semisolpr12

# semisolpr12 - Semiconductor Physics and Devices Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 12 Solutions Manual Problem Solutions 185 Chapter 12 Problem Solutions 12.1 (a) I V V D GS t = () L N M O Q P 10 2.1 15 exp For VV GS = 05 ., I D =⇒ ( ) L N M O Q P 10 2.1 0 0259 15 exp . . Ix A D = 9.83 10 12 For GS = 07 A D = 388 10 10 . For GS = 09 A D = 154 10 8 . Then the total current is: II Total D = 10 6 bg For GS = IA Total = 9.83 µ For GS = . , Im A Total = 0388 . For GS = A Total = 154 . (b) Power: PI V Total DD =⋅ Then For GS = PW = 49.2 For GS = . , Pm W = 194 . For GS = W = 77 12.2 We have ∆∆ L eN Vs a t V a fp DS DS = ⋅+ + 2 φ −+ () fp DS a t where fp t a i V N nx == F H G I K J F H G I K J ln . ln . 00259 10 15 10 16 10 or fp V = 0347 . We find 2 2117 88510 16 10 10 14 19 16 12 = L N M O Q P eN x x a .. . / b gb g = 0360 ./ / mV We have a t V V DS GS T =− (a) For V s a t V GS DS = 54 . 2 5 Then L =+ + 0347 5 0347 4.25 . or Lm = 00606 . If L is 10%of L , then = 0606 . (b) For V s a t V DS GS DS ⇒= 52 1 2 5 ,. Then L + 0347 125 . . or = 0377 . Now if L is 10% of L , then = 377 . 12.3 L eN a a fp DS DS = + 2 fp DS a t where fp t a i V N n x x F H G I K J F H G I K J ln . ln . 41 0 16 10 or fp V = 0383 and x eN dT fp a = L N M O Q P 4 / = ( ) L N M O Q P 4117 88510 4 10 14 19 16 . . / x xx b g or xm dT = 0157 . Then = Qe N x SD a dT max

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 12 Solutions Manual Problem Solutions 186 = −− 16 10 4 10 0157 10 19 16 4 .. xx x bg b g b g or = () QC c m SD max / 10 72 Now VQ Q t TS D S S ox ox ms fp = ++ F H G I K J max af φφ 2 so that V x x T = 10 3 10 400 10 39 885 10 71 9 1 0 8 14 . b g b g 02 0 3 8 3 . or VV T = 187 Now Vs a t V V V DS GS T =− = = 51 8 7 3 1 3 We find 2 2117 88510 4 10 14 19 16 12 = L N M O Q P eN x a . / b gb g = 180 10 5 . x Now ∆∆ Lx V DS =⋅ + + 180 10 0 383 313 5 . −+ 0383 313 or V DS =+ 180 10 3513 5 . We obtain V DS Lm µ 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 a t V DS = 1 , We have Ε sat DS a t L = We find Ε sat Vc m / 3 1 05 025 013 333 10 3 . x 10 4 21 0 4 x 41 0 4 x 7.69 10 4 x (b) Assume n cm V s 500 2 / , we have v ns a t = Ε Then For = 3 , vx c m s = 167 10 6 ./ For = 1 , vxc m s = 510 6 / For . , vc m s 10 7 / 12.8 We have IL LL I DD 1 We may write g I V L I L V O D DS D DS = ∂ ′ −∂ ()( ) F H G I K J 1 2 = ⋅⋅ L I L V D DS 2 We have L eN s a t a fp DS fp DS = ⋅+ + 2 We find = + L Ve N V DS a fp DS 2 φ (a) For GS DS == , , and a V DS GS T = −= 20 81 2 Also s a t V V DS DS DS = + = 12 1 2.2 and fp x x V F H G I K J 00259 310 15 10 0376 16 10 .l n . Now
Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 12 Solutions Manual Problem Solutions 187 2 2117 88510 16 10 3 10 14 19 16 12 = () L N M O Q P eN x xx a .. . / bg b g = 02077 ./ / µ mV We find L =+ + 0376 2.2 0376 12 . . = 00726 . m Then =⋅ + L V DS 2 1 . . = 00647 From the previous problem, Im A L m D == 048 2 ., Then gx O = 2 2 0 0726 0 48 10 0 0647 2 3 . or S O = 167 10 5 . so that r g k O O 1 59.8 (b) If Lm = 1 , then from the previous problem, we would have A D = 096 . , so that O = 1 1 0 0726 0 96 10 0 0647 2 3 .

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semisolpr12 - Semiconductor Physics and Devices Basic...

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