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Solutions Manual to Accompany SEMICONDUCTOR DEVICES Physics and Technology rd 3 Edition S. M. SZE Etron Chair Professor College of Electrical and Computer Engineering National Chaio Tung University Hsinchu, Taiwan M. K. LEE Department of Electrical Engineering National Sun Yat-sen University Kaohsiung, Taiwan John Wiley and Sons, Inc New York. Chicester / Weinheim / Brisband / Singapore / Toronto 1 Contents Ch.0 Introduction---------------------------------------------------------------------------- 0 Ch.1 Energy Bands and Carrier Concentration in Thermal Equilibrium ------------ 1 Ch.2 Carrier Transport Phenomena ------------------------------------------------------- 9 Ch.3 p-n Junction ---------------------------------------------------------------------------18 Ch.4 Bipolar Transistor and Related Devices -------------------------------------------35 Ch.5 MOS Capacitor and MOSFET------------------------------------------------------52 Ch.6 Advanced MOSFET and Related Devices ----------------------------------------62 Ch.7 MESFET and Related Devices -----------------------------------------------------68 Ch.8 Microwave Diode, Quantum-Effect and Hot-Electron Devices ----------------76 Ch.9 Light Emitting Diodes and Lasers -------------------------------------------------81 Ch.10 Photodetectors and Solar Cells-----------------------------------------------------88 Ch.11 Crystal Growth and Epitaxy -------------------------------------------------------96 Ch.12 Film Formation ----------------------------------------------------------------------105 Ch.13 Lithography and Etching -----------------------------------------------------------112 Ch.14 Impurity Doping---------------------------------------------------------------------118 Ch.15 Integrated Devices-------------------------------------------------------------------126 0 CHAPTER 1 1. (a) From Fig. 11a, the atom at the center of the cube is surround by four equidistant nearest neighbors that lie at the corners of a tetrahedron. Therefore the distance between nearest neighbors in silicon (a = 5.43 Å) is 1/2 [(a/2)2 + ( 2a /2)2]1/2 = 3a /4 = 2.35 Å. (b) For the (100) plane, there are two atoms (one central atom and 4 corner atoms each contributing 1/4 of an atom for a total of two atoms as shown in Fig. 4a) for an area of a2, therefore we have 2/ a2 = 2/ (5.43 × 10-8)2 = 6.78 × 1014 atoms / cm2 Similarly we have for (110) plane (Fig. 4a and Fig. 6) (2 + 2 ×1/2 + 4 ×1/4) / 2a 2 = 9.6 × 1015 atoms / cm2, and for (111) plane (Fig. 4a and Fig. 6) ⎛3⎞ ⎝2⎠ (3 × 1/2 + 3 × 1/6) / 1/2( 2a )( ⎜ ⎟ a ) = ⎛ ⎜ ⎜ ⎝ 2 3 ⎞⎟ 2 a 2 ⎟⎠ = 7.83 × 1014 atoms / cm2. 2. The heights at X, Y, and Z point are 3 4, 1 4, and 3 . 4 3. (a) For the simple cubic, a unit cell contains 1/8 of a sphere at each of the eight corners for a total of one sphere. ∴ Maximum fraction of cell filled = no. of sphere × volume of each sphere / unit cell volume = 1 × 4π(a/2)3 / a3 = 52 % (b) For a face-centered cubic, a unit cell contains 1/8 of a sphere at each of the eight corners for a total of one sphere. The fcc also contains half a sphere at each of the six faces for a total of three spheres. 1 The nearest neighbor distance is 1/2(a 2 ). Therefore the radius of each sphere is 1/4 (a 2 ). ∴ Maximum fraction of cell filled = (1 + 3) {4π[(a/2) / 4 ]3 / 3} / a3 = 74 %. (c) For a diamond lattice, a unit cell contains 1/8 of a sphere at each of the eight corners for a total of one sphere, 1/2 of a sphere at each of the six faces for a total of three spheres, and 4 spheres inside the cell. The diagonal distance between (1/2, 0, 0) and (1/4, 1/4, 1/4) shown in Fig. 9a is 2 2 ⎛a⎞ ⎛a⎞ ⎛a⎞ ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ ⎝2⎠ ⎝2⎠ ⎝2⎠ 2 = a 3 4 The radius of the sphere is D/2 = a 3 8 1 D= 2 ∴ Maximum fraction of cell filled ⎡ 4π ⎣ 3 = (1 + 3 + 4) ⎢ 3 ⎛a ⎞⎤ 3 3 ⎟⎥ / a ⎜ 8 ⎝ ⎠⎦ = π 3 / 16 = 34 %. This is a relatively low percentage compared to other lattice structures. 4. = d2 = d3 = d4 = d d1 d1 + d 2 + d 3 + d 4 = 0 d1 • ( d1 + d 2 + d 3 + d 4 ) = d1 • 0 = 0 2 d 1 + d1 • d 2 + d1 • d 3 + d1 • d 4 = 0 ∴d2+ d2 cosθ12 + d2cosθ13 + d2cosθ14 = d2 +3 d2 cosθ= 0 ∴ cosθ = θ= cos-1 ( −1 3 −1 ) = 109.470 . 3 5. Taking the reciprocals of these intercepts we get 1/2, 1/3 and 1/4. three integers having the same ratio are 6, 4, and 3. The smallest The plane is referred to as (643) plane. 6. (a) The lattice constant for GaAs is 5.65 Å, and the atomic weights of Ga and As are 69.72 and 74.92 g/mole, respectively. There are four gallium atoms and 2 four arsenic atoms per unit cell, therefore 4/a3 = 4/ (5.65 × 10-8)3 = 2.22 × 1022 Ga or As atoms/cm2, Density = (no. of atoms/cm3 × atomic weight) / Avogadro constant = 2.22 × 1022(69.72 + 74.92) / 6.02 × 1023 = 5.33 g / cm3. (b) If GaAs is doped with Sn and Sn atoms displace Ga atoms, donors are formed, because Sn has four valence electrons while Ga has only three. The resulting semiconductor is n-type. 7. Eg (T) = 1.17 – 4.73x10 −4 T 2 for Si (T + 636) ∴ Eg ( 100 K) = 1.163 eV , and Eg (600 K) = 1.032 eV Eg(T) = 1.519 – 5.405x10 −4 T 2 for GaAs (T + 204) ∴Eg( 100 K) = 1.501 eV, and Eg (600 K) = 1.277 eV . 8. The density of holes in the valence band is given by integrating the product N(E)[1-F(E)]dE from top of the valence band ( EV taken to be E = 0) to the bottom of the valence band Ebottom: p= ∫ { [ where 1 –F(E) = 1 − 1 / E bottom 0 1 + e N(E)[1 – F(E)]dE (E − E F )/kT ]} = [1 + e (1) ] ( E − E F ) / kT −1 If EF – E >> kT then 1 – F(E) ~ exp [− (E F − E ) kT ] (2) Then from Appendix H and , Eqs. 1 and 2 we obtain p = 4π[2mp / h2]3/2 ∫ E bottom 0 E1/2 exp [-(EF – E) / kT ]dE Let x ≣ E / kT , and let Ebottom = − ∞ , Eq. 3 becomes p = 4π(2mp / h2)3/2 (kT)3/2 exp [-(EF / kT)] ∫ −∞ 0 where the integral on the right is of the standard form and equals ∴ x1/2exdx π / 2. p = 2[2πmp kT / h2]3/2 exp [-(EF / kT)] By referring to the top of the valence band as EV instead of E = 0 we have, p = 2(2πmp kT / h2)3/2 exp [-(EF – EV) / kT ] 3 (3) or p = NV exp [-(EF –EV) / kT ] NV = 2 (2πmp kT / h2)3 . where 9. From Eq. 18 NV = 2(2πmp kT / h2)3/2 The effective mass of holes in Si is mp = (NV / 2) 2/3 ( h2 / 2πkT ) ⎛ 2.66 × 1019 × 10 6 m − 3 ⎞ ⎟ = ⎜ 2 ⎝ ⎠ 2 3 (6.625 × 10 ) 2π (1.38 × 10 )(300 ) −34 2 − 23 = 9.4 × 10-31 kg = 1.03 m0. Similarly, we have for GaAs mp = 3.9 × 10-31 kg = 0.43 m0. 10. Using Eq. 19 ( 2 )ln (N E i = ( E C + EV ) 2 + kT V NC ) 2 = (EC+ EV)/ 2 + (3kT / 4) ln ⎡( m p m n )(6) 3 ⎤ ⎢⎣ ⎥⎦ (1) At 77 K Ei = (1.16/2) + (3 × 1.38 × 10-23T) / (4 × 1.6 × 10-19) ln(1.0/0.62) = 0.58 + 3.29 × 10-5 T = 0.58 + 2.54 × 10-3 = 0.583 eV. At 300 K Ei = (1.12/2) + (3.29 × 10-5)(300) = 0.56 + 0.009 = 0.569 eV. At 373 K Ei = (1.09/2) + (3.29 × 10-5)(373) = 0.545 + 0.012 = 0.557 eV. Because the second term on the right-hand side of the Eq.1 is much smaller compared to the first term, over the above temperature range, it is reasonable to assume that Ei is in the center of the forbidden gap. ∫ (E − E ) E − E E−E e ∫ E top 11. KE = C EC E top EC C C e − ( E − E F )/kT dE − ( E − E F ) / kT dE 4 x ≡ ( E − EC ) ∫ = kT ∫ ∞ 0 ∞ 0 3 x 2 e − x dx 1 x 2 e − x dx = ⎛5⎞ Γ⎜ ⎟ 1.5 × 0.5 × π 2 = kT ⎝ ⎠ = kT ⎛3⎞ 0.5 π Γ⎜ ⎟ 2 ⎝ ⎠ 3 kT . 2 12. (a) p = mv = 9.109 × 10-31 ×105 = 9.109 × 10-26 kg–m/s λ = (b) λ n = 6.626 × 10 −34 h = 7.27 × 10-9 m = 72.7 Å = − 26 p 9.109 × 10 m0 1 × 72.7 = 1154 Å . λ = 0.063 mp 13. From Fig. 22 when ni = 1015 cm-3, the corresponding temperature is 1000 / T = 1.8. So that T = 1000/1.8 = 555 K or 282 ℃. 14. From Ec – EF = kT ln [NC / (ND – NA)] which can be rewritten as ND – NA = NC exp [–(EC – EF) / kT ] Then ND – NA = 2.86 × 1019 exp(–0.20 / 0.0259) = 1.26 × 1016 cm-3 or ND= 1.26 × 1016 + NA = 2.26 × 1016 cm-3 A compensated semiconductor can be fabricated to provide a specific Fermi energy level. 15. From Fig. 28a we can draw the following energy-band diagrams: 5 AT 77K - Ec(0.59eV) -- EF(0.53) ·- ·- ·- ·- ·- · -· - ·- · - · Ej(O) Ev(- 0.59) AT 300K Ec(0.56 eV ) EF(0.38) Ej (0 ) ;·I - - - - - -- - -- -- Ev(-O.SS) AT 600K -·-·- 16. Ec(0.50 eV) .. - ·- ·- ·- ·- ·-·- · EF: Ej(O) - - - - - - - -- - - Ev(-0.50) (a) The ionization energy for boron in Si is 0.045 eV impurities are ionized. Thus pp = NA = 1015 cm-3 np = 11?I nA = (9.65 x At 300 K, all boron 109i I 1015 = 9.3 x 104 cm-3 . The Fe1m i level measured from the top of the valence band is given by: Ep - Ev = kTln(NvfND) = 0.0259 ln (2.66 x 1019 I 10 15) = 0.26 eV (b) The boron atoms compensate the arsenic atoins; we have PP = NA - ND= 3 x 1016 - 2.9 x 10 16 = 1015 cm-3 Since pP is the same as given in (a), the values for np and Ep are the same as in (a). However, the mobilities and resistivities for these two samples are different. 17. Since ND >> n;, we can approximate no = ND and Po = n? I no= 9.3 x 10 19 I 1017 = 9.3 From no = n;exp E F ( x 102 cm- 3 -E) kT i , we have Ep - E; = kT ln (n 0 1n;) =0.0259ln(10 11 19.65 x 109)= 0.42eV 6 The resulting flat band diagram is : ----------0.42eV ------- 1.12eV - - Ej t~, ----------- 18. ' -:' .Ev From Eq. 28 n = l /2[Nn - NA +~(ND - 2 NA) +4n/ J =112[ 2.5 x1013 +~(2.5x10 13 ) 2 +4(2.5x1013 / J = 4.04x1013 20. Assuming complete ionization, the Fenni level measm ed from the intrinsic Fenni level is 0.35 eV for 1015 cm-3 , 0.45 eV for 1017 cm-3 , and 0.54 eV for 1019 The number of electrons that are ionized is given by Using the Fenni levels given above, we obtain the number of ionized donors as n = 10 15 cm-3 for ND = 1015 cm-3 n = 0.93 x 10 17 cm-3 for ND = 1017 cm-3 n = 0.27 x 10 19 cm-3 Therefore, the assumption of complete ionization is valid only for the case of 1015 cm-3 . 7 21. ND+ = = 1016 1 + e −( ED − EF ) / kT = 1016 1 + e −0.135 1016 = 5.33 × 1015 cm-3 1 1+ 1.145 The neutral donor = 1016 – 5.33 ×1015 cm-3 = 4.67 × 1015 cm-3 N DO 4.76 ∴ The ratio of = = 0.876 . + 5.33 ND 8 CHAPTER 2 1. (a) For intrinsic Si, μn = 1450, μp = 505, and n = p = ni = 9.65×109 We have ρ = 1 1 = = 3.31 × 10 5 qnμ n + qpμ p qni ( μ n + μ p ) Ω-cm (b) Similarly for GaAs, μn = 9200, μp = 320, and n = p = ni = 2.25×106 We have ρ = 2. 1 1 = = 2.92 × 10 8 qnμ n + qpμ p qni ( μ n + μ p ) Ω-cm. For lattice scattering, μn ∝ T-3/2 200 −3 / 2 = 2388 cm2/V-s T = 200 K, μn = 1300× −3 / 2 300 T = 400 K, μn = 1300× 3. Since 1 = μ ∴ 1 μ 1 μ1 = + 1 μ2 1 1 + 250 500 4. (a) p = 5×1015 400 −3 / 2 = 844 cm2/V-s. −3 / 2 300 μ = 167 cm2/V-s. cm-3, n = ni2/p = (9.65×109)2/5×1015 = 1.86×104 cm-3 μp = 410 cm2/V-s, μn = 1300 cm2/V-s ρ= 1 1 = 3 Ω-cm ≈ qμ n n + qμ p p qμ p p (b) p = NA – ND = 2×1016 – 1.5×1016 = 5×1015 cm-3, n = 1.86×104 cm-3 μp = μp (NA + ND) = μp (3.5×1016) = 290 cm2/V-s, μn = μn (NA + ND) = 1000 cm2/V-s ρ= 1 1 ≈ = 4.3 Ω-cm qμ n n + qμ p p qμ p p (c) p = NA (Boron) – ND + NA (Gallium) = 5×1015 cm-3, n = 1.86×104 cm-3 μp = μp (NA + ND+ NA) = μp (2.05×1017) = 150 cm2/V-s, 9 μn = μn (NA + ND+ NA) = 520 cm2/V-s ρ = 8.3 Ω-cm. 5. Assume ND − NA>> ni, the conductivity is given by σ ≈ qnμn = qμn(ND − NA) We have that 16 = (1.6×10-19)μn(ND − 1017) Since mobility is a function of the ionized impurity concentration, we can use Fig. 3 along with trial and error to determine μn and ND . For example, if we choose ND = 2×1017, then NI = ND+ + NA- = 3×1017, so that μn ≈ 510 cm2/V-s which gives σ = 8.16. Further trial and error yields ND≈ 3.5×1017 cm-3 and μn ≈ 400 cm2/V-s which gives σ ≈ 16 (Ω-cm)-1. 6. σ = q( μ n n + μ p p) = qμ p (bn + ni2 / n) From the condition dσ/dn = 0, we obtain n = ni / b Therefore 1 ρ m qμ p (bni / b + b ni ) b + 1 = = . 1 ρi 2 b qμ p ni (b + 1) 7. At the limit when d >> s, CF = π ln 2 = 4.53. Then from Eq. 16 10 10 × 10 −3 V ρ = × W × CF = × 50 × 10 −4 × 4.53 = 0.226 −3 I 1 × 10 Ω-cm From Fig. 6, CF = 4.2 (d/s = 10); using the a/d = 1 curve we obtain 0.226 × 10 −3 V = ρ ⋅ I /(W ⋅ CF ) = = 10.78 50 × 10 − 4 × 4.2 8. mV. Hall coefficient, VH A 10 × 10 −3 × 1.6 × 10 −3 = RH = = 426.7 IB zW 2.5 × 10 −3 × (30 × 10 −9 × 10 4 ) × 0.05 cm3/C Since the sign of RH is positive, the carriers are holes. From Eq. 22 p= 1 1 = = 1.46 × 1016 −19 qRH 1.6 × 10 × 426.7 cm-3 Assuming NA ≈ p, from Fig. 7 we obtain ρ = 1.1 Ω-cm The mobility μp is given by Eq. 15b μp = 9. 1 1 = = 380 −19 qpρ 1.6 × 10 × 1.46 × 1016 × 1.1 cm2/V-s. 1 1 , hence R ∝ qnμ n + qpμ p nμ n + pμ p Since R ∝ ρ and ρ = From Einstein relation D ∝ μ μ n / μ p = Dn / D p = 50 1 R1 = 0.5R1 N D μn 1 N D μn + N Aμ p We have NA = 50 ND . 10. The electric potential φ is related to electron potential energy by the charge (− q) 1 q φ = + (EF − Ei) The electric field for the one-dimensional situation is defined as E(x) = − 11 dφ 1 dE i = dx q dx ⎛ E F − Ei ⎞ ⎟ = ND(x) ⎝ kT ⎠ n = ni exp ⎜ Hence ⎛ ND( x ) ⎞ ⎟⎟ ⎝ ni ⎠ EF − Ei = kT ln ⎜⎜ ⎛ kT ⎞ 1 dN D ( x ) ⎟ E (x) = − ⎜ . dx ⎝ q ⎠ ND( x ) 11. (a) From Eq. 31, Jn = 0 and E( x) = - Dn μn dn − ax dx = - kT N 0 ( − a )e = + kT a n q N 0 e −ax q (b) E (x) = 0.0259 (104) = 259 V/cm. 12. At thermal and electric equilibria, J n = qμ n n( x )E + qDn E( x) = − =− L V =∫ 0 13. dn( x ) =0 dx D 1 dn( x ) 1 =− n μ n n( x ) dx μ n N 0 + ( N L − N 0 )( x NL − N0 μ n LN 0 + ( N L − N 0 ) x Dn NL − N0 D N = − n ln L . μ LN 0 + ( N L − N 0 ) x μn N 0 Dn Δn = Δp = τ p G L = 10 × 10 −6 × 1016 = 1011 cm-3 n = n no + Δn = N D + Δn = 1015 + 1011 ≈ 1015 cm-3 p= NL − N0 L) L Dn n i2 (9.65 × 10 9 ) 2 + Δp = + 1011 ≈ 1011 cm -3 . ND 1015 14. (a) τ p ≈ 1 σ pν th N t = 5 × 10 −15 1 = 10 −8 7 15 × 10 × 2 × 10 L p = D pτ p = 9 × 10 −8 = 3 × 10 −4 cm S lr = ν thσ s N sts = 10 7 × 2 × 10 −16 × 1010 = 20 12 s cm/s (b) The hole concentration at the surface is given by Eq. 67 ⎛ τ p S lr pn (0) = pno + τ p GL ⎜1 − ⎜ L p + τ p S lr ⎝ ⎞ ⎟ ⎟ ⎠ ⎞ (9.65 × 109 ) 2 10 −8 × 20 17 ⎛ −8 ⎟ = + 10 × 10 ⎜1 − 16 −4 −8 2 × 10 ⎝ 3 × 10 + 10 × 20 ⎠ ≈ 10 9 cm -3 . 15. σ = qnμ n + qpμ p Before illumination nn = nno , p n = p no After illumination nn = nno + Δn = nno + τ p G, p n = p no + Δp = p no + τ p G Δσ = [ qμ n ( nno + Δn ) + qμ p ( pno + Δp )] − ( qμ n nno + qμ p pno ) = q( μ n + μ p )τ p G . 16. (a) J p , diff = −qD p dp dx = − 1.6×10-19×12× 1 ×1015exp(-x/12) 12 × 10 − 4 = 1.6exp(-x/12) A/cm2 (b) J n , drift = J total − J p , diff = 4.8 − 1.6exp(-x/12) A/cm2 (c) ∵ J n , drift = qnμ nE ∴ 4.8 − 1.6exp(-x/12) = 1.6×10-19×1016×1000×E E = 3 − exp(-x/12) V/cm. 17. For E = 0 we have p − pno ∂ 2 pn ∂p =− n + Dp =0 ∂t ∂x 2 τp 13 at steady state, the boundary conditions are pn (x = 0) = pn (0) and pn (x = W) = pno. Therefore p n ( x) = p no J p ( x = 0) = − qD p J p ( x = W ) = − qD p ⎡ ⎛W − x ⎞⎤ ⎟⎥ ⎢ sinh ⎜ ⎜ ⎟ L ⎢ p ⎠⎥ ⎝ + [ p n (0) − p no ] ⎢ ⎥ ⎢ sinh ⎛⎜ W ⎞⎟ ⎥ ⎢ ⎜ Lp ⎟ ⎥ ⎝ ⎠ ⎦ ⎣ ∂p n ∂x ∂p n ∂x = q[ p n (0) − p no ] ⎛W coth ⎜ ⎜ Lp Lp ⎝ Dp x =0 = q[ p n (0) − p no ] x =W ⎞ ⎟ ⎟ ⎠ Dp 1 Lp ⎛W sinh ⎜ ⎜ Lp ⎝ ⎞ ⎟ ⎟ ⎠ . 18. The portion of injection current that reaches the opposite surface by diffusion is given by α0 = J p (W ) = J p (0) 1 cosh(W / L p ) L p ≡ D pτ p = 50 × 50 × 10 −6 = 5 × 10 −2 ∴ α0 = cm 1 = 0.98 cosh(10 / 5 × 10 − 2 ) −2 Therefore, 98% of the injected current can reach the opposite surface. 19. In steady state, the recombination rate at the surface and in the bulk is equal Δp n , bulk τ p , bulk = Δp n , surface τ p , surface so that the excess minority carrier concentration at the surface Δpn, surface = 1014⋅ 10 −7 =1013 10 −6 cm-3 The generation rate can be determined from the steady-state conditions in the bulk G= 1014 = 1020 −6 10 14 cm-3s-1 From Eq. 62, we can write Dp Δp ∂ 2 Δp +G− =0 2 τp ∂x The boundary conditions are Δp(x = ∞ ) = 1014 cm-3 and Δp(x = 0) = 1013 cm-3 Hence Δp(x) = 1014( 1 − 0.9e − x / L ) where Lp = p 10 ⋅ 10 −6 = 31.6 μm. 20. The potential barrier height φ B = φ m − χ = 4.2 − 4.0 = 0.2 volts. 21. The number of electrons occupying the energy level between E and E+dE is dn = N(E)F(E)dE where N(E) is the density-of-state function, and F(E) is Fermi-Dirac distribution function. Since only electrons with an energy greater than E F + qφ m and having a velocity component normal to the surface can escape the solid, the thermionic current density is J = ∫ qv x = ∫ ∞ E F + qφ m 4π ( 2m ) h3 3 2 1 v x E 2 e −( E − EF ) kT dE where v x is the component of velocity normal to the surface of the metal. Since the energy-momentum relationship E= P2 1 = ( p x2 + p 2y + p z2 ) 2m 2m Differentiation leads to dE = PdP m By changing the momentum component to rectangular coordinates, 4πP 2 dP = dp x dp y dp z ∞ 2q ∞ ∞ − ( p x2 + p 2y + p z2 − 2 mE f ) / 2 mkT p e dp x dp y dp z x mh 3 ∫ p x 0 ∫ p y = −∞ ∫ p = −∞ ∞ ∞ 2 2 q ∞ − ( p x2 − 2 mE f ) 2 mkT − p 2 2 mkT = e p x dp x ∫ e y dp y ∫ e − p z 3 ∫p −∞ −∞ x0 mh J= Hence where p x20 = 2m( E F + qφ m ). 15 2 mkT dp z ∫ Since ∞ −∞ e − ax 2 ⎛π ⎞ dx = ⎜ ⎟ ⎝a⎠ 1 2 , the last two integrals yield (2πmkT) 1 2 . p x2 − 2mE F =u . 2mkT The first integral is evaluated by setting Therefore we have du = p x dp x mkT The lower limit of the first integral can be written as 2m( E F + qφ m ) − 2mE F qφ m = 2mkT kT so that the first integral becomes mkT ∫ Hence J = 22. 4πqmk 2 2 −qφm T e h3 kT ∞ qφ m / kt e −u du = mkT e − qφ m kT ⎛ − qφ m ⎞ ⎟⎟ . = A*T 2 exp⎜⎜ ⎝ kT ⎠ Equation 79 is the tunneling probability β= 2mn ( qV0 − E ) 2 = 2(9.11 × 10 −31 )( 20 − 2)(1.6 × 10 −19 ) = 2.17 × 1010 m −1 −34 2 (1.054 × 10 ) [ ⎧⎪ 20 × sinh(2.17 × 10 0 × 3 × 10 −10 T = ⎨1 + 4 × 2 × ( 20 − 2) ⎩ ] ⎫⎪ 2 −1 = 3.19 × 10 −6 . ⎬ ⎭ 23. Equation 79 is the tunneling probability β= 2mn ( qV0 − E ) 2 = 2(9.11 × 10 −31 )(6 − 2.2)(1.6 × 10 −19 ) = 9.99 × 10 9 m −1 −34 2 (1.054 × 10 ) [ ] 2 ⎧ 6 × sinh(9.99 × 10 9 × 10 −10 ) ⎫ −10 T (10 ) = ⎨1 + ⎬ 4 × 2.2 × ( 6 − 2.2 ) ⎪⎩ ⎪⎭ [ ] 2 ⎧⎪ 6 × sinh (9.99 × 10 9 × 10 −9 ) ⎫⎪ T (10 ) = ⎨1 + ⎬ 4 × 2.2 × (6 − 2.2 ) ⎩ ⎭ −9 16 −1 = 0.403 −1 = 7.8 × 10 −9 . 24. From Fig. 22 As E = 103 V/s νd ≈ 1.3×106 cm/s (Si) and νd ≈ 8.7×106 cm/s (GaAs) t ≈ 77 ps (Si) and t ≈ 11.5 ps (GaAs) As E = 5×104 V/s νd ≈ 107 cm/s (Si) and νd ≈ 8.2×106 cm/s (GaAs) t ≈ 10 ps (Si) and t ≈ 12.2 ps (GaAs). 25. Thermal velocity vth = 2 Eth = m0 2kT m0 2 × 1.38 × 10-23 × 300 9.1 × 10 −31 = 9.5 × 104 m/s = 9.5 × 106 cm/s = For electric field of 100 V/cm, drift velocity v d = μ n E = 1350 × 100 = 1.35 × 10 5 cm/s << v th For electric field of 104 V/cm. μ n E = 1350 × 10 4 = 1.35 × 10 7 cm/s ≈ v th . The value is comparable to the thermal velocity, the linear relationship between drift velocity and the electric field is not valid. 17 CHAPTER3 1. The impmity profile is, x (J.lm) The overall space charge neutrality of the semiconductor requires that the total negative space charge per unit area in the p -side must equal the total positive space charge per unit area in the n-side, thus we can obtain the depletion layer width in the n-side region: 0.8 X 8 X 10 14 - - - - - = Wn x3x 1014 2 Hence, then-side depletion layer width is: The total depletion layer width is 1.867 J.lm. We use the Poisson 's equation for calculation of the electric field E(x) . In the n-side region, 18 q dE q = N D ⇒ E( x n ) = NDx + K dx ε s εs E( x n = 1.067 μm ) = 0 ⇒ K = − ∴E ( x n ) = q εs q εs N D × 1.067 × 10 − 4 × 3 × 1014 ( x − 1.067 × 10 − 4 ) Emax = E( x n = 0 ) = −4.86 × 10 3 V/cm In the p-side region, the electrical field is: q dE q N A ⇒ E( x p ) = = × ax 2 + K ' dx ε s 2ε s E( x p = −0.8μm ) = 0 ⇒ K ' = − ( ) ∴E x p = ( q × a × 0.8 × 10 − 4 2ε s ( ) 2 ) 2 q × a × ⎡ x 2 − 0.8 × 10 − 4 ⎤ ⎦ ⎣ 2ε s Emax = E( x p = 0 ) = −4.86 × 10 3 V/cm The built-in potential is: Vbi = − ∫ 2. xn −xp E ( x )dx = − ∫ 0 −xp E ( x )dx − ∫ E (x )dx xn p − side 0 n − side = 0.52 V . From Vbi = − ∫ E ( x )dx , the potential distribution can be obtained With zero potential in the neutral p-region as a reference, the potential in the p-side depletion region is V p ( x ) = − ∫ E ( x )dx = − ∫ x 0 [ ] 2 q qa × a × x 2 − (0.8 × 10 − 4 ) dx = − 0 2ε 2ε s s x 2 3⎤ 2 ⎡1 = −7.596 × 1011 × ⎢ x 3 − (0.8 × 10 − 4 ) x − (0.8 × 10 − 4 ) ⎥ 3 ⎣3 ⎦ With the condition Vp(0)=Vn(0), the potential in the n-region is 19 2 ⎡1 3 −4 2 −4 3 ⎢⎣ 3 x − (0.8 × 10 ) x − 3 (0.8 × 10 ) 7( 2 3 4 = - 4.56 x l0 x 1 x -1. 067 x l 0- x - 0.8 -- x l o- 2 9 The potential distribution is Distance -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.067 p-regwn n-regwn 0.000 0.006 0.022 0.048 0.081 0.120 0.164 0.211 0.259 0.259413333 0.305788533 0.347603733 0.384858933 0.417554133 0.445689333 0.469264533 0.488279733 0.502734933 0.512630133 0.517965333 0.518988825 20 7) Potential Distribution v.~v v.Jvv v.-.vv V•J' .L / / / ~ -- /.~vv ___./ v V•'VV v.vvv - .(.5 05 -v . .vv Distance (urn) 21 I 5 3. The int1insic caniers density in Si at different temperatures can be obtained by using Fig.22 in Chapter 2 : Temperature (K) Inuinsic catTier density (n;) 250 1.50x108 300 9.65x109 350 2.00x10 11 400 8.50xl0 12 450 9.00xl0 13 500 2.20xl0 14 The Vbi can be obtained by using Eq. 12, and the results ru·e listed in the following table. T Vbi (V) ni 250 1.500E+08 0.777 300 9.65E+9 0.717 350 400 2.00E+11 0.653 8.50E+12 0.488 450 9.00E+l3 500 2.20E+14 0.366 0.329 Thus, the built-in potential is decreased as the temperan1re is increased. The depletion layer width and the maximum field at 300 K are W= 2ssVbi = qND 14 2 X 11.9 X 8 .85 X 10 - ~ 0.717 = 0. 9715 1.6x 10 - 19 x 10b flill 19 5 15 ~ = qNDW = 1.6 x 10- x 10 x 9.7 15 x 10- =1.476 x 104 V /cm . 14 &5 11.9 X 8.85 X 10- 22 ⎡ 2qVR ≈⎢ ⎣ εs ⎛ N A N D ⎞⎤ ⎜⎜ ⎟⎟⎥ 4. .Emax ⎝ N A + N D ⎠⎦ ND ⇒ 1.755 × 1016 = ND 1 + 18 10 1/ 2 ⎡ 2 × 1.6 × 10 −19 × 30 ⎛ 1018 N D ⎜ 18 ⇒ 4 × 10 = ⎢ −14 ⎜ ⎣ 11.9 × 8.85 × 10 ⎝ 10 + N D 5 ⎞⎤ ⎟⎟⎥ ⎠⎦ 1/ 2 We can select n-type doping concentration of ND = 1.755×1016 cm-3 for the junction. 5. From Eq. 12 and Eq. 35, we can obtain the 1/C2 versus V relationship for doping concentration of 1015, 1016, or 1017 cm-3, respectively. For ND=1015 cm-3, 1 Cj 2 = 2(Vbi − V ) 2 × (0.837 − V ) = = 1.187 × 1016 (0.837 − V ) −19 −14 15 qε s N B 1.6 × 10 × 11.9 × 8.85 × 10 × 10 For ND=1016 cm-3, 1 Cj 2 = 2(Vbi − V ) 2 × (0.896 − V ) = = 1.187 × 1015 (0.896 − V ) −19 −14 16 qε s N B 1.6 × 10 × 11.9 × 8.85 × 10 × 10 For ND =1017 cm-3, 1 Cj 2 = 2(Vbi − V ) 2 × (0.956 − V ) = = 1.187 × 1014 (0.956 − V ) −19 −14 17 qε s N B 1.6 × 10 × 11.9 × 8.85 × 10 × 10 When the reversed bias is applied, we summarize a table of 1/ C 2j vs V for various ND values as following, 23 v No= lEIS No=1E16 No=1E17 -4 5.741E+16 5.812E+l5 5.883E+l4 -3.5 5.148E+16 5.218E+l5 5.289E+l4 -3 -2.5 4.555E+16 4.625E+15 4.696E+l4 3.961E+16 4.031E+l5 4.102E+l4 -2 3.368E+l6 3.438E+l5 3.509E+l4 -1.5 2.774E+16 2.844E+15 2.915E+l4 -1 2.181E+l6 2.251E+l5 2.322E+14 -0.5 1.587E+l6 1.657E+l5 1.728E+l4 0 9.935E+l5 1.064E+15 1.134E+l4 Hence, we obtain a seiies of curves of 1/C versus Vas following, 1/C"2 VS v <'-'T>V -~ VuT>V ~~ ~ ~ -~ JuT>V ~ ~~ ''"' ,.v ~ ~ J'-' ' ~ ~ ~'"' •v ,.v ~ ''-'T>V -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -I -0.5 0 Applied Voltage The slopes of the curves is positive prop01tional to the values of the doping concentration. The interceptions give the built-in potential of the p-n junctions. 24 6. The built-in potential is ⎛ 10 20 × 10 20 × 11.9 × 8.85 × 10 −14 × 0.0259 ⎞ 2 kT ⎛ a 2 ε s kT ⎞ 2 ⎜ ⎟ ⎟ ln⎜⎜ = × 0 . 0259 × ln Vbi = 9 3 −19 ⎜ ⎟ 3 q ⎝ 8q 2 ni3 ⎟⎠ 3 8 × 1.6 × 10 × 9.65 × 10 ⎝ ⎠ = 0.5686 V ( ) From Eq. 38, the junction capacitance can be obtained ⎡ qaε s 2 ⎤ Cj = =⎢ ⎥ W ⎣12(Vbi − V R ) ⎦ εs 1/ 3 ( ⎡1.6 × 10 -19 × 10 20 × 11.9 × 8.85 × 10 −14 =⎢ 12(0.5686 − V R ) ⎣⎢ ) 2 ⎤ ⎥ ⎦⎥ 1/ 3 At reverse bias of 4V, the junction capacitance is 6.866×10-9 F/cm2. 7. From Eq. 35, we can obtain 1 Cj 2 = 2(Vbi − V ) 2(Vbi − V R ) 2 ⇒ ND = Cj qε s N B qε s ∵V R >> Vbi ⇒ N D ≅ 2(V R ) 2 2×4 Cj = × (0.85 × 10 −8 ) 2 −19 −14 qε s 1.6 × 10 × 11.9 × 8.85 × 10 ⇒ N d = 3.43 × 1015 cm −3 We can select the n-type doping concentration of 3.43×1015cm-3. 8. From Eq. 56, ⎤ ⎡ ⎥ ⎢ σ pσ nυ th N t ⎥n G = −U = ⎢ i ⎢ ⎛ Et − Ei ⎞ ⎛ Ei − Et ⎞ ⎥ σ σ exp exp + ⎜ ⎟ ⎜ ⎟⎥ ⎢ n p ⎝ kT ⎠ ⎝ kT ⎠ ⎦ ⎣ ⎤ ⎡ ⎥ ⎢ 7 15 −15 −15 10 × 10 × 10 × 10 ⎥ × 9.65 × 10 9 = 3.89 × 1016 =⎢ 0 . 02 − ⎥ ⎢ −15 ⎛ ⎞ ⎛ 0.02 ⎞ −15 ⎢10 exp⎜ 0.0259 ⎟ + 10 exp⎜ 0.0259 ⎟ ⎥ ⎝ ⎠⎦ ⎝ ⎠ ⎣ 25 and 2ε s (Vbi + V ) 2 × 11.9 × 8.85 × 10 −14 × (0.717 + 0.5) W= = = 12.66 × 10 −5 cm = 1.266 μm −19 15 qN A 1.6 × 10 × 10 Thus J gen = qGW = 1.6 × 1019 × 3.89 × 1016 × 12.66 × 10 −5 = 7.879 × 10 −7 A/cm 2 . 2 9. From Eq. 49, and p no n = i ND We can obtain the hole concentration at the edge of the space charge region, 08 ) ni ( 0 0259 (9.65 × 10 9 ) e ⎜⎝ 0 0259 ⎟⎠ = 2.42 × 1017 cm −3 . pn = e = ND 1016 2 2 10. ⎛ 08 ⎞ J = J p ( x n ) + J n (− x p ) = J s (e qV / kT − 1) V J ⇒ = e 0 0259 − 1 Js V ⇒ 0.95 = e 0 0259 − 1 ⇒ V = 0.017 V. 11. The parameters are ni= 9.65×109 cm-3 Dn= 21 cm2/sec Dp=10 cm2/sec τp0=τn0=5×10-7 sec From Eq. 52 and Eq. 54 26 J p (xn ) = qD p p no Lp (e qV / kT − 1) = q Dp τ po 2 ⎡ ⎛⎜ qVa ⎞⎟ ⎤ ni × × ⎢e ⎝ kT ⎠ − 1⎥ ND ⎢ ⎥⎦ ⎣ ⇒ 7 = 1.6 × 10 2 ⎛ 07 ⎞ ( 10 9.65 × 10 9 ) ⎡ ⎜⎝ 0 0259 ⎟⎠ ⎤ × × × ⎢e − 1⎥ ND 5 × 10 −7 ⎢⎣ ⎥⎦ −19 ⇒ N D = 5.2 × 1015 cm −3 J n (− x p ) = qDn n po Ln (e qV / kT − 1) = q Dn τ no 2 ⎡ ⎛⎜ qVa ⎞⎟ ⎤ ni × × ⎢e ⎝ kT ⎠ − 1⎥ NA ⎢ ⎥⎦ ⎣ ⇒ 25 = 1.6 × 10 −19 2 ⎛ 07 ⎞ ( 21 9.65 × 10 9 ) ⎡ ⎜⎝ 0 0259 ⎟⎠ ⎤ × × × ⎢e − 1⎥ NA 5 × 10 −7 ⎢⎣ ⎥⎦ ⇒ N A = 5.278 × 1016 cm −3 We can select a p-n diode with the conditions of NA= 5.278×1016cm-3 and ND = 5.4×1015cm-3. 12. Assumeτg =τp =τn = 10-6 s, Dn= 21 cm2/sec, and Dp = 10 cm2/sec (a) The saturation current calculation. From Eq. 55a and L p = D pτ p , we can obtain Js = qD p p n 0 Lp + qDn n p 0 Ln ⎛ 1 = qni2 ⎜ ⎜ ND ⎝ 2⎛ 1 = 1.6 × 10 −19 × 9.65 × 10 9 ⎜⎜ 18 ⎝ 10 = 6.87 × 10 −12 A/cm 2 ( ) Dp τ p0 + 1 NA Dn ⎞⎟ τ n0 ⎟ ⎠ 10 1 + 16 −6 10 10 21 10 −6 ⎞ ⎟ ⎟ ⎠ And from the cross-sectional area A = 1.2×10-5 cm2, we obtain 27 I s = A × J s = 1.2 × 10 −5 × 6.87 × 10 −12 = 8.244 × 10 −17 A . (b) The total current density is ⎛ qV ⎞ = J J s ⎜⎜ e kt − 1⎟⎟ ⎝ ⎠ Thus I 0 7V = 8.244 × 10 −17 7 ⎛ 0 00259 ⎞ ⎜e − 1⎟⎟ = 8.244 × 10 −17 × 5.47 × 1011 = 4.51 × 10 −5 A ⎜ ⎝ ⎠ ⎛ −0 7 ⎞ I − 0 7V = 8.244 × 10 −17 ⎜⎜ e 0 0259 − 1⎟⎟ = 8.244 × 10 −17 A. ⎝ ⎠ ⎛ qV ⎞ 13. From J = J s ⎜⎜ e kt − 1⎟⎟ ⎝ ⎠ we can obtain ⎡⎛ J V = ln ⎢⎜⎜ 0.0259 ⎣⎝ J s ⎡⎛ ⎞ ⎤ ⎞ ⎤ 10 −3 ⎟⎟ + 1⎥ ⇒ V = 0.0259 × ln ⎢⎜⎜ ⎟ + 1⎥ = 0.78 V . −17 ⎟ 8 . 244 10 × ⎠ ⎦ ⎠ ⎦ ⎣⎝ 14. From Eq. 59, and assume Dp= 10 cm2/sec, we can obtain JR ≅ q D p ni 2 qniW + τ p ND τg = 1.6 × 10 −19 ( 10 9.65 × 10 9 10 −6 1015 Vbi = 0.0259ln 1019 × 1015 (9.65 × 10 ) 9 2 ) 2 + 1.6 × 10 −19 × 9.65 × 10 9 10 −6 = 0.834 V Thus J R = 5.26 × 10 −11 + 1.872 × 10 −7 0.834 + V R 28 2 × 11.9 × 8.85 × 10 −14 × (Vbi + V R ) 1.6 × 10 −19 × 1015 Js 1.713E-07 1.755E-07 1.941E-07 2,129E-07 2.316E-07 2.503E-07 2.691E-07 2.878E-07 3.065E-07 3.252E-07 3.439E-07 VR 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 J.£ -01 , . -- - - - - - - - - - - - - - - - - - - - - - . l.£-07 1-----------~-""'-------1 2£ -07 v----- .i:l ~ ~ l .£-01 1.£ -07 1--- - - - - - - - - - - - - - - - - - - ----1 5.£-08 1--- - - - - - - - - - - - - - - - - - - ----1 O.E•OO .__--~---~---~--~---~------' o.z 0 0.< 0.6 Applied Voltaa.e vbi 10 17 )2 = 0.953 v \9.65 X 10 9 = o.0259ln ~ 10 19 X 29 0.8 1.2 J R = 5.26 x 1o-n + 1.872 x 1o-s Jo.956 + vR Current Density 3.00E-08 . . - - - - - - - - - - - - - - - - - - - - - - - - - - . 2.50E-08 f---- - - --------- ---=---=:::! -=- - - - - - ----l ~ 2.00E-08 ~ i! 8 ~ 1.50E-08 I.OOE-08 1---- - - - - - - - - - - - - - - - - - - - - - --1 5.00E-09 f---- - - - - - - - - - - - - - - - - - - - - - --1 O.OOE+OO .___ _ _......__ _ __.___ _ __.__ _ __.___ _ __.__ _ __, 0.2 0 0.4 0.6 0.8 1.2 Applied Voltage 15. From Eq. 39, _ f oo Pno\e ( qV ! kT - q _ x. 1) e - (x- x. )ILpdX The hole diffusion length is larger than the length of neutral region. _ f x~ - q x. ( qV ! kT _ Pno\e ) - (x- x. )ILPd 1 e X =L6 x Io-" x <9 6 ~~,;o')' (- 5 x 10-•{ e'~" - IXe~ - e ~) = 8.784 x 10-3 C/cm 2 . 16. From Fig. 26, the critical field at breakdown for a Si one-sided abmpt 30 junction is about 2.8 ×105 V/cm. Then from Eq. 85, we obtain V B (breakdown voltage) = = EcW ε s Ec = 2 2q ( 2 (N B )−1 11.9 × 8.85 × 10 −14 × 2.8 × 10 5 2 × 1.6 × 10 −19 ) (10 ) 2 15 −1 = 258 V W= 2ε s (Vbi − V ) ≅ qN B 2 × 11.9 × 8.85 × 10 −14 × 258 = 1.843 × 10 −3 cm = 18.43μm 1.6 × 10 −19 × 1015 When the n-region is reduced to 5μm, the punch-through will take place first. From Eq. 87, we can obtain W V B ' shaded area in Fig. 29 insert ⎛ W ⎞⎛ ⎟⎟ 2 − = = ⎜⎜ (Ec Wm )/2 VB Wm ⎝ Wm ⎠⎝ ⎛W V B ' = V B ⎜⎜ ⎝ Wm ⎞⎛ W ⎟⎟⎜⎜ 2 − Wm ⎠⎝ ⎞ ⎟⎟ ⎠ ⎞ 5 ⎞ ⎛ 5 ⎞⎛ ⎟⎟ = 258 × ⎜ ⎟⎜ 2 − ⎟ = 121 V 18.43 ⎠ ⎝ 18.43 ⎠⎝ ⎠ Compared to Fig. 29, the calculated result is the same as the value under the conditions of W = 5 μm and NB = 101 5 cm-3. 17. We can use following equations to determine the parameters of the diode. D p ni 2 qV / kT qWni qV / 2 kT D p ni 2 qV / kT JF = q e e + ≅q e τ p ND 2τ r τ p ND VB = EcW ε sEc = 2 2q 2 (N D )−1 ⇒ 31 ( D p ni 2 qV / kT D p 9.65 × 10 9 AJ F = Aq e ⇒ A × 1.6 × 10 −19 × ND τ p ND 10 −7 VB = EcW ε s Ec = 2 2q 2 ( N D )−1 ⇒ 130 = 11.9 × 8.85 × 10 −14 Ec 2 × 1.6 × 10 −19 2 ) 2 0.7 e 0.0259 = 2.2 × 10 −3 ( N D )−1 Let Ec= 4×105 V/cm, we can obtain ND = 4.05×1015 cm-3. The mobility of minority carrier hole is about 500 at ND = 4.05×1015 ∴Dp= 0.0259×500=12.95 cm2/s Thus, the cross-sectional area A is 8.6×10-5 cm2. 18. As the temperature increases, the total reverse current also increases. That is, the total electron current increases. The impact ionization takes place when the electron gains enough energy from the electrical field to create an electron-hole pair. When the temperature increases, total number of electron increases resulting in easy to lose their energy by collision with other electron before breaking the lattice bonds. This need higher breakdown voltage. 19. (a) The i-layer is easy to deplete, and assume the field in the depletion region is constant. From Eq. 84, we can obtain. ∫ W 0 6 6 1 ⎛ E ⎞ ⎛ E ⎞ −3 5 5 10 ⎜ dx = 1 ⇒ 10 4 ⎜ ⎟ × 10 = 1 ⇒ Ecritical = 4 × 10 × (10) 6 = 5.87 × 10 V/cm 5 ⎟ 4 105 × 4 10 × ⎝ ⎠ ⎝ ⎠ 4 ∴ V B = 5.87 × 10 5 × 10 −3 = 587 V (b) From Fig. 26, the critical field is 5 × 105 V/cm. 32 V B (breakdown voltage) = EcW ε s Ec = 2 2q 2 (N B )−1 12.4 × 8.85 × 10 −14 × (5 × 10 5 ) (2 × 1016 )−1 −19 2 × 1.6 × 10 2 = = 42.8 V. 20. a= 2 × 1018 = 10 22 cm − 4 2 × 10 − 4 3/ 2 2EW 4Ec = VB = 3 3 = 4E c 3/ 2 3 ⎡ 2ε s ⎤ ⎢ ⎥ ⎣ q ⎦ 1/ 2 (a )−1 / 2 ⎡ 2 × 11.9 × 8.85 × 10 −14 ⎤ ⎢ ⎥ 1.6 × 10 −19 ⎣ ⎦ = 4.84 × 10 −8 Ec 1/ 2 ( × 10 22 ) −1 / 2 3/ 2 The breakdown voltage can be determined by a selected Ec. 21. To calculate the results with applied voltage of V = 0.5 V , we can use a similar calculation in Example 10 with (1.6-0.5) replacing 1.6 for the voltage. The obtained electrostatic potentials are 1.1 V and 3.4 × 10 −4 V , respectively. The depletion widths are 3.821 × 10 −5 cm and 1.274 × 10 −8 cm , respectively. Also, by substituting V = −5 V to Eqs. 90 and 91, the electrostatic potentials are 6.6 V and 20.3 × 10 −4 V , and the depletion widths are 9.359 × 10 −5 cm and 3.12 × 10 −8 cm , respectively. The total depletion width will be reduced when the heterojunction is forward-biased from the thermal equilibrium condition. On the other hand, when the heterojunction is reverse-biased, the total depletion width will be increased. 22. Eg(0.3) = 1.424 + 1.247 × 0.3 = 1.789 eV Vbi = Eg2 q − ΔE C − (E F 2 − EV 2 ) / q − (E C1 − E F 1 ) / q q 33 = 1.789 – 0.21– kT 4.7 × 1017 kT 7 × 1018 – = 1.273 V ln ln q q 5 × 1015 5 × 1015 ⎡ ⎤ 2 N A ε 1ε 2 vbi x1 = ⎢ ⎥ ⎣ qN D (ε 1 N D + ε 2 N A ) ⎦ 1 2 ⎡ 2 × 12.4 × 11.46 × 8.85 × 10 −14 × 1.273 ⎤ =⎢ ⎥ 15 −19 ⎣ 1.6 × 10 × 5 × 10 (12.4 + 11.46) ⎦ = 4.1 × 10 −5 cm. Since N D x1 = N A x 2 ∴ x1 = x 2 ∴W = 2 x1 = 8.2 × 10 −5 cm = 0.82 μm . 34 1 2 CHAPTER 4 1. (a) The common-base and common-emitter current gains is given by α 0 = γα T = 0.997 × 0.998 = 0.995 α 0.995 β0 = 0 = 1 − α 0 1 − 0.995 = 199 . (b) Since I B = 0 and I Cp = 10 × 10 −9 A , then I CBO is 10 × 10 −9 A . The emitter current is I CEO = (1 + β 0 )I CBO = (1 + 199 ) ⋅ 10 × 10 −9 = 2 × 10 −6 A . 2. For an ideal transistor, α 0 = γ = 0.999 α0 = 999 . β0 = 1 − α0 I CBO is known and equals to 10 × 10 −6 A . Therefore, I CEO = (1 + β 0 )I CBO = (1 + 999) ⋅ 10 × 10 −6 = 10 mA . 3. (a) The emitter-base junction is forward biased. From Chapter 3 we obtain Vbi = kT ⎛⎜ N A N D ln q ⎜⎝ ni 2 18 17 ⎤ ⎡ ⎞ ⎟ = 0.0259 ln ⎢ 5 × 10 ⋅ 2 × 10 ⎥ = 0.956 V . ⎟ 9 2 ⎠ ⎣⎢ 9.65 × 10 ⎦⎥ ( The depletion-layer width in the base is 35 ) ⎛ NA W1 = ⎜⎜ ⎝ NA + ND ⎞ ⎟⎟ (Total depletion - layer width of the emitter - base junction) ⎠ = 2ε s q = 2 ⋅ 1.05 × 10 −12 1.6 × 10 −19 ⎛ NA ⎜⎜ ⎝ ND ⎞⎛ 1 ⎟⎟⎜⎜ ⎠⎝ N A + N D ⎞ ⎟⎟(Vbi − V ) ⎠ ⎛ 5 × 1018 ⎜⎜ 17 ⎝ 2 × 10 ⎞⎛ 1 ⎞ ⎟⎟⎜ (0.956 − 0.5) 18 17 ⎟ 5 × 10 + 2 × 10 ⎠ ⎠⎝ = 5.364 × 10 -6 cm = 5.364 × 10 -2 μm . Similarly we obtain for the base-collector function ⎡ 2 × 1017 ⋅ 1016 ⎤ Vbi = 0.0259 ln ⎢ = 0.795 V . 2 ⎥ ⎢⎣ 9.65 × 10 9 ⎥⎦ ( ) and W2 = 2 ⋅ 1.05 × 10 −12 1.6 × 10 −19 ⎛ 1016 ⎜⎜ 17 ⎝ 2 × 10 ⎞⎛ 1 ⎞ ⎟⎟⎜ 16 (0.795 + 5) 17 ⎟ ⎠⎝ 10 + 2 × 10 ⎠ = 4.254 × 10 -6 cm = 4.254 × 10 -2 μm . Therefore the neutral base width is W = WB − W1 − W2 = 1 − 5.364 × 10 −2 − 4.254 × 10 -2 = 0.904 μm . (b) Using Eq. 13a pn (0) = pno e qVEB kT 2 = ni qVEB e ND kT = (9.65 × 10 ) 9 2 2 × 1017 e 0 5 0 0259 = 2.543 × 1011 cm -3 . 4. In the emitter region L E = 52 ⋅ 10 −8 = 0.721 × 10 −3 cm D E = 52 cm s (9.65 ×10 ) = 9 2 n EO 5 × 1018 = 18.625 . In the base region 36 L p = D pτ p = 40 ⋅ 10 −7 = 2 × 10 −3 cm D p = 40 cm s ( 2 pno n 9.65 × 10 9 = i = ND 2 × 1017 ) 2 = 465.613 . In the collector region DC = 115 cm s nCO = (9.65 × 10 ) LC = 115 ⋅ 10 −6 = 10.724 × 10 −3 cm 9 2 16 10 = 9.312 × 10 3 . The current components are given by Eqs. 20, 21, 22, and 23: 1.6 × 10 −19 ⋅ 0.2 × 10 −2 ⋅ 40 ⋅ 465.613 0 5 0 0259 e = 1.596 × 10 -5 A −4 0.904 × 10 ≅ I Ep = 1.596 × 10 -5 A I Ep = I Cp I En I Cn I BB 5. (a) 1.6 × 10 −19 ⋅ 0.2 × 10 − 2 ⋅ 52 ⋅ 18.625 0 5 0 0259 (e = − 1) = 1.041 × 10 -7 A −3 0.721 × 10 −19 1.6 × 10 ⋅ 0.2 × 10 − 2 ⋅ 115 ⋅ 9.312 × 10 3 = = 3.196 × 10 -14 A −3 10.724 × 10 = I Ep − I Cp = 0 . The emitter, collector, and base currents are given by I E = I Ep + I En = 1.606 × 10 -5 A I C = I Cp + I Cn = 1.596 × 10 -5 A I B = I En + I BB − I Cn = 1.041 × 10 -7 A . (b) We can obtain the emitter efficiency and the base transport factor: γ= I Ep IE αT = = I Cp I Ep 1.596 × 10 −5 = 0.9938 1.606 × 10 −5 = 1.596 × 10 −5 =1 . 1.596 × 10 −5 Hence, the common-base and common-emitter current gains are α 0 = γα T = 0.9938 α0 = 160.3 . β0 = 1 − α0 37 (c) To improve γ , the emitter has to be doped much heavier than the base. To improve α T , we can make the base width narrower. 6. We can sketch p n ( x) p n (0) curves by using a computer program: 1.0 0.8 pn(x)/pn(0) 0.6 W/Lp<0.1 1 0.4 2 5 0.2 10 20 0.0 0.0 0.2 0.4 0.6 0.8 1.0 DISTANCE x In the figure, we can see when W L p < 0.1 ( W L p = 0.05 in this case), the minority carrier distribution approaches a straight line and can be simplified to Eq. 15. 7. Using Eq.14, I Ep is given by 38 ⎛ ⎞ dp ⎟ I Ep = A ⎜⎜ − qD p n ⎟ dx x =0 ⎠ ⎝ ⎧ ⎪ ⎪ = A(− qD p ) ⎨ p no ( e qVEB ⎪ ⎪ ⎩ = qA = qA D p p no Lp D p p no Lp ⎧ ⎪ ⎪ qVEB ⎨( e ⎪ ⎪ ⎩ ⎡ 1 ⎛W − x ⎞⎤ ⎟ cosh⎜ ⎢− ⎜ L p ⎟ ⎥⎥ L p ⎢ kT ⎝ ⎠ +p − 1) ⎢ no ⎥ ⎛ ⎞ W ⎟ ⎥ ⎢ ⎜ sinh ⎜ Lp ⎟ ⎥ ⎢ ⎝ ⎠ ⎦ ⎣ ⎡ ⎛ x 1 cosh⎜ ⎢ − ⎜ Lp ⎢ Lp ⎝ ⎢ ⎛ W ⎞ ⎢ sinh⎜ ⎟ ⎜ Lp ⎟ ⎢ ⎝ ⎠ ⎣ ⎞ ⎤⎫ ⎟ ⎥⎪ ⎟ ⎥⎪ ⎠ ⎥⎬ ⎥⎪ ⎥⎪ ⎦ ⎭ x =0 ⎤⎫ ⎡ ⎛ W ⎞⎤ ⎡ ⎥⎪ ⎢ cosh⎜⎜ ⎟⎟ ⎥ ⎢ L 1 ⎥⎪ p ⎢ ⎥ ⎢ kT ⎝ ⎠ + − 1) ⎢ ⎥⎬ ⎢ ⎥ ⎛ ⎞ ⎛ ⎞ W W ⎢ sinh⎜ ⎟ ⎥ ⎢ sinh⎜ ⎟ ⎥ ⎪ ⎜ Lp ⎟ ⎥⎪ ⎜ Lp ⎟ ⎥ ⎢ ⎢ ⎝ ⎠ ⎦⎭ ⎝ ⎠⎦ ⎣ ⎣ ⎛W coth⎜ ⎜ Lp ⎝ ⎡ ⎢ ⎢ qVEB ⎢e ⎢ ⎢ ⎣ ⎞ ⎟ ⎟ ⎠ ( kT ⎤ ⎥ 1 ⎥ . −1 + ⎛ W ⎞ ⎥⎥ cosh⎜ ⎟ ⎜ Lp ⎟ ⎥ ⎝ ⎠⎦ ) Similarly, we can obtain I Cp : ⎛ dp I Cp = A ⎜⎜ − qD p n dx ⎝ x =W ⎞ ⎟⎟ ⎠ ⎧ ⎪ ⎪ = A(− qD p ) ⎨ pno ( e qVEB ⎪ ⎪ ⎩ = qA = qA 8. D p pno Lp D p pno Lp ⎧ ⎪ ⎪ qVEB ⎨( e ⎪ ⎪ ⎩ ⎡ 1 ⎡ ⎛W − x ⎞⎤ ⎛ x 1 ⎟⎥ − cosh⎜ cosh⎜ ⎢− ⎢ ⎜ L ⎟⎥ ⎜L Lp ⎢ Lp p ⎝ ⎠ +p ⎢ ⎝ p kT − 1) ⎢ no ⎢ ⎥ ⎛W ⎞ ⎛W ⎞ ⎢ ⎥ ⎢ sinh⎜ ⎟ sinh⎜ ⎟ ⎜L ⎟ ⎜L ⎟ ⎢ ⎥ ⎢ ⎝ p⎠ ⎝ p⎠ ⎣ ⎦ ⎣ ⎡ ⎢ 1 ⎢ kT − 1) ⎢ ⎢ sinh⎛⎜ W ⎜L ⎢ ⎝ p ⎣ 1 ⎛W sinh⎜ ⎜ Lp ⎝ ⎡ qV ⎢ e EB ⎞ ⎢⎣ ⎟ ⎟ ⎠ ( kT ⎤ ⎡ ⎛ W ⎞ ⎤⎫ ⎥ ⎢ cosh⎜⎜ ⎟⎟ ⎥ ⎪ ⎥ ⎢ ⎝ Lp ⎠ ⎥⎪ + ⎬ ⎞⎥ ⎢ ⎛ W ⎞ ⎥⎪ ⎥ ⎢ ⎥ ⎟ sinh⎜ ⎟ ⎪ ⎟⎥ ⎢ ⎜L ⎟⎥ ⎠⎦ ⎣ ⎝ p ⎠ ⎦⎭ ⎛W − 1 + cosh⎜ ⎜ Lp ⎝ ) ⎞⎤ ⎟⎥ . ⎟⎥ ⎠⎦ The total excess minority carrier charge can be expressed by 39 ⎞ ⎤⎫ ⎟ ⎥⎪ ⎟ ⎥⎪ ⎠ ⎥⎬ ⎥⎪ ⎥⎪ ⎦ ⎭ x =W Q B = qA = qA W ∫ [ p ( x) − p ] dx n 0 ∫ W 0 no ⎡ qVEB ⎢⎣ pno e kT (1 − x ⎤ ) dx W ⎥⎦ W = qApno e qVEB kT qAWpno e qVEB 2 qAWpn (0) = . 2 = x2 (x − ) 2W 0 kT From Fig. 6, the triangular area in the base region is Wpn (0) . 2 By multiplying this value by q and the cross-sectional area A, we can obtain the same expression as Q B . In Problem 3, 1.6 × 10 −19 ⋅ 0.2 × 10 −2 ⋅ 0.904 × 10 −4 ⋅ 2.543 × 1011 2 -15 = 3.678 × 10 C . QB = 9. In Eq. 27, ( I C = a 21 e qVEB ≅ kT ) − 1 + a 22 qAD p p n (0) W 2 D p qAQp n (0) = 2 W2 2D p = QB . W2 Therefore, the collector current is directly proportional to the minority carrier charge stored in the base. 10. The base transport factor is 40 αT ≅ I Cp I Ep ⎡ qV kT ⎛ W ⎞⎤ ⎢ e EB − 1 + cosh⎜⎜ ⎟⎟⎥ ⎞ ⎢⎣ ⎝ L p ⎠⎥⎦ ⎟ ⎟ ⎠ . ⎡ ⎤ ⎢ ⎥ ⎛ W ⎞ ⎢ qV kT 1 ⎥ coth⎜ ⎟ ⎢ e EB − 1 + ⎜ Lp ⎟ ⎛ W ⎞ ⎥⎥ ⎝ ⎠ ⎢ cosh⎜ ⎟ ⎜ Lp ⎟ ⎥ ⎢ ⎝ ⎠⎦ ⎣ 1 ⎛W sinh⎜ ⎜ Lp ⎝ = ( ) ( ) For W L p << 1 , cosh(W L p ) ≅ 1 . Thus, αT = 1 ⎛W sinh⎜ ⎜ Lp ⎝ ⎞ ⎛ ⎟ ⋅ coth⎜ W ⎟ ⎜ Lp ⎠ ⎝ ⎞ ⎟ ⎟ ⎠ ⎞ ⎟ ⎟ ⎠ ⎛W = sech ⎜ ⎜ Lp ⎝ 1⎛W =1− ⎜ 2 ⎜⎝ L p ( ⎞ ⎟ ⎟ ⎠ 2 = 1 − W 2 2 Lp 2 ). 11. The common-emitter current gain is given by β0 ≡ α0 γα T = . 1 − α 0 1 − γα T β0 ≅ αT 1 − αT Since γ ≅ 1 , = ( 1 − W 2 2 Lp [ ( 2 ) 1 − 1 − W 2 2 Lp ( = 2Lp 2 ) 2 )] W 2 −1 . If W L p << 1 , then β 0 ≅ 2 L p W 2 . 2 12. L p = D pτ p = 100 ⋅ 3 × 10 −7 = 5.477 × 10 −3 cm = 54.77 μm Therefore, the common-emitter current gain is 41 2(54.77 × 10 −4 ) 2 β 0 ≅ 2L p 2 W 2 = (2 × 10 ) −4 2 = 1500 . 13. In the emitter region, μ pE = 54.3 + 407 1 + 0.374 × 10 −17 ⋅ 3 × 1018 = 87.6 D E = 0.0259 ⋅ 87.6 = 2.26 cm s . In the base region, μ n = 88 + 1252 1 + 0.698 × 10 −17 ⋅ 2 × 1016 = 1186.63 D p = 0.0259 ⋅1186.63 = 30.73 cm s . In the collector region, μ pC = 54.3 + 407 1 + 0.374 × 10 −17 ⋅ 5 × 1015 = 453.82 DC = 0.0259 ⋅ 453.82 = 11.75 cm s . 14. In the emitter region, LE = DEτ E = 2.269 ⋅ 10 −6 = 1.506 × 10 −3 cm 2 p EO = ( ni 9.65 × 10 9 = NE 3 × 1018 ) 2 = 31.04 cm -3 . In the base region, Ln = 30.734 ⋅ 10 −6 = 5.544 × 10 −3 cm n po = (9.65 ×10 ) 9 2 2 × 1016 = 4656.13 cm -3 . In the collector region, LC = 11.754 ⋅ 10 −6 = 3.428 × 10 −3 cm 42 (9.65 ×10 ) = 9 2 pCO 5 × 1015 = 18624.5 cm -3 . The emitter current components are given by I En = 1.6 × 10 −19 ⋅ 10 −4 ⋅ 30.734 ⋅ 4656.13 0 6 0 0259 e = 526.83 × 10 −6 A −4 0.5 × 10 I Ep = 1.6 × 10 −19 ⋅ 10 −4 ⋅ 2.269 ⋅ 31.04 0 6 0 0259 e − 1 = 8.609 × 10 −9 A . 1.506 × 10 −3 ( ) Hence, the emitter current is I E = I En + I Ep = 526.839 × 10 −6 A . And the collector current components are given by I Cn = 1.6 × 10 −19 ⋅ 10 −4 ⋅ 30.734 ⋅ 4656.13 0 6 0 0259 e = 526.83 × 10 −6 A −4 0.5 × 10 I Cp = 1.6 × 10 −19 ⋅ 10 −4 ⋅ 11.754 ⋅ 18624.5 = 1.022 × 10 −14 A . 3.428 × 10 −3 Therefore, the collector current is obtained by I C = I Cn + I Cp = 5.268 × 10 −4 A . 15. The emitter efficiency can be obtained by γ = I En 526.83 × 10 −6 = = 0.99998 . I E 526.839 × 10 −6 The base transport factor is αT = I Cn 526.83 × 10 −6 = =1 . I En 526.83 × 10 −6 Therefore, the common-base current gain is obtained by α 0 = γα T = 1 × 0.99998 = 0.99998 . The value is very close to unity. The common-emitter current gain is 43 β0 = 16. (a) α0 0.99998 = ≅ 50000 . 1 − α 0 1 − 0.99998 The total number of impurities in the neutral base region is QG = ∫ 0 W ( N AO e − x l dx = N AO l 1 − e −W ( = 2 × 1018 ⋅ 3 × 10 −5 1 − e −8×10 −5 l ) 3×10 −5 ) = 5.583 × 10 13 cm -2 . (b) Average impurity concentration is QG 5.583 × 1013 = W 8 × 10 −5 = 6.979 × 1017 cm -3 . 17. For N A = 6.979 × 1017 cm -3 , Dn = 7.77 cm 3 s , and Ln = Dnτ n = 7.77 ⋅ 10 −6 = 2.787 × 10 −3 cm αT ≅ 1 − W2 2 Ln = 1− 2 1 γ = 1+ DE QG Dn N E L E = (8 ×10 ) −5 2 ( 2 2.787 × 10 −3 ) 2 = 0.999588 1 = 0.99287 . 1 5.583 × 1013 1+ ⋅ 7.77 1019 ⋅ 10 −4 Therefore, α 0 = γα T = 0.99246 β0 = 18. α0 = 131.6 . 1−α0 The mobility of an average impurity concentration of 6.979 × 1017 cm -3 is about 300 cm 2 Vs . The average base resistivity ρ B is given by ρB = 1 = 0.0299 Ω - cm . qμ n (QG W ) 44 Therefore, ( ) RB = 5 × 10 −3 (ρ B W ) = 5 × 10 −3 ⋅ 0.0299 8 × 10 −5 = 1.869 Ω . For a voltage drop of kT q , IB = kT = 0.0139 A . qRB Therefore, I C = β 0 I B = 131.6 ⋅ 0.0139 = 1.83 A . 19. From Fig. 10b and Eq. 35, we obtain I B (μA ) I C (mA ) 0 0.20 --- 5 0.95 150 10 2.00 210 15 3.10 220 20 4.00 180 25 4.70 140 β0 = ΔI C ΔI B β 0 is not a constant. At low I B , because of generation-recombination current, β 0 increases with increasing I B . At high I B , VEB increases with I B , this in turn causes a reduction of VBC since VEB + VBC = VEC = 5 V . The reduction of VBC causes a widening of the neutral base region, therefore β 0 decreases. The following chart shows β 0 as function of I B . It is obvious that β 0 is not a constant. 45 250 β0 200 150 100 0 5 10 15 20 25 IB (μA) 20. Comparing the equations with Eq. 32 gives I FO = a11 , α R I RO = a12 α F I FO = a 21 , and I RO = a 22 . Hence, αF = αR = 21. a 21 = a11 a12 = a 22 1 W D E n EO 1+ ⋅ ⋅ LE D p p no 1 . W DC nCO ⋅ ⋅ 1+ LC D p p no In the collector region, LC = DCτ C = 2 ⋅ 10 −6 = 1.414 × 10 −3 cm nCO = ni 2 ( N C = 9.65 × 10 9 ) 2 From Problem 20, we have 46 5 × 1015 = 1.863 × 10 4 cm -3 . 30 1 1 = − 4 W DE n EO 0.5 × 10 1 9.31 1+ ⋅ ⋅ 1+ ⋅ ⋅ − 3 LE D p pno 10 9.31 × 10 −2 10 = 0.99995 αF = 1 1 = −4 W DC nCO 0 . 5 × 10 2 1.863 × 10 4 1+ ⋅ ⋅ 1+ ⋅ ⋅ LC D p pno 1.414 × 10 −3 10 9.31 × 10 −2 = 0.876 αR = ⎞ ⎟⎟ ⎠ ⎛ D p pno DE n EO + I FO = a11 = qA⎜⎜ LE ⎝ W ⎛ 10 ⋅ 9.31 × 10 2 1 ⋅ 9.31 ⎞ ⎟ = 1.6 × 10 −19 ⋅ 5 × 10 −4 ⋅ ⎜⎜ + −4 10 −4 ⎟⎠ ⎝ 0.5 × 10 = 1.49 × 10 −14 A ⎛ D p p no DC nCO + I RO = a 22 = qA⎜⎜ LC ⎝ W ⎞ ⎟⎟ ⎠ ⎛ 10 ⋅ 9.31 × 10 2 2 ⋅ 1.863 × 10 4 = 1.6 × 10 −19 ⋅ 5 × 10 −4 ⋅ ⎜⎜ + −4 1.414 × 10 −3 ⎝ 0.5 × 10 ⎞ ⎟⎟ ⎠ = 1.7 × 10 −14 A . The emitter and collector currents are ( I E = I FO e qVEB kT ) − 1 + α R I RO = 1.715 × 10 −4 A ( I C = α F I FO e qVEB = 1.715 × 10 −4 kT ) − 1 + I RO A . Note that these currents are almost the same (no base current) for W L p << 1 . 22. Referring Eq. 11, the field-free steady-state continuity equation in the collector region is ⎡ d 2 nC ( x' ) ⎤ nC (x' ) − n po = 0. DC ⎢ ⎥− 2 τC ⎣ dx' ⎦ The solution is given by ( LC = DCτ C ) nC (x' ) = C1e x' LC + C 2 e − x' LC Applying the boundary condition at x' = ∞ yields 47 . C1e ∞ LC + C 2 e −∞ LC = 0 . Hence C1 = 0 . In addition, for the boundary condition at x' = 0 , C 2 e −0 LC = C 2 = nC (0) ( nC (0). = nCO e qVCB kT ) −1 . The solution is ( nC ( x ) = nCO e qVCB kT ) − 1 e − x' LC . The collector current can be expressed as ⎛ ⎞ ⎛ ⎞ dp dn ⎟ + A ⎜ − qDC C ⎟ I C = A ⎜⎜ − qD p n ⎟ ⎜ ⎟ dx dx ' x =W ⎠ x' = 0 ⎠ ⎝ ⎝ D p p no qVEB kT ⎛ D p p no DC nCO e = qA − 1 − qA ⎜⎜ + W LC ⎝ W ( ( = a 21 e qVEB 23. ) kT − 1 − a 22 ) (e qVCB kT ) ( ⎞ qVCB ⎟ e ⎟ ⎠ kT ) −1 −1 . Using Eq. 44, the base transit time is given by τ B = W 2 2D p = (0.5 × 10 ) −4 2 2 × 10 = 1.25 × 10 -10 s . We can obtain the cutoff frequency : f T ≅ 1 2πτ B = 1.27 GHz . From Eq. 41, the common-base cutoff frequency is given by: fα ≅ f T α 0 = 1.27 × 10 9 = 1.275 GHz . 0.998 The common-emitter cutoff frequency is f β = (1 − α 0 ) f α = (1 − 0.998) × 1.275 × 10 9 = 2.55 MHz . Note that fβ f β = (1 − α 0 ) f α = (1 − α 0 ) α 0 × f T = can 1 β0 be fT . 48 expressed by 24. Neglect the time delays of emitter and collector, the base transit time is given by τB = 1 1 = = 31.83 × 10 −12 s . 2πf T 2π × 5 × 10 9 From Eq. 44, W can be expressed by W = 2 D pτ B . Therefore, W = 2 × 10 × 31.83 × 10 −12 = 2.52 × 10 -5 cm = 0.252 μm . The neutral base width should be 0.252 µm. 25. ΔE g =9.8% ×1.12 ≅ 110 meV. ⎛ ΔE g ⎞ ⎟ ⎟ kT ⎠ ⎝ β o ~exp ⎜⎜ ∴ 26. β o (100 °C ) ⎛ 110 meV 110 meV ⎞ = exp ⎜ − ⎟ = 0.29. β o (0 °C ) 273 k ⎠ ⎝ 373 k ⎛ E gE − E gB β 0 ( HBT ) = exp⎜⎜ β 0 ( BJT ) kT ⎝ ⎞ ⎡ E ( x ) − 1.424 ⎤ ⎟⎟ = exp ⎢ gE ⎥ 0.0259 ⎠ ⎣ ⎦ where E gE ( x ) = 1.424 + 1.247 x, x ≤ 0.45 = 1.9 + 0.125 x + 0.143x, 0.45 < x ≤ 1 . The plot of β 0 ( HBT ) β 0 ( BJT ) is shown in the following graph. 49 1E12 1E11 1EIO IE9 ~1~ ~ ~ 1 .424+1~47X IOOCOO 10000 1000 100 10 1~-r~~,-~--r-~-,--T--r~ M U ~ U M ~ M U M ~ ~ Note that /30 (HBT) increases exponentially when x increases. 27. The impmity concentration of the n1 region is 1014 cm-3 . The avalanche breakdown voltage (for W > Wm) is larger than 1500 V (Wm > 100 Jlm). For a reverse block voltage of 120 V, we can choose a width such that punch-through occm·s, 1.e., Thus, When switching occms, That is, a1 = = osJ*In( ~ J 1- 0.4 = 0.6 hl( .:!_J = 0.6 25 X 10-4 J0 0.5 39.6 x10-4 = 1.51 . 50 Therefore, J ≅ 4.5 J 0 = 2.25 × 10 −5 A cm 2 Area = Is 1 × 10 −3 = = 44.4 cm 2 . J 2.25 × 10 −5 51 CHAPTER 5 1. VG = VT EC EF Ei EV ψS = 2ψB METAL OXIDE 52 n-TYPE SEMICONDUCTOR 2. EC Ei EF EV EF EC Ei EV n+ POLYSILICON OXIDE p-TYPE SEMICONDUCTOR 3. VG = VFB = φms EC Ei EF EV φms n+ POLYSILICON OXIDE 53 p-TYPE SEMICONDUCTOR 4. Ps(x) Qp qND -d (a) X E (x) A -d 0 w .... (b) X '1/(X) -d 0 w (c) X 54 5. Wm = 2 ⎛ NA ⎝ ni ε s kTln⎜⎜ ⎞ ⎟⎟ ⎠ q2N A ⎛ 5 × 1016 11.9 × 8.85 × 10 −14 × 0.026ln⎜⎜ 9 ⎝ 9.65 × 10 =2 1.6 × 10 −19 × 5 × 1016 ⎞ ⎟ ⎟ ⎠ = 1.5 ×10-5 cm = 0.15 μ m. 6. C min = ε ox d + (ε ox / ε s )Wm Wm = 0.15μ m ∴ C min = 7. ψB = Es = W = From Prob. 5 3.9 × 8.85 × 10 −14 = 6.03 × 10 −8 F/cm2 . 3 .9 −7 −5 × 1.5 × 10 8 × 10 + 11.9 kT N A 1017 = 0.026ln = 0.42 V ln q ni 9.65 × 10 9 qN AW εs at intrinsic ψ s = ψ B 2ε sψ s = qN A 2 × 11.9 × 8.85 × 10 −14 × 0.42 = 0.74 × 10 −5 cm −19 17 1.6 × 10 × 10 Es = Eo = Es ε s ε ox = 1.11 1.6 × 10 −19 × 1017 × 0.74 × 10 −5 = 1.11 × 10 5 V/cm −14 11.9 × 8.85 × 10 ×105× 11.9 = 3.38 ×105 V/cm 3 .9 V = Vo+ψ s = Eod + ψ s = (3.38 × 10 5 × 5 × 10 −7 ) +0.42. = 0.59 V. 8. At the onset of strong inversion, ψ s =2ψ B ⇒ VG=VT thus, VG = qN AWm +2ψ B Co ⎛ 5 × 1016 9 ⎝ 9.65 × 10 From Prob. 5, Wm = 0.15 μm , ψ B = 0.026ln ⎜⎜ 55 ⎞ ⎟ = 0.4 V ⎟ ⎠ Co = 3.9 × 8.85 × 10 −14 = 3.45×10-7 F/cm2 10 −6 ∴VG = 9. Qot = 1.6 × 10 −19 1 [ × 1017 × (10 −6 ) 2 ] −6 ∫0 2 10 = 8×10-9 C/cm2 Q 8 × 10 −9 = ot = = 2.32 × 10 − 2 V. −7 C o 3.45 × 10 1 d ΔVFB 10. 1.6 × 10 −19 × 5 × 1016 × 1.5 × 10 −5 +0.8 = 0.35 + 0.8 = 1.15V. 3.45 × 10 −7 Qot = 10 −6 1 d ∫ d 0 yq(1017 )dy = yρ ot ( y )dy ⎧∞, ρ ot = q × 5 × 10 δ ( x) , where δ ( x) = ⎨ ⎩ 0, 11 y = 5 × 10 −7 y ≠ 5 × 10 −7 1 × 5 × 1011 × 5 × 10 −7 × 1.6 × 10 −19 10 −6 ∴ Qot = = 4×10-8 C/cm2 Q 4 × 10 −8 ∴ ΔVFB = ot = = 0.12 V C o 3.45 × 10 −7 = 11. Qot = 1 1 × 1.6 × 10 −19 × × 5 × 10 23 (10 − 6 ) 3 . −6 3 10 1 d ∫ 10 −6 0 y ( q × 5 × 10 23 × y )dy = 2.67×10-8 C/cm2 Qot 2.67 × 10 −8 ∴ ΔVFB = = = 7.74 × 10 −2 V. −7 C o 3.45 × 10 12. ΔV FB = Qot q d = × × N m , where Nm is the area density of Qm. Co Co d ΔVFB × C o 0.3 × 3.45 × 10 −7 ⇒ Nm = = = 6.47 × 1011 cm-2 . −19 q 1.6 × 10 56 13. Since VD << (VG − VT ) , the first term in Eq. 35 can be approximated as V Z μn Co (VG − 2ψ B − D )VD . 2 L Performing Taylor’s expansion on the 2nd term in Eq. 35, we obtain ( V D + 2ψ B )3 / 2 − ( 2ψ B )3 / 2 ≅ ( 2ψ B )3 / 2 + 3 3 ( 2ψ B )1 / 2 V D − ( 2ψ B )3 / 2 = ( 2ψ B )1 / 2 V D 2 2 Equation 33 can now be re-written as ID ≅ ⎡ ⎤ V Z 2 2ε s qN A 3 × μ n Co ⎢(VG − 2ψ B − D )VD − 2ψ B VD ⎥ L Co 2 3 2 ⎢⎣ ⎥⎦ ⎧⎪ ⎡ 2ε s qN A (2ψ B ) ⎤ VD ⎫⎪ Z μ n Co ⎨VG − ⎢ 2ψ B + ⎥ − ⎬VD L Co ⎢⎣ ⎥⎦ 2 ⎪⎭ ⎪⎩ V Z ≅ ( ) μn Co (VG − VT − D )VD L 2 ≅ where VT = 2ψ B + 14. 2ε s qN A ( 2ψ B ) Co . When the drain and gate are connected together, VG = VD and the MOSFET is operated in saturation (VD > VDsat ) . I D can be obtained by substituting VD = VDsat in Eq. 35 ID VD =VG = ⎧⎪⎛ V Z 2 ⎛ 2ε s qN A ⎞ μ n C o ⎨⎜ Dsat − 2ψ B ⎟V Dsat − ⎜ 2 L Co 3 ⎜⎝ ⎠ ⎩⎪⎝ ⎞ ⎟ (VDsat + 2ψ B ) 3 / 2 − 2(ψ B ) 3 / 2 ⎟ ⎠ [ where V Dsat is given by Eq. 40. Inserting the condition Qn ( y = L) = 0 into Eq. 29 yields V Dsat = 1 Co 2ε s qN A ( V Dsat + 2ψ B ) + 2ψ B − VG = 0 For V Dsat << 2ϕ B , the above equation reduces to VG = V D = 2ε s qN A ( 2ψ B ) Co + 2ψ B = VT . Therefore a linear extrapolation from the low current region to I D = 0 will yield the threshold voltage value. 15. ψ= ⎡ 5 × 1016 ⎤ N kT = 0.40 V ln( A ) = 0.026ln ⎢ −9 ⎥ ni q ⎣ 9.65 × 10 ⎦ 57 ⎫ ]⎪⎬ ⎪⎭ K≡ ε s qN A Co 11.9 × 8.85 × 10 −14 × 1.6 × 10 −19 × 5 × 10 −16 3.45 × 10 −7 = = 0.27 ⎛ 2V ∴VDsat ≅ VG − 2ψ B + K 2 ⎜1 − 1 + G2 ⎜ K ⎝ = 5 − 0.8 + (0.27) 2 [1 − 1 + ⎞ ⎟ ⎟ ⎠ 10 ] (0.27) 2 = 3.42 V I Dsat = Z μ n Co 10 × 800 × 3.45 × 10−7 (VG − VT )2 = (5 − 0.7)2 2L 2 ×1 = 2.55×10-2 A. 16. The device is operated in linear region, since V D = 0.1 V < (VG − VT ) = 0.5 V Therefore, g d = ∂I D ∂V D VG = const = Z μ n C o (VG − VT ) L = 5 × 500 × 3.45 × 10 −7 × 0.5 0.25 = 1.72×10-3 S. 17. gm = ∂I D ∂VG = V D = const . = Z μ n C oVD L 5 × 500 × 3.45 × 10 −7 × 0.1 0.25 =3.45×10-4 S. 18. VT = V FB + 2ψ B + V FB = φ ms − Qf Co 2 ε s qN Aψ B =− Co Eg 2 , ψ B = 0.026 ln( −ψ B − 1017 9.65 × 10 9 ) 1.6 × 10 −19 × 5 × 1010 3.45 × 10 −7 = − 0.56 − 0.42 − 0.02 = −1 V 2 11.9 × 8.85 × 10 −14 × 1017 × 0.42 × 1.6 × 10 −19 3.45 × 10 −7 = − 1 + 0.84 + 0.49 ∴VT = −1 + 0.84 + 58 = 0.33 V ( φ ms can also be obtained from Fig. 8 to be – 0.98 V). 19. 0.7 = 0.33 + FB = 20. qFB 3.45 × 10 −7 0.37 × 3.45 × 10 −7 = 8 × 1011 cm-2 . −19 1.6 × 10 φ ms = − Eg 2 VT = φ ms − + ψ B = −0.56 + 0.42 = −0.14 V Qf Co − 2ψ B − 2 ε s qN Dψ B = −0.14 − 0.02 − 0.84 − Co 2 11.9 × 8.85 × 10 −14 × 1.6 × 10 −19 × 1017 × 0.42 3.45 × 10 −7 = − 1.49 V. 21. − 0.7 = −1.49 + FB = 22. qFB 3.45 × 10 −7 0.79 × 3.45 × 10 −7 = 1.7 × 1012 cm-2 . −19 1.6 × 10 The bandgap in degenerately doped Si is around 1eV due to bandgap-narrowing effect. Therefore, φ ms = −0.14 + 1 = 0.86 V ∴VT = 0.86 − 0.02 − 0.84 − 0.49 = −0.49 V. 59 23. ⎛ ⎞ ⎟ = 0.42 V ⎟ 9 . 65 × 10 ⎝ ⎠ 1017 ψ B = 0.026 ln⎜⎜ VT = φ ms − 9 qQ f Co + 2ψ B + 2 ε s qN Aψ B Co = −0.98 − 1.6 × 10 −19 × 1011 2 11.9 × 8.85 × 10 −14 × 1017 × 0.42 × 1.6 × 10 −19 + 0.84 + Co Co = −0.14 + 15.2 × 10 −8 Co 3.9 × 8.85 × 10 −14 3.45 × 10 −13 Co = = d d d × 15.2 × 10 −8 > 20.14 VT > 20 ⇒ 3.45 × 10 −13 ∴ d > 4.57 × 10 −5 cm = 0.457 μm. 24. VT = 0.5 V at I d = 0.1μA ⎛ log I D Subthreshold swing = ⎜ ⎜ ⎝ 0.1 = VG =VT − log I D VG = 0 VT − 0 ⎞ ⎟ ⎟ ⎠ −1 0.5 − 7 − log I D log I D VG = 0 = −12 VG = 0 ∴ID VG = 0 = 1 × 10 −12 A . 25. ΔVT = 2 qε s N A Co ( 2ψ B + V BS − 2ψ B ) 1017 ψ B = 0.026 ln( ) = 0.42 V 9.65 × 10 9 3.9 × 8.85 × 10 −14 Co = 6.9 × 10 −7 F/cm2 −7 5 × 10 ΔVT = 0.1 V if we want to reduce ID at VG = 0 by one order of magnitude, since the subthreshold swing is 100 mV/decade. 60 0.1 = ∴V BS 2 × 1.6 × 10 −19 × 4.9 × 8.85 × 10 −14 × 1017 6.9 × 10 −7 = 0.83 V. 61 ( 0.84 + V BS − 0.84 ) CHAPTER 6 1. Scaling factor κ =10 1 Switching energy = (C ⋅ A)V 2 2 C′ = A′ = V′ = ε ox d A = κC κ2 V κ ∴ scaling factor for switching energy : κ ⋅ 1 κ 2 ⋅ 1 κ 2 = 1 κ 3 = 1 1000 A reduction of one thousand times. 1 2 (a) ∴ Switching energy = (C ⋅ A)V 2 = (b) Pτ ~ 1 2000 1 1 1 1 ×1 = 3 ×1 = × 1 = 1 mJ k2 k k 1000 2. From Fig. 2 we have (r j + Δ) 2 = (r j + Wm ) 2 − Wm 2 ∴ Δ2 + 2Δr j − 2W m r j = 0 Δ = − r j + r j + 2W m r j L' = L − 2Δ rj L + L' 2 L − 2Δ Δ = = 1− = 1− L L 2L 2L ⎛ ⎞ ⎜ 1 + 2W m − 1⎟ ⎜ ⎟ rj ⎝ ⎠ From Eq. 17, Ch. 5 we have ΔVT = = (space charge in the trapezaidal region − space charge in the rectangular region) Co ⎞ qN W r ⎛ 2W qN AWm L + L ' qN AWm ( )− = − A m j ⎜ 1 + m − 1⎟ . ⎟ 2L Co Co Co L ⎜⎝ rj ⎠ 62 3. 4. Pros: 1. 2. Cons: 1. 2. Higher operation speed. High device density More complicated fabrication flow. High manufacturing cost. 5. (a) (b) 63 (c) At point C, the voltage Va= Vin (input voltage) , the NMOS is just becoming saturated from the linear region. Since NMOS is in the linear region and PMOS in the saturation region, the drain current of NMOS is equal to the drain current of the PMOS, that is IDN= IDP Therefore, 6. The maximum width of the surface depletion region for bulk MOS Wm = 2 ε s kTln ( N A / ni ) q2N A 11.9 × 8.85 × 10 −14 × 0.026 × ln(5 × 1017 / 9.65 × 10 9 ) 1.6 × 10 −19 × 5 × 1017 = 4.9 × 10 −6 cm =2 = 49 nm For FD-SOI, d si ≤ Wm = 49 nm . 7. VT = VFB + 2ψ B + qN A d si Co 64 Eg V FB = φ ms NA ⎛ 5 × 10 17 kT 1.12 ln( )=− =− − − 0.026ln⎜⎜ 9 q ni 2 2 ⎝ 9.65 × 10 2ψ B = 2 × N kT ln( A ) = 0.92 V q ni ⎞ ⎟⎟ = −1.02 V ⎠ 3.9 × 8.85 × 10 −14 = 8.63 × 10 − 7 F/cm 2 4 × 10 − 7 1.6 × 10 −19 × 5 × 1017 × 3 × 10 − 6 ∴VT = −1.02 + 0.92 + 8.63 × 10 − 7 = −0.1 + 0.28 Co = = 0.18 V. 8. ΔVT = qN A Δd si Co 1.6 × 10 −19 × 5 × 1017 × 5 × 10 −7 8.63 × 10 −7 = 46 mV = Thus, the range of VT is from (0.18-0.046)= 0.134 V to (0.18+0.046) = 0.226 V. 9. (a) It is fully depleted with a SOI structure, the leakage current is low. (b) Source and drain to body junction capacitance is low, the speed is fast. (c) The Si is very thin, there is no leakage path far from the gate. (d) Low vertical field and less impurity scattering, the mobility is higher. 10. The planar capacitor (1 × 10 −4 ) 2 3.9 × 8.86 × 10 −14 ε ox = 3.45 × 10 −15 F = C= A −6 d 1 × 10 For the trench capacitor A = 4× 7 µm2+ 1 µm2 = 29 µm2 C = 29 ×3.45 ×10-15 F = 100 × 10-15 F. 11. C= ε ox tox = 3.9 × 8.85 × 10−14 = 6.9 × 10−7 F/cm 2 −8 50 × 10 65 Area = (0.5 × 0.5) + 4(0.5 × l ) = 0.25 + 2l μm 2 Q 105 × 1.6 × 10−19 = = 8 × 10−15 F V 2 8 × 10−15 = 6.9 × 10−7 (0.25 + 2l ) × 10−8 = 6.9 × 10−15 (0.25 + 2l ) 1.16 − 0.25 ∴ l= = 0.46 μm 2 Q = CV 12. 13. I =C C= dV 2.5 = 5 × 10−14 × = 3.1 × 10−11 A dt 4 × 10−3 Z μ p C i (Vo − VT ) L 4 × 10-5 = A ( − 5 − VT ) go = ∴VT = 7 V -5 1 × 10 = A( − 5 + 2 ) ΔV = 7 − ( −2) = 9 V. 14. E1 = VG Q + . d1 + d 2 (ε1 / ε 2 ) ε1 + ε 2 ( d1 / d 2 ) ⎧ ⎫ ⎪⎪ 10 × 107 ⎪⎪ Q 5 J1 = σ E1 = 10−7 ⎨ + ⎬ = 0.2 − 2.26 × 10 Q 4 10 ⎤ ⎪ 10 + 100( ) ⎡ 4 + 10( )8.85 × 10−14 ⎥ ⎪ ⎢⎣ 10 100 ⎪⎩ ⎦ ⎪⎭ 0.2 when J1 = 0 Q = = 8.84 × 107 Couls 2.26 × 105 Q 8.85 × 107 ΔVT = = = 10 V ε 2 / d 2 10 × 8.85 × 10−4 /10−5 15. QFG = CFG iCG × ΔV nFG = CFG i CG × ΔV 2.59 × 10−15 × 0.5 = = 8094 electrons. q 1.6 × 10−19 16. Cell size Write one byte rate Rewrite cycle Keep data without power SRAM Large Fast unlimited No DRAM Mid Fast unlimited No Flash small Slow limited Yes 66 Applications Embedded in ligic chips Stand-alone chips and enbeded Nonvoltatile storage stand-alone 17. VT = V FB + 2ψ B + ε s qN Aψ B C0 8.854 × 10 −14 C 0 = 3.9 × = 3.45 × 10 −8 F/cm 2 −5 10 ⎛ 1017 ⎞ ⎟ = 0.42 V ψ B = 0.026 ln⎜ 9 9 . 65 × 10 ⎝ ⎠ V FB = φ ms − Qf 2 Eg −ψ B − 0 2 = −0.56 − 0.42 = −0.98 V =− 2 11.9 × 8.85 × 10 −14 × 1017 × 0.42 × 1.6 × 10 −19 ∴ VT = −0.98 + 0.42 + 3.45 × 10 −8 = − 0.98 + 0.42 + 4.9 = 4.34 V 18. ΔVFB 5 × 1011 × 1.6 × 10−19 =− =− = − 2.32 V C0 3.45 × 10−8 Qf VT = 4.34 − 2.32 = 2.02 V. 67 CHAPTER 7 1. From Eq.1, the theoretical barrier height is φ Bn = φ m − χ = 4.55 − 4.01 = 0.54 eV We can calculate Vn as Vn = 2.86 × 1019 kT N C ln = 0.0259 ln( ) = 0.188 V q ND 2 × 1016 Therefore, the built-in potential is Vbi = φ Bn − Vn = 0.54 − 0.188 = 0.352 V . 2. (a) From Eq.11 d (1 / C 2 ) (6.2 − 4.6) × 1014 = = −2.3 × 1014 (cm 2 /F) 2 /V dV −2−0 ND = Vn = 2 qε s ⎤ ⎡ −1 16 -3 ⎥ = 4.7 × 10 cm ⎢ 2 ⎣ d (1 / C ) / dV ⎦ ⎛ 4.7 × 1017 kT N C ln = 0.0259 ln⎜⎜ 16 q ND ⎝ 4.7 × 10 ⎞ ⎟ = 0.06 V ⎟ ⎠ From Fig.6, the intercept of the GaAs contact is the built-in potential Vbi, which is equal to 0.7 V. Then, the barrier height is φ Bn = Vbi + Vn = 0.76 V (b) J s = 5 × 1017 A/cm 2 A* = 8 A/K2-cm2 for n– type GaAs J s = A*T 2 e − qφ Bn / kT φ Bn ⎡ 8 × (300) 2 ⎤ kT A*T 2 ln( ) = 0.0259 ln ⎢ = 0.72 eV = −7 ⎥ q Js ⎣ 5 × 10 ⎦ The barrier height from capacitance is 0.04 V or 5% larger. 68 (c) For V = –1 V W= 2ε s (Vbi + V R ) 2 × 1.09 × 10 −12 (0.7 + 1) = qN D 1.6 × 10 −19 × 4.7 × 1016 = 2.22 × 10 −5 cm = 0.222 μm Em = C= 3. qN DW εs W εs = 1.43 × 10 5 V/cm = 5.22 × 10 −8 F/cm 2 . The barrier height is φ Bn = φ m − χ = 4.65 − 4.01 = 0.64 V Vn = ⎛ 2.86 × 1019 kT N C ln = 0.0259 × ln⎜⎜ 16 q ND ⎝ 3 × 10 ⎞ ⎟ = 0.177 V ⎟ ⎠ The built-in potential is Vbi = φ Bn − Vn = 0.64 − 0.177 = 0.463 V The depletion width is W = 2ε s (Vbi − V ) = qN D 2 × 11.9 × 8.85 × 10 −14 × 0.463 = 0.142 μm 1.6 × 10 −19 × 3 × 1016 The maximum electric field is Em = E ( x = 0) = qN D εs W= (1.6 × 1019 ) × (3 × 1016 ) × (1.42 × 10 −5 ) = 6.54 × 10 4 V/cm. −14 11.9 × (8.85 × 10 ) 4. The unit of C needs to be changed from μF to F/cm2, so 1/C2 = 1.74×1015 – 2.12×1015 Va (cm2/F)2 Therefore, we obtain the built-in potential at 1/C2 = 0 Vbi = 1.57 × 1015 = 0.74 V 2.12 × 1015 69 From the given relationship between C and Va, we obtain ⎛ 1 ⎞ d⎜ 2 ⎟ ⎝ C ⎠ = −2.12 × 1015 dVa (cm2/F)2/V From Eq.11 ⎤ ⎡ −1 ⎥ ⎢ 2 ⎣ d (1 / C ) / dV ⎦ 2 1 ⎞ ⎛ = −19 −14 ⎜ 15 ⎟ 1.6 × 10 × 11.9 × 8.85 × 10 ⎝ 2.12 × 10 ⎠ ND = 2 qε s = 5.6 × 1015 cm -3 Vn = ⎛ 2.86 × 1019 kT N C ln = 0.0259 ln⎜⎜ 16 q ND ⎝ 5.6 × 10 ⎞ ⎟ = 0.161 V ⎟ ⎠ We can obtain the barrier height φ Bn = Vbi + Vn = 0.74 + 0.161 = 0.901 V. 5. The built-in potential is Vbi = φ Bn − kT N C ln q ND ⎛ 2.86 × 1019 = 0.8 − 0.0259 ln⎜⎜ 16 ⎝ 1.5 × 10 = 0.8 − 0.195 ⎞ ⎟ ⎟ ⎠ = 0.605 V Then, the work function is φ m = φ Bn + χ = 0.8 + 4.01 = 4.81 V. 6. The saturation current density is 70 ⎛ − qφ Bn ⎞ J s = A*T 2 exp⎜ ⎟ ⎝ kT ⎠ ⎛ − 0 .8 ⎞ = 110 × (300) 2 × exp⎜ ⎟ ⎝ 0.0259 ⎠ = 3.81 × 10 −7 A/cm 2 The injected hole current density is J po = qD p ni 2 Lp N D = 1.6 × 10 −19 × 12 × (9.65 × 10 9 ) 2 = 1.19 × 10 −11 A/cm 2 −3 16 1 × 10 × 1.5 × 10 J po (e qV / kT − 1) Hole current = Electron current J s (e qV / kT − 1) J po 1.18 × 10 −11 = 3 × 10 −5. = = Js 3.81 × 10 − 3 7. The difference between the conduction band and the Fermi level is given by ⎛ 4.7 × 1017 Vn = 0.0259 ln⎜⎜ 17 ⎝ 1 × 10 ⎞ ⎟ = 0.04 V. ⎟ ⎠ The built-in potential barrier is then Vbi = 0.9 − 0.04 = 0.86 V For a depletion mode operation, VT is negative. Therefore, From Eq.38a VT = 0.86 − V P < 0 qa 2 N D 1.6 × 10 −19 a 2 × 1017 VP = > 0.86 = 2ε s 2 × 12.4 × 8.85 × 10 −14 1.6 × 10 − 2 2 a > 0.86 2.19 × 10 −12 a > 1.08 × 10 −5 cm = 0.108 μ m. 71 8. From Eq.33 we obtain gm = = IP 2V p2 VP VD VG + Vbi Zμ n ε s aL VP VD Vbi qN D a 2 1.6 × 10 −19 × 7 × 1016 × (3 × 10 −5 ) 2 VP = = 4.62 V = 2ε s 2 × 12.4 × 8.85 × 10 −14 ∴ gm = 5 × 10 −4 × 4500 × 12.4 × 8.85 × 10 −14 0.3 × 10 − 4 × 1.5 × 10 − 4 4.62 ×1 0.84 = 1.28×10-3 S = 1.28 mS. 9. (a) The built-in voltage is ⎛ 4.7 × 1017 Vbi = φ Bn − Vn = 0.9 − 0.025 ln⎜⎜ 17 ⎝ 10 ⎞ ⎟⎟ = 0.86 V ⎠ At zero bias, the width of the depletion layer is W = 2ε sVbi 2 × 1.09 × 10 −12 × 0.86 = qN D 1.6 × 10 −19 × 1017 = 1.07 × 10-5 cm = 0.107 µm Since W is smaller than 0.2 µm, it is a depletion-mode device. (b) The pinch-off voltage is qN D a 2 1.6 × 10 −19 × 1017 (2 × 10 −5 ) VP = = 2.92 V = 2ε s 2 × 12.4 × 8.85 × 10 −14 and the threshold voltage is VT = Vbi – VP = 0.86 – 2.92 = –2.06 V. 10. From Eq.31b, the pinch-off voltage is 72 qN D a 2 1.6 × 10 −19 × 1017 (2 × 10 −5 ) VP = = 0.364 V = 2ε s 2 × 12.4 × 8.85 × 10 −14 The threshold voltage is VT = Vbi – Vp = 0.8 – 0.364 = 0.436 V and the saturation current is given by Eq. 39 I Dsat = = 11. Zε s μ n (VG − VT )2 2aL 50 × 10 −4 × 12.4 × 8.85 × 10 −14 × 4500 (0 − 0.436) 2 = 4.7 × 10 − 4 A. −4 −4 2 × (0.5 × 10 ) × (1 × 10 ) ⎛ 4.7 × 1017 0.85 – 0.0259ln ⎜⎜ ⎝ ND ⎞ 1.6 × 10 −19 × N D ⎟⎟ − a2 = 0 −14 2 × 12 . 4 × 8 . 85 × 10 ⎠ For ND = 4.7 × 1016 cm-3 ⎛ 4.7 × 1017 a = ⎜⎜ 0.85 − 0.0259 ln ND ⎝ ⎞ ⎟⎟ ⎠ 1 2 (3.7 × 10 3 ) ND = 1.52 ×10-5 cm = 0.152 µm For ND = 4.7 × 1017 cm-3 a = 0.496 × 10-5 cm = 0.0496 µm. 12. From Eq.48 the pinch-off voltage is V P = φ Bn − ΔEC − VT q = 0.89 – 0.23 – (–0.5) = 0.62 V and then, VP = qN D d12 1.6 × 10 −19 × 3 × 1018 × d12 = 0.62 V = 2ε s 2 × 12.3 × 8.85 × 10 −14 73 d1 = 1.68 × 10-6 cm d1= 16.8 nm Therefore, this thickness of the doped AlGaAs layer is 16.8 nm. 13. The pinch-off voltage is VP = qN D d12 1.6 × 10 −19 × 1018 × (50 × 10 −7 ) = 1.84 V = 2ε s 2 × 12.3 × 8.85 × 10 −4 The threshold voltage is VT = φ Bn − ΔEC − VP q = 0.89 – 0.23 – 1.84 = –1.18 V When ns = 1.25× 1012 cm-2, we obtain ns = 12.3 × 8.85 × 10 −14 × [0 − (−1.18)] = 1.25 × 1012 −19 −7 1.6 × 10 × (50 + d 0 + 8) × 10 and then d0 +58.5 = 64.3 d0 = 5.8 nm The thickness of the undoped spacer is 5.8 nm. 14. The pinch-off voltage is VP = qN D d12 1.6 × 10 −19 × 5 × 1017 × (50 × 10 −7 ) 2 = 1.84 V = 2ε s 2 × 12.3 × 8.85 × 10 −14 The barrier height is φ Bn = VT + ΔEC + VP = −1.3 + 0.25 + 1.84 = 0.79 V q 74 The 2DEG concentration is 12.3 × 8.85 × 10 −14 ns = × [0 − (−1.3)] = 1.29 × 1012 cm-2. −19 −7 1.6 × 10 × (50 + 10 + 8) × 10 15. The pinch-off voltage is VP = qN D d12 1.6 × 10 −19 × 1 × 1018 × d12 = 1.5 = 2ε s 2 × 12.3 × 8.85 × 10 − 4 The thickness of the doped AlGaAs is 1.5 × 2 × 12.3 × 8.85 × 10 −14 d1 = = 4.45×10-8 cm = 44.5 nm −19 18 1.6 × 10 × 10 VT = φ Bn − 16. ΔE C − V P = 0.8 − 0.23 − 1.5 = −0.93 V. q The pinch-off voltage is VP = = qN D 2 d1 2ε s 2 1.6 × 10 −19 × 3 × 1018 ( 35 × 10 −7 ) = 2.7 V −14 2 × 12.3 × 8.85 × 10 the threshold voltage is VT = φ Bn − ΔEC −Vp q = 0.89 – 0.24 – 2.7 = –2.05 V Therefore, the two-dimensional electron gas is ns = 12.3 × 8.85 × 10 −14 × [0 − (−2.05)] = 3.2 × 1012 cm-2. −19 −7 1.6 × 10 × (35 + 8) × 10 75 CHAPTER 8 Vbi ≅ ( E g / q) + Vn + V p = 1.42+0.03+0.03 = 1.48 V 1. 2ε S q W = = ⎛ NA + ND ⎜⎜ ⎝ N AND 2 × 1.16 × 10 −12 1.6 × 10 −19 ⎞ ⎟⎟(Vbi + V ) ⎠ ⎛ 1019 + 1019 ⎞ ⎟(1.48 − 0.25) ⎜ 19 19 ⎝ 10 × 10 ⎠ = 1.89 × 10 −6 cm = 18.9 nm C= 2. ⎛V I = I p ⎜⎜ ⎝ VP εS W = 1.16 × 10 −12 = 6.13 × 10 −7 F/cm 2 . 1.89 × 10 −6 ⎞ ⎛ V ⎟⎟ exp⎜⎜1 − ⎠ ⎝ VP ⎞ ⎛ qV ⎞ ⎟⎟ + I 0 exp⎜ ⎟ ⎝ kT ⎠ ⎠ From Fig. 4, We note that the largest negative differential resistance occurs between Vp<V<VV. The corresponding voltage can be obtained from the condition d2I/d2V = 0. By neglecting the second term in Eq. 5, we obtain dI ⎛ I P I PV ≅⎜ − dV ⎜⎝ VP VP2 ⎞ ⎛ V ⎞ ⎟⎟ exp ⎜⎜1 − ⎟⎟ ⎠ ⎝ VP ⎠ ⎛ d 2 I ⎛ − 2 I P I PV ⎞ V ⎞ = ⎜⎜ 2 + 3 ⎟⎟ exp ⎜⎜1 − ⎟⎟ = 0 2 dV VP ⎠ ⎝ VP ⎝ VP ⎠ ∴V = 2VP = 2 × 0.1 = 0.2V dI dV ⎡10 −2 10 −2 × 0.2 ⎤ ⎛ 0.2 ⎞ = − ⎟ = −0.0367 0 2V ⎢ ⎥ exp⎜1 − 2 (0.1) ⎦ ⎝ 0.1 ⎠ ⎣ 0.1 ⎛ dI R=⎜ ⎝ dV 3. (a) RSC = −1 ⎞ = −27.2 Ω . 0 2V ⎟ ⎠ 1 W 1 W Ix W2 Δ E = = dx dx I ∫0 I ∫ 0 Aε S v S 2 Aε S v S 76 = (12 × 10 ) −4 2 2(5 × 10 − 4 ) 1.05 × 10 −12 × 10 7 = 137 Ω (b) The breakdown voltage for ND = 1015 cm-3 and W = 12 μm is 250 V (Refer to Chapter 4). The voltage due to RSC is ( ) IRSC = 10 3 × 5 × 10 −4 × 137 = 68.5 V The total applied voltage is then 250 + 68.5 = 318.5 V. 4. (a) The dc input power is 100V(10-1A) = 10W. For 25% efficiency, the power dissipated as heat is 10W(1-25%) = 7.5 W. ΔT = 7.5W × (10 o C/W) = 75 o C (b) ΔVB = (60mV/ o C) × 75 o C = 4.5 V The breakdown voltage at room temperature is (100-4.5) = 95.5V . 5. (a) For a uniform breakdown in the avalanche region, the maximum electric field is Em = 4.4 ×105 V/cm. The total voltage at breakdown across the diode is ⎛ qQ ⎞ VB = Em xA + ⎜⎜Em − ⎟⎟ (W − xA ) εS ⎠ ⎝ ⎛ 1.6 ×10−19 ×1.5 ×1012 ⎞ ⎟ (3 - 0.4) ×10−4 = 4.4 ×105 (0.4 ×10−4 ) + ⎜ 4.4 ×105 − −12 1 . 09 10 × ⎝ ⎠ = 17.6 + 57.2 = 74.8 V (b) The average field in the drift region is 77 57.2 (3 − 0.4) × 10 −4 = 2.2 × 10 5 V/cm This field is high enough to maintain velocity saturation in the drift region. (c) f = vS 10 7 = = 19 GHz . 2(W − x A ) 2(3 − 0.4 )10 − 4 6. (a) In the p layer E1(x) = Em − E2(x) = Em − qN 1 x εs qN 1b εs 0 ≤ x ≤ b = 3 µm b ≤ x ≤ W = 12 µm E2(x) should be larger than 105 V/cm for velocity saturation ∴ Em − qN 1b εs ≥ 10 5 or Em ≥ 10 5 + qN 1b εs = 10 5 + 4.66 × 10 −11 N1. This equation coupled the plot of Em versus N in Chapter 3 gives N1 = 7 × 1015 cm-3 for Em = 4.2 × 105 V/cm ∴VB = (Em − E2 )b (4.2 − 1) × 10 5 × 3 × 10 −4 + E2 W = + 10 5 × 9 × 10 − 4 2 2 = 138 V (b) Transit time t = W − b (12 − 3) × 10 −4 = 9×10-15 = 90 ps . = 7 vs 10 7. (a) For transit-time mode, we require n0L ≥ 1012 cm-2. n0 ≈ 1012 / L = 1012 / 1 × 10 −4 = 1016 cm −3 (b) t = L/v = 10-4 / 107 = 10-11 s = 10 ps 78 (c) The threshold field for InP is 10.5 kV/cm; the corresponding applied voltage is ⎛ 10.5 × 10 3 ⎞ ⎟⎟(1 × 10 − 4 ) = 0.525 V V = ⎜⎜ 2 ⎝ ⎠ The current is ( ) I = JA = (qμ n n0E )A = 1.6 × 10−19 × 4600× 1016 × 5.25 × 103 × 10−4 = 3.86 A The power dissipated in the device is then P = IV = 2.02 W . 8. (a) Referring to Chapter 2, we have N CU ⎛ 2πm nU kT ⎞ = 2⎜ ⎟ 2 ⎝ ⎠ 3 2 ⎛m = N CL ⎜⎜ nU ⎝ m nL ⎛ 1.2m 0 = 4.7 × 1017 ⎜⎜ ⎝ 0.07m 0 ⎞ ⎟⎟ ⎠ 3 2 ⎞ ⎟⎟ ⎠ 3 2 = 4.7 × 1017 × 71 = 3.33 × 1019 cm −3 (b) For Te = 300 K N CU ⎛ − 0.31eV ⎞ exp(−ΔE / kTe ) = 71 × exp⎜ ⎟ = 71 × exp (− 11.97 ) N CL ⎝ 0.0259 ⎠ = 4.4 × 10 − 4 (c) For Te = 1500 K N CU ⎛ ⎞ − 0.31eV ⎟ = 71× exp (− 2.394) exp(−ΔE / kTe ) = 71× exp⎜ N CL ⎝ 0.0259 (1500 / 300) ⎠ = 6.5 Therefore, at Te = 300 K most electrons are in the lower valley. However, at Te = 1500 K , 87%, i.e., 6.5/(6.5+1), of the electrons are in the upper valley. 9. The energy En for infinitely deep quantum well is 79 En = h2 * 2 n2 8m L ( ) ΔE n h2n2 (− 2) L−3 = 2 E1 = * ΔL L 8m 2 E1 ΔL L ∴ ΔE1 = 3 meV , ΔE n = ΔE 2 = 11 meV . 10. From Fig. 14 we find that the first excited energy is at 280 meV and the width is 0.8 meV. For same energy but a width of 8 meV, we use the same well thickness of 6.78 nm for GaAs, but the barrier thickness must be reduced to 1.25 nm for AlAs. The resonant-tunneling current is related to the integrated flux of electrons whose energy is in the range where the transmission coefficient is large. Therefore, the current is proportional to the width ΔE n , and sufficiently thin barriers are required to achieve a high current density. 80 CHAPTER 9 1. hv (0.6μm) = 1.24/0.6 = 2.07 eV (From Eq.9) α (0.6 μm) = 3×104 cm-1 The net incident power on the sample is the total incident power minus the reflected power, or 10 mW. ( 10 −2 1 − e −3 ×10 4 W ) = 5 × 10 −3 W = 0.231 μm. The portion of each photon’s energy that is converted to heat is hν − E g hν = 2.07 − 1.42 = 31.4% 2.07 The amount of thermal energy dissipated to the lattice per second is 31.4%×5 = 1.57 mW. 2. P(x) = Po(1－R)exp(－αx) x = ( − 1/α)ln[P(x)/Po(1 − R)] = ( − 1/4×104)ln[1/(2×0.9)] = 0.15 μm. 3. From Eqs. 11 and 12, the spectral half-width of the intensity is 2Δλ = (2/hc) λ2kT = (2/hc)(550×10-9)2×k ×293 = 18 nm. 4. For λ = 0.898 μm, the corresponding photon energy is E = 1.24 λ = 1.38 eV From Fig. 18, we obtain n 2 (Al0.3Ga0.7As) = 3.38 81 sin θ c = Efficiency = n1 1 = = 0.2958 ⇒ θ c = 17 o12 ′. n 2 3.38 4n1 n 2 (1 − cos θ c ) (n1 + n 2 ) 2 = [ ] 4 × 1 × 3.38 1 − cos(17 o12 ′) (1 + 3.38) 2 = 0.0315 = 3.15% . fT = 5. 1 2πτ 1019 1 + = 1010 + 107 = 1010 τ 109 10−7 τ = 10−10 s 1 = fT = 6. 1 = 1.6 GHz 2π × 10−10 From Eq. 13 P(ω ) = p(0) 1 + (ωτ )2 P(ω ) 1 1 = = 2 p(0) 2 1 + (ωτ ) 4 = 1 + (ωτ )2 3 = (2π f τ ) 2 f = 3 2πτ = 3 = 0.28 × 1010 Hz −10 2π × 10 7. The Fresnel transmission coefficient from Eq. 22 T = 1 − R = 4 n1 n2 /( n1 + n2 )2 = 4×3.38 /(1+3.38)2 = 0.70 8. 80×1.8×0.716=103.1 mW. 0.5×103.1×exp(-5×105×3×106)=11.5 mW 9. 11.5× 4 n1 n2 /( n1 + n2 )2 × 4 n2 n3 /( n2 + n3 )2 = 11.5×4×3.38×1.6/(1.6+3.38)2 ×4×1.6 /(1+1.6)2= 9.5 mW. 10. From Eq. 1 82 λ= c hc 1.24 = = μm, v hv E g (eV) dλ hc ⎛ dE ⎞ 2.8 =− 2 ⎜ g ⎟=− nm / o K = −0.28 × 10−3 μm / o K dT E g ⎝ dT ⎠ 10 dE g dT 11. = E g 2 ⎛ d λ ⎞ (1.42)2 (0.28 × 10−3 ) = 4.55 × 10 −4 eV/K ⎜ ⎟= hc ⎝ dT ⎠ 1.24 (a) For GaAs, n2 = 3.66 at λ = 0.8 μm and for air, n1 = 1.0 . The critical angle is θ C = sin −1 FG n IJ = sin F 1 I = 15.9° H n K H 3.66K −1 1 2 The fraction of photons that will not experience total internal reflection is 2 × 15.9 = 0.0883 = 8.83% 360 (b) Fresnel loss: F n − n IJ = F 3.66 − 1I R=G H n + n K H 3.66 + 1K 2 2 1 2 1 2 = 0.326 0.0883(1 − 0.326) = 0.0595 = 5.95% 12. From Eq. 22 ⎛ 3.39 − 1 ⎞ ⎟ R = ⎜⎜ ⎟ ⎝ 3.39 + 1 ⎠ 2 = 0.296 (a) The mirror loss 1 ⎛1⎞ 1 ⎛ 1 ⎞ -1 ln⎜ ⎟ = ln⎜ ⎟ = 40.58 cm . −4 0 . 296 L ⎝ R ⎠ 300 × 10 ⎝ ⎠ (b) The threshold current reduction is 83 J th (R = 0.296) − J th (R1 = 0.296, R 2 = 0.9 ) J th (R = 0.296) 1 ⎛ 1 ⎞⎤ ⎡ 1 ⎛ 1 ⎡ ⎢α + L ln⎜ R ⎟⎥ − ⎢α + 2 L ln⎜⎜ R R ⎝ ⎠⎦ ⎣ ⎣ ⎝ 1 2 ≈ 1 ⎛1⎞ α + ln⎜ ⎟ L ⎝R⎠ 40.58 − = ⎞⎤ ⎟⎟⎥ ⎠⎦ 1 1 ⎛ ⎞ ln⎜ ⎟ −4 2 ⋅ 300 × 10 ⎝ 0.296 ⋅ 0.90 ⎠ 10 + 40.58 = 36.6%. 13. (a) From Eq. 30 Δλ ≅ Δλ ≅ λ2 2nL . (1300 nm)2 = 0.828 nm 2 × 3.4 × 300 × 103 nm (b) v = c λ dv c = dλ λ 2 c λ2 c = 2 2 2 λ λ λ 2nL 2nL c 3 × 1010 cm/s Δv = = = 147 GHz 2nL 2 × 3.4 × 300 × 10−4 cm Δv = 14. c ΔΔv = c Δλ = We assume a conventional p-n junction laser and a stripe DH laser have the same active area. The area is 300×10-4×20×10-4 = 6×10-5 cm2. For a conventional p-n junction laser, the current is 5×104 A/cm2×6×10-5 cm2 = 3 A. It is impractical. For a stripe DH laser, the current is 103 cm2×6×10-5 A/cm2=6×10-2 A=60 mA. 15. From Eq. 23 sin θ c = n1 ⇒ n1 = n 2 ⋅ sin θ c n2 84 From Eq. 26 ( Γ ≅ 1 − exp(− CΔn d ) = 1 − exp − 8 × 10 5 ⋅ 3.6(1 − sin θ c ) ⋅ 1 × 10 −4 For θc = 84o ) n1 = 3.58 Γ1 = 0.794 θc = 78o n2 = 3.52 Γ2 = 0.998 . 16. From Eq. 28 we have mλ = 2 n L . Differentiating the above equation with respect to λ, we obtain λ dm dn + m = 2L . dλ dλ Substituting 2nL λ for m and letting dm/dλ = - Δm/Δλ, yield dn ⎛ − Δm ⎞ 2n L = 2L ⎟+ λ dλ ⎝ Δλ ⎠ 2 λ Δm ∴ Δλ = ⎡ ⎛ λ ⎞⎛ dn ⎞⎤ 2n L ⎢1 − ⎜ ⎟ ⎟⎥ ⎣ ⎝ n ⎠⎝ dλ ⎠⎦ λ⎜ and Δλ = (0.89)2 × 1 ⎡ ⎛ 0.89 ⎞ ⎤ 2(3.58)300⎢1 − ⎜ ⎟(2.5)⎥ ⎣ ⎝ 3.58 ⎠ ⎦ = 9.7 × 10 − 4 μm = 0.97 nm. 17. From Eq. 34 ⎛ n − 1⎞ 1⎡ 1 ⎛ 1 ⎞⎤ ⎟ g = ⎢α + ln⎜ ⎟⎥ and Eq. 27 R = ⎜⎜ ⎟ Γ⎣ L ⎝ R ⎠⎦ ⎝ n + 1⎠ R1 = 0.317, R2 = 0.311 85 2 g 84 o = ( ) 1 ⎡ 1 ⎛ 1 ⎞⎤ 100 + ln⎜ ⎟⎥ = 270 ⎢ −4 0.794 ⎣ 100 × 10 ⎝ 0.317 ⎠⎦ ( ) 1 ⎡ 1 ⎛ 1 ⎞⎤ 100 + ln⎜ ⎟⎥ = 217 ⎢ −4 0.998 ⎣ 100 × 10 ⎝ 0.311 ⎠⎦ g 78 o = 1 ⎛ 1 ⎞ 1 ⎛1⎞ ln⎜ ln⎜ ⎟ ⎟= −4 2 L′ ⎝ R ⋅ 0.99 ⎠ 100 × 10 ⎝R⎠ For R1 = 0.317 1 ⎛ 1 1 ⎞ ⎛ 1 ⎞ ln⎜ ln⎜ ⎟= ⎟ ⇒ L ′ = 50 μm −4 2 L ′ ⎝ 0.317 ⋅ 0.99 ⎠ 100 × 10 ⎝ 0.317 ⎠ For R2 = 0.311 1 1 1 ⎛ ⎞ ⎛ 1 ⎞ ln⎜ ln⎜ ⎟= ⎟ ⇒ L2′ = 50.43 μm. −4 2 L2′ ⎝ 0.311 ⋅ 0.99 ⎠ 100 × 10 ⎝ 0.311 ⎠ 18. From Eq. 35a For R = 0.317 ⎡ 1 1 ⎛ ⎞⎤ −2 ln⎜ J th = 10 × ⎢100 + ⎟⎥ = 1000 A cm −4 2 × 100 × 10 ⎝ 0.317 × 0.99 ⎠⎦ ⎣ and so I th = 1000 × 100 × 10 −4 × 5 × 10 −4 = 5 mA For R = 0.311 β ∝Γ ⇒ β2 = 0.1×0.998/0.764 = 0.13 ⎡ 1 1 ⎛ ⎞⎤ −2 ln⎜ J th = 7.66 × ⎢100 + ⎟⎥ = 766 A cm −4 2 × 100 × 10 ⎝ 0.311 × 0.99 ⎠⎦ ⎣ and so I th = 766 × 100 × 10 −4 × 5 × 10 −4 = 3.83 mA . 19. From the equation, we have for m = 0: λ2B ± 4n Lλ B ∓ 4n Lλo = 0 which can be solved as 86 (25a) ⎛ − 4n L ± 16n 2 L2 + 16n Lλ o λ B = ±⎜ ⎜ 2 ⎝ ⎞ ⎟ ⎟ ⎠ (25b) There are several variations of ± in this solutions. Take the solution which is the only practical one, i.e., λB ≈ λo, gives λB = 1.3296 or 1.3304 μm. Λ≅ 1.33 = 0.196 μ m 2 × 3.4 . 20. The threshold current in Fig. 38b is given by Ith = I0 exp (T/110). Therefore ξ≡ 1 dI th 1 = = 0.0091 I th dT 110 ( C) o −1 . If T0 = 50 oC, the temperature coefficient becomes ξ= 1 = 0.02 50 ( C) o −1 . which is larger than that for T0 = 110 oC. Therefore the laser with T0 = 50 oC is worse for high-temperature operation. 87 CHAPTER 10 1. λ = 0.8 μm R(ideal) =(100%)×0.8×10-4/1.24=0.645 A/W The photon energy for 0.8 μm is 1.24/0.8 = 1.55 eV (l) GaAs bandgap is 1.42 eV, since photon energy is larger than the bandgap, we have η= 0.645 A/W. (2) For A10.34Ga0.66As the bandgap is 1.424 + 1.247×0.34 = 1.85 eV (Eq. 58 in Ch. 12). The photons with 1.55 eV energy have insufficient energy to generate electrons and holes ∴η = 0 (3) For the heterojunction cell, photons will be absorbed in the 1.42 eV region, but will not be absorbed in the 1.85 eV regions ∴ η = 0.645 A/W (4) For the tandem cell, photons wiI1 not be absorbed in the upper 1.85 eV cell, but will be absorbed in the lower 1.42 eV cell. However, since the cells are in series, the top cell wiI1 block the current generated by the bottom cell once the current is larger than the top cell's dark saturation current va1ue. The bottom cell wiI1 be forced to open-circuit，dissipating the generated current ∴η =0 2. (a) ΔI = q (μn + μp) ΔnEA ∴ Δn = electron − hole pairs = = ΔI q(μ n + μ p )EA 2.83 × 10 − 3 = 1013 cm − 3 −19 −2 1.6 × 10 (3600 + 1700)(10 0.6) 2 × 1 × 10 ( (b) τ = 2.83 × 10 23.6 −3 ) = 120 μs ⎛ 10 −3 ⎞ ⎟ = 2.5 × 10 9 cm-3 . (c) Δn(t ) = Δn exp(− t τ ) = 1013 exp⎜ − −4 × 1 . 2 10 ⎝ ⎠ 3. From Eq. 6 ⎛ 10 −6 ⎞ ⎛ 3000 ⋅ 6 × 10 −10 ⋅ 5000 ⎞ ⎟⋅⎜ ⎟⎟ I p = q⎜⎜ 0.85 ⋅ 3 × q ⎟⎠ ⎜⎝ 10 × 10 − 4 ⎝ ⎠ = 2.55 × 10 −6 = 2.55 μA and from Eq. 8 88 Gain = 4. 5. μnτ E L = 3000 ⋅ 6 × 10−10 ⋅ 5000 =9. 10 × 10−4 ηext = 0.9×0.8×[exp(-106×10-6)－exp(-106×2×10-6)] = 0.168. From Eq. 9 ⎛Ip η = ⎜⎜ ⎝ q ⎞ ⎛ Popt ⎟⋅⎜ ⎟ ⎜ hν ⎠ ⎝ ⎞ ⎟ ⎟ ⎠ −1 ⎛ Ip =⎜ ⎜ Popt ⎝ ⎞ ⎛ hν ⎟⋅⎜ ⎟ ⎜⎝ q ⎠ ⎞ ⎛ hν ⎟⎟ = R ⋅ ⎜⎜ ⎠ ⎝ q ⎞ ⎟⎟ ⎠ The wavelength λ of light is related to its frequency ν by ν = c/λ, where c is the velocity of light in vacuum. Therefore hν /q = hc / λq and h = 6.625×10-34 J-s, c = 3.0×1010 cm/s, q = 1.6×10–19 coul, 1 eV = 1.6×10–19 J. Therefore, h v/q = 1.24 / λ (μm) Thus, η = (R×1.24) / λ and R = (ηλ) / 1.24. 6. R = ηλ/1.24 = 0.168 × 1.1/1.24 = 0.149 A/W. Iopt = 0.149×5×10-6 =0.745 μA. 7. The electric field in the p-layer is given by qN x E1 ( x ) = Em − 1 0 ≤ x ≤ b, εs where Em is the maximum field. In the p layer, the field is essentially a constant given by E2 ( x ) = Em − qN1b εs b < x ≤W The electric field required to maintain velocity saturation of holes is ~ 105 V/cm. Therefore Em − qN1b εs ≥ 105 89 or Em ≥ 105 + qN1b εs = 105 + 4.66 × 10−11 N1 . From the plot of the critical field versus doping, the corresponding Em are obtained : N1 = 7×1015 cm-3 Em = 4.2×105 V/cm The biasing voltage is given by 3.2 × 105 × ( 3 × 10−4 ) Em − E2 ) b ( = +EW = + 105 VB 2 2 2 (9 ×10 ) −4 = 138 V. The transit time is t≈ 8. (W − b ) = 9 × 10 −4 νs 10 7 = 9 × 10 −11 = 90 ps. From Fig. 5, Ch. 9, the absorption coefficient at 900 nm is ~3 × 102 cm-1, and the absorption depth is ~33 μm. We can assume that absorption and hence photogeneration occurs over the entire i layer. The field in the i-Si layer is E = (100V)/(20 × 10-4 cm) = 5 × 104 V cm-1 At this field the electron drift velocity is very near its saturation at 107 cm/s, whereas the hole drift velocity is about 7 × 106 cm/s as shown in Fig. 22, Ch. 2. The transit time tr of holes across the i-Si layer is tr = (20 × 10-4 cm)/(7 × 106 cm/s)= 2.86 × 10-10 s or 0.3 ns. 9. The optimum occurs when the depletion layer is chosen so that the transit time is of the order of one-half the modulation period τ 2 =t= w υ If υs = 107 cm/s 90 1 107 ⎛τ ⎞ = = 5 × 10−4 = 5 μm. W =υ⎜ ⎟ =υ 2 f 2 × 10 × 109 ⎝2⎠ 10. Energy at earth’s mean distance/unit area/unit time = (rs/des)2 energy at sun’s surface/unit area/unit time = (rs/des)2σT4 Absorbing energy of the Earth = radiating energy of the Earth (rs/des)2σTs4πre2=σTe44πre2 Te=Ts(rs/des)1/2/21/2=289.38 K (b) The absorbing energy of the Earth decreases to 0.7, and the radiating energy of the Earth decreases to 0.6. The earth is hotter. 0.7(rs/des)2σTs4πre2=0.6σTe’44πre2 Te’= Te(0.7/0.6)1/4=300.75 K (c) The increase of 2 oC is roughly from a doubling of atmospheric CO2. 302.75= Te(0.7/ε’)1/4 ε’ = 0.584 11. (a) For a photodiode, only a narrow wavelength range centered at the optical signal wavelength is important; whereas for a solar cell, high spectral response over a broad solar wavelength range are required. (b) Photodiode are small to minimize junction capacitance, while solar cells are large-area device. (c) An important figure of merit for photodiodes is the quantum efficiency (number of electron-hole pairs generated by incident photon), whereas the main concern for solar cells is the power conversion efficiency (power delivered to the load per incident solar 91 energy). 12. Pmax=(0.8)(90)(0.75)=54 mW 13. From Fig. 15, we draw a load line with a slope 1/5Ω to intersect the I-V curve at about I = 80mA and V =0.4V. The power delivered to the load is Pload = 0.08×0.4= 0.032 W The input power is Pin = 600×4×10-4=0.24 W The efficiency is η= 0.032/0.24 = 0.13 = 13%. ⎛ 1 (a ) I s = AqN C N V ⎜ ⎜ NA ⎝ 14. Dn τn + 1 ND D p ⎞ −Eg ⎟e τ p ⎟⎠ kT = 2(1.6 × 10 −19 )(2.86 × 1019 )(2.66 × 1019 ) × ⎛ 1 9.3 1 2.5 ⎜ ⎜ 1.7 × 1016 10 −5 + 5 × 1019 5 × 10 −7 ⎝ = 2.43 × 10 20 (5.67 × 10 −14 )e − 43 2 ⎞ −1 12 eV ⎟×e ⎟ ⎠ kT = 2.28 × 10 −12 A I = I s (e qV kT )− I I = I L − I s (e V I s (e IL 0 qV kT − 1) 0 L qV kT − 1) 0.1 0.2 - 0.3 -6 0.4 0.5 - - 0.6 0.65 1.1×10 5×10 2.5×10 1.1×10 0.55 26.8 179 7 4 95 95 95 (b) VOC = kT ln⎛⎜ I L q ⎜⎝ I s ( (c) P = I sV e qV kT 0.7 (V) 1520 (mA) 2 95 95 ⎛ 95 × 10 −3 ⎞ ⎟⎟ = 0.0259 ln⎜⎜ −12 ⎠ ⎝ 2.28 × 10 ) − 1 − I LV 92 94.5 68.2 -84 -1425 ⎞ ⎟ = 0.68 V ⎟ ⎠ (mA) dP dV ( = 0 = I s e qV ≈ I s e qV ∴ e qV kT kT kT ) − 1 + Is (1 + V ) − qV kT e qV kT − IL IL IL = I s (1 + V ) Vm = 0.64 V ⎡ kT ⎛ qVn ⎞ kT ⎤ ln⎜1 + Pm = I mVm = I L ⎢VOC − ⎟− q ⎝ kT ⎠ q ⎥⎦ ⎣ = 95 × 10 −3 [0.68 − 0.0259 ln (1 + 24.7 ) − 0.0259] = 52 mW I (mA) 100 50 0 0 0.2 0.4 0.6 0.8 V (V) 15. From Eq. 38 and 39 ( I = I s e qV / kT ) − 1 − IL and Voc = ⎞ kT ⎛ I L ⎞ kT ⎛ I L ln⎜⎜ + 1⎟⎟ ≅ ln⎜⎜ ⎟⎟ q ⎝ Is ⎠ q ⎝ Is ⎠ I s = I L e − qV kT = 3 ⋅ e − 0.6 0.02585 = 2.493 × 10 − 10 I = 3 − 2.493 × 10 −10 ⋅ eV 0 02585 and P = I V 93 A v 0 0 .1 0.2 0 .3 0.4 0.5 0.5 1 0.52 0.53 0.54 0.55 0 .6 I p 3.00 3.00 3.00 3.00 3.00 2 .94 2 .91 2 .86 2 .80 2.71 2 .57 0.00 0 .00 0.30 0.60 0 .90 1.20 1.47 1.48 1.49 1.48 1.46 1 .4 1 0.00 2 .0 0 ,.-..__ r:/) 1. 50 t: ro ~ ...._., 0.., 1.0 0 0.50 0 .0 0 0 0.5 1 V (vo lts) .·.Maximum power output = 1.49 W Fill Factor FF = lmVm =~= 1.49 JLVoc JLVoc 0.6 x 3 = 0.83. 16. From Fig. 40 The output powers for Rs = 0 and Rs = 5 n can be obtained from the area P1 (Rs = 0) = 95 mAx0.375V = 35.6 mW, P2(Rs = 5 Q) = 50 mAx0.18V = 9.0 mW For Rs = 5 n P2/ P1 = 9/35.6 = 25.3%. 17. From Eq. 8, Ch. 9, - 1 <P(W) - 1 -5 For am01phous Si, W =-In- - = - ln(0. 1) = 2.3 x 10 em = 0.23 ~un. a <P0 a For CIGS, the thickness is 2.3 Jlm. 94 18. The efficiencies are 14.2% (1 sun), 16.2% (10-sun), 17.8% (100-sun), and 18.5% (1000-sun). Solar cells needed under 1-sun condition = η (concentration ) × Pin (concentration ) η (1 - sun ) × Pin (1 - sun ) 16.2% × 10 = 11.4 cells for 10 - sun 14.2% × 1 17.8% × 10 = 125 cells for 100 - sun = 14.2% × 1 18.5% × 10 = 1300 cells for 1000 - sun. = 14.2% × 1 = 95 CHAPTER 11 ko(As in Si) = 0.3 Cs= k{)Co(l - MIMo)kO-l X 0 0.2 0.4 0 .6 0.8 0 .9 l (em) 0 10 20 30 40 45 Cs (cm-3) 3x l0 16 3.5 x l0 16 4 .28xl0 16 5.68x l0 16 1.07x l0 17 1.5x l0 17 16 14 12 '7 e 10 :2 8 0 ...... 6 :z. 4 2 0 I rl (.) _.-.---- Q 0 10 20 / ./ I I 30 40 I (em) 2. (a) The radius of a silicon atom can be expressed as J3 a r =- 8 J3 x5.43 = 1.175A so r = - 8 (b) The numbers of Si atom in its diamond structure are 8. So the density of silicon atoms is 8 8 22 3 n=- 3 = A 3 =5 .0x l0 atoms/cm a (5.43 ) 96 • 50 (c) The density of Si is ρ = 3. M / 6.02 × 10 23 1/ n = 28.09 × 5 × 10 22 6.02 × 10 23 g / cm 3 = 2.33 g / cm3. k0 = 0.8 for boron in silicon M / M0 = 0.5 The density of Si is 2.33 g / cm3. The acceptor concentration for ρ = 0.01 Ω–cm is 9×1018 cm-3. The doping concentration CS is given by C s = k 0 C 0 (1 − M k0 −1 ) M0 Therefore Cs 9 × 1018 = C0 = M k 0 −1 0.8(1 − 0.5) −0 2 ) k 0 (1 − M0 = 9.8 × 1018 cm −3 The amount of boron required for a 10 kg charge is 10,000 × 9.8 × 1018 = 4.2 × 10 22 boron atoms 2.338 So that 10.8g/mole × 4.2 × 10 22 atoms = 0.75g boron . 6.02 × 10 23 atoms/mole 4. (a) The molecular weight of boron is 10.81. The boron concentration can be given as 97 number of boron atoms volume of silicon wafer 5.41 × 10 −3 g / 10.81g × 6.02 × 10 23 = 10.0 2 × 3.14 × 0.1 = 9.78 × 1018 atoms/cm 3 nb = (b) The average occupied volume of everyone boron atoms in the wafer is V = 1 1 = cm 3 nb 9.78 × 1018 We assume the volume is a sphere, so the radius of the sphere ( r ) is the average distance between two boron atoms. Then r= 3V = 2.9 × 10 −7 cm . 4π 5. The cross-sectional area of the seed is 2 ⎛ 0.55 ⎞ 2 π⎜ ⎟ = 0.24 cm 2 ⎝ ⎠ The maximum weight that can be supported by the seed equals the product of the critical yield strength and the seed’s cross-sectional area: (2 × 10 6 ) × 0.24 = 4.8 × 10 5 g = 480 kg The corresponding weight of a 200-mm-diameter ingot with length l is 2 ⎛ 20.0 ⎞ ( 2.33g/cm 3 )π ⎜ ⎟ l = 480000 g ⎝ 2 ⎠ ∴ l = 656 cm = 6.56 m. 6. We have 98 0 021 / !:.2 u 011 __. ~ 001 0 02 04 06 08 Fraction Solidified ko-1 C, IC0 =k0 ( 1-:, ) Fractional 0 0.2 0.4 0.6 0.8 0.05 0.06 0.08 0.12 0.23 1.0 solidified 7. The segregation coefficient of boron in silicon is 0.72. It is smaller than unity, so the solubility of B in Si under solid phase is smaller than that of the melt. Therefore, the excess B atoms will be thrown-off into the melt, then the concentration of B in the melt will be increased. The tail-end of the crystal is the last to solidify. Therefore, the concentration of B in the tail-end of grown c1ystal will be higher than that of seed-end. 8. The reason is that the solubility in the melt is prop01tional to the temperature, and the temperature is higher in the center prut than at the perimeter. Therefore, the solubility is higher in the center prui, causing a higher impmity concentration there. 9. The segregation coefficient of Ga in Si is 8 x 1o-3 FromEq. 18 We have 99 x= L ⎛ 1− k ln⎜ k ⎜⎝ 1 − C s / C 0 ⎞ ⎟⎟ ⎠ ⎛ ⎞ 2 1 − 8 × 10 − 3 ⎟ = ln⎜ -3 15 16 8 × 10 ⎝ 1 − 5 × 10 / 5 × 10 ⎠ = 250 ln(1.102) = 24 cm. 10. We have from Eq.18 Cs = C0[1 − (1 − ke ) exp(−ke x / L)] So the ratio Cs / C0 = [1 − (1 − ke ) exp(−ke x / L)] = 1 − (1 − 0.3) • exp(−0.3 × 1) = 0.52 = 0.38 at x / L = 1 at x/L = 2. 11. For the conventionally-doped silicon, the resistivity varies from 120 Ω-cm to 155 Ω-cm. The corresponding doping concentration varies from 2.5×1013 to 4×1013 cm-3. Therefore the range of breakdown voltages of p+ - n junctions is given by VB ≅ = ε s Ec 2 2q ( N B ) −1 1.05 × 10 −12 × (3 × 10 5 ) 2 ( N B ) −1 = 2.9 × 1017 / N B = 7250 to 11600 V −19 2 × 1.6 × 10 ΔV B = 11600 − 7250 = 4350 V ⎛ ΔV ⎞ ∴ ⎜ B ⎟ / 7250 = ±30% ⎝ 2 ⎠ For the neutron irradiated silicon, ρ = 148 ± 1.5 Ω-cm. The doping concentration is 3×1013 (±1%). The range of breakdown voltage is 100 V B = 1.3 × 10 17 / N B = 2.9 × 10 17 / 3 × 10 13 ( ±1%) = 9570 to 9762 V . ΔV B = 9762 − 9570 = 192 V ⎛ ΔV ⎞ ∴ ⎜ B ⎟ / 9570 = ±1% . ⎝ 2 ⎠ 12. We have M s weight of GaAs at Tb C m − C l s = = = M l weight of liquid at Tb C s − C m l Therefore, the fraction of liquid remained f can be obtained as following f = Ml l 30 = ≈ = 0.65 . M s + M l s + l 16 + 30 13. From the Fig.11, we find the vapor pressure of As is much higher than that of the Ga. Therefore, the As content will be lost when the temperature is increased. Thus the composition of liquid GaAs always becomes gallium rich. 14. ⎡ − 88.8 ⎤ n s = N exp(− E s / kT ) = 5 × 10 22 exp(2.3 eV / kT ) = 5 × 10 22 exp ⎢ ⎥ ⎣ (T / 300) ⎦ = 1.23 × 10 −16 cm −3 ≈ 0 at 27 0 C = 300 K 15. = 6.7 × 1012 cm −3 at 900 0 C = 1173 K = 6.7 × 1014 cm −3 at 1200 0 C = 1473 K . n f = NN ` exp(− E f / 2kT ) = 5 × 10 22 × 1 × 10 27 e −3 8eV / kT × e −1 1eV / 2 kT = 7.07 × 10 24 × e −94 7 /(T / 300) = 5.27 × 10 −17 at 27oC = 300 K 101 =2.14x10 14 16. at 900°C = 1173 K. 37 x 4 = 148 chips In tenns of litho-stepper considerations, there are 500 )..till space tolerance between the mask boundaty of two dice. We divide the wafer into four symmetrical patts for convenient dicing, and discard the perimeter parts of the wafer. Usually the quality of the perimeter patts is the worst due to the edge effects. ~ / I \ \ J 300 ~ / 17. 3 2 Where f. =4-(-M-) v .[; 2kT ' MM Toi:ol Dies: 148 2 l v 2 exp( -Mv -2kT 102 M: Molecular mass k: Boltzmann constant = 1.38×10-23 J/k T: The absolute temperature ν: Speed of molecular So that ν av = 18. λ= π 0.66 cm P( in Pa ) ∴P = 19. 2 × 1.38 × 10 −23 × 300 = 468 m/sec = 4.68 × 10 4 cm/sec . − 27 29 × 1.67 × 10 2 0.66 λ = 0.66 = 4.4 × 10 −3 Pa . 150 For close-packing arrange, there are 3 pie shaped sections in the equilateral triangle. Each section corresponds to 1/6 of an atom. Therefore 1 number of atoms contained in the triangle 6 = Ns = area of the triangle 1 3 d× d 2 2 3× = 2 3d 2 = 2 3 (4.68 × 10 −8 ) 2 = 5.27 × 1014 atoms/cm 2 . dd 20. (a) The pressure at 970°C (=1243K) is 2.9×10-1 Pa for Ga and 13 Pa for As2. The arrival rate is given by the product of the impringement rate and A/πL2 : 103 ⎛ P ⎞⎛ A ⎞ Arrival rate = 2.64×1020 ⎜ ⎟ 2 ⎟ ⎝ MT ⎠⎝ πL ⎠ ⎛ 2.9 × 10 −1 ⎞⎛ 5 ⎞ ⎟⎜ = 2.64×1020 ⎜ 2 ⎟ ⎝ 69.72 × 1243 ⎠⎝ π × 12 ⎠ = 2.9×1015 Ga molecules/cm2 –s The growth rate is determined by the Ga arrival rate and is given by (2.9×1015)×2.8/(6×1014) = 13.5 Å/s = 810 Å/min . (b) The pressure at 700ºC for tin is 2.66×10-6 Pa. The molecular weight is 118.69. Therefore the arrival rate is ⎛ 2.66 × 10 −6 ⎞⎛ 5 ⎞ ⎟⎟⎜ 2.64 × 10 20 ⎜⎜ = 2.28 × 1010 molecular/cm 2 ⋅ s 2 ⎟ π × 12 ⎠ ⎝ 118.69 × 973 ⎠⎝ If Sn atoms are fully incorporated and active in the Ga sublattice of GaAs, we have an electron concentration of ⎛ 2.28 × 1010 ⎜⎜ 15 ⎝ 2.9 × 10 ⎞⎛ 4.42 × 10 22 ⎟⎟ 2 ⎠⎝ ⎞ ⎟⎟ = 1.74 × 1017 cm -3 . ⎠ 21. The x value is about 0.25, which is obtained from Fig. 26. 22. The lattice constants for InAs, GaAs, Si and Ge are 6.05, 5.65,5.43, and 5.65 Å, respectively (Appendix F). Therefore, the f value for InAs-GaAs system is f = (5.65 − 6.05) 6.05 = −0.066 And for Ge-Si system is f = (5.43 − 5.65) 5.65 = −0.39 . 104 CHAPTER 12 1. From Eq. 11 (with τ=0) x2+Ax = Bt From Figs. 6 and 7, we obtain B/A =1.5 µm /hr, B=0.47 µm 2/hr, therefore A= 0.31 µm. The time required to grow 0.45µm oxide is t= 2. 1 2 1 (x + Ax) = (0.45 2 + 0.31 × 0.45) = 0.72 hr = 44 min . B 0.47 After a window is opened in the oxide for a second oxidation, the rate constants are B = 0.01 µm 2/hr, A= 0.116 µm (B/A = 6 ×10-2 µm /hr). If the initial oxide thickness is 20 nm = 0.02 µm for dry oxidation, the value ofτ can be obtained as followed: (0.02)2 + 0.166(0.02) = 0.01 (0 +τ) or τ= 0.372 hr. For an oxidation time of 20 min (=1/3 hr), the oxide thickness in the window area is x2+ 0.166x = 0.01(0.333+0.372) = 0.007 or x = 0.0350 µm = 35 nm (gate oxide). For the field oxide with an original thickness 0.45 µm, the effectiveτis given by τ= 1 2 1 ( x + Ax) = (0.45 2 + 0.166 × 0.45) = 27.72 hr. B 0.01 x2+ 0.166x = 0.01(0.333+27.72) = 0.28053 or x = 0.4530 µm (an increase of 0.003µm only for the field oxide). 3. x2 + Ax = B (t + τ ) (x + A 2 A2 ) − = B (t + τ ) 2 4 105 (x + ⎡ A2 ⎤ A 2 ) =B⎢ + (t + τ )⎥ 2 ⎣ 4B ⎦ when t >> τ , t >> A2 , 4B then, x2 = Bt similarly, when t >> τ , t >> then, x = A2 , 4B B (t + τ ) A 4. At 980℃(=1253K) and 1 atm, B = 8.5×10-3 µm 2/hr, B/A = 4×10-2 µm /hr (from Figs. 6 and 7). Since A ≣2D/k , B/A = kC0/C1, C0 = 5.2×1016 molecules/cm3 and C1 = 2.2×1022 cm-3 , the diffusion coefficient is given by D= Ak A ⎛ B C1 = ⎜⎜ ⋅ 2 2 ⎝ A C0 ⎞ B ⎛ C1 ⎟⎟ = ⎜⎜ ⎠ 2 ⎝ C0 ⎞ ⎟⎟ ⎠ 8.5 × 10 −3 2.2 × 10 22 μm 2 / hr 16 2 5.2 × 10 3 = 1.79 × 10 μm 2 / hr = = 4.79 × 10 -9 cm 2 / s. 5. (a) For SiNxHy Si 1 = = 1.2 N x ∴ x = 0.83 atomic % H = 100 y = 20 1 + 0.83 + y ∴ y = 0.46 The empirical formula is SiN0.83H0.46. (b) ρ= 5× 1028e-33 3×1 2 = 2× 1011 Ω-cm As the Si/N ratio increases, the resistivity decreases exponentially. 106 6. Set Ta2O5 thickness = 3t, ε1 = 25 SiO2 thickness = t, ε2 = 3.9 Si3N4 thickness = t, ε3 = 7.6, area = A then C Ta 2O5 = 1 t + t + t ε 2ε 0 A ε 3ε 0 A ε 2ε 0 A εεε A = 2 3 0 (ε 2 + 2ε 3 )t C ONO C Ta 2 O5 7. 3t = C ONO C ONO ε 1ε 0 A = ε 1 (ε 2 + 2ε 3 ) 25(3.9 + 2 × 7.6) = = 5.37 . 3ε 2 ε 3 3 × 3.9 × 7.6 Set BST thickness = 3t, ε1 = 500, area = A1 SiO2 thickness = t, ε2 = 3.9, area = A2 Si3N4 thickness = t, ε3 = 7.6, area = A2 then ε 1ε 0 A1 3t = ε 2 ε 3ε 0 A2 (ε 2 + 2ε 3 )t A1 = 0.0093. A2 8. Let Ta2O5 thickness = 3t, ε1 = 25 SiO2 thickness = t, ε2 = 3.9 Si3N4 thickness = t, ε3 = 7.6 area = A then ε 1ε 0 A 3t d= = 3ε 2 t ε1 ε 2ε 0 A d = 0.468t. 107 9. The deposition rate can be expressed as r = r0 exp (-Ea/kT) where Ea = 0.6 eV for silane-oxygen reaction. Therefore for T1 = 698 K ⎡ ⎛ 1 r (T2 ) 1 ⎞ ⎤ ⎟⎟ ⎥ = 2 = exp ⎢0.6⎜⎜ − r (T1 ) ⎣ ⎝ kT1 kT2 ⎠ ⎦ ln 2 = 0.6 ⎡⎛ 300 300 ⎞ ⎤ ⎟ ⎥ − ⎢⎜ 0.0259 ⎣⎜⎝ 698 T2 ⎟⎠ ⎦ ∴ T2 =1030 K= 757 ℃. 10. We can use energy-enhanced CVD methods such as using a focused energy source or UV lamp. Another method is to use boron doped P-glass which will reflow at temperatures less than 900℃. 11. Moderately low temperatures are usually used for polysilicon deposition, and silane decomposition occurs at lower temperatures than that for chloride reactions. In addition, silane is used for better coverage over amorphous materials such SiO2. 12. There are two reasons. One is to minimize the thermal budget of the wafer, reducing dopant diffusion and material degradation. In addition, fewer gas phase reactions occur at lower temperatures, resulting in smoother and better adhering films. Another reason is that the polysilicon will have small grains. The finer grains are easier to mask and etch to give smooth and uniform edges. However, for temperatures less than 575 ºC the deposition rate is too low. 13. Atomic density = (Avogadro’s No.)(density/g/mole) = 6 × 1023 (molecules/mole) [(3g/cm3)/(102g/mole)] The surface density (#molecules/cm2) is just (molecules/cm3)2/3 approximately 7×1014 molecules/cm2. 14. 3600/4 = 900 cycles. 0.34nm/cycle × 900 cycles/hr = 306 nm/hr. 15. The flat-band voltage shift is 108 or Q ot C0 ΔVFB = 0.5 V ~ C0 = ε ox d = 3.9 × 8.85 × 10 −14 = 6.9 × 10 −8 F/cm − 2 . −8 500 × 10 ∴ Number of fixed oxide charge is 0.5C 0 0.5 × 6.9 × 10 −8 = = 2.1 × 1011 cm -2 −19 q 1.6 × 10 To remove these charges, a 450℃ heat treatment in hydrogen for about 30 minutes is required. 16. 20/0.25 = 80 sqs. Therefore, the resistance of the metal line is 5×50 = 400 Ω . 17. For TiSi2 30 × 2.37 = 71.1nm For CoSi2 30 × 3.56 = 106.8nm. 18. For TiSi2: Advantage: low resistivity It can reduce native-oxide layers TiSi2 on the gate electrode is more resistant to high-field-induced hot-electron degradation. Disadvantage: bridging effect occurs. Larger Si consumption during formation of TiSi2 Less thermal stability For CoSi2: Advantage: low resistivity High temperature stability No bridging effect A selective chemical etch exits Low shear forces Disadvantage: not a good candidate for polycides 109 19. (a) L 1 = 2.67 × 10 −6 × = 3.2 × 10 3 Ω −4 −4 A 0.28 × 10 × 0.3 × 10 R=ρ C= εA εTL = d S 3.9 × 8.85 × 10 −14 × 0.3 × 10 −4 × 1 × 10 4 × 10 −6 = 2.9 × 10 −13 F 0.36 × 10 − 4 = RC = 3.2 × 10 5 × 2.9 × 10 −15 = 0.93 ns (b) L 1 = 1.7 × 10 −6 × = 2 × 10 3 Ω −4 A 0.28 × 10 × 0.3 × 10 − 4 R=ρ εA εTL 2.8 × 8.85 × 10 −14 × 0.3 × 10 −4 × 1 = 2.1 × 10 −13 F −4 d S 0.36 × 10 3 −13 RC = 2 × 10 × 2.1 × 10 = 0.42 ns C= = = (c) We can decrease the RC delay by 55%. Ratio = 20. (a) L 1 = 2.67 × 10 −6 × = 3.2 × 10 3 Ω 4 4 A 0.28 × 10 × 0.3 × 10 R=ρ C= 0.42 = 0.45. 093 εA d = εTL S = 3.9 × 8.85 × 10 −14 × 0.3 × 10 −4 × 1 × 3 = 8.7 × 10 −13 F 4 0.36 × 10 . RC = 3.2 ×103 ×8.7 × 10-13 = 2.8 ns. (b) R = ρ L 1 = 1.7 × 10 −6 × = 2 × 10 3 Ω 4 A 0.28 × 10 × 0.3 × 10 4 C= εA d = εTL S = 2.8 × 8.85 × 10 −14 × 0.3 × 10 −4 × 1 × 3 = 6.3 × 10 −13 F 4 0.36 × 10 RC = 2 × 10 3 × 8.7 × 10 −13 = 2.5 ns RC = 3.2 × 10 3 × 8.7 × 10 −13 = 2.5 ns. 21. (a) The aluminum runner can be considered as two segments connected in series: 20% (or 0.4 mm) of the length is half thickness (0.5 µm) and the remaining 1.6 mm is full thickness (1µm). The total resistance is ⎡ ⎤ ⎡ 0.16 ⎤ 0.04 + −4 R = ρ ⎢ 1 + 2 ⎥ = 3 × 10 −6 ⎢ − 4 −4 −4 ⎥ 10 × (0.5 × 10 ) ⎦ ⎣10 × 10 ⎣ A1 A2 ⎦ = 72 Ω. The limiting current I is given by the maximum allowed current density times 110 cross-sectional area of the thinner conductor sections: I = 5×105 A/cm2× (10-4×0.5×10-4) = 2.5×10-3 A = 2.5 mA. The voltage drop across the whole conductor is then V = RI = 72Ω × 2.5 × 10 −3 A = 0.18V. 22. 0.5 μm 0.5 μm 40 nm Cu = Al 60 nm h: height , W : width , t : thickness, assume that the resistivities of the cladding layer and TiN are much larger than ρ A and ρ Cu R Al = ρ Al × RCu = ρ Cu × h ×W h ×W = 2.7 = 1.7 (0.5 − 0.1) × 0.5 (0.5 − 2t ) × (0.5 − 2t ) When R Al = RCu Then ⇒ 2.7 1.7 = 0.4 × 0.5 (0.5 − 2t ) 2 t = 0.073 μm = 73 nm . 111 CHAPTER 13 1. With reference to Fig. 2 for class 100 clean room we have a total of 3500 particle sizes ≥ 0.5 µm particles/m3 with 21 × 3500 = 735 particles/m2 with particle sizes ≥ 1.0 µm 100 4.5 × 3500 = 157 particles/m2 with particle sizes ≥ 2.0 µm 100 Therefore, (a) 3500-735 = 2765 particles/m3 between 0.5 and 1 µm (b) 735-157 = 578 particles/m3 between 1 and 2 µm (c) 157 particles/m3 above 2 µm. 2. 9 Y = Π e − D1 A n =1 A = 50 mm2 = 0.5 cm2 Y = e −4( 0 1×0 5) × e −4( 0 25×0 5) × e −1(1×0 5) = e −1 2 = 30.1% . 3. The available exposure energy in an hour is 0.3 mW2/cm2 × 3600 s =1080 mJ/cm2 For positive resist, the throughput is 1080 = 7 wafers/hr 140 For negative resist, the throughput is 1080 = 120 wafers/hr . 9 4. (a) The resolution of a projection system is given by l m = k1 λ NA DOF = k 2 (b) = 0 .6 × 0.193μm = 0.178 µm 0.65 λ ⎡ 0.193μm ⎤ = 0 .5 ⎢ = 0.228 µm 2 ⎥ ( NA) ⎣ (0.65) ⎦ 2 We can increase NA to improve the resolution. We can adopt resolution enhancement techniques (RET) such as optical proximity correction (OPC) and phase-shifting Masks 112 (PSM). We can also develop new resists that provide lower k1 and higher k2 for better resolution and depth of focus. (c) PSM technique changes k1 to improve resolution. 5. (a) Using resists with high γ value can result in a more vertical profile but throughput decreases. (b) Conventional resists can not be used in deep UV lithography process because these resists have high absorption and require high dose to be exposed in deep UV. This raises the concern of damage to stepper lens, lower exposure speed and reduced throughput. 6. (a) A shaped beam system enables the size and shape of the beam to be varied, thereby minimizing the number of flashes required for exposing a given area to be patterned. Therefore, a shaped beam can save time and increase throughput compared to a Gaussian beam. (b) We can make alignment marks on wafers using e-beam and etch the exposed marks. We can then use them to do alignment with e-beam radiation and obtain the signal from these marks for wafer alignment. X-ray lithography is a proximity printing lithography. Its accuracy requirement is very high, therefore alignment is difficult. (c) X-ray lithography using synchrotron radiation has a high exposure flux so X-ray has better throughput than e-beam. 7. (a) To avoid the mask damage problem associated with shadow printing, projection printing exposure tools have been developed to project an image from the mask. With a 1:1 projection printing system is much more difficult to produce defect-free masks than it is with a 5:1 reduction step-and-repeat system. (b) It is not possible. The main reason is that X-rays cannot be focused by an optical lens. When it is through the reticle. So we can not build a step-and-scan X-ray lithography 113 system. 8. As shown in the figure, the profile for each case is a segment of a circle with origin at the initial mask-film edge. As overetching proceeds the radius of curvature increases so that the profile tends to a vertical line. 9. (a) 20 sec 0.6 × 20/60 = 0.2 µm…..(100) plane 0.6/16 × 20/60 = 0.0125 µm……..(110) plane 0.6/100 × 20/60 = 0.002 µm…….(111) plane Wb = W0 − 2l = 1.5 − 2 × 0.2 = 1.22 µm (b) 40 sec 0.6 × 40/60 = 0.4 µm….(100)plane 0.6/16 × 40/60 = 0.025 µm…. (110) plane 0.6/100 × 40/60 = 0.004 µm…..(111) plane Wb = W0 − 2l = 1.5 − 2 × 0.4 = 0.93 µm (c) 60 sec 0.6 ×1 = 0.6 µm….(100)plane 0.6/16 ×1 = 0.0375 µm…. (110) plane 0.6/100 ×1= 0.006 µm…..(111) plane Wb = W0 − 2l = 1.5 − 2 × 0.6 = 0.65 µm. 10. Using the data in Prob. 9, the etched pattern profiles on <100>-Si are shown in below. (a) 20 sec l = 0.012 µm, W0 = Wb = 1.5 µm 114 (b) 40 sec l = 0.025 µm, W0 = Wb = 1.5 µm (c) 60 sec l = 0.0375 µm W0 = Wb = 1.5 µm. 11. If we protect the IC chip areas (e.g. with Si3N4 layer) and etch the wafer from the top, the width of the bottom surface is W = W1 + 2l = 1000 + 2 × 625 = 1884 µm The fraction of surface area that is lost is (W 2 − W12 ) / W 2 × 100%=(18842-10002) /18842× 100% = 71.8 % In terms of the wafer area, we have lost 71.8 % × π (15 / 2) 2 =127 cm2 Another method is to define masking areas on the backside and etch from the back. The width of each square mask centered with respect of IC chip is given by W = W1 − 2l = 1000 − 2 × 625 = 116 µm Using this method, the fraction of the top surface area that is lost can be negligibly small. 12. 1 Pa = 7.52 m Torr PV = nRT 7.52 /760 × 10-3 = n/V ×0.082 × 273 n/V = 4.42 × 10-7 mole/liter = 4.42 × 10-7 × 6.02 × 1023/1000 =2.7 ×1014 cm-3 mean–free–path λ = 5 × 10 −3 / P cm = 5× 10-3 ×1000/ 7.52 = 0.6649 cm = 6649 µm 150Pa = 1128 m Torr PV = nRT 115 1128/ 760 × 10-3 = n/V × 0.082 × 273 n/V = 6.63 × 10-5 mole/liter = 6.63 ×10-5×6.02×1023/1000 = 4 × 1016 cm-3 mean-free-path λ = 5 × 10 −3 / P cm = 5× 10-3 ×1000/1128 = 0.0044 cm = 44 µm. 13. Si Etch Rate (nm/min) = 2.86 × 10-13 × n F × T -13 = 2.86 × 10 1 2 ×e − Ea RT 15 ×3×10 × (298) 1 2 ×e −2 48×103 1 987×298 = 224.7 nm/min. 14. SiO2 Etch Rate (nm/min) = 0.614× 10 -13 15 ×3×10 × (298) 1 2 ×e −3 76×103 1 987×298 Etch selectivity of SiO2 over Si = 5.6 = 0.025 224.7 Or etch rate (SiO2)/etch rate (Si) = 0.614 ( −3 76+ 2 48)1 987×298 ×e = 0.025 . 2.86 = 5.6 nm/min 15. A three–step process is required for polysilicon gate etching. Step 1 is a nonselective etch process that is used to remove any native oxide on the polysilicon surface. Step 2 is a high polysilicon etch rate process which etches polysilicon with an anisotropic etch profile. Step 3 is a highly selective polysilicon to oxide process which usually has a low polysilicon etch rate. 16. If the etch rate can be controlled to within 10 %, the polysilicon may be etched 10 % longer or for an equivalent thickness of 40 nm. The selectivity is therefore 40 nm/1 nm = 40. 17. Assuming a 30% overetching, and that the selectivity of Al over the photoresist maintains 3. The minimum photoresist thickness required is 116 (1+ 30%) × 1 µm/3 = 0.433 µm = 433.3 nm. 18. ωe = qB me 2π × 2.45 × 10 9 = 1.6 × 10 −19 × B 9.1 × 10 −31 B = 8.75 × 10-2(tesla) = 875 (gauss). 19. Traditional RIE generates low-density plasma (109 cm-3) with high ion energy. ECR and ICP generate high-density plasma (1011 to 1012 cm-3) with low ion energy. Advantages of ECR and ICP are low etch damage, low microloading, low aspect-ratio dependent etching effect, and simple chemistry. However, ECR and ICP systems are more complicated than traditional RIE systems. 20. The corrosion reaction requires the presence of moisture to proceed. Therefore, the first line of defense in controlling corrosion is controlling humidity. Low humidity is essential,. especially if copper containing alloys are being etched. Second is to remove as much chlorine as possible from the wafers before the wafers are exposed to air. Finally, gases such as CF4 and SF6 can be used for fluorine/chlorine exchange reactions and polymeric encapsulation. Thus, Al-Cl bonds are replaced by Al-F bonds. Whereas Al-Cl bonds will react with ambient moisture and start the corrosion process , Al-F bonds are very stable and do not react. Furthermore, fluorine will not catalyze any corrosion reactions. 117 CHAPTER 14 1. Ea(boron) = 3.46 eV, D0 = 0.76 cm2/sec From Eq. 6, D = D0 exp( − Ea − 3.46 ⎛ ⎞ −15 2 ) = 0.76 exp⎜ ⎟ = 4.142 × 10 cm /s −5 kT ⎝ 8.614 × 10 × 1223 ⎠ L = Dt = 4.142 × 10 −15 × 1800 = 2.73 × 10 −6 cm From Eq. 9, C ( x ) = C s erfc( x x ⎛ ⎞ ) = 1.8 × 10 20 erfc ⎜ −6 ⎟ 2L ⎝ 5.46 × 10 ⎠ If x = 0, C (0) = 1.8 × 10 20 atoms /cm 3 ; x = 0.05 ×10-4, C(5× 10-6) = 3.6 × 1019 atoms/cm3; x = 0.075 ×10-4 , C(7.5×10-6) = 9.4 ×1018 atoms/cm3; x = 0.1×10-4, C(10-5) = 1.8 × 1018 atoms/cm3; x = 0.15× 10-4, C(1.5×10-5) = 1.8× 1016 atoms/cm3. The x j = 2 Dt (erfc -1 C sub ) = 0.15μm Cs Total amount of dopant introduced = Q(t) = 2. 2 π C s L = 5.54 × 1014 atoms/cm2. − 3.46 ⎛ − Ea ⎞ ⎛ ⎞ −14 2 D = D0 exp⎜ ⎟ = 0.76 exp⎜ ⎟ = 4.96 × 10 cm /s −5 ⎝ 8.614 × 10 × 1323 ⎠ ⎝ kT ⎠ From Eq. 15, C S = C (0, t ) = S πDt = 2.342 × 1019 atoms/cm 3 x ⎛ x ⎞ ⎛ ⎞ C ( x) = C S erfc⎜ ⎟ = 2.342 × 1019 erfc⎜ −5 ⎟ ⎝ 2L ⎠ ⎝ 2.673 × 10 ⎠ If x = 0, C(0) = 2.342 × 1019 atoms/cm3; x = 0.1×10-4, C(10-5) = 1.41×1019 atoms/cm3; 118 x = 0.2×10-4, C(2×10-5) = 6.79×1018 atoms/cm3; x = 0.3×10-4, C(3×10-5) = 2.65×1018 atoms/cm3; x = 0.4×10-4, C(4×10-5) = 9.37×1017 atoms/cm3; x = 0.5×10-4, C(5×10-5) = 1.87×1017 atoms/cm3; x = 0.6×10-4, C(6×10-5) = 3.51×1016 atoms/cm3; x = 0.7×10-4, C(7×10-5) = 7.03×1015 atoms/cm3; x = 0.8×10-4, C(8×10-5) = 5.62×1014 atoms/cm3. The x j = 4 Dt ln 3. S C B πDt = 0.72 μm . ⎛ ⎞ 10 −8 ⎟ 1 × 1015 = 1 × 1018 exp⎜⎜ −13 ⎟ ⎝ 4 × 2.3 × 10 t ⎠ t = 1573 s = 26 min For the constant-total-dopant diffusion case, Eq. 15 gives C S = S πDt S = 1 × 1018 π × 2.3 × 10 −13 × 1573 = 3.4 × 1013 atoms/cm 2 . 4. The process is called the ramping of a diffusion furnace. For the ramp-down situation, the furnace temperature T is given by T = T0 - rt where T0 is the initial temperature and r is the linear ramp rate. The effective Dt product during a ramp-down time of t1 is given by ( Dt ) eff = ∫ t1 0 D (t )dt In a typical diffusion process, ramping is carried out until the diffusivity is negligibly small. Thus the upper limit t1 can be taken as infinity: 119 1 T = 1 T0 − rt ≈ 1 T0 (1 + rt T0 + ...) and − Ea ⎞ = D exp ⎡ − E a (1 + rt + ...)⎤ = D (exp − E a )(exp − rE a t ...) ≈ D (T ) exp − rE a t D = D0 exp⎛⎜ 0 0 0 ⎥ ⎢ kT 2 2 kT ⎟⎠ ⎝ T0 kT0 kT0 kT0 ⎦ ⎣ 0 where D(T0) is the diffusion coefficient at T0. Substituting the above equation into the expression for the effective Dt product gives ∞ ( Dt ) eff ≈ ∫ D (T0 ) exp 0 − rE a t kT0 2 2 dt = D (T0 ) kT0 rE a Thus the ramp-down process results in an effective additional time equal to kT02/rEa at the initial diffusion temperature T0. For phosphorus diffusion in silicon at 1000°C, we have from Fig. 4: D(T0) = D (1273 K) = 2× 10-14 cm2/s 1273 − 773 r= 20 × 60 = 0.417 K / s Ea = 3.66 eV Therefore, the effective diffusion time for the ramp-down process is kT0 rE 5. 2 a = 1.38 × 10 −23 (1273) 0.417( 3.66 × 1.6 × 10 2 −19 ) = 91s ≈ 1.5 min . For low-concentration drive-in diffusion, the diffusion is given by Gaussian distribution. The surface concentration is then C (0, t ) = S S ⎛ E ⎞ exp⎜ a ⎟ = πDt πD0 t ⎝ 2kT ⎠ 120 dC S ⎛ E ⎞⎛ − t 3 / 2 = exp⎜ a ⎟⎜⎜ dt πD 0 ⎝ 2kT ⎠⎝ 2 or dC C = −0.5 × ⎞ C ⎟⎟ = −0.5 × t ⎠ dt t which means 1% change in diffusion time will induce 0.5% change in surface concentration. dC S ⎛ E ⎞⎛ − E a exp⎜ a ⎟⎜ = 2 dT πD0 t ⎝ 2kT ⎠⎝ 2kT or dC C = − Ea 2 kT × dT T = E ⎞ ⎟ = −C a 2 2kT ⎠ − 3.6 × 1.6 × 10 −19 2 × 1.38 × 10 − 23 × 1273 × dT T = −16.9 × dT T which means 1% change in diffusion temperature will cause 16.9% change in surface concentration. 6. At 1100°C, ni = 6×1018 cm-3. Therefore, the doping profile for a surface concentration of 4 × 1018 cm-3 is given by the “intrinsic” diffusion process: ⎛ x ⎞ C ( x, t ) = C s erfc⎜ ⎟ ⎝ 2 Dt ⎠ where Cs = 4× 1018 cm-3, t = 3 hr = 10800 s, and D = 5x10-14 cm2/s. The diffusion length is then Dt = 2.32 × 10 −5 cm = 0.232μm x ⎛ ⎞ The distribution of arsenic is C ( x) = 4 × 1018 erfc⎜ −5 ⎟ ⎝ 4.64 × 10 ⎠ The junction depth can be obtained as follows xj ⎞ ⎛ ⎟ 1015 = 4 × 1018 erfc⎜⎜ −5 ⎟ 4 . 64 10 × ⎠ ⎝ xj = 1.2× 10-4 cm = 1.2 μm. 121 7. At 900°C, ni = 2× 1018 cm-3. For a surface concentration of 4×1018 cm-3, given by the “extrinsic” diffusion process D = D0 e − Ea kT −4 05×1 6×10 −19 n 4 × 1018 − 23 × = 45.8e 1 38×10 ×1173 × = 3.77 × 10 −16 cm 2 /s ni 2 × 1018 x j = 1.6 Dt = 1.6 3.77 × 10 −16 × 10800 = 3.23 × 10 −6 cm = 32.3 nm . 8. Intrinsic diffusion is for dopant concentration lower than the intrinsic carrier concentration ni at the diffusion temperature. Extrinsic diffusion is for dopant concentration higher than ni. 9. For impurity in the oxidation process of silicon, segregation coefficeint = equilibrium concentration of impurity in silicon . equilibrium concentration of impurity in SiO 2 3 × 1011 3 = = 0.006 . 13 500 5 × 10 10. κ= 11. –0.5 = –1.1 + qFB/Ci C ox = FB = 10 εs d = 3.9 × 8.85 × 10 −14 = 3.45 × 10 − 7 −6 10 3.45 × 10 −7 × 0.6 = 1.3 × 1012 cm − 2 −19 1.6 × 10 −5 π (10.16) 2 t = 1.3 × 1012 × 1.6 × 10 −19 the implant time t = 6.7 s. 12. The ion dose per unit area is It 10 × 10 −6 × 5 × 60 N q 1.6 × 10 −19 = = = 2.38 × 1012 ions/cm 2 10 A A π × ( )2 2 122 From Eq. 25 and Example 3, the peak ion concentration is at x = Rp. Figure. 17 indicates the σp is 20 nm. Therefore, the ion concentration is S σ p 2π = 2.38 × 1012 20 × 10 −7 2π = 4.74 × 1017 cm −3 . From Fig. 17, the Rp = 230 nm, and σp = 62 nm. 13. The peak concentration is S σ p 2π = 2 × 1015 62 × 10 −7 2π = 1.29 × 10 20 cm −3 From Eq. 25, ⎡ − (x j − Rp )2 ⎤ 10 = 1.29 × 10 exp ⎢ ⎥ 2 2σ p ⎢⎣ ⎥⎦ 15 20 xj = 0.53 μm. 14. Dose per unit area = Q C 0 ΔVT 3.9 × 8.85 × 10 −14 × 1 = = = 8.6 × 1011 cm − 2 −8 −19 q q 250 × 10 × 1.6 × 10 From Fig. 17 and Example 3, the peak concentration occurs at 140 nm from the surface. Also, it is at (140-25) = 115 nm from the Si-SiO2 interface. 15. The total implanted dose is integrated from Eq. 25 QT = ∫ S ∞ 0 σp ⎡ − (x − Rp )2 ⎤ R p ⎤ ⎫⎪ S S ⎧⎪ ⎡ S 1 1 erfc ( )⎥ ⎬ = [2 − erfc( 2.3)] = × 1.9989 exp ⎢ = + − dx ⎢ ⎥ ⎨ 2 2 ⎪ ⎢⎣ 2 σ p 2 ⎥⎦ ⎪⎭ 2 2π ⎢⎣ 2σ p ⎥⎦ ⎩ The total dose in silicon is as follows (d = 25 nm): Q Si = ∫ S ∞ d σp ⎡− (x − Rp )2 ⎤ R p − d ⎤ ⎫⎪ S S ⎧⎪ ⎡ S 1 1 erfc ( )⎥ ⎬ = [2 − erfc(1.87)] = × 1.9918 dx exp ⎢ = + − ⎢ ⎥ ⎨ 2 2 ⎪ ⎢⎣ 2 σ p 2 ⎥⎦ ⎪⎭ 2 2π ⎢⎣ 2σ p ⎥⎦ ⎩ 123 the ratio of dose in the silicon = QSi/QT = 99.6%. 16. The projected range is 150 nm (see Fig. 17). The average nuclear energy loss over the range is 60 eV/nm (Fig. 16). 60× 0.25 = 15 eV (energy loss of boron ion per each lattice plane) the damage volume = VD = π (2.5 nm)2(150 nm) = 3× 10-18 cm3 total damage layer = 150/0.25 = 600 displaced atom for one layer = 15/15 = 1 damage density = 600/VD = 2×1020 cm-3 2×1020/5.02×1022 = 0.4%. 17. The higher the temperature, the faster defects anneal out. Also, the solubility of electrically active dopant atoms increases with temperature. 18. ΔV t = 1 V = Q1 C ox where Q1 is the additional charge added just below the oxide-semiconductor surface by ion implantation. COX is a parallel-plate capacitance per unit area given by C ox = εs d (d is the oxide thickness, ε s is the permittivity of the semiconductor) Q1 = ΔVt C ox = 1V × 3.9 × 8.85 × 10 −14 F/cm C = 8.63 × 10 −7 −6 cm 2 0.4 × 10 cm 8.63 × 10 −7 = 5.4 ×1012 ions/cm2 −19 1.6 × 10 Total implant dose = 5.4 × 1012 = 1.2 × 1013 ions/cm2. 45% 124 19. The discussion should mention much of Section 13.6. Diffusion from a surface film avoids problems of channeling. Tilted beams cannot be used because of shadowing problems. If low energy implantation is used, perhaps with preamorphization by silicon, then to keep the junctions shallow, RTA is also necessary. 20. From Eq.35 Sd 1 ⎛ 0.4 − 0.6 ⎞ = erfc⎜ ⎟ = 0.84 2 S ⎝ 0.2 2 ⎠ The effectiveness of the photoresist mask is only 16%. Sd 1 ⎛ 1 − 0.6 ⎞ = erfc⎜ ⎟ = 0.023 2 S ⎝ 0.2 2 ⎠ The effectiveness of the photoresist mask is 97.7%. 21. 2 e -u = 10 −5 T= 2 π u 1 ∴u = 3.02 d = Rp + 4.27 σ p = 0.53 + 4.27 × 0.093 = 0.927 µm. 125 CHAPTER15 1. Each U-shape section (refer to the figure) has an area of2500 ~m x 8 ~m = 2 x 104 ~2 . Therefore, there are (2500)2/2 x 104 = 312.5 U-shaped section. Each section contains 2 long lines with 1248 squares each, 4 comer squares, 1 bottom square, and 2 half squares at the top. Therefore the resistance for each section is 1 k:Q I[] (1248 x2 + 4x0.65 +2) = 2500.6 k:Q The maximum resistance is then 312.5 x2500.6 = 7.81 X 108 Q = 781 MQ 2. The area required on the chip is 126 = 4.35 × 103 μm2 = 66 × 66 μm Refer to Fig.4a and using negative photoresist of all levels (a) Ion implantation mask (for p+ implantation and gate oxide) (b) Contact windows (2×10 μm) (c) Metallization mask (using Al to form ohmic contact in the contact window and form the MOS capacitor). Because of the registration errors, an additional 2 μm is incorporated in all critical dimensions. 127 ( 0) (b) (c ) 73 3. If the space between lines is 2 Jlm, then there is 4 Jlm for each tum (i.e., 2xn, for one tum). Assume there are n tmns, from Eq.6, L ~ jl()n2r ~ 1.2 x 10-6n2r, where r can be replaced by 2 x n. Then, we can obtain that n is 13. 128 4. (a) Metal 1, (b) contact hole, (c) Metal 2. (a) Metal 1, (b) contact hole, (c) Metal 2. 129 5. The circuit diagram and device cross-section of a d amped transistor are shown in (a) and (b), respectively. SCHOTTKY DIODE BASE (0) COL L ECTOR~------~-J SCHOTTKY DIODE -----o EMITTER B E (b) p- SUBSTRATE 6. OHMIC CONTACT (a) The undoped polysilicon is used for isolation. (b) The polysilicon 1 is used as a solid-phase diffusion somce to f01m the extrinsic base region and the base electrode. (c) The polysilicon 2 is used as a solid-phase diffusion som ce to fonn the emitter region and the emitter electrode. 7. (a) For 30 keV boron, Rp = 100 nm and Mp = 34 nm. Assmni.ng that Rp and Mp 130 for boron are the same in Si and SiO2 the peak concentration is given by S 2π ΔR p 8 × 1011 = 2π (34 × 10 ) −7 = 9.4 × 1016 cm −3 The amount of boron ions in the silicon is ⎡ (x − R p )2 ∞ Q S =∫ exp ⎢ − d q 2 ΔR p2 2π ΔR p ⎣⎢ = ⎛ Rp − d S⎡ ⎢2 − erfc⎜ ⎜ 2 ΔR 2⎢ p ⎝ ⎣ = 8 × 1011 2 ⎤ ⎥dx ⎦⎥ ⎞⎤ ⎟⎥ ⎟⎥ ⎠⎦ ⎡ ⎛ 750 ⎞⎤ ⎟⎥ ⎢2 − erfc⎜ ⎝ 2 × 340 ⎠⎦ ⎣ = 7.88 × 1011 cm − 2 Assume that the implanted boron ions form a negative sheet charge near the Si-SiO2 interface, then ⎛Q ⎞ 1.6 × 10 −19 × (7.88 × 1011 ) = 0.91 V ΔVT = q⎜ ⎟ / C ox = 3.9 × 8.85 × 10 −14 / (25 × 10 − 7 ) ⎝q⎠ (b) For 80 keV arsenic implantation, Rp = 49 nm and Δ Rp = 18 nm. The peak arsenic concentration is S 2π ΔR p = 1016 π × (18 × 10 ) 131 −7 = 2.21 × 10 21 cm − 3 . Rp =490 A(As) .~. I ..' /L 2~p\ . I \ARSENIC (UPPER SCALE) I \ \ \ \ \ \ \ \ \ Rp=1000 A (B) 1015~~------~~l~--~~c=~~1o=n=-=c=m==<=10=0;>~ 0 d ---!Si0 21-- 250 A 3000 X(Al - FOR CHANNEL REGION 8. (a) Because (100)-oriented silicon has lower (~ one tenth) interface-trapped charge and a lower fixed oxide charge. (b) If the field oxide is too thin, it may not provide a large enough threshold voltage for adequate isolation between neighboring MOSFETs. (c) The typical sheet resistance of heavily doped polysilicon gate is 20 to 30 n /0 , which is adequate for MOSFETs with gate lengths larger than 3 Jlm. For sh01t er gates, the sheet resistance of polysilicon is too high and will cause 132 large RC delays. We can use refractory metals (e.g., Mo) or silicides as the gate material to reduce the sheet resistance to about 1 Ω /□. (d) A self-aligned gate can be obtained by first defining the MOS gate structure, then using the gate electrode as a mask for the source/drain implantation. The self-aligned gate can minimize parasitic capacitance caused by the source/drain regions extending underneath the gate electrode (due to diffusion or misalignment). (e) P-glass can be used for insulation between conducting layers, for diffusion and ion implantation masks, and for passivation to protect devices from impurities, moisture, and scratches. 9. The lower insulator has a dielectric constant ε1/ε0 = 4 and a thickness d1= 10 nm The upper insulator has a dielectric constant ε2/ε0 = 10 and a thickness d2 = 100 nm. Upon application of a positive voltage VG to the external gate, electric field E1 and E2 are established in the d1 and d2 respectively. We have, from Gauss’ law, that ε1E1 = ε2E2 +Q and VG = E1d1 + E2d2 where Q is the stored charge on the floating gate. From these above two equations, we obtain E1 = VG Q + d1 + d 2 (ε 1 / ε 2 ) ε 1 + ε 2 (d1 / d 2 ) 133 ⎧ ⎪ 10 × 10 7 −7 ⎪ = = 10 J σE1 + ⎨ ⎪10 + 100⎛⎜ 4 ⎞⎟ ⎪⎩ ⎝ 10 ⎠ (a) ⎫ ⎪ Q ⎪ 5 ⎬ = 0.2 − 2.26 × 10 Q ⎡ ⎛ 10 ⎞⎤ −14 ⎪ ⎢4 + 10⎜ 100 ⎟⎥ × 8.85 × 10 ⎪ ⎝ ⎠⎦ ⎣ ⎭ If the stored charge does not reduce E1 by a significant amount (i.e., 0.2 >> 2.26×105 ⎥Q⎥, we can write Q = ∫ σE1 dt ' ≈ 0.2Δt = 0.2 × (0.25 × 10 − 6 ) = 5 × 10 −8 C t 0 ΔVT = (b) Q 5 × 10 −8 = 0.565 V = C2 10 × 8.85 × 10 −14 / 100 × 10 −7 ( )( ) when t → ∞, J → 0 we have Q → 0.2 / 2.26 × 10 5 ≅ 8.84×10-7 C. Then ΔVT = Q 8.84 × 10 −7 = 9.98 V. = C2 10 × 8.85 × 10 −14 / 10 −5 ( ) 10. 134 + + (o) p- TUB + + + + (b) POLYSILICON GATE + + + + (c) n-TYPE DIFFUSION . 1L - - - -_ _ _ _ . + + 135 + + (d) p-TYPE DtFFUSION D + + + + D D D (e ) CONTACT WINDOWS 0 0 0 D + + + {f) METALLIZATION 11 . The oxide capacitance per unit area is given by and the maximum cunent supplied by the device is I DS 2 z ..!_ W C (V -V )2 = 1 SJ.Im 3.5x 10- 7 (V - V ) z5mA 2 L J1 ox G T 2 0.5J111l G T 136 and the maximum allowable wire resistance is 0.1 V/5 mA, or 20Ω. Then, the length of the wire must be L≤ R × Area ρ = 20Ω × 10 −8 cm 2 = 0.074 cm 2.7 × 10 −8 Ω − cm or 740 μm. This is a long distance compared to most device spacing. When driving signals between widely spaced logic blocks however, minimum feature sized lines would not be appropriate. 12. 137 X p X l ~~TUB 1 ( 0) v-EPITAXY B SiOz .......................... (b) -----------------p-TUB I __________ n-TUB ' t--------',._. .._ _______ --v (c) --------~ J j FIELD OXIDE (d) ---------------RESIST (e) -------------------II p-~ASS ,___n__P_-_ru_e_ __, ___ n.- T.1JB __: --------- .-...- ------.-II 138 (f ) 13. To solve the short-channel effect of devices. 14. The device performance will be degraded from the boron penetration. There are methods to reduce this effect: (1) using rapid thermal annealing to reduce the time at high temperatures, consequently reduces the diffusion of boron, (2) using nitrided oxide to suppress the boron penetration, since boron can easily combine with nitrogen and becomes less mobile, (3) making a multi-layer of polysilicon to trap the boron atoms at the interface of each layer. 15. Total capacitance of the stacked gate structure is : C= ε1 d1 × ε2 d2 ⎛ ε1 ε 2 ⎞ 7 25 ⎛ 7 25 ⎞ ⎜ + ⎟ = × + ⎟ = 2.12 ⎜ 0.5 10 ⎝ 0.5 10 ⎠ ⎝ d1 d 2 ⎠ 3.9 = 2.12 d ∴d = 3.9 =1.84 nm. 2.12 16. Disadvantages of LOCOS: (1) high temperature and long oxidation time cause VT shift, (2) bird’s beak, (3) not a planar surface, (4) exhibits oxide thinning effect. Advantages of shallow trench isolation: (1) planar surface, (2) no high temperature processing and long oxidation time, (3) no oxide thinning effect, (4) no bird’s beak. 17. For isolation between the metal and the substrate. 18. GaAs lacks of high-quality insulating film. 139 19. (a) 1 A⎞ ⎛ ⎛ L ⎞⎛ RC = ⎜ ρ ⎟ ε ox ⎟ = ⎜10 −5 × d⎠ ⎝ 1 × 0.5 × 10 −8 ⎝ A ⎠⎝ 1 × (1 × 10 −4 )⎤ ⎞⎡ −14 ⎟ ⎢3.9 × 8.85 × 10 × ⎥ 0.5 × 10 − 4 ⎦ ⎠⎣ = 2000 × (69.03 × 10 −14 ) = 1.38 × 10 − 9 s = 1.38 ns. (b) For a polysilicon runner L ⎞⎛ A⎞ ⎛ RC = ⎜ Rsquare ⎟⎜ ε ox ⎟ W ⎠⎝ d⎠ ⎝ ⎛ 1 ⎞ = 30⎜ − 4 ⎟ 69.03 × 10 −14 = 2.07 × 10 −7 s ⎝ 10 ⎠ = 207 ns ( ) Therefore the polysilicon runner’s RC time constant is 150 times larger than the aluminum runner. 22. When we combine the logic circuits and memory on the chip, we need multiple supply voltages. For reliability issue, different oxide thicknesses are needed for different supply voltages. 21. (a) 1 C total = hence EOT 1 C Ta 2 O5 3.9 + 1 = 75 25 C nitride + 10 = 17.3 Å 7 (b) EOT = 16.7 Å. 140