2 - Solutions to Chapter 2 Problems ‘Fundamentals of...

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Unformatted text preview: Solutions to Chapter 2 Problems ‘Fundamentals of III-V Devices, by W. Liu 1. (a) For the (M AlwgagfiAs/ (m QaAs heterojunction: According to-Example 1-3, NC for A10.35Ga0,65As is 3.72 X 1017 cm'3. n/Nc = 2/3.72 :4 0.54. From Fig. 1—21,'we locate 0.54 from the y—axis. The corresponding x—axis value is — 0.4. Therefore, Ef- EC = 4 0.4 X 0.0258 = - 0.01 eV. Therefore, QDN = 0.01 eV. - For GaAs. NV = 4.7 X 1018 cm‘3. p/N‘. = 5 /4.7 =1.06. From Fig. 1-21, we locate 1.06 from the y—axis. The corresponding x—axis value is 0.4. Therefore, Ev - Ef: 0.4 X 0.0258 ' =/0.01 eV. Therefore, (1),, = — 0.01 eV. From Eq. (2—14), $1,,- = 1.668 eV; 0N0: 1.6115 V; 4),,0: 0.0564 V; XNO = 0.104 um; Xpo = 4.15 x 10’3 um. Band Energy (eV) 2 I _ -0.05 0 0.05 0.1 0.15 0.2 Position (pm) We shall use the following parameters for InGaAs: Eg = 0.75 eV, M, = 6 x 1013 cm'3; 6, , = 14; and for InP: E8 = 1.35 eV, NC = 5.8 x 1017 cm‘3; E, = 12.6. The valence band discontinuity is 0.37 eV. For the InP, n/Nc = 2/5.8 = 0.345. From Fig. 1-21, we locate 0.345 from the y—axis. The corresponding x-axis value is -"0.9. Therefore, Ef— EC = - 0.9 x 0.0258 = - 0.023 eV. Therefore, GDN = 0.023 eV. For the -InGaAs, p/NV = 5 / 6 = 0.83. From Fig. 1-21, we locate 0.83-from the y-axis. The corresponding x-axis value is 0.2. Therefore, Ev - Ef: 0.2 x 0.0258 = 0.005 eV. Therefore, 45p = — 0.005 eV. From Eq. (2- 14), 0w =. 0.962 V; ¢No= 0.9285 V; ¢p0= 0.0335 V; XNO = 0.0805 mm; X,» = 3.22 x 10'3 pm. - = ,. ,, Band Energy (6V)- POsition (pm) 0)) .6 WM ‘ i - ‘ For the AlGaAs, n c = 1 /3.72 = 2.7. From Fig. 1-21, we locate 2.7 from the y- axis. The corresponding x-axis value is 2. Therefore, Ef~ EC = 2 x 0.0258 = 0.0516 eV. Therefore, CDN = ~ 0.0516 eV. For GaAs, p/Nv = 100/ 4.7 = 21.3. From Fig. 1-21, we IQCate 21.3 from the y—axis. The corresponding x-axis value is 9.2. Therefore, Ev - Ef: 9.2 X 0.0258 x 0.237 eV. Therefore, q, = — 0.237 eV. From Eq. (2-14), $1": = 1.9566 V; ¢No= 1.9388 V; 01,0 = 0.01788 V; XNO = 0.0509 gum; Xpo = 5.09 x 10'4 um. Band Energy (eV) —0.02 ’ _ 0 0.02 . 0.04 0.06 0.08 Position (urn) Wmfiagiqmmgm - i For the 112?, n/Nc = 1015. = 1.72. From Fig. 1-2-1, we locate 1.72 from the y-axis. The cerresponding x-axis value is 1.2., Therefore, Ef— EC = 1.2 x 0.0258 = 0.031 eV. Therefore, 451v = - 0.031 eV. For the InGaAs, p/Nv =100 / 6 = 16.7. From Fig. 1-21, we locate 16.7 from the y—axis. The corresponding x-axis value is 7.8. Therefore, Ev - Ef: 7.8 x 0.0258 = 0.20 eV. Therefore, ch = - 0.20 eV. From Eq. (2.14), (151,; = 1.211 V; 210 ¢N0= 1.20 V; (FPO = 0.0108 V; XNO = 0.0409 Mm; XNO = 4.09 X 10'4 um. Band Energy (CV) l -2 - -0.02-0.01 0 0.01 0.02 0.03 0.04 0.05 0.06 Position (pm) 2. (a) Em the (M AleaMAs/ 12-! Gags hetergjunction: Although pi = n;2 holds only for non—degenerate semiconductors, we will continue to use it for all samples. According to Example l-2, n,- for Alo,35Gao.65As is 3.6 X 102 cm'3. . Also, n,- for GaAs is 1.79 x 106 cmi3. At x = XNO, n = 2 x 1017 cm-3 and p = niZ/n = (3.6 x 102)2/(2 x 1017) = 6.5 X 10'13 cm'3. Atx = 0*, we use Eq. (2-17) to find n, knowing that «pm: 1.6115 V. So, It at x = 0+ is equal to 2 x 1017 x7exp( — ' 1.6115/00258) _= 1.5 X 10'10 cm’3. Further, we use Eq. (2—18) to find p at x = 0+, which ' is equal to 6.5 X 10'13 X exp(1.6115/0.0258) = 8.7 X 1017 cm'3. Now, let us find the p-side. At x = - Xpo, p = 5 X 1018 cm‘3 and n = nizlp = (1.79 x 105)2/(5 x 1013) = 6.2 X 10'7 cm‘3, At x = 0‘, we (use Eq. (2-19) to find p, knowing that ¢po= 0.0564 V. So, p at x = 0‘ is equal to 5 X 1013 X exp( - 0.0564/0.0258) = 5.6 X 1017 cm'3. Finally, ‘we are ready to find n at x = 0' . We use n = ni2/p = (1.79'x 106)2/(5.6 x ‘ 1017) = 5.7 x 10‘6 cm'3. We tabulate the results: ‘ , ‘ nlx=0+ Plx=0+ nlxr-XNO plx=XNoni (AlGaAS) p|x=0— n|x=0- P|x=-Xpo nlx=‘Xpo ni (GaAS) (cm'3) 1.5E-10 8.7E17 2El7 6.5E-13 3.6E2 , 5.6E17 5.7E-6 5E18 6.2E-7 1.79E6 loglo -9.8 17.94 17.3} -12.1 2.56 17.75 -5.24 18.7 —6.2 6.25 5n - ' [0 (carrier concentration) hole 420 gm electron “1 10 ,l - — — - — - - —| L . _ _ _ _ _ _ _ _ . _ .. . I 0 I ,l # electron ‘ - 10:. ______—+—*——-————-—————_|———. ‘ X Xpo o I . No F91 Lhe 1M lnfl (m IanaMAs heterojunction: ‘ .n,- for InP is 1.05 X 107 cm'3. Also, 11,- for InGaAs is 6.31 X 1011 em'3. At x = XNO, n = 2 x 1.017 cm'3 andp = nizln = 5.5 X 10“4 cm'3. At x = 0*, we use Eq. (2-17) to find n, knowing that we: 0.9285 V. So, n at x = 0+ is equal to 2 x 1017 x exp( - ' 0.9285/0.0258) = 4.7 x 101 cm’3. Further, we use Eq. (2-18) to find p at x = 0+, which is . 7 equal to 5.5 X 10"4 X exp(0.9285/0.0258) = 2.3 X 1012 cm‘3. Now, let us find the p-side. At x = - Xpo, p = 5 x 1018 cm-3 and n = ni2/p = (6.31 x 10“)2/(5 x 1013) ‘ 8.0x 104 cm'3. Atx = 0', we use Eq. (2-19) to find p, knowing that ' ¢po= ¢b,- — m, = 0.0335 V. So, p at x = o— is equal to 5 x 1018 x exp( — 0.0335/0.0258) =, 1.4 x 1018 cm‘3. Finally, we are ready to find 72 atx = 0‘ . We use n = nizlp = (6.31 x 1011)2/(1.4:x 1018)='2.8 x 105 cm-3. We tabulate the results: ‘ ' . nlx=0+ plx=0+ nlx=XNo Plx=XNo ni (InP) Plx=o nlx=o p|x=-Xpo nlxé-Xponi (InGaA: (cm‘3) 4.7E1 2.3E12 2E17 5.5E-4 1.1E7 1.4E18 2.8E5 5E18 8E4 6.3E11 loglo 1.67. 12.4 17.3 -3.25 7.0 18.1 5.4 18.7 4.9 11.8 n '. lo (carrier concentration) hole , 4 gm 20 electron n------—- i ‘10I electron Xp0 0 (b) For the (121 1319356 _' As/ 12) GaAs hetegojunctjon: J96: . . ’ n,- for Alo.3sGao,65As is 3.6 X 102 cm‘3. Also, ni for GaAs is 1.79 X 106 cm‘3. At x = XNO, =1x 1018 cm'3 and p = nizln = (3.6 x102)2/(1X 1018) = 1.3 x 10-13 cm-3. At x = 0*, we use Eq. (2-17) to find n, knowing that (two: 1.939 V. So, n at x = 0+ is equal to 1 X 1018 X exp( - 1.939/0.0258) = 2.3 X 10'15 cm‘3. Further, we use'Eq. (2—18) to findp at x = 0* is nizln = (3.6 X 102)2/(2.3 X 1015) = 5.6 X 1019 cm‘3. ' Now, let us find the p—side. At x = - Xpo, p = 1 X 1020 cm'3 and n = ni2/p = (1.79 X 106)2/(1 X 1020) = 3.2 x 10'8 cm’3. At x = 0', we use Eq. '(2—19) to find p, knowing that ¢po= 0.01778 V. So, p at x = 0‘ is equal to 1 X 1020 X exp( — 0.01778/0.0258) = 5.02 X 1019 cm'3. Finally, we are ready to find n atx = 0’ . We use n = niZ/p = (1.79 X 106)2/(5.02 X 1019) = 6.38 X 10‘8 cm'3. We tabulate the results: nlx=0+ Plx=0+ nlx=XNo P|x=XNoni (AlGaAS) Plx=0— n|x=0- .P|x=-Xpo n|x=-Xpo ni (GaAs) » ' (cm'3) 2.3E—15 5.6E19 1E18 1.3E-13 3.6E2 5.02E19 6.38E-8 1E20 3.2E-8 1.79E6 loglo -l4.6 19.75 18 -12.9 2.56 19.7 —7.2 20 -7.49 6.25 l(3 - electron 3. (a) For the (m AlmgagééAs/ 12) GaAs heterojgnctign: Let's calculate for Va = 0.5 V first. From the solutions of Problem 1a, we see that (DN = 0.01 eV and dip = — 0.01 eV. Valp is calculated to be 0.0177 V. VaN = Va - Vap = 0.4823 V, The built-in voltages at the tWO sides‘of the junction, which are critical to the construction of the band diagram, are the following. (1),, = «ppo - Vap = 0.0413 V; ¢N = ¢N0 ' - VaN = 1.127 V. The depletion thicknesses, according to Eqs. (2-41) and (2-42) are, Xpr = 3.47 x to? cm‘and XN = 8.68 x 10-6 cm. Band Energy (eV) -l .5 , -0.05 r 0 0.05 0.1 015 Position (pm) Let's calculate for Va == - 5 V. Va? is calculated to be - 0.177 V. VaN = Va - Vap = -4.823 V. The builbin voltages at the two sides of the junction, which are critical to the construction of the band diagram, are the following. ¢p = ope - Vap = 0.236 V; (M = ¢No - VaN = 6.432 V. The depletion thicknesses, according to Eqs. (2—41) and (2-42) are, Xp = 8.30 x 10-7 cm and XN = 2.74 x 10-5 cm. I IS N .1 J . ,1 o r— - ~15 -2 ~ *1 5? r “g -6 L . V - Ev L ' ,8; E ___1__ _L I I _.___l -O.l 0 0.1 0.2 0.3 0.4. ‘ * , Position (pm) The Fermi levels are nearly on top of the valence band in GaAs and the conduction band in AlGaAs. ‘ For the gm Igfl (Q) InQflQgMAs hetergjunction: ' Let's calculate for Va = 0.5 V first. From the solutions of Problem 121, we see that CDN = 0.023 eV and (15;, = - 0.005 eV. Vap is calculated to be 0.01737 V. VdN = Va - Vap = 0.4826 V. The built-in voltages at the two sides of the junction, which are critical to the construction of the band diagram, are the following. (1),, = ¢p0 — Vap = 0.01605 V; (M = (two - VaN = 0.4459 V. The depletion thicknesses, according to Eqs. (2-41) and (2-42) are, Xp = 2.23 x ‘10’7 cm and XN = 5.57 x 10'6 cm. ' A ' Band Energy (6‘!) I -1.5 -0.05 0 0.05 , 0.1 0.15 Position (um) Let's calculate for Va = — 5 V. Vap is calculated to be - 0.1737 V. VaN = Va - Vap = -4.826 V. The built-in voltages at the two sides of the junction, which are critical to the construction of the band diagram, are the following. «7),, = (ppo - Vap = 0.207 V; (M = ¢N0 - ‘ VaN = 5.755 V. The depletion thicknesses, according to Eqs. (2-41) and (2-42) are, Xp = 9M: 8.01 x 10‘7 cm and XN = 2.00 X 10'5 cm. 0 LI! ._ [m1 > U) iii: m L_~2L: Band lincrgy (CV) ~-0.02 0 0.02 0.04 0.06 0.08 0.1 Position (um) (b) For the (m AlmQQQfiAS/ gm GaAs heteroju- notion: Let's calculate for Va = 0.5 V first. From the solutions of Problem la, we see that (DN = -0.057 eV and (Pp = - 0.237 eV. Vap is calculated to be 0.00454 V. VaN = Va - Vap = 0.4955 V. The built—in voltages at the two sides of the junction, which are critical to the construction of the band diagram, are the following. 4),, = 45150 -_ Vap = 0.01324 V; ¢N = _ ¢N0 - VaN = 1;4433 V. The depletion thicknesses, according to Eqs. (2-41) and (2-42) are, Xp = 4.39 x 10'7 cm and XN = 4.39 x 10?6 cm. Band Energy (eV) -2 . . V . -0.02 0 002 ' 0.04 0.06 0.08 Position (um) Let's calculate for Va = - 5 V. 'Vap is calculated to be ~ 0.0454‘V. VaN = Va - Vap = -4.955 V.‘ The built-in voltages at'the two sides of the junCtion, which are critical to the cOnstruction of the band diagram, are the following. «pp = $1” — Vap = 0.0632 V; ¢N = ¢N0 - V'aN = 6.893 V. The depletion thicknesses, according to Eqs. (2-41) and (2-42) are, Xp = 9.60 X 10‘8 cm and XNE 9.60 x 10'6 cm. at] Band Energy (eV) Position (pm) For the (M 132/ 12) IanMAs heterojunction: ' Let's calculate for Va = 0.5 V first. From the solution/sof Problem 1a, we see that @N = —0.031 eV and Q, = - 0.20 eV. Vap is calculated to be 0.00446 V. VaN = Va - Val, = 0.4955V. The built‘in voltages at the two sides of the junction, which are critical to the construction of the band diagram, are the following. ¢,, = 0P0 - Vap = 0.00634 V; (M = $100 - VaN =~0.7047 V. The depletion thicknesses, according to Eqs. (2-41) and (2—42) are, Xp = 3.13 x 10‘7 cm and XN = 3.13 x 10‘6 cm. Band Energy (eV) .5 v " .002 -001 0- 0.01 0.02 0.03 0.04 0.05 Position (um) Let's calculate for Va 2 - 5 V. Val, is calculated to be - 0.0446 V. VaN = Va - Vap = -4.955 ' , V. The built—in voltages at the two sides of the junction, which are critical to the construction of the band diagram, are the following. «0,, = ¢p0 — Vap = 0.0554 V; ¢N = 0N0 - VaN= 6.156 V. The depletion thicknesses, according to Eqs. (2-41) and (2-42) are, X}, = 9.26 x 10'8 cm and XN = 9.26 X 10'6 cm. 1013 V- 9 8 >. -2 E.“ » U 1 :5 ‘3 ;» 1 _. g ‘4 ; ': 5Q Z 5 i ._6 -7 _L_ ‘ __:I ~0.05 0 0.05 - 0.1 0.15 Position (um) 4. Because the'doping profile of this problem is unmodified from that of Example 2-6,‘ the Poisson potential is still that given in Eq. (2—86). However, we the new aluminum composition grading given in the problem, the energy gap potential will be modified from that of Eq. (2-99).' In particular, let us consider the region 0 S x S Xgmding. With fimax translated tolAEanax, we obtain: ’ ' ’Q‘Pg : Esp + AEcmax ‘ c..max -'—l"‘( ngrading ' x2 ) 2 X grading 2e 2 V; ESP + AECJDKX ' Xgrading ' x2 ) where 26 s Xgrading = AEC max qud Now, in this same region (0 S x .<_ Xgmding), (ppoisson was given in Eq. (2-86). We sum up qwrois‘son and mag. and get: ‘ . 2 I 2N p} Ec(x )= q Nd( XNo - x )2 - q¢~o+Egp+ AEC,max -.__q d ( xgzrading-x2) 265 S I . 2N = X1310 ‘ _2XN0x + x2 ' ngrading + x2 )' Q¢No + Egp + AEc,max es . ' It is clear that the x2 dependence does not vanish. (NOTE: in Eq. (2-100), which is the linear conduction band profile from using the correct parabolic grading, the term inside the parantheses would be, (XNO " x? ' (Xgrading ' x)2 V: X160 ' ngrading ‘ 2(Xgrading 4 XNOV We see that the x2 term disappears. 5. From Example 2-5, NC for Alo,3sGao,65As is 3.72 x 1017 cm'3 and Nv for GaAs is v4.7«x 1018 cm‘3. From Fig. 1-21, we find that for Nd/Nc = 0.267, Ef— EC = -1.2'kT = - 0.031 eV and Na/Nv = 21.3, E, - Ef: 8.9-kT =- 0.23 eV. Hence, (DN = 0.031 eV and (Pp = - 0.23 eV. AEC max = 1247-an = 0.244 eV. ¢N0, taken to be (phi, is calculated from Eq. (2-14) to be 1.424 + 0.244 - 0.031 + 0.23 = 1.867 V. In order for the parabolic 1‘1 grading scheme to work, the grading distance needs to be that given in Eq. (2-96): - F—‘—"25 2-13.18-8.85 x10-14 Xgrading 2, AEC max = ~ ' 1.6 x 10~19~_1 x 1017 The aluminum gnole fraction should be varied from 0 at the junction interface to 0.35 at 0.244 = 6.0 x 10-6 cm. Xgmdmg = 600 A in a manner given by Eq. (2-97): . = .35— .-5 ———X—u——2. fix) 0 03(1600A) I This way, the conduction band profile is given by Eq. (2—100). We plot the result for Va = l and 1.2 V: _ I - ' 1.4' ‘ ‘13 (eV) 1.2 E 0.9 0.8 ' 9 0 200 400 600 800 1000 1200 Position away from the Junction, x (A) 6. This is an abrupt heterojunction because the ideality differs significantly from unity. A i graded heterojunction would have an ideality factor very clOse to 1. According to Eq. (2- 109), this ideality factor is simply I + K. K is thus equal to 0.2. If we neglect the differences in the dielectric constants of the two hetero-materials, we establish from Eq. (2- 110) that: N . e a a Note that, as discussed in the textbook, this ratio is obtained with the assumption that the thermionic emission dominates the current flow. When tunneling is an significant part of . the overall current conduction, then the ideality factor can. approach values like 1.2 although the Na greatly exceeds'Nd by, for example, 2 order of magnitude. 7. (21) During the reverse bias, the total current (consisting of the ideal p-n junction diode current and the generation current in the space-charge region) is the sum of Eqs. (2-128) and (2-157). Therefore, we write; , DP "'2 Dn "‘2 [ (an ) ] qADni exp H, -1- 27 (XN+Xp) .. Dp ":2 Bu "1‘2 ) qADnt‘ I qAD(Lde+LnNa 21. (XN+Xpl _ The term due to space-charge generation is the same whether the more heavily doped side is the n-type or p-type. Therefore, we shall make the decision based entirely from the first term; Rev'vrite the first term, which is the reverse saturation current (10), we obtain: 'uzo ' ' ' a- 15), (15,21 = 0.16 eV. Now for the right hand side whose doping is 1 x 1017 cm'3. This deping level is heavy enough that we need to use the Joyce-Dixon approximation. n/NC = 1/4.7 = 0.213. From Fig. 1-21, we locate 0.213 from the y-axis. The corresponding x- axis value is - 1.5. Therefore, Ef— EC = - 1.5 x 0.0258 = ~ 0.039 eV. Therefore, 49,12 = 0.039 eV. ' $1,,- = (15,,1- CD”; = 0.16 — 0.039 = 0.121 eV. 11. (a) NC for Si is 2.8 X 1019 cm'3. For the n—GaAs layer doped at 5 X 1016 cm'3, we find Ef- EC =_ 0.0258 x ln(5/2800) = - 0.16 eV. Therefore, from Eq. (2-15), 4),,0 = 0.16 /eV. The band diagram is similar to Fig. 2—21: / Ec 0.16 eV (b) When the electron hole pairs are generated within the depletion region, the holes are swept across the junction and move toward the emitter side. The electrons are repelled by the potential barrier. Therefore, the current moves from the semiconductor toward the metal. 0.45 eV W/ O . (c) The thermionic Current of the, junction .is given by} Eq. (2-102a), . . * V I , V ‘ ’rhermionic=AvA T2“? lexP(%rl'1l=’°lexP(i—il‘ll . where AD is the area of the metal-semiconductor diode area. The metal is taken to be the positive polarityside. Therefore, the current Ib- h, Of part (b) is considered a negative ' current. Hence, the total device current is equ to: I=Io[exp(-%l:)- 1]- Ilight -' When I is made zero by the open-circuit condition, there is a positive voltage formed at the 13.2% junction so that the thermionic current cancels out lug/1,. This voltage is solved by forcing I to be zero in the above equation. V high, = Io [exp ( 3:7 ) — 1] Therefore, , I - V=1‘—Iln[ "‘9’" +1 - ‘1 10 12. The equation shown in problem 12 is more accurate than Eq. (2—133). However, the difference will be very small. 13. When there is no external bias, the total current should be zero. The reason that 'Eq. (2-158) suggests a nonzero current as bias goes to zero is the following. The equation assumes that the generation‘rate U is given by Eq. (2—156). This generation rate, in turn, assumes that the p-n‘ junction is reverse biased such that the p-n product, equal to niZ-exp(an/kT), is approximately zero. In reality when the bias goes to zero, U is zero, and the generation current is zero. . 9 ...
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This note was uploaded on 03/13/2010 for the course EECS 277B taught by Professor Peterburke during the Spring '09 term at UC Irvine.

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2 - Solutions to Chapter 2 Problems ‘Fundamentals of...

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