半导体物理与器件第三版(尼曼)12章答案

Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 185 Chapter 12 Problem Solutions 12.1 (a) I V VD GS t = − ( ) L NM O QP10 2.1 15 exp For V VGS = 0 5. , I D = ⇒ − ( )( ) L NM O QP10 0 5 2.1 0 0259 15 exp . . I x AD = −9.83 10 12 For V VGS = 0 7. , I x AD = −3 88 10 10. For V VGS = 0 9. , I x AD = −154 10 8. Then the total current is: I ITotal D= 10 6b g For V VGS = 0 5. , I ATotal = 9.83 µ For V VGS = 0 7. , I mATotal = 0 388. For V VGS = 0 9. , I mATotal = 15 4. (b) Power: P I VTotal DD= ⋅ Then For V VGS = 0 5. , P W= 49.2 µ For V VGS = 0 7. , P mW= 1 94. For V VGS = 0 9. , P mW= 77 12.2 We have ∆ ∆L eN V sat V a fp DS DS= ∈ ⋅ + +( )2 φ − + ( )φ fp DSV sat where φ fp t a i V N n x = = F HG I KJ ( ) F HG I KJln . ln .0 0259 10 15 10 16 10 or φ fp V= 0 347. We find 2 2 11 7 8 85 10 1 6 10 10 14 19 16 1 2 ∈ = ( )L NM O QP − −eN x xa . . . /b g b gb g = 0 360 1 2. / /µm V We have V sat V VDS GS T( ) = − (a) For V V V sat VGS DS= ⇒ =( )5 4.25 Then ∆L = + − +0 360 0 347 5 0 347 4.25. . . or ∆L m= 0 0606. µ If ∆L is 10%of L , then L m= 0 606. µ (b) For V V V V V sat VDS GS DS= = ⇒ =( )5 2 1 25, . Then ∆L = + − +0 360 0 347 5 0 347 1 25. . . . or ∆L m= 0 377. µ Now if ∆L is 10% of L, then L m= 3 77. µ 12.3 ∆ ∆L eN V sat V a fp DS DS= ∈ ⋅ + +( )2 φ − + ( )φ fp DSV sat where φ fp t a i V N n x x = = F HG I KJ ( ) F HG I KJln . ln .0 0259 4 10 15 10 16 10 or φ fp V= 0 383. and x eNdT fp a = ∈L NM O QP 4 1 2φ / = ( ) ( )L NM O QP − − 4 11 7 8 85 10 0 383 1 6 10 4 10 14 19 16 1 2 . . . . / x x x b g b gb g or x mdT = 0 157. µ Then ′ =( )Q eN xSD a dTmax Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 186 = − −16 10 4 10 0157 1019 16 4. .x x xb gb gb g or ′ =( ) −Q C cmSD max /10 7 2 Now V Q Q t T SD SS ox ox ms fp= ′ − ′ ∈ + +( ) FHG I KJmaxa f φ φ2 so that V x x x xT = − − − − −( ) 10 1 6 10 3 10 400 10 3 9 8 85 10 7 19 10 8 14 . . . b gb g b g b g + + ( )0 2 0 383. or V VT = 187. Now V sat V V VDS GS T( ) = − = − =5 187 313. . We find 2 2 11 7 8 85 10 1 6 10 4 10 14 19 16 1 2 ∈ = ( )L NM O QP − −eN x x xa . . . /b g b gb g = −180 10 5. x Now ∆ ∆L x VDS= ⋅ + + −180 10 0 383 3135. . . − +0 383 313. . or ∆ ∆L x VDS= + − −180 10 3 513 3 5135. . . We obtain ∆VDS ∆L mµ( ) 0 1 2 3 4 5 0 0.0451 0.0853 0.122 0.156 0.188 12.4 Computer plot 12.5 Plot 12.6 Plot 12.7 (a) Assume V sat VDS ( ) = 1 , We have Ε sat DSV sat L = ( ) We find L mµ( ) Ε sat V cm/( ) 3 1 0 5. 0 25. 0 13. 3 33 103. x 104 2 104x 4 104x 7.69 104x (b) Assume µ n cm V s= −500 2 / , we have v n sat= µ Ε Then For L m= 3 µ , v x cm s= 1 67 106. / For L m= 1 µ , v x cm s= 5 106 / For L m≤ 0 5. µ , v cm s≈ 107 / 12.8 We have ′ = −( )−I L L L ID D∆ 1 We may write g I V L L L I L VO D DS D DS = ∂ ′ ∂ = − − −∂ ∂ ( ) ( ) ( )FHG I KJ −1 2∆ ∆ = − ⋅ ⋅ ∂ ∂( ) ( )L L L I L VD DS∆ ∆ 2 We have ∆L eN V V sat a fp DS fp DS= ∈ ⋅ + − + ( )2 φ φ We find ∂ ∂ = ∈ ⋅ + ( )∆L V eN VDS a fp DS 2 1 2 φ (a) For V V V VGS DS= =2 1, ∆ , and V sat V V VDS GS T( ) = − = − =2 0 8 1 2. . Also V V sat V VDS DS DS= + = + =( ) ∆ 1 2 1 2.2. and φ fp x x V= =( ) FHG I KJ0 0259 3 10 15 10 0 376 16 10 . ln . . Now Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 187 2 2 11 7 8 85 10 1 6 10 3 10 14 19 16 1 2 ∈ = ( )L NM O QP − −eN x x xa . . . /b g b gb g = 0 2077 1 2. / /µm V We find ∆L = + − +0 2077 0 376 2.2 0 376 1 2. . . . = 0 0726. µm Then ∂ ∂ = ⋅ + ( )∆L VDS 0 2077 2 1 0 376 2.2 . . = 0 0647. /µm V From the previous problem, I mA L mD = =0 48 2. , µ Then g xO = −( ) ( ) − 2 2 0 0726 0 48 10 0 0647 2 3 . . .b g or g x SO = −1 67 10 5. so that r g kO O = = 1 59.8 Ω (b) If L m= 1 µ , then from the previous problem, we would have I mAD = 0 96. , so that g xO = −( ) ( ) − 1 1 0 0726 0 96 10 0 0647 2 3 . . .b g or g x SO = −7.22 10 5 so that r g kO O = = 1 13 8. Ω 12.9 (a) I sat W C L V VD n ox GS T( ) = − µ 2 2a f = −FH I K( ) − 10 2 500 6 9 10 18 2. x VGSb ga f or I sat V mAD GS( ) ( )= −0 173 1 2. a f and I sat V mAD GS( ) ( )= −0173 1 1 2. /a f (b) Let µ µeff O eff C = F HG I KJ − Ε Ε 1 3/ Where µ O cm V s= −1000 2 / and ΕC x V cm= 2.5 10 4 / . Let Ε eff GS ox V t = We find C t t C x xox ox ox ox ox ox = ∈ ⇒ = ∈ = ( ) − − 3 9 8 85 10 6 9 10 14 8 . . . b g or t Aox = °500 Then VGS Ε eff µ eff I satD( ) 1 2 3 4 5 -- 4E5 6E5 8E5 10E5 -- 397 347 315 292 0 0.370 0.692 0.989 1.27 (c) The slope of the variable mobility curve is not constant, but is continually decreasing. 12.10 Plot 12.11 V V Q CT FB SD ox fp= + ′ + ( )max 2φ We find φ fp t a i V N n x x = = F HG I KJ ( ) F HG I KJln . ln .0 0259 5 10 15 10 16 10 or φ fp V= 0 389. and x eNdT fp a = ∈L NM O QP 4 1 2φ / = ( ) ( )L NM O QP − − 4 11 7 8 85 10 0 389 1 6 10 5 10 14 19 16 1 2 . . . . / x x x b g b gb g or x mdT = 0 142. µ Now Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 188 ′ =( )Q eN xSD a dTmax = − −16 10 5 10 0142 1019 16 4. .x x xb gb gb g or ′ =( ) −Q x C cmSD max . /114 10 7 2 Also C t x x xox ox ox F cm= ∈ = = ( ) − − − 3 9 8 85 10 400 10 8 63 10 14 8 8 2 . . . / b g Then V x xT = − + + − − ( )112 114 10 8 63 10 2 0 389 7 8 . . . . or V VT = +0 90. (a) I W C L V V V VD n ox GS T DS DS= − − µ 2 2 2a f and V sat V VDS GS T( ) = − We have I xD = FH IKFH IK( ) − 20 2 1 2 400 8 63 10 8.b g × − −2 2V V V VGS T DS DSa f or I V V V V mAD GS T DS DS= − − ( )0173 2 2. a f For V V sat V V VDS DS GS T= = − =( ) 1 , I sat mAD ( ) = 0 173. For V V sat V V VDS DS GS T= = − =( ) 2 , I sat mAD ( ) = 0 692. (b) For V V cm V sDS n≤ = = −1 25 400 2. , /µ µ . The curve for V V VGS T− = 1 is unchanged. For V V VGS T− = 2 and 0 1 25≤ ≤V VDS . , the curve in unchanged. For V VDS ≥ 1 25. , the current is constant at I mAD = − =( )( ) ( )0 173 2 2 1 25 1 25 0 5952. . . . When velocity saturation occurs, V sat VDS ( ) = 1 25. for the case of V V VGS T− = 2 . 12.12 Plot 12.13 (a) Non-saturation region I C W L V V V VD n ox GS T DS DS= − − FH IK 1 2 2 2µ a f We have C t C kox ox ox ox = ∈ ⇒ and W kW L kL⇒ ⇒, also V kV V kVGS GS DS DS⇒ ⇒, So I C k kW kL kV V kV kVD n ox GS T DS DS= − − FH IKFH IK 1 2 2 2µ a f a f Then I kID D⇒ ≈ In the saturation region I C k kW kL kV VD n ox GS T= − FH IKFH IK 1 2 2µ Then I kID D⇒ ≈ (b) P I V kI kV k PD DD D DD= ⇒ ⇒a fa f 2 12.14 I sat WC V V vD ox GS T sat( ) = −a f ⇒ −( )FH IKkW C k kV V vox GS T sata f or I sat kI satD D( ) ( )≈ 12.15 (a) (i) I K V VD n GS T= − = −( )( )a f2 20 1 5 0 8. . or I mAD = 1 764. (ii) I D = − FH IK ( )( ) 0 1 0 6 0 6 5 0 8 2 . . . . or I mAD = 0 807. (b) (i) P = ⇒( )( )1 764 5. P mW= 8 82. (ii) P = ⇒( )( )( )0 807 0 6 5. . P mW= 2.42 Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 189 (c) Current: Ratio = = 0 807 1 764 0 457 . . . Power: Ratio = = 2.42 8 82 0 274 . . 12.16 ∆V eN x C r L x rT a dT ox j dT j = − + − L NM O QP RST UVW 1 2 1 Now φ fp t a i V N n x = = F HG I KJ ( ) F HG I KJln . ln .0 0259 10 15 10 16 10 or φ fp V= 0 347. and x eNdT fp a = ∈L NM O QP 4 1 2φ / = ( ) ( )L NM O QP − − 4 11 7 8 85 10 0 347 1 6 10 10 14 19 16 1 2 . . . . / x x b g b gb g or x mdT = 0 30. µ Also C t x xox ox ox = ∈ = ( ) − − 3 9 8 85 10 450 10 14 8 . .b g = −7.67 10 8 2x F cm/ Then ∆V x x xT = − − − − 1 6 10 10 0 3 10 7.67 10 19 16 4 8 . .b gb gb g × + − ( )L NM O QP RST UVW 0 3 1 1 2 0 3 0 3 1 . . . or ∆V VT = −0 137. 12.17 ∆V eN x C r L x rT a dT ox j dT j = − + − L NM O QP RST UVW 1 2 1 Now φ fp x x V= =( ) FHG I KJ0 0259 3 10 15 10 0 376 16 10 . ln . . and x eNdT fp a = ∈L NM O QP 4 1 2φ / = ( ) ( )L NM O QP − − 4 11 7 8 85 10 0 376 1 6 10 3 10 14 19 16 1 2 . . . . / x x x b g b gb g or x mdT = 0 180. µ Also C t x xox ox ox = ∈ = ( ) − − 3 9 8 85 10 800 10 14 8 . .b g or C x F cmox = −4.31 10 8 2/ Then ∆V x x x xT = − = − − − 0 20 1 6 10 3 10 0 18 10 4.31 10 19 16 4 8 . . .b gb gb g × + − ( )L NM O QP RST UVW 0 6 1 2 0 18 0 6 1 . . .L or = − = −0 20 0 319 . . L which yields L m= 159. µ 12.18 We have ′ = − +( )L L a b and from the geometry (1) a r x r xj dT j dS+ + = +b g b g2 2 2 and (2) b r x r xj dT j dD+ + = +b g b g2 2 2 From (1), a r r x xj j dS dT+ = + −b g b g2 2 2 so that a r x x rj dS dT j= + − −b g2 2 which can be written as a r x r x rj dS j dT j = + − − F HG I KJ F HG I KJ L N MM O Q PP1 1 2 2 or Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 190 a r x r x r x rj dS j dS j dT j = + + − − F HG I KJ F HG I KJ L N MM O Q PP1 2 1 2 2 Define α 2 2 2 2 = −x x r dS dT j We can then write a r x rj dS j = + + − L NM O QP 1 2 12α Similarly from (2), we will have b r x rj dD j = + + − L NM O QP 1 2 12β where β 2 2 2 2 = −x x r dD dT j The average bulk charge in the trapezoid (per unit area) is ′ ⋅ = + ′FH IKQ L eN x L L B a dT 2 or ′ = + ′FH IKQ eN x L L LB a dT 2 We can write L L L L L L L a b + ′ = + ′ = + − +( ) 2 1 2 2 1 2 1 2 which is = − +( ) 1 2 a b L Then ′ = − +( )L NM O QPQ eN x a b LB a dT 1 2 Now ′QB replaces ′ ( )QSD max in the threshold equation. Then ∆V Q C Q CT B ox SD ox = ′ − ′ ( )max = − + − ( )L NM O QP eN x C a b L eN x C a dT ox a dT ox 1 2 or ∆V eN x C a b LT a dT ox = − ⋅ +( ) 2 Then substituting, we obtain ∆V eN x C r L x rT a dT ox j dS j = − ⋅ + + − L NM O QP RST2 1 2 12α + + + − L NM O QP UVW 1 2 12 x r dD j β Note that if x x xdS dD dT= = , then α β= = 0 and the expression for ∆VT reduces to that given in the text. 12.19 We have ′ =L 0 , so Equation (12.25) becomes L L L L L r L x r j dT j + ′ ⇒ = = − + − L NM O QP RST UVW2 2 1 2 1 1 2 1 or r L x r j dT j 1 2 1 1 2 + − = L NM O QP Then Equation (12.26) is ′ = FH IKQ eN xB a dT 1 2 The change in the threshold voltage is ∆V Q C Q CT B ox SD ox = ′ − ′ ( )max or ∆V eN x C eN x CT a dT ox a dT ox = − 1 2a fa f a f or ∆V eN x CT a dT ox = − FH IK 1 2 a f 12.20 Computer plot 12.21 Computer plot 12.22 ∆V eN x C r L x rT a dT ox j dT j = − + − L NM O QP RST UVW 1 2 1 Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 191 ⇒ − + − FH IK FH IK L NM O QP RST UVW e N k kx C k kr kL kx kr a dT ox j dT j a f 1 2 1 or ∆ ∆V k VT T= 12.23 ∆V eN x C x WT a dT ox dT = FH IK ξ We find φ fp x V= =( ) FHG I KJ0 0259 10 15 10 0 347 16 10 . ln . . and x eNdT fp a = ∈L NM O QP 4 1 2φ / = ( ) ( )L NM O QP − − 4 11 7 8 85 10 0 347 1 6 10 10 14 19 16 1 2 . . . . / x x b g b gb g or x mdT = 0 30. µ Also C t x xox ox ox = ∈ = ( ) − − 3 9 8 85 10 450 10 14 8 . .b g or C x F cmox = −7.67 10 8 2/ Then ∆V x x xT = − − − 1 6 10 10 0 3 10 7.67 10 19 16 4 8 . .b gb gb g × − − L NM O QP π 2 0 3 10 2.5 10 4 4 a fb g. x x or ∆V VT = +0 118. 12.24 Additional bulk charge due to the ends: ∆Q eN L x eN Lx xB a dT a dT dT= ⋅ = FH IK 1 2 22 ξa f where ξ = 1 . Then ∆V eN x C WT a dT ox = 2 We find φ fp x x V= =( ) FHG I KJ0 0259 3 10 15 10 0 376 16 10 . ln . . and x eNdT fp a = ∈L NM O QP 4 1 2φ / = ( ) ( )L NM O QP − − 4 11 7 8 85 10 0 376 1 6 10 3 10 14 19 16 1 2 . . . . / x x x b g b gb g or x mdT = 0 180. µ Also C t x xox ox ox = ∈ = ( ) − − 3 9 8 85 10 800 10 14 8 . .b g or C x F cmox = −4.31 10 8 2/ Now, we can write W eN x C V a dT ox T = 2 ∆a f = − − − ( ) 1 6 10 3 10 0 18 10 4.31 10 0 25 19 16 4 2 8 . . . x x x x b gb gb g b g or W m= 1 44. µ 12.25 Computer plot 12.26 ∆V eN x C x WT a dT ox dT = FH IK ξ Assume that ξ is a constant ⇒ ⋅ FH IK FH IK F H I K e N k kx C k kx kW a dT ox dT a f ξ or ∆ ∆V k VT T= 12.27 (a) V x t x xBD ox= = −6 10 6 10 250 106 6 8b g b gb g or Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 192 V VBD = 15 (b) With a safety factor of 3, VBD = ⋅ ⇒ 1 3 15 V VBD = 5 12.28 We want V VG = 20 . With a safety factor of 3, then V VBD = 60 , so that 60 6 106= ⇒x toxb g t Aox = °1000 12.29 Snapback breakdown means αM = 1 , where α = ( ) FH IK−0 18 3 1010 9. log I x D and M V V CE BD m = − F HG I KJ 1 1 Let V V mBD = =15 3, . Now when α α M VCE = = − FH IK 1 1 15 3 we can write this as 1 15 15 1 3 3 − = ⇒ = − F H I K V VCE CEα α Now I D α VCE E-8 E-7 E-6 E-5 E-4 E-3 0.0941 0.274 0.454 0.634 0.814 0.994 14.5 13.5 12.3 10.7 8.6 2.7 12.30 One Debye length is L kT e eND a = ∈L NM O QP a f 1 2/ = ( ) ( )L NM O QP − − 11 7 8 85 10 0 0259 1 6 10 10 14 19 16 1 2 . . . . / x x b g b gb g or L x cmD = −4.09 10 6 Six Debye lengths: 6 4.09 10 0 2466x m− =b g . µ From Example 12.4, we have x mdO = 0 336. µ , which is the zero-biased source-substrate junction width. At near punch-through, we will have x L x LdO D d+ + =6 where xd is the reverse-biased drain-substrate junction width. Now 0 336 0 246 1 2 0 618. . . .+ + = ⇒ =x x md d µ at near punch-through. We have x V V eNd bi DS a = ∈ +L NM O QP 2 1 2a f / or V V x eN bi DS d a+ = ∈ 2 2 = − − −( ) 0 618 10 1 6 10 10 2 11 7 8 85 10 4 2 19 16 14 . . . . x x x b g b gb g b g which yields V V Vbi DS+ = 2.95 From Example 12.4, we have V Vbi = 0 874. , so that V VDS = 2.08 which is the near punch-through voltage. The ideal punch-through voltage was V VDS = 4.9 12.31 V x x Vbi = =( ) L N MM O Q PP0 0259 10 3 10 15 10 0 902 19 16 10 2 . ln . . b gb g b g The zero-biased source-substrate junction width: x V eNdO bi a = ∈L NM O QP 2 1 2/ = ( ) ( )L NM O QP − − 2 11 7 8 85 10 0 902 1 6 10 3 10 14 19 16 1 2 . . . . / x x x b g b gb g or x mdO = 0 197. µ The Debye length is Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 12 Solutions Manual Problem Solutions 193 L kT e eND a = ∈L NM O QP a f 1 2/ = ( ) ( )L NM O QP − − 11 7 8 85 10 0 0259 1 6 10 3 10 14 19 16 1 2 . . . . / x x x b g b gb g or L x cmD = −2.36 10 6 so that 6 6 2.36 10 01426L x mD = = −b g . µ Now x L x LdO D d+ + =6 We have for V VDS = 5 , x V V eNd bi DS a = ∈ +L NM O QP 2 1 2a f / = +( ) ( )L NM O QP − − 2 117 8 85 10 0 902 5 1 6 10 3 10 14 19 16 1 2 . . . . / x x x b g b gb g or x md = 0 505. µ Then L = + +0197 0142 0 505. . . or L m= 0 844. µ 12.32 With a source-to-substrate voltage of 2 volts, x V V eNdO bi SB a = ∈ +L NM O QP 2 1 2a f / = +( ) ( )L NM O QP − − 2 11 7 8 85 10 0 902 2 1 6 10 3 10 14 19 16 1 2 . . . . / x x x b g b gb g or x mdO = 0 354. µ We have 6 0 142L mD = . µ from the previous problem. Now x V V V eNd bi DS SB a = ∈ + +L NM O QP 2 1 2a f / = + +( ) ( )L NM O QP − − 2 117 8 85 10 0 902 5 2 1 6 10 3 10 14 19 16 1 2 . . . . / x x x b g b gb g or x md = 0 584. µ Then L x L xdO D d= + +6 = + +0 354 0142 0 584. . . or L m= 1 08. µ 12.33 (a) φ fp x x V= =( ) FHG I KJ0 0259 2 10 15 10 0 306 15 10 . ln . . an