~ 15), we have 2 0. 35 3:33;;71? ﬁr Cr Given that there are 6 X minima along the<100 > directions, Ngoge  (o. { ELL—‘5
"r 4 0.30 \
0.065 ﬁom Eq.(3 ~16b) we get :
30)3/ 2 Nap x @095)” L\ l 0.35 ekT Whennr=nx,weobtaih 99.5.
e” =6x [ 0.30
0.065 3
'2' That is: kT == 0.08576V or, T = 988K mi). 3.5
For the given b and structure, calculate and sketch how the conductivity varies ﬁ‘om low T to high Tandﬁnd the ratio of the conductivities at 1000°C and 300°C. n = nr +nL z nr[l+(15 0.30
)3/2 6 k1 = constant, independent of temperature according to the problem From Eq.(3  15), we have : 932
2.4:.“ng k?
"r Nor
EVE:
nercre kT
EVE“ i . 3/2 45, 5
15 "“ ‘4‘
nlchLe kr .ekT =[_m‘ﬂ "FekT =(15)3/2,,ren
r
20 20 6 = qD’rﬂr + "£1111: Q‘D‘I‘Hr + "L /2 LEE = anﬂr 1+ ’1
=q——————————oh30 pf 1+ 2+(15)3’2e77 :2 .939
1+g§2e *7 ‘ 9"“1" 50 0.30 1+(15)3’2e k7 [ 0.30 1+———5:(')1 e 1‘7
WWI“? 0.30 [1+58‘1xe *7 T<<%—, 6 =60 =qnul E 1+1.16
J = = T»k’° °°" q""r(1+ss.1
6(1000°C) 6 (300°C) : 0.254 9.523;“?
50 111' 50 0.30 ) 21 21 If the F to L separation is assumed to be 0.35 eV instead of 0.30 6V, we get: 581 l+———e “' 6 = qwr 1+58.1><e'F 6(1000°C) _
6(300°C) 0.322 i:
‘2: 22 22 my. 3.6
Find ngbr Si ﬁ'om F ig. 31 7. g 111n§1=h1 r , / E 1
Inan =1“ N'ng ‘— ]
2 'FmSi, an .1108 mi: 4x10'3 n,’1=3x10” at —1—=Zx10'3
T1 1 1 Eg = ZkN’Iil/"iﬂ/(E‘g‘ﬁ‘) 2(8.62x10‘5) 1n(3x10‘4/108)
(4— 2) 10—3 =1.3eV This result is only approximate, since we neglect the temperature dependencias och, Ny,
, and Es. 23 23 W Show that Eq.(3—25) results from Eqs.(315) and (319) and ﬁnd lhe position of zhe Fermi
level relative to E; at 300K EC
Emem—m _..._. 0.347eV «— —————— —————Ei EV no = Nce'(Ee“ErV"T = Ncg'erO/WewrE,)IkT
= "f e(£,E,)lk1‘
p0 : "‘2 me = nie(E,—E,)/k7‘ no =1016 =LSX10‘0 Xe(E;—E¢)/0‘0259 E,— ~ Ei = 0.0259 x1n(6.667 x105) = 0.347ev Prob. 3.8
Find the displacement of E; ﬁom the middle of Eg for Si. E; is not exactly in the middle of the gap because the density of states NC and N y differ.
Equating qu.(321) and (3—23). Nee(a—EJ/kr = Mia/21:1
Eg/z ~ (EC — 12,) = len(Nv/Nc)1"2 = kT1n(m;/m;)3/‘ For Si at 300K,
Eg/Z — (EC ~ E1) = 0.0259 xi 1n (0.56/1.1)= —0.013 eV Thus, E, is about kT/2 below the center of the gap. 24 24 b. 3.9
(a) Explain why holes appear at the top of the valence band. Electron energy is plotted “up” in band diagrams such as Fig. 35. Thus conduction band
electrons relax to the bottom of the conduction band. Holes, having positive charge, have
energies which increase oppositely to that of negatively charged electrons. Thai is, hole
energy would be plotted “ wn” on an electron energy diagram such as Fig.3—S. Holes therefore relax to the lowest hole energy available to them; Le. the “mp” of the valence
band. (b) Explain why Si doped with 10“ cm” donors is n—type at 400K, but Ge is not. According Fig. 317, the intrinsic concemmtion n; at 400K is
m (400K) ~ 10” cm" for Ge
~ 10” cm'3 for Si Thus at this temperature, Nd » n; for Si, N, « n, for Ge. Prob. 3.12.
For 5: with M Na 2 4 x 10” cm", ﬁnd Epanng. no =Nd — Na =4x10‘5 = niemF‘E‘mT .— 0.0259ln(4><10‘5/1.5X101°)= 0.324eV R” = —(qno)“' = —(1.6x10“‘9 x4x10‘5r‘ = 4562.5 cm’lc 25 25 Prob. 3.11.
(a) F ind the value ofnojbr minimum conductivity 6 = «nun +pup) = q(nun + upnf/n) do
'8'”— : (IOLn "' Rang/"2) Setting this equal to zero and deﬁning n”. as the electron concentration for minimum
conductivity, we have "3. = nfup/um nm = rim/Rpm” (b) What is am? 6min = Mum/Hymn + um/un/up ) = anrxiunup (c) Calculate cm. and 0'. for Si For Si,
om = 20.6x10"")(15><10“’)(1350><4:30)”2 = 3.9x10‘5(Q cm)"
at. = mm" + up): (1.6x10'19)(15x101°)(1830)
= 4.4x10“(Qcm “ or, take the reciprocal of p; in Appendix III. 26 26 2 1‘ ar 0.1 cm long and 100 [MW2 in cross sectional area is doped with 10" cm'3
antimony. Find the current at 300K with 10V applied. Fm Fig.323, 1.1,. a 700 cmz/Vs §=qﬁnno =1.6x10“9 x700x10” =11.2(Q~cm)'l = p"
' =0.08939.~cm R pL/A = 0.0893 ><0.1/10‘6 = 8.93 x103 :2
>1: V/R =10/(s.93 x103) = 1.12 mA apcat for a length of 111m. w 8 = lOV/lO“ cm = 105 V/cm, which is in the velocity saturation regime. From
319.24, V; = 107 cm/s , = (1.6 x10“9)(10")(10”)ao") = 0.16 A (I!) How long does it take an average electron to (in)? 1 gm in pure Si at an electric ﬁeld
of100 V/cm? Repeatjbr 105 V/cm. From Appendix m, p, = 1350 cmzNs ,ow ﬁcld: vd= 9,8 = 1350 x 100 = 1.35 x 105 cm/s t= L/vd =1o4/(1.3s x 105) = 7.4 x 104° 5 = 0.74 as high ﬁeld: seaminglimited velocity v, ——~ 107 cm/s (Fig. 324)
=1o4/107 =10'“s = 10 ps 27 27 Pgoh. 3:13
A perfect ['11 Vsemiconductar is doped with column VI and 11 impurities. For the given 31", p1,, calculate the energy levels inncalmed in the bandgap. qt . =1.5x10"9x10"3 “n a t '7" ma m2
"’n 100000—47)
cm m=l.6x10"3]kg=0.176mo m; = 1.408»;0 From Eq.(3  8) and using the value of the ground state energy of a H atom
(l Rydberg)=l3.6ev: ‘ 13.61% ED = 62"“) =14.2meV below EC
7' E A = 113.6meV above EV [399. 3,14
Find V” with Hall probes misaligned. Displacement of the probes by an amount 5 gives a small IR drop V5 in addition to V”.
The Hall voltage reverses when .6 is reversed; however, V3; is insensitive m the direction of the magnetic ﬁeld. Thus,
wifhfapositive: VA}, = VH + V; negative: VXB = —VH + V5 subtracting, VXB  VXB = 2V“
We obtain the true Hall voltage t‘mtn VH = g(VA'B — VKB). 28 28 diheposition ofthe Fermi Ievelfor 11 electrons in an inﬁnite 1D potential well 1001i
he probability of exciting a carrier to the ﬁrst excited state. 6219112
E6 =EmZ2—, where m= ﬁeeelcctron mass, L =100A 52(0.00120) eV = 0.0432 eV mLz =0‘0588cV l 1 3115, I + e(0.05880.0432)/0.0259
l+e "T At 300K, this is 0.354. 29 29 ob.6 Electron mobility (cmzNs) 0 100 200 300 400 500
Tcmpaam(K)
£mb.3.l7
i
'25.
E
5
Lu O 1 0 0 2 O O 3 O 0 4 O 0 5 0 0
Temperature (K)
When freezeout occurs, ionized impurity scattering disappears, and only the phonon scattering remains. In real Si, other mechanisms, including neutral impurity scattenng,
contribute to mobility. 30 30 Prgp. 3.; 8 Find the hole concentration and mobility with Hall measurement on a 1743722
semiconductor bar. The voltage measured is the Hall voltage‘plus the ohmic drop. The Sign of V” changes
with the magnetic ﬁeld, but the ohmic voltage does not. VH, ‘ V” True, VHaI; = z 3 3 mV Thus the ohmic drop is 3.2  3.0 = 0.2 mV
From Eq.(3 — 50)
(3xlO'3AX10x10‘5Wb/cm2) = 3.125x10‘7cm‘3 1’” “ q(20><10”4cm/)(3 x 1041/)
0.2 mV
p: 3"“ =0.033Q~cm= l
ﬁn Qﬂppo
50051:» x 20m
1 1 =———=————————————=600cm2/V~s
up qppo 1.6x10"9(o.033)(3.125x10”) ( ) rob. 3.19 Calculate the conductivity of a hypothetical semiconductor at 600K.
The intrinsic conductivity is given as 5.  V'NCNve M x2000=4x10'5(52cm)" 9 6i zqniq‘l'ﬂ +up)=q 5 4x10'6 = 1.6x1049(10‘9)(2000)e"51F
As T goes ﬁom 300K to 600K, Eg, NC. N, do not change, and I? E! ..__L. ..
e 3”» increases to e “(273), wherer, = 300K. Therefore,
E _ . —6
e 2k(2x):\!___~.f‘_§i(l___._=3.s4x10‘5
1.6x10"9><10‘9><2000 49—— = 0.11352  cm)‘l 6 = __5
3.54X10 31 Prob. §,20 Calculate the number of electrons, holes, and n; in the unknown semiconductor with 151.
0.25eV below Ea;
Ec 0.2SeV  “26"
ED Incomplete ionization : fwd): Jim = 0.1267
l+eW9 n =(1— f)Nd = 8.733x 1014cnf‘3 Ec~EP Also, I: = Nee "T E —E
M
NC = "3 k7 = 8.733X1014 x 3025103259 =1.359x10‘9cm"3 = Nv
_Er'Ev
p = me k7 = 1.359 x10” Xe‘(“‘“°~25)’°'°259 = 7.591 x 104cm‘3 n, = np = 8.142X109cm“3 W
Referring to Fig. 3.25, ﬁnd the type, concentration and mobility of the majorin carrier. Given,
3, = 10‘4 Wb/cmz
From the sign of VAR, we can see the majon'ty carriers are electrons. —3 4
=_Lx§z._=__——_(_‘_1%l}3L)—__T:3stx1o”cm‘3
qt(—VAB) 1.6x10 (10 )(2x10‘) 3
p=i=kﬂa=~ﬂlﬂﬁ=aoom~cm
L/wt L/wt 0.5/0.01x10 1 ﬂ 1
”” " pqm0 (o.ooz)(1.6x10“9)(3.125x10”) "o = 10,000 cm2(V  8Y4 32 32 ...
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This note was uploaded on 03/19/2008 for the course EE 339 taught by Professor Banjeree during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Banjeree

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