3rd edition neamen solutions

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3rd edition neamen solutions

  1. 1. Ramey Soft Innovation with Excellence. Date:- ______________ There’s more to a product than just code. We understand there is more that goes into a project than coding. Your software design, strategy, tools, and processes all have an impact on how successful your custom software project or product will be. Our software consulting services are broken into three offers to assist with those key areas of success. Exceed client expectations by going beyond software to provide solutions that transform data into knowledge, enabling them to solve problems and better serve our clients. Also counseling in graduate and post graduate projects. All kinds of software source codes available. Our expertise’s are: Web Services Data Base Design Web Development Managements Systems (Hospitals, Warehouses, Commercial businesses, Super Stores and so on) Desktop Application Development Android Application Development Java, Asp.net, C/C++, PHP, C#, SQL, MySQL, Graphics, Logo Design Skype: rameysoft +92-311-1379796 +92-334-7595788 Contact Details : Skype: rameysoft Email: rameysoft@gmail.com Contact No. +92-311-1379796 +92-334-7595788
  2. 2. Chapter 1 Exercise Problems EX1.1 ⎛ − Eg ⎞ ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2kT ⎠ GaAs: ni = ( 2.1× 1014 ) ( 300 ) Ge: ni = (1.66 × 1013 ) ( 300 ) 3/ 2 3/ 2 ⎛ ⎞ −1.4 ⎟ or ni = 1.8 × 106 cm −3 exp ⎜ ⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟ ⎝ ⎠ ⎛ ⎞ −0.66 ⎟ or ni = 2.40 × 1013 cm −3 exp ⎜ ⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟ ⎝ ⎠ EX1.2 (a) majority carrier: holes, po = 1017 cm −3 minority carrier: electrons, n 2 (1.5 × 10 no = i = 1017 po ) 10 2 = 2.25 × 103 cm −3 (b) majority carrier: electrons, no = 5 × 1015 cm −3 minority carrier: holes, n 2 (1.5 × 10 ) = 4.5 × 104 cm −3 po = i = 5 × 1015 no 10 2 EX1.3 For n-type, drift current density J ≅ eμn nE or 200 = (1.6 × 10−19 ) ( 7000 ) (1016 ) E which yields E = 17.9 V / cm EX1.4 Diffusion current density due to holes: dp J p = −eD p dx ⎛ −1 ⎞ ⎛ −x ⎞ = −eD p (1016 ) ⎜ ⎟ exp ⎜ ⎟ ⎜L ⎟ ⎜L ⎟ ⎝ p⎠ ⎝ p⎠ (a) At x = 0 (1.6 ×10 ) (10 ) (10 ) = 16 A / cm = −19 Jp 16 2 10−3 −3 (b) At x = 10 cm ⎛ −10−3 ⎞ J p = 16 exp ⎜ −3 ⎟ = 5.89 A / cm 2 ⎝ 10 ⎠ EX1.5 ⎡N N Vbi = VT ln ⎢ a 2 d ⎣ ni ⎡ (1016 )(1017 ) ⎤ ⎤ ⎥ or Vbi = 1.23 V = ( 0.026 ) ln ⎢ ⎥ ⎢ (1.8 × 106 )2 ⎥ ⎦ ⎣ ⎦ EX1.6 ⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠ and −1/ 2
  3. 3. ⎡N N ⎤ Vbi = VT ln ⎢ a 2 d ⎥ ⎣ ni ⎦ ⎡ (1017 )(1016 ) ⎤ ⎥ = 0.757 V = ( 0.026 ) ln ⎢ ⎢ (1.5 × 1010 )2 ⎥ ⎣ ⎦ 5 ⎞ ⎛ Then 0.8 = C jo ⎜ 1 + ⎟ ⎝ 0.757 ⎠ or C jo = 2.21 pF −1/ 2 = C jo ( 7.61) −1/ 2 EX1.7 ⎡ ⎛v ⎞ ⎤ iD = I S ⎢exp ⎜ D ⎟ − 1⎥ ⎢ ⎝ VT ⎠ ⎥ ⎣ ⎦ ⎡ ⎛ v ⎞ ⎤ so 10−3 = (10−13 ) ⎢ exp ⎜ D ⎟ − 1⎥ ⎝ 0.026 ⎠ ⎦ ⎣ ⎡ 10−3 ⎤ Solving for the diode voltage, we find vD = ( 0.026 ) ln ⎢ −13 + 1⎥ ⎣10 ⎦ or vD ≅ ( 0.026 ) ln (1010 ) which yields vD = 0.599 V EX1.8 ⎛V ⎞ VPS = I D R + VD and I D ≅ I S exp ⎜ D ⎟ ⎝ VT ⎠ ( 4 − VD ) so 4 = I D ( 4 ×103 ) + VD ⇒ I D = 4 ×103 and ⎛ V ⎞ I D = (10 −12 ) exp ⎜ D ⎟ ⎝ 0.026 ⎠ By trial and error, we find I D ≅ 0.864 mA and VD ≅ 0.535 V EX1.9 (a) ID = (b) ID = Then R = (c) VPS − Vγ R VPS − Vγ R 5 − 0.7 ⇒ I D = 1.08 mA 4 VPS − Vγ ⇒R= ID = 8 − 0.7 = 6.79 kΩ 1.075
  4. 4. ID(mA) Diode curve 1.25 1.08 Load lines (b) (a) 0 0.7 2 4 VD(v) 6 8 EX1.10 PSpice analysis EX1.11 Quiescent diode current I DQ = VPS − Vγ = 10 − 0.7 = 0.465 mA 20 R Time-varying diode current: V 0.026 We find that rd = T = = 0.0559 kΩ I DQ 0.465 Then id = vI 0.2sin ω t (V ) = ⋅ or id = 9.97sin ω t ( μ A) rd + R 0.0559 + 20 ( kΩ ) EX1.12 ⎛I ⎞ ⎛ 1.2 × 10−3 ⎞ or VD = 0.6871 V For the pn junction diode, VD ≅ VT ln ⎜ D ⎟ = ( 0.026 ) ln ⎜ −15 ⎟ ⎝ 4 × 10 ⎠ ⎝ IS ⎠ The Schottky diode voltage will be smaller, so VD = 0.6871 − 0.265 = 0.4221 V ⎛V ⎞ Now I D ≅ I S exp ⎜ D ⎟ ⎝ VT ⎠ or 1.2 × 10−3 IS = ⇒ I S = 1.07 × 10−10 A 0.4221 ⎞ ⎛ exp ⎜ ⎟ ⎝ 0.026 ⎠ EX1.13 P = I ⋅ VZ ⇒ 10 = I ( 5.6 ) ⇒ I = 1.79 mA Also I = 10 − 5.6 = 1.79 ⇒ R = 2.46 kΩ R Test Your Understanding Exercises TYU1.1 (a) T = 400K ⎛ − Eg ⎞ Si: ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2kT ⎠ ni = ( 5.23 × 1015 ) ( 400 ) or ni = 4.76 × 1012 cm −3 3/ 2 ⎡ ⎤ −1.1 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) ( 400 ) ⎥ ⎣ ⎦
  5. 5. Ge: ni = (1.66 × 1015 ) ( 400 ) 3/ 2 ⎡ ⎤ −0.66 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) ( 400 ) ⎥ ⎣ ⎦ or ni = 9.06 × 1014 cm −3 GaAs: ni = ( 2.1× 1014 ) ( 400 ) 3/ 2 ⎡ ⎤ −1.4 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) ( 400 ) ⎥ ⎣ ⎦ or ni = 2.44 × 109 cm −3 (b) T = 250 K Si: ni = ( 5.23 × 1015 ) ( 250 ) 3/ 2 ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢ 2 ( 86 × 10−6 ) ( 250 ) ⎥ ⎣ ⎦ or ni = 1.61× 108 cm −3 Ge: ni = (1.66 × 1015 ) ( 250 ) 3/ 2 ⎡ ⎤ −0.66 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) ( 250 ) ⎥ ⎣ ⎦ or ni = 1.42 × 1012 cm −3 GaAs: ni = ( 2.10 × 1014 ) ( 250 ) 3/ 2 ⎡ ⎤ −1.4 ⎥ exp ⎢ ⎢ 2 ( 86 × 10−6 ) ( 250 ) ⎥ ⎣ ⎦ or ni = 6.02 × 103 cm −3 TYU1.2 (a) n = 5 × 1016 cm −3 , p <<< n, so σ ≅ eμ n n = (1.6 × 10 −19 ) (1350 ) ( 5 × 1016 ) or σ = 10.8 ( Ω − cm ) (b) −1 p = 5 × 1016 cm −3 , n <<< p, so σ ≅ eμ p p = (1.6 × 10−19 ) ( 480 ) ( 5 × 1016 ) or σ = 3.84 ( Ω − cm ) −1 TYU1.3 J = σ E = (10 )(15 ) or J = 150 A / cm2 TYU1.4 (a) J n = eDn ⎛ 1015 − 1016 ⎞ dn Δn so J n = 1.6 × 10−19 ( 35 ) ⎜ = eDn −4 ⎟ dx Δx ⎝ 0 − 2.5 × 10 ⎠ ( ) or J n = 202 A / cm 2 (b) J p = −eD p or J p = −24.5 A / cm2 TYU1.5 ⎛ 1014 − 5 × 1015 ⎞ dp Δp so J p = − 1.6 × 10−19 (12.5 ) ⎜ = −eD p −4 ⎟ dx Δx ⎝ 0 − 4 × 10 ⎠ ( )
  6. 6. no = N d = 8 × 1015 cm −3 (a) 10 n 2 (1.5 × 10 ) po = i = = 2.81× 10 4 cm −3 no 8 × 1015 2 (b) n = no + δ n = 8 × 1015 + 0.1× 1015 or n = 8.1×1015 cm−3 p = po + δ p = 2.81 × 10 4 + 1014 or p ≅ 1014 cm −3 TYU1.6 (a) ⎡ (1015 )(1017 ) ⎤ ⎡N N ⎤ ⎥ = 0.697 V Vbi = VT ln ⎢ a 2 d ⎥ so Vbi = ( 0.026 ) ln ⎢ ⎢ (1.5 × 1010 )2 ⎥ ⎣ ni ⎦ ⎣ ⎦ (b) ⎡ (1017 )(1017 ) ⎤ ⎥ = 0.817 V Vbi = ( 0.026 ) ln ⎢ ⎢ (1.5 × 1010 )2 ⎥ ⎣ ⎦ TYU1.7 ⎡ ⎛V I D = I S ⎢exp ⎜ D ⎢ ⎝ VT ⎣ (a) ⎞ ⎤ ⎟ − 1⎥ ⎠ ⎥ ⎦ ⎛ 0.5 ⎞ I D ≅ 10−14 exp ⎜ ⎟ ⎝ 0.026 ⎠ Then, for VD = 0.5 V, I D = 2.25 μ A VD = 0.6 V, I D = 0.105 mA VD = 0.7 V, I D = 4.93 mA (b) I D ≅ − I S = −10 −14 A for both cases. TYU1.8 ΔT = 100C so ΔVD ≅ 2 × 100 = 200 mV Then VD = 0.650 − 0.20 = 0.450 V TYU1.9 ID(mA) VD Diode 1.0 ഠ0.87 Load line 0 1 ഠ0.54v 2 VD(v) ID 0.45 0.50 0.55 0.033 0.225 1.54 3 TYU1.10 P = I DVD ⇒ 1.05 = I D ( 0.7 ) so I D = 1.5 mA 4
  7. 7. Now R = VPS − Vγ ID = 10 − 0.7 ⇒ R = 6.2 kΩ 1.5 TYU1.11 I 0.8 gd = D = = 30.8 mS VT 0.026 TYU1.12 V 0.026 0.026 rd = T ⇒ 50 = ⇒ ID = ID ID 50 or I D = 0.52 mA TYU1.13 For the pn junction diode, 4 − 0.7 ID = = 0.825 mA 4 For the Schottky diode, I D = 4 − 0.3 = 0.925 mA 4 TYU1.14 Vz = Vzo + I z rz ⇒ Vzo = Vz − I z rz so Vzo = 5.20 − (10 −3 ) ( 20 ) = 5.18 V Then Vz = 5.18 + (10 × 10−3 ) ( 20 ) ⇒ Vz = 5.38 V
  8. 8. Chapter 1 Problem Solutions 1.1 − E / 2 kT ni = BT 3 / 2 e g (a) Silicon ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢ 2 ( 86 × 10−6 ) ( 250 ) ⎥ ⎣ ⎦ = 2.067 × 1019 exp [ −25.58] ni = 1.61× 108 cm −3 (i) ni = ( 5.23 × 1015 ) ( 250 ) (ii) ni = ( 5.23 × 1015 ) ( 350 ) (b) GaAs (i) ni = ( 2.10 × 1014 ) ( 250 ) 3/ 2 ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢ 2 ( 86 × 10−6 ) ( 350 ) ⎥ ⎣ ⎦ = 3.425 × 1019 exp [ −18.27 ] ni = 3.97 ×1011 cm −3 3/ 2 3/ 2 ⎡ ⎤ −1.4 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) ( 250 ) ⎥ ⎣ ⎦ = ( 8.301× 1017 ) exp [ −32.56] ni = 6.02 × 103 cm −3 (ii) ni = ( 2.10 × 1014 ) ( 350 ) 3/ 2 ⎡ ⎤ −1.4 ⎥ exp ⎢ ⎢ 2 ( 86 × 10−6 ) ( 350 ) ⎥ ⎣ ⎦ = (1.375 × 1018 ) exp [ −23.26] ni = 1.09 × 108 cm −3 1.2 ⎛ − Eg ⎞ ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2kT ⎠ ⎛ ⎞ −1.1 1012 = 5.23 × 1015 T 3 / 2 exp ⎜ ⎟ 2(86 × 10−6 )(T ) ⎠ ⎝ a. ⎛ 6.40 × 103 ⎞ 1.91× 10−4 = T 3 / 2 exp ⎜ − ⎟ T ⎝ ⎠ By trial and error, T ≈ 368 K b. ni = 109 cm −3 ⎛ ⎞ −1.1 ⎟ 109 = 5.23 × 1015 T 3 / 2 exp ⎜ ⎜ 2 ( 86 × 10−6 ) (T ) ⎟ ⎝ ⎠ ⎛ 6.40 × 103 ⎞ 1.91× 10−7 = T 3 / 2 exp ⎜ − ⎟ T ⎝ ⎠ By trial and error, T ≈ 268° K 1.3 Silicon
  9. 9. ni = ( 5.23 × 1015 ) (100 ) (a) 3/ 2 ⎡ ⎤ −1.1 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) (100 ) ⎥ ⎣ ⎦ = ( 5.23 × 1018 ) exp [ −63.95] ni = 8.79 ×10−10 cm −3 ni = ( 5.23 × 1015 ) ( 300 ) (b) 3/ 2 ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢ 2 ( 86 × 10−6 ) ( 300 ) ⎥ ⎣ ⎦ = ( 2.718 × 1019 ) exp [ −21.32] ni = 1.5 × 1010 cm −3 ni = ( 5.23 × 1015 ) ( 500 ) (c) 3/ 2 ⎡ ⎤ −1.1 ⎥ exp ⎢ −6 ⎢ 2 ( 86 × 10 ) ( 500 ) ⎥ ⎣ ⎦ = ( 5.847 × 1019 ) exp [ −12.79] ni = 1.63 × 1014 cm −3 Germanium. ni = (1.66 × 1015 ) (100 ) (a) 3/ 2 ⎡ ⎤ −0.66 ⎥ = (1.66 × 1018 ) exp [ −38.37 ] exp ⎢ ⎢ 2 ( 86 × 10−6 ) (100 ) ⎥ ⎣ ⎦ ni = 35.9 cm −3 ni = (1.66 × 1015 ) ( 300 ) (b) 3/ 2 ⎡ ⎤ −0.66 ⎥ = ( 8.626 × 1018 ) exp [ −12.79] exp ⎢ ⎢ 2 ( 86 × 10−6 ) ( 300 ) ⎥ ⎣ ⎦ ni = 2.40 × 1013 cm −3 ni = (1.66 × 1015 ) ( 500 ) (c) 3/ 2 ⎡ ⎤ −0.66 ⎥ = (1.856 × 1019 ) exp [ −7.674] exp ⎢ ⎢ 2 ( 86 × 10−6 ) ( 500 ) ⎥ ⎣ ⎦ ni = 8.62 ×1015 cm −3 1.4 a. N d = 5 × 1015 cm −3 ⇒ n − type n0 = N d = 5 × 1015 cm −3 n 2 (1.5 × 10 ) ⇒ p0 = 4.5 × 10 4 cm −3 p0 = i = 5 × 1015 n0 10 2 N d = 5 × 1015 cm −3 ⇒ n − type b. no = N d = 5 × 1015 cm −3 ni = ( 2.10 × 1014 ) ( 300 ) = ( 2.10 × 1014 ) ( 300 ) 3/ 2 3/ 2 ⎛ ⎞ −1.4 exp ⎜ ⎟ −6 ⎝ 2(86 × 10 )(300) ⎠ (1.65 ×10 ) −12 = 1.80 × 106 cm −3 6 ni2 (1.8 × 10 ) = ⇒ p0 = 6.48 × 10 −4 cm −3 p0 = 15 5 × 10 n0 2 1.5
  10. 10. (a) (b) n-type no = N d = 5 × 1016 cm −3 10 n 2 (1.5 × 10 ) po = i = = 4.5 × 103 cm −3 no 5 × 1016 2 no = N d = 5 × 1016 cm −3 (c) From Problem 1.1(a)(ii) ni = 3.97 × 1011 cm −3 ( 3.97 × 10 ) = 11 2 po = 3.15 × 106 cm −3 5 × 1016 1.6 a. N a = 1016 cm −3 ⇒ p − type p0 = N a = 1016 cm −3 n 2 (1.5 × 10 ) ⇒ n0 = 2.25 × 10 4 cm −3 n0 = i = 1016 p0 b. Germanium N a = 1016 cm −3 ⇒ p − type 10 2 p0 = N a = 1016 cm −3 ni = (1.66 × 1015 ) ( 300 ) = (1.66 × 1015 ) ( 300 ) 3/ 2 3/ 2 ⎛ ⎞ −0.66 ⎟ exp ⎜ ⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟ ⎝ ⎠ ( 2.79 × 10 ) −6 = 2.4 × 1013 cm −3 13 n 2 ( 2.4 × 10 ) n0 = i = ⇒ n0 = 5.76 × 1010 cm −3 p0 1016 2 1.7 (a) (b) p-type po = N a = 2 × 1017 cm −3 10 ni2 (1.5 × 10 ) no = = = 1.125 × 103 cm −3 po 2 × 1017 2 (c) po = 2 × 1017 cm −3 From Problem 1.1(a)(i) ni = 1.61 × 108 cm −3 (1.61×10 ) = 8 2 no 1.8 (a) 2 × 10 17 = 0.130 cm −3 no = 5 × 1015 cm −3 10 ni2 (1.5 × 10 ) po = = ⇒ po = 4.5 × 104 cm −3 no 5 × 1015 2 po ⇒ n-type (b) no (c) no ≅ N d = 5 × 1015 cm −3 1.9 Add Donors a. N d = 7 × 1015 cm −3
  11. 11. Want po = 106 cm −3 = ni2 / N d b. So ni2 = (106 )( 7 × 1015 ) = 7 × 10 21 ⎛ − Eg ⎞ = B 2T 3 exp ⎜ ⎟ ⎝ kT ⎠ ⎛ ⎞ 2 −1.1 ⎟ 7 × 1021 = ( 5.23 × 1015 ) T 3 exp ⎜ −6 ⎜ ( 86 × 10 ) (T ) ⎟ ⎝ ⎠ By trial and error, T ≈ 324° K 1.10 I = J ⋅ A = σ EA I = ( 2.2 )(15 ) (10−4 ) ⇒ I = 3.3 mA 1.11 J 85 = E 12 σ = 7.08 (ohm − cm) −1 J =σE ⇒σ = 1.12 g≈ 1 1 1 ⇒ Na = = eμ p N a eμ p g (1.6 × 10 −19 ) ( 480 )( 0.80 ) N a = 1.63 × 10 16 cm −3 1.13 σ = eμ n N d Nd = σ eμ n = ( 0.5) (1.6 ×10 ) (1350 ) −19 N d = 2.31× 1015 cm −3 1.14 (a) For n-type, σ ≅ eμ n N d = (1.6 × 10 −19 ) ( 8500 ) N d For 1015 ≤ N d ≤ 1019 cm −3 ⇒ 1.36 ≤ σ ≤ 1.36 × 104 ( Ω − cm ) (b) J = σ E = σ ( 0.1) ⇒ 0.136 ≤ J ≤ 1.36 × 103 A / cm2 1.15 J n = eDn dn Δn = eDn dx Δx ⎡10 15 −10 2 ⎤ = (1.6 × 10 −19 ) (180 ) ⎢ −4 ⎥ ⎣ 0.5 × 10 ⎦ J n = 576 A/cm 2 1.16 −1
  12. 12. J p = −eD p dp dx ⎛ −1 ⎞ ⎛ −x ⎞ = −eD p (10 15 ) ⎜ ⎟ exp ⎜ ⎟ ⎜ Lp ⎟ ⎜ Lp ⎟ ⎝ ⎠ ⎝ ⎠ (1.6 ×10 ) (15) (10 ) exp ⎛ − x ⎞ −19 Jp = 15 ⎜ ⎟ ⎜L ⎟ ⎝ p⎠ 10 × 10 −4 J p = 2.4 e − x / Lp J p = 2.4 A/cm2 (a) x=0 (b) x = 10 μ m J p = 2.4 e−1 = 0.883 A/cm 2 (c) x = 30 μ m J p = 2.4 e−3 = 0.119 A/cm 2 1.17 a. N a = 1017 cm −3 ⇒ po = 1017 cm −3 6 ni2 (1.8 × 10 ) no = = ⇒ no = 3.24 × 10−5 cm −3 po 1017 2 b. n = no + δ n = 3.24 × 10 −5 + 1015 ⇒ n = 1015 cm −3 p = po + δ p = 1017 + 1015 ⇒ p = 1.01× 1017 cm −3 1.18 (a) ⎛N N ⎞ Vbi = VT ln ⎜ a 2 d ⎟ ⎝ ni ⎠ ⎡ (10 16 )(10 16 ) ⎤ ⎥ = 0.697 V = ( 0.026 ) ln ⎢ ⎢ (1.5 × 10 10 )2 ⎥ ⎣ ⎦ (b) ⎡ (10 18 )(10 16 ) ⎤ ⎥ = 0.817 V Vbi = ( 0.026 ) ln ⎢ ⎢ (1.5 × 10 10 )2 ⎥ ⎣ ⎦ (c) ⎡ (10 18 )(10 18 ) ⎤ ⎥ = 0.937 V Vbi = ( 0.026 ) ln ⎢ ⎢ (1.5 × 10 10 )2 ⎥ ⎣ ⎦ 1.19 ⎛N N ⎞ Vbi = VT ln ⎜ a 2 d ⎟ ⎝ ni ⎠ a. ⎡ (1016 )(1016 ) ⎤ ⎥ ⇒ Vbi = 1.17 V Vbi = ( 0.026 ) ln ⎢ 6 2 ⎢ (1.8 × 10 ) ⎥ ⎣ ⎦ b. ⎡ (1018 )(1016 ) ⎤ ⎥ ⇒ Vbi = 1.29 V Vbi = ( 0.026 ) ln ⎢ 6 2 ⎢ (1.8 × 10 ) ⎥ ⎣ ⎦ c. ⎡ (1018 )(1018 ) ⎤ ⎥ ⇒ Vbi = 1.41 V Vbi = ( 0.026 ) ln ⎢ 6 2 ⎢ (1.8 × 10 ) ⎥ ⎣ ⎦ 1.20 ⎡ N a (1016 ) ⎤ ⎛ Na Nd ⎞ ⎥ Vbi = VT ln ⎜ 2 ⎟ = ( 0.026 ) ln ⎢ 10 2 ⎢ (1.5 × 10 ) ⎥ ⎝ ni ⎠ ⎣ ⎦
  13. 13. For N a = 1015 cm −3 , Vbi = 0.637 V For N a = 1018 cm −3 , Vbi = 0.817 V Vbi (V) 0.817 0.637 1015 1016 1018 Na(cmϪ3) 1017 1.21 ⎛ T ⎞ kT = (0.026) ⎜ ⎟ ⎝ 300 ⎠ T kT (T)3/2 200 0.01733 2828.4 250 0.02167 3952.8 300 0.026 5196.2 350 0.03033 6547.9 400 0.03467 8000.0 450 0.0390 9545.9 500 0.04333 11,180.3 ⎛ ⎞ −1.4 ⎟ ni = ( 2.1× 1014 )(T 3 / 2 ) exp ⎜ −6 ⎜ 2 ( 86 × 10 ) (T ) ⎟ ⎝ ⎠ ⎛N N ⎞ Vbi = VT ln ⎜ a 2 d ⎟ ⎝ ni ⎠ T 200 250 300 350 400 450 500 ni 1.256 6.02 × 103 1.80 × 106 1.09 × 108 2.44 × 109 2.80 × 1010 2.00 × 1011 Vbi 1.405 1.389 1.370 1.349 1.327 1.302 1.277 Vbi (V) 1.45 1.35 1.25 200 250 300 1.22 ⎛ V ⎞ C j = C jo ⎜ 1 + R ⎟ ⎝ Vbi ⎠ −1/ 2 350 400 450 500 T(ЊC)
  14. 14. ⎡ (1.5 × 10 16 )( 4 × 10 15 ) ⎤ ⎥ = 0.684 V Vbi = ( 0.026 ) ln ⎢ ⎢ (1.5 ×10 10 ) 2 ⎥ ⎣ ⎦ 1 ⎞ ⎛ C j = ( 0.4 ) ⎜ 1 + ⎟ ⎝ 0.684 ⎠ −1/ 2 (a) 3 ⎞ ⎛ C j = ( 0.4 ) ⎜ 1 + ⎟ ⎝ 0.684 ⎠ −1/ 2 (b) 5 ⎞ ⎛ C j = ( 0.4 ) ⎜ 1 + ⎟ 0.684 ⎠ ⎝ −1/ 2 (c) = 0.255 pF = 0.172 pF = 0.139 pF 1.23 (a) ⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠ −1 / 2 5 ⎞ ⎛ For VR = 5 V, C j = (0.02) ⎜ 1 + ⎟ ⎝ 0. 8 ⎠ −1 / 2 = 0.00743 pF −1 / 2 ⎛ 1. 5 ⎞ For VR = 1.5 V, C j = (0.02) ⎜1 + = 0.0118 pF ⎟ ⎝ 0. 8 ⎠ 0.00743 + 0.0118 C j (avg ) = = 0.00962 pF 2 vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ where τ = RC = RC j (avg ) = (47 × 103 )(0.00962 × 10−12 ) or τ = 4.52 ×10−10 s Then vC ( t ) = 1.5 = 0 + ( 5 − 0 ) e − ti / τ 5 + r /τ ⎛ 5 ⎞ = e 1 ⇒ t1 = τ ln ⎜ ⎟ 1.5 ⎝ 1.5 ⎠ −10 t1 = 5.44 × 10 s For VR = 0 V, Cj = Cjo = 0.02 pF −1/ 2 ⎛ 3.5 ⎞ For VR = 3.5 V, C j = ( 0.02 ) ⎜ 1 + = 0.00863 pF ⎟ ⎝ 0.8 ⎠ 0.02 + 0.00863 C j (avg ) = = 0.0143 pF 2 τ = RC j ( avg ) = 6.72 ×10−10 s (b) vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ ( 3.5 = 5 + (0 − 5)e − t2 /τ = 5 1 − e − t2 /τ so that t2 = 8.09 × 10 −10 ) s 1.24 ⎡ (1018 )(1015 ) ⎤ ⎥ = 0.757 V Vbi = ( 0.026 ) ln ⎢ ⎢ (1.5 × 1010 )2 ⎥ ⎣ ⎦ a. VR = 1 V 1 ⎞ ⎛ C j = (0.25) ⎜ 1 + ⎟ ⎝ 0.757 ⎠ −1/ 2 = 0.164 pF
  15. 15. f0 = 1 2π LC = 1 ( 2.2 ×10 )( 0.164 ×10 ) 2π −3 −12 f 0 = 8.38 MHz b. VR = 10 V −1/ 2 10 ⎞ ⎛ = 0.0663 pF C j = (0.25) ⎜ 1 + ⎟ ⎝ 0.757 ⎠ 1 f0 = −3 2π ( 2.2 × 10 )( 0.0663 × 10−12 ) f 0 = 13.2 MHz 1.25 ⎡ ⎛V I = I S ⎢ exp ⎜ D ⎝ VT ⎣ a. ⎞ ⎤ ⎛ VD ⎟ − 1⎥ − 0.90 = exp ⎜ ⎠ ⎦ ⎝ VT ⎛V ⎞ exp ⎜ D ⎟ = 1 − 0.90 = 0.10 ⎝ VT ⎠ VD = VT ln ( 0.10 ) ⇒ VD = −0.0599 V b. IF IR ⎡ ⎛ VF ⎢ exp ⎜ I ⎝ VT = S ⋅⎣ IS ⎡ ⎛ VR ⎢exp ⎜ ⎝ VT ⎣ = ⎞ ⎤ ⎛ 0.2 ⎞ ⎟ − 1⎥ exp ⎜ ⎟ −1 ⎠ ⎦ ⎝ 0.026 ⎠ = ⎛ −0.2 ⎞ ⎞ ⎤ ⎟ − 1⎥ exp ⎜ 0.026 ⎟ − 1 ⎝ ⎠ ⎠ ⎦ 2190 −1 IF = 2190 IR 1.26 a. ⎛ 0.5 ⎞ I ≅ (10−11 ) exp ⎜ ⎟ ⇒ I = 2.25 mA ⎝ 0.026 ⎠ ⎛ 0.6 ⎞ I = (10−11 ) exp ⎜ ⎟ ⇒ I = 0.105 A ⎝ 0.026 ⎠ ⎛ 0.7 ⎞ I = (10−11 ) exp ⎜ ⎟ ⇒ I = 4.93 A ⎝ 0.026 ⎠ b. ⎛ 0.5 ⎞ I ≅ (10−13 ) exp ⎜ ⎟ ⇒ I = 22.5 μ A ⎝ 0.026 ⎠ ⎛ 0.6 ⎞ I = (10−13 ) exp ⎜ ⎟ ⇒ I = 1.05 mA ⎝ 0.026 ⎠ ⎛ 0.7 ⎞ I = (10−13 ) exp ⎜ ⎟ ⇒ I = 49.3 mA ⎝ 0.026 ⎠ 1.27 (a) 150 × 10 ( ) I = I S eVD / VT − 1 −6 = 10 −11 (e VD / VT ) − 1 ≅ 10−11 eVD / VT ⎞ ⎟ −1 ⎠
  16. 16. ⎛ 150 × 10−6 Then VD = VT ln ⎜ −11 ⎝ 10 Or VD = 0.430 V ⎞ ⎛ 150 × 10−6 ⎞ ⎟ = (0.026) ln ⎜ ⎟ −11 ⎠ ⎝ 10 ⎠ (b) ⎛ 150 × 10−6 ⎞ VD = VT ln ⎜ ⎟ −13 ⎝ 10 ⎠ Or VD = 0.549 V 1.28 ⎛ 0.7 ⎞ 10−3 = I S exp ⎜ ⎟ ⎝ 0.026 ⎠ I S = 2.03 × 10 −15 A (b) VD I D ( A ) ( n = 1) I D ( A )( n = 2 ) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 9.50 ×10 −14 4.45 ×10 −12 2.08 ×10 −10 9.75 ×10 −9 4.56 ×10 −7 2.14 ×10 −5 10 −3 1.39 ×10 −14 9.50 ×10 −14 6.50 ×10 −13 4.45 ×10 −12 3.04 ×10 −11 2.08 ×10 −10 1.42 ×10 −9 ID(A) 4.68 ×10−11 2.19 ×10−9 1.03 ×10−7 4.80 ×10−6 2.25 ×10−4 1.05 ×10−2 4.93 ×10−1 log10ID −10.3 −8.66 −6.99 −5.32 −3.65 −1.98 −0.307 ID(A) 4.68 ×10−13 2.19 ×10−11 1.03 ×10−9 4.80 ×10−8 2.25 ×10−6 1.05 ×10−4 4.93 ×10−3 log10ID −12.3 −10.66 −8.99 −7.32 −5.65 −3.98 −2.31 (a) 1.29 (a) I S = 10 −12 A VD(v) 0.10 0.20 0.30 0.40 0.50 0.60 0.70 (b) I S = 10 −14 A VD(v) 0.10 0.20 0.30 0.40 0.50 0.60 0.70 1.30 a. ⎛ V − VD1 ⎞ ID2 = 10 = exp ⎜ D 2 ⎟ I D1 ⎝ VT ⎠ ΔVD = VT ln (10) ⇒ ΔVD = 59.9 mV ≈ 60 mV
  17. 17. b. ΔVD = VT ln (100 ) ⇒ ΔVD = 119.7 mV ≈ 120 mV 1.31 (a) (i) (ii) ⎛I VD = Vt ln ⎜ D ⎝ IS VD = 0.669 V ⎞ ⎛ 150 × 10−6 ⎞ ⎟ = ( 0.026 ) ln ⎜ ⎟ −15 ⎝ 10 ⎠ ⎠ ⎛ 25 × 10−6 ⎞ VD = ( 0.026)ln ⎜ ⎟ −15 ⎝ 10 ⎠ VD = 0.622 V (ii) ⎛ 0.2 ⎞ −12 I D = (10−15 )exp ⎜ ⎟ = 2.19 × 10 A ⎝ 0.026 ⎠ ID = 0 (iii) I D = −10 −15 A (iv) I D = −10 −15 A (b) (i) 1.32 ⎛I VD = Vt ln ⎜ D ⎝ IS ⎞ ⎛ 2 × 10−3 ⎞ = (0.026) ln ⎜ = 0.6347 V ⎟ −14 ⎟ ⎝ 5 × 10 ⎠ ⎠ ⎛ 2 × 10−3 ⎞ = 0.5150 V VD = (0.026) ln ⎜ −12 ⎟ ⎝ 5 × 10 ⎠ 0.5150 ≤ VD ≤ 0.6347 V 1.33 ⎛V ⎞ I D = I S exp ⎜ D ⎟ ⎝ Vt ⎠ ⎛ 1.10 ⎞ −21 12 ×10−3 = I S exp ⎜ ⎟ ⇒ I S = 5.07 × 10 A ⎝ 0.026 ⎠ (a) (b) ⎛ 1.0 ⎞ I D = ( 5.07 × 10−21 ) exp ⎜ ⎟ ⎝ 0.026 ⎠ I D = 2.56 × 10−4 A = 0.256 mA 1.34 (a) (b) (c) ⎛ 1.0 ⎞ −7 I D = 10−23 exp ⎜ ⎟ = 5.05 × 10 A ⎝ 0.026 ⎠ ⎛ 1.1 ⎞ −5 I D = 10−23 exp ⎜ ⎟ = 2.37 × 10 A 0.026 ⎠ ⎝ ⎛ 1.2 ⎞ −3 I D = 10−23 exp ⎜ ⎟ = 1.11× 10 A ⎝ 0.026 ⎠ 1.35 IS doubles for every 5C increase in temperature. I S = 10 −12 A at T = 300K For I S = 0.5 × 10 −12 A ⇒ T = 295 K For I S = 50 × 10 −12 A, (2) n = 50 ⇒ n = 5.64 Where n equals number of 5C increases. Then ΔT = ( 5.64 )( 5 ) = 28.2 K So 295 ≤ T ≤ 328.2 K
  18. 18. 1.36 I S (T ) = 2ΔT / 5 , ΔT = 155° C I S (−55) I S (100) = 2155 / 5 = 2.147 × 109 I S (−55) VT @100°C ⇒ 373°K ⇒ VT = 0.03220 VT @− 55°C ⇒ 216°K ⇒ VT = 0.01865 I D (100) = (2.147 × 109 ) × I D (−55) ⎛ 0.6 ⎞ exp ⎜ ⎟ ⎝ 0.0322 ⎠ ⎛ 0.6 ⎞ exp ⎜ ⎟ ⎝ 0.01865 ⎠ ( 2.147 ×10 )(1.237 ×10 ) ( 9.374 ×10 ) 9 = 8 13 I D (100) = 2.83 × 103 I D (−55) 1.37 3.5 = ID (105) + VD ⎛ V ⎞ ⎛ ID ⎞ I D = 5 ×10−9 exp ⎜ D ⎟ ⇒ VD = 0.026 ln ⎜ −9 ⎟ ⎝ 0.026 ⎠ ⎝ 5 × 10 ⎠ Trial and error. VD ID VD −5 0.50 0.226 3 ×10 0.40 0.227 3.1×10−5 −5 0.250 0.228 3.25 ×10 −5 0.229 0.2284 3.271×10 −5 0.2285 0.2284 3.2715 ×10 (a) So VD ≅ 0.2285 V I D ≅ 3.272 × 10−5 A (b) I D = I S = 5 × 10−9 A VR = ( 5 × 10−9 )(105 ) = 5 × 10−4 V VD = 3.4995 V 1.38 ⎛ I ⎞ 10 = I D ( 2 × 10 4 ) + VD and VD = ( 0.026 ) ln ⎜ D12 ⎟ − ⎝ 10 ⎠ Trial and error. VD(v) ID(A) VD(v) 0.50 0.5194 4.75 ×10−4 −4 0.517 0.5194 4.7415 ×10 0.5194 0.5194 4.740 ×10−4 VD = 0.5194 V I D = 0.4740 mA 1.39
  19. 19. I s = 5 × 10 −13 A R1 ϭ 50 K ϩ 1.2 V Ϫ ϩ VD Ϫ R2 ϭ 30 K ID RTH ϭ R1 ͉͉ R2 ϭ 18.75 K ϩ VTH Ϫ ϩ VD Ϫ ID ⎛ R2 VTH = ⎜ ⎝ R1 + R2 ⎞ ⎛ 30 ⎞ ⎟ (1.2) = ⎜ ⎟ (1.2) = 0.45 V ⎝ 80 ⎠ ⎠ ⎛I ⎞ 0.45 = I D RTH + VD , VD = VT ln ⎜ D ⎟ ⎝ IS ⎠ By trial and error: I D = 2.56 μ A, VD = 0.402 V 1.40 ϩ VDϪ ϩ VDϪ ϩ VI Ϫ I1 1K IR I2 ϩ V0 Ϫ I S = 2 × 10 −13 A V0 = 0.60 V ⎛V ⎞ ⎛ 0.60 ⎞ I 2 = I S exp ⎜ 0 ⎟ = ( 2 × 10−13 ) exp ⎜ ⎟ ⎝ 0.026 ⎠ ⎝ VT ⎠ = 2.105 mA 0.6 = 0.60 mA 1K I1 = I 2 + I R = 2.705 mA IR = ⎛I VD = VT ln ⎜ 1 ⎝ IS = 0.6065 ⎞ ⎛ 2.705 × 10−3 ⎞ = (0.026) ln ⎜ ⎟ ⎟ −13 ⎝ 2 × 10 ⎠ ⎠ VI = 2VD + V0 ⇒ VI = 1.81 V 1.41 (a) Assume diode is conducting. Then, VD = Vγ = 0.7 V So that I R 2 = 0. 7 ⇒ 23.3 μ A 30
  20. 20. 1.2 − 0.7 ⇒ 50 μ A 10 Then I D = I R1 − I R 2 = 50 − 23.3 I R1 = Or I D = 26.7 μ A (b) Let R1 = 50 k Ω Diode is cutoff. 30 ⋅ (1.2) = 0.45 V 30 + 50 Since VD < Vγ , I D = 0 VD = 1.42 ϩ5 V 3 k⍀ 2 k⍀ ID VB ϭVAϪVr VA 2 k⍀ 2 k⍀ A&VA: 5 − VA V = ID + A 2 2 A& VA − Vr (1) (2) 5 − (VA − Vr ) + ID = (VA − Vr ) 2 2 5 − (VA − Vr ) ⎡ 5 − VA VA ⎤ VA − Vr +⎢ − ⎥= So 3 2⎦ 2 ⎣ 2 Multiply by 6: 10 − 2 (VA − Vr ) + 15 − 6VA = 3 (VA − Vr ) 25 + 2Vr + 3Vr = 11VA Vr = 0.6 V (a) 11VA = 25 + 5 ( 0.6 ) = 28 ⇒ VA = 2.545 V 5 − VA VA − = 2.5 − VA ⇒ I D Neg. ⇒ I D = 0 2 2 Both (a), (b) I D = 0 From (1) I D = VA = 2.5, VB = 2 ⋅ 5 = 2 V ⇒ VD = 0.50 V 5 1.43 Minimum diode current for VPS (min) I D (min) = 2 mA, VD = 0.7 V I2 = 0.7 5 − 0.7 4.3 , I1 = = R2 R1 R1 We have I1 = I 2 + I D
  21. 21. 4.3 0.7 = +2 R1 R2 Maximum diode current for VPS (max) P = I DVD 10 = I D ( 0.7 ) ⇒ I D = 14.3 mA so (1) I1 = I 2 + I D or 9.3 0.7 = + 14.3 (2) R1 R2 Using Eq. (1), 9.3 4.3 = − 2 + 14.3 ⇒ R1 R1 R1 = 0.41 kΩ Then R2 = 82.5Ω 82.5Ω 1.44 (a) Vo = 0.7 V 5 − 0.7 I= ⇒ I = 0.215 mA 20 10 − 0.7 (b) I= ⇒ I = 0.2325 mA 20 + 20 Vo = I (20 K) − 5 ⇒ Vo = −0.35 V (c) (d) 10 − 0.7 ⇒ I = 0.372 mA 5 + 20 Vo = 0.7 + I (20) − 8 ⇒ Vo = +0.14 V I =0 Vo = I (20) − 5 ⇒ Vo = −5 V I= 1.45 (a) VD 0.6 0.482 (b) Vo 0.5 0.484 (c) 5 = I ( 2 × 109 ) + VD → ID → 2.2 ×10−4 2.259 ×10−4 10 = I ( 4 × 10 4 ) + VD → I → I ⎛ ⎞ VD = ( 0.026 ) ln ⎜ −12 ⎟ ⎝ 2 ×10 ⎠ VD Vo = VD = 0.482 V 0.481 0.482 I = 0.226 mA I ⎛ ⎞ VD = ( 0.026 ) ln ⎜ −12 ⎟ ⎝ 2 ×10 ⎠ VD VD = 0.483 V 2.375 ×10−4 2.379 ×10−4 I = 0.238 mA Vo = −0.24 V I ⎛ ⎞ VD = ( 0.026 ) ln ⎜ ⎟ 2 ×10−12 ⎠ ⎝ VD VD = 0.496 V I → −4 0.496 I = 0.380 mA 3.808 ×10 −4 0.496 Vo = −0.10 V 3.802 ×10 10 = I ( 2.5 × 10 4 ) + VD Vo 0.480 0.496 → (d) I = − I S ⇒ I = 2 × 10−12 A Vo ≅ −5 V 1.46 (a) 0.4834 0.4834 Diode forward biased VD = 0.7 V
  22. 22. 5 = (0.4)(4.7) + 0.7 + V ⇒ V = 2.42 V P = I ⋅ VD = (0.4)(0.7) ⇒ P = 0.28 mω (b) 1.47 0.65 = 0.65 mA = I D1 1 = 2(0.65) = 1.30 mA I R 2 = I D1 = (a) ID2 ID2 = VI − 2Vr − V0 5 − 3(0.65) = = 1.30 ⇒ R1 = 2.35 K R1 R1 0.65 = 0.65 mA 1 8 − 3(0.65) ID2 = ⇒ I D 2 = 3.025 mA 2 I D1 = I D 2 − I R 2 = 3.025 − 0.65 I D1 = 2.375 mA IR2 = (b) 1.48 τd = a. VT (0.026) = = 0.026 kΩ = 26Ω 1 I DQ id = 0.05 I DQ = 50 μ A peak-to-peak vd = idτ d = (26)(50) μ A ⇒ vd = 1.30 mV peak-to-peak For I DQ = 0.1 mA ⇒ τ d = b. (0.026) = 260Ω 0. 1 id = 0.05 I DQ = 5 μ A peak-to-peak vd = idτ d = (260)(5) μ V ⇒ vd = 1.30 mV peak-to-peak 1.49 RS ␯S a. ϩ ␯d Ϫ ϩ Ϫ diode resistance rd = VT /I ⎛ ⎜ VT /I ⎛ rd ⎞ vd = ⎜ ⎟ vS = ⎜ V ⎝ rd + RS ⎠ ⎜ T + RS ⎜ ⎝ I ⎛ VT ⎞ vd = ⎜ ⎟ vs = vo ⎝ VT + IRS ⎠ b. RS = 260Ω ⎞ ⎟ ⎟ vS ⎟ ⎟ ⎠
  23. 23. ⎞ v0 ⎛ VT v 0.026 =⎜ ⇒ 0 = 0.0909 ⎟= vS ⎝ VT + IRS ⎠ 0.026 + (1)(0.26) vS v v 0.026 I = 0.1 mA, 0 = ⇒ 0 = 0.50 vs 0.026 + ( 0.1)( 0.26 ) vS I = 1 mA, I = 0.01 mA. v0 v 0.026 = ⇒ 0 = 0.909 vS 0.026 + (0.01)(0.26) vS 1.50 ⎛V I ≅ I S exp ⎜ a ⎝ VT ⎛ I ⎞ ⎞ ⎟ , Va = VT ln ⎜ ⎟ ⎠ ⎝ IS ⎠ ⎛ 100 × 10−6 ⎞ pn junction, Va = (0.026) ln ⎜ ⎟ −14 ⎝ 10 ⎠ Va = 0.599 V ⎛ 100 × 10−6 ⎞ Schottky diode, Va = (0.026) ln ⎜ ⎟ −9 ⎝ 10 ⎠ Va = 0.299 V 1.51 Schottky pn junction I ⎛V ⎞ Schottky: I ≅ I S exp ⎜ a ⎟ ⎝ VT ⎠ ⎛ I ⎞ ⎛ 0.5 × 10−3 ⎞ Va = VT ln ⎜ ⎟ = (0.026) ln ⎜ −7 ⎟ ⎝ 5 × 10 ⎠ ⎝ IS ⎠ = 0.1796 V Then Va of pn junction = 0.1796 + 0.30 = 0.4796 I 0.5 × 10−3 = ⎛V ⎞ ⎛ 0.4796 ⎞ exp ⎜ a ⎟ exp ⎜ ⎟ ⎝ 0.026 ⎠ ⎝ VT ⎠ I S = 4.87 × 10 −12 A IS = 1.52 (a) ϩ VD Ϫ I1 0.5 mA I2 I1 + I 2 = 0.5 × 10 −3
  24. 24. ⎛V 5 × 10−8 exp ⎜ D ⎝ VT ⎞ ⎛ VD ⎞ −12 −3 ⎟ + 10 exp ⎜ ⎟ = 0.5 × 10 VT ⎠ ⎠ ⎝ ⎛V ⎞ 5.0001× 10 −8 exp ⎜ D ⎟ = 0.5 × 10 −3 ⎝ VT ⎠ ⎛ 0.5 × 10−3 ⎞ ⇒ VD = 0.2395 VD = (0.026) ln ⎜ −8 ⎟ ⎝ 5.0001× 10 ⎠ Schottky diode, I 2 = 0.49999 mA pn junction, I1 = 0.00001 mA (b) ϩ VD1 Ϫ I ϩ ϩ VD2 Ϫ Ϫ 0.90 V ⎛V ⎞ ⎛V ⎞ I = 10 −12 exp ⎜ D1 ⎟ = 5 × 10−8 exp ⎜ D 2 ⎟ ⎝ VT ⎠ ⎝ VT ⎠ VD1 + VD 2 = 0.9 ⎛V ⎞ ⎛ 0.9 − VD1 ⎞ 10−12 exp ⎜ D1 ⎟ = 5 × 10−8 exp ⎜ ⎟ ⎝ VT ⎠ ⎝ VT ⎠ ⎛ 0.9 ⎞ ⎛ −VD1 ⎞ = 5 ×10−8 exp ⎜ ⎟ ⎟ exp ⎜ ⎝ VT ⎠ ⎝ VT ⎠ ⎛ 2V ⎞ ⎛ 5 × 10−8 ⎞ ⎛ 0.9 ⎞ exp ⎜ D1 ⎟ = ⎜ exp ⎜ ⎟ −12 ⎟ ⎝ 0.026 ⎠ ⎠ ⎝ VT ⎠ ⎝ 10 ⎛ 5 × 10−8 ⎞ + 0.9 = 1.1813 2VD1 = VT ln ⎜ −12 ⎟ ⎝ 10 ⎠ VD1 = 0.5907 pn junction VD 2 = 0.3093 Schottky diode ⎛ 0.5907 ⎞ I = 10−12 exp ⎜ ⎟ ⇒ I = 7.35 mA ⎝ 0.026 ⎠ 1.53 R ϭ 0.5 K V0 ϩ VPS ϭ 10 V Ϫ I ϩ VZ Ϫ IZ VZ = VZ 0 = 5.6 V at I Z = 0.1 mA rZ = 10Ω I Z rZ = ( 0.1)(10 ) = 1 mV VZ0 = 5.599 a. RL → ∞ ⇒ IZ = 10 − 5.599 4.401 = = 8.63 mA R + rZ 0.50 + 0.01 VZ = VZ 0 + I Z rZ = 5.599 + ( 0.00863)(10 ) VZ = V0 = 5.685 V RL IL
  25. 25. 11 − 5.599 = 10.59 mA 0.51 VZ = V0 = 5.599 + ( 0.01059 )(10 ) = 5.7049 V b. VPS = 11 V ⇒ I Z = 9 − 5.599 = 6.669 mA 0.51 VZ = V0 = 5.599 + ( 0.006669 )(10 ) = 5.66569 V VPS = 9 V ⇒ I Z = ΔV0 = 5.7049 − 5.66569 ⇒ ΔV0 = 0.0392 V c. I = IZ + IL V0 V − V0 V − VZ 0 IL = , I = PS , IZ = 0 RL R rZ 10 − V0 V0 − 5.599 V0 = + 0.50 0.010 2 10 5.599 1 1⎤ ⎡ 1 + = V0 ⎢ + + ⎥ 0.50 0.010 ⎣ 0.50 0.010 2 ⎦ 20.0 + 559.9 = V0 (102.5) V0 = 5.658 V 1.54 a. 9 − 6.8 ⇒ I Z = 11 mA 0.2 PZ = (11)( 6.8 ) ⇒ PZ = 74.8 mW IZ = 12 − 6.8 ⇒ I Z = 26 mA 0.2 b. 26 − 11 %= × 100 ⇒ 136% 11 PZ = ( 26 )( 6.8 ) = 176.8 mW IZ = %= 176.8 − 74.8 × 100 ⇒ 136% 74.8 1.55 I Z rZ = ( 0.1)( 20 ) = 2 mV VZ 0 = 6.8 − 0.002 = 6.798 V a. RL = ∞ 10 − 6.798 ⇒ I Z = 6.158 mA 0.5 + 0.02 V0 = VZ = VZ 0 + I Z rZ = 6.798 + ( 0.006158)( 20 ) IZ = V0 = 6.921 V b. I = IZ + IL 10 − V0 V0 − 6.798 V0 = + 0.50 0.020 1 10 6.798 1 1⎤ ⎡ 1 + = V0 ⎢ + + 0.30 0.020 0.50 0.020 1⎥ ⎣ ⎦ 359.9 = V0 (53) V0 = 6.791 V ΔV0 = 6.791 − 6.921 ΔV0 = −0.13 V 1.56
  26. 26. For VD = 0, I SC = 0.1 A ⎛ 0.2 ⎞ + 1⎟ For ID = 0 VD = VT ln ⎜ −14 ⎝ 5 × 10 ⎠ VD = VDC = 0.754 V
  27. 27. Chapter 2 Exercise Problems EX2.1 30 − 12 − 0.6 = 87 mA 0.2 vR ( max ) = 30 + 12 = 42 V iD ( peak ) = ⎛ 12.6 ⎞ ⎟ ⇒ 24.8° ⎝ 30 ⎠ By symmetry ω t2 = 180 − 24.8 = 155.2° ω t1 = sin −1 ⎜ % time = 155.2 − 24.8 × 100% = 36.2% 360 EX2.2 (a) vO = 12sin θ1 − 1.4 = 0 1.4 = 0.1166 12 which yields θ1 = 6.7° or sin θ1 = By symmetry, θ 2 = 180 − 6.7 = 173.3° 173.3 − 6.7 × 100% = 46.3% 360 1.4 sinθ1 = (b) = 0.35 4 which yields θ1 = 20.5° Then % time = By symmetry, θ 2 = 180 − 20.5 = 159.5° Then % time = 159.5 − 20.5 × 100% = 38.6% 360 EX2.3 VM 24 C= = ⇒ or C = 500 μ F 2 fRVr 2 ( 60 ) (103 ) ( 0.4 ) EX2.4 Vr = VM VM 75 or R = ⇒R= f RC f CVr 60 ) ( 50 × 10−6 ) ( 4 ) ( Then R = 6.25 k Ω EX2.5 10 ≤ VPS ≤ 14 V , VZ = 5.6 V , 20 ≤ RL ≤ 100 Ω 5.6 = 0.28 A, 20 5.6 I L ( min ) = = 0.056 A 100 ⎡VPS ( max ) − VZ ⎤ ⋅ I L ( max ) ⎡VPS ( min ) − VZ ⎤ ⋅ I L ( min ) ⎦ ⎣ ⎦ − I Z (max) = ⎣ VPS ( min ) − 0.9VZ − 0.1VPS ( max ) VPS ( min ) − 0.9VZ − 0.1VPS ( max ) I L ( max ) = or I L ( max ) = (14 − 5.6 )( 280 ) − (10 − 5.6 )( 56 ) 10 − ( 0.9 )( 5.6 ) − ( 0.1)(14 )
  28. 28. or I L ( max ) = 591.5 mA Power(min) = I Z ( max ) ⋅ VZ = ( 0.5915)( 5.6 ) So Power(min) = 3.31 W VPS ( max ) − VZ 14 − 5.6 = Now Ri = or Ri ≅ 13 Ω I Z ( max ) + I L ( min ) 0.5915 + 0.056 EX2.6 13.6 − 9 = 0.2383 A 15.3 + 4 = 9 + ( 4 )( 0.2383) = 9.9532 V For vPS = 13.6 V , I Z = vL ,max 11 − 9 = 0.1036 A 15.3 + 4 = 9 + ( 4 )( 0.1036 ) = 9.4144 V For vPS = 11 V , I Z = vL ,min ΔvL 9.9532 − 9.4144 × 100% = × 100% ΔvPS 13.6 − 11 or Source Reg = 20.7% Source Reg = 13.6 − 9 = 0.2383 A 15.3 + 4 = 9 + ( 4 )( 0.2383) = 9.9532 V For I L = 0, I Z = vL , noload For I L = 100 mA, I Z = 13.6 − ⎡9 + I Z ( 4 ) ⎤ ⎣ ⎦ 15.3 which yields I Z = 0.1591 A vL , full load = 9 + ( 4 )( 0.1591A ) = 9.6363 V vL , noload − vL , full load × 100% Load Reg = vL , full load 9.9532 − 9.6363 × 100% 9.6363 or Load Reg = 3.29% = EX2.7 R1 VO ␯I ϩ Ϫ D1 ϩ V1 Ϫ For vI < 5 V , D2 on ⇒ VO = −5 V Then, V2 = 4.3 V. D1 turns on when v1 = 2.5 V, Then, V1 = 1.8 V. D2 Ϫ V2 ϩ − 0.10
  29. 29. For vI > 2.5 V , ΔvO 1 R2 1 = ⇒ = ΔvI 3 R1 + R2 3 So that R1 = 2R2 EX2.8 For Vγ = 0, vO ( max ) = −2 V Now, ΔvO = 8 V , so that vO ( min ) = −10 V EX2.9 10 − 4.4 = 0.5895 mA 9.5 Set I = ID1, then vI = 4.4 − 0.6 − ( 0.5895 )( 0.5 ) = 3.505 V vO = 4.4 V , I = Summary: For 0 ≤ vI ≤ 3.5 V , vO = 4.4 V For vI > 3.5 V , D2 turns on and when vI ≥ 9.4 V , vO = 10 V ␯O(V) 10 4.4 0 3.505 9.4 ␯I (V) EX2.10 VO = −0.6 V , I D1 = 0 ID2 = I = −0.6 − ( −10 ) 2.2 ⇒ I D 2 = I = 4.27 mA EX2.11 At the VA node: 15 − VA V (1) = ID2 + A 5 15 At the VB node: 15 − (VB + 0.7 ) V + I D 2 = B (2) 5 10 We see that VB = VA − 0.7, so Equation (2) becomes 15 − VA V − 0.7 (2’) + ID2 = A 5 10 Solving for ID2 from Equation (1) and substituting into Equation (2’), we find ⎛ 15 − VA ⎞ VA VA − 0.7 2⎜ = ⎟− 10 ⎝ 5 ⎠ 15 Then VA = 10.71 V and VB = 10.01 V 15 − 10.71 Solving for the diode currents, we obtain I D1 = ⇒ I D1 = 0.858 mA 5
  30. 30. Also I D 2 = 15 − 10.71 10.71 − ⇒ I D 2 = 0.144 mA 5 15 EX2.12 Assuming all diodes are conducting, we have VB = −0.7 V and VA = 0 Summing currents, we have 5 − VA (1) = I D1 + I D 2 5 V − ( −5 ) I D 2 + I D3 = B (2) 5 V − ( −10 ) 10 − 0.7 = I D1 = B (3) 10 10 From ( 3) , we find I D1 = 0.93 mA From (1) , we obtain I D 2 = 0.07 mA From ( 2 ) , we have I D 3 = 0.79 mA EX2.13 (a) I ph = η eΦ ⎡ 6.4 × 10 −2 ⎤ ⎥ ( 0.5 ) so I = ( 0.8 ) (1.6 × 10−19 ) ⎢ ⎢ ( 2 ) (1.6 × 10−19 ) ⎥ ⎣ ⎦ or I ph = 12.8 mA (b) We have vO = (12.8)(1) = 12.8 V . The diode must be reverse biased so that VPS > 12.8 V EX2.14 The equivalent circuit is ϩ5 V ϩ Vr ϭ 1.7 V Ϫ rf ϭ 15 ⍀ I R 0.2 V So I = 5 − 1.7 − 0.2 = 15 mA rf + R 15 − 1.7 − 0.2 3.1 = = 0.207 kΩ 15 15 Then R = 207 − 15 ⇒ R = 192 Ω 2.15 40 − 12 IZ = = 0.233 A (a) 120 P = ( 0.233)(12 ) = 2.8 W Or rf + R =
  31. 31. I R = 0.233 A, I L = ( 0.9 )( 0.233) = 0.21 A (b) So 0.21 = 12 ⇒ RL = 57.1 Ω RL P = ( 0.1)( 0.233)(12 ) ⇒ P = 0.28 W (c) Test Your Understanding Exercises TYU2.1 vI = 120sin ( 2π 60t ) , Vγ = 0.7 V , and R = 2.5 kΩ Full-wave rectifier: Turns ratio 1:2 so that vS = v I VM = 120 − 0.7 = 119.3 V Vr = 119.3 − 100 = 19.3 V So C = VM 119.3 = or C = 20.6 μ F 2 f RVr 2 ( 60 ) ( 2.5 x103 ) (19.3) TYU2.2 vI = 50sin ( 2π 60t ) , Vγ = 0.7 V , and R = 10 kΩ. Full-wave rectifier C= ( 50 − 1.4 ) VM = 2 f RVr 2 ( 60 ) (10 × 103 ) ( 2 ) or C = 20.3 μ F TYU2.3 Using Equation (2.10) (a) ω Δt = 2Vr = VM 2 ( 4) 75 = 0.327 ⎛ 0.327 ⎞ Percent time = ⎜ ⎟ × 100% = 5.2% ⎝ 2π ⎠ (b) ω Δt = 2Vr = VM 2 (19.3) 119.3 = 0.569 ⎛ 0.569 ⎞ Percent time = ⎜ ⎟ × 100% = 18.1% ⎝ π ⎠ (c) ω Δt = 2Vr = VM 2 ( 2) 48.6 = 0.287 ⎛ 0.287 ⎞ Percent time = ⎜ ⎟ × 100% = 9.14% ⎝ π ⎠ TYU2.4 V − VZ I Z = PS − IL Ri 11 − 9 − 0.1 = 0 (Minimum Zener current is zero.) 20 13.6 − 9 For VPS (max) and IL (min), then I Z ( max ) = − 0 ⇒ I Z ( max ) = 230 mA 20 The characteristic of the minimum Zener current being one-tenth of the maximum value is violated. The proper circuit operation is questionable. For VPS (min) and IL (max), then I Z ( min ) = TYU2.5
  32. 32. I Z ( min ) = so 30 = VPS ( min ) − VZ Ri − I L ( max ) 10 − 9 − I L ( max ) which yields I L ( max ) = 35.4 mA 0.0153 TYU2.6 ␯O(V) 4.35 2.7 Ϫ4.7 2.7 6 ␯I (V) Ϫ4.7 TYU2.7 As vS goes negative, D turns on and vO = +5 V . As vS goes positive, D turns off. Output is a square wave oscillating between +5 and +35 volts. TYU2.8 ␯O(V) 4.4 0 0.6 5 ␯I (V) 5 ␯I (V) (a) ␯O(V) 4.4 2.4 0 (b) 3
  33. 33. TYU2.9 (a) VO = 0.6 V for all V1. VO (V) 3.6 0.6 0 (b) PSpice Exercises 3 VI (V)
  34. 34. Chapter 2 Problem Solutions 2.1 ␯I ␯0 Rϭ1K ␯I 10 0 Ϫ10 Vγ = 0.6 V, rf = 20 Ω ⎛ R ⎞ For vI = 10 V, v0 = ⎜ 10 − 0.6 ) ⎜ R + r ⎟( ⎟ f ⎠ ⎝ ⎛ 1 ⎞ =⎜ ⎟ ( 9.4 ) ⎝ 1 + 0.02 ⎠ v0 = 9.22 10 0.6 ␯0 9.22 2.2 ␯I ϩ ␯D Ϫ ␯0 D R v0 = vI − vD ⎛i ⎞ v vD = VT ln ⎜ D ⎟ and iD = 0 R ⎝ IS ⎠ ⎛ v ⎞ v0 = vI − VT ln ⎜ 0 ⎟ ⎝ IS R ⎠ 2.3 80sin ω t = 13.33sin ω t 6 13.33 Peak diode current id (max) = R (a) vs =
  35. 35. (b) PIV = vs (max) = 13.3 V (c) vo ( avg ) = 1 To To 1 π ∫ v (t )dt = 2π ∫ 13.33sin × dt o o o 13.33 13.33 13.33 π = ⎡ − ( −1 − 1) ⎤ = [ − cos x ]o = ⎣ ⎦ 2π 2π π vo ( avg ) = 4.24 V (d) 50% 2.4 v0 = vS − 2Vγ ⇒ vS ( max ) = v0 ( max ) + 2Vγ a. For v0 ( max ) = 25 V ⇒ vS ( max ) = 25 + 2 ( 0.7 ) = 26.4 V N1 160 N = ⇒ 1 = 6.06 N 2 26.4 N2 b. For v0 ( max ) = 100 V ⇒ vS ( max ) = 101.4 V N1 N 160 = ⇒ 1 = 1.58 N 2 101.4 N2 From part (a) PIV = 2vS ( max ) − Vγ = 2 ( 26.4 ) − 0.7 or PIV = 52.1 V or, from part (b) PIV = 2 (101.4 ) − 0.7 or PIV = 202.1 V 2.5 (a) vs (max) = 12 + 2(0.7) = 13.4 V 13.4 vs ( rms ) = ⇒ vs (rms) = 9.48 V 2 (b) VM VM Vr = ⇒C = 2 f RC 2 f Vr R C= 12 ⇒ C = 2222 μ F 2 ( 60 )( 0.3)(150 ) (c) VM ⎡ 2VM ⎤ ⎢1 + π ⎥ R ⎢ Vr ⎥ ⎣ ⎦ 2 (12 ) ⎤ 12 ⎡ ⎢1 + π ⎥ = 150 ⎢ 0.3 ⎥ ⎣ ⎦ id , peak = 2.33 A id , peak = 2.6 (a) vS ( max ) = 12 + 0.7 = 12.7 V vS ( rms ) = (b) vS ( max ) Vr = 2 ⇒ vS ( rms ) = 8.98 V VM V 12 or C = 4444 μ F ⇒C = M = fRC fRVr ( 60 )(150 )( 0.3)
  36. 36. For the half-wave rectifier iD , max = (c) VM R ⎛ VM ⎜ 1 + 4π ⎜ 2Vr ⎝ iD , max = 4.58 A 2.8 (a) vs ( peak ) = 15 + 2 ( 0.7 ) = 16.4 V 16.4 vs ( rms ) = = 11.6 V 2 VM 15 = = 2857 μ F C= (b) 2 f RVr 2 ( 60 )(125 )( 0.35 ) 2.9 ␯S 26 0.6 Ϫ0.6 Ϫ26 25.4 ϩ ␯O 0 0 Ϫ ␯O Ϫ25.4 2.10 R iD ϩ VB ϭ 12 V Ϫ ␯S ϭ ϩ Ϫ ␯S iD x2 x1 ␻t vS ( t ) = 24sin ω t Now iD ( avg ) = T 1 iD ( t ) dt T∫ 0 24sin x − 12.7 R To find x1 and x2, 24sin x1 = 12.7 We have for x1 ≤ ω t ≤ x2 iD = x1 = 0.558 rad x2 = π − 0.558 = 2.584 rad Then ⎞ 12 ⎛ 12 ⎞ ⎜ 1 + 4π ⎟ or ⎟= ⎟ 150 ⎜ 2 ( 0.3) ⎟ ⎠ ⎝ ⎠
  37. 37. 1 2π iD ( avg ) = 2 = x2 ⎡ 24sin x − 12.7 ⎤ ⎥dx R ⎦ x1 ∫⎢ ⎣ 1 ⎛ 24 ⎞ 1 ⎛ 12.7 ⎞ x2 6.482 4.095 x2 − ⇒ R = 1.19Ω ⎜ ⎟ ( − cos x ) x1 − ⎜ ⎟ ⋅ xx or 2 = 2π ⎝ R ⎠ 2π ⎝ R ⎠ 1 R R x −x 2.584 − 0.558 Fraction of time diode is conducting = 2 1 × 100% = × 100% or Fraction = 32.2% 2π 2π Power rating = = x2 1 2 x R 2 R 2 ⎡ 24sin x − 12.7 ⎤ ∫ iD dt = 2π x∫ ⎢ ⎥ dx T 0 R ⎣ ⎦ 1 T 2 Pavg = R ⋅ irms = ⎡( 24 ) 2π R ∫ ⎣ x1 2 2 sin 2 x − 2 (12.7 )( 24 ) sin x + (12.7 ) ⎤dx ⎦ x = 1 ⎡ sin 2 x ⎤ 2 2 x 2 x2 x2 ⎤ ( 24 ) ⎡ − ⎢2 ⎥ − 2 (12.7 )( 24 )( − cos x ) x1 + (12.7 ) xx1 ⎦ 2π R ⎣ 4 ⎦x ⎣ 1 For R = 1.19 Ω, then Pavg = 17.9 W 2.11 (a) 15 = 150Ω 0.1 vS ( max ) = vo ( max ) + Vγ = 15 + 0.7 or vS ( max ) = 15.7 V R= Then vS ( rms ) = Now 15.7 2 = 11.1 V N1 120 N = ⇒ 1 = 10.8 N 2 11.1 N2 VM VM 15 ⇒C = = or C = 2083 μ F 2 fRC 2 fRVr 2 ( 60 )(150 )( 0.4 ) (b) Vr = (c) PIV = 2vS ( max ) − Vγ = 2 (15.7 ) − 0.7 or PIV = 30.7 V 2.12 ␯i ϩ Ϫ ␯0 D2 R2 RL R1 ␯0 ␯i ϩ Ϫ RL R2 R1 For vi > 0 Vγ = 0 Voltage across RL + R1 = vi ⎛ RL ⎞ 1 Voltage Divider ⇒ v0 = ⎜ ⎟ vi = vi 2 ⎝ RL + R1 ⎠
  38. 38. ␯0 20 2.13 For vi > 0, (Vγ = 0 ) ␯i R1 ϩ Ϫ ␯0 R2 RL a. ⎛ R2 || RL ⎞ v0 = ⎜ ⎟ vi ⎝ R2 || RL + R1 ⎠ R2 || RL = 2.2 || 6.8 = 1.66 kΩ ␯0 ⎛ 1.66 ⎞ v0 = ⎜ ⎟ vi = 0.43 vi ⎝ 1.66 + 2.2 ⎠ 4.3 v0 ( rms ) = b. v0 ( max ) 2 ⇒ v0 ( rms ) = 3.04 V 2.14 3.9 ⇒ I 2 = 0.975 mA 4 20 − 3.9 IR = = 1.3417 mA 12 I Z = 1.3417 − 0.975 ⇒ I Z = 0.367 mA IL = PT = I Z ⋅ VZ = ( 0.367 )( 3.9 ) ⇒ PT = 1.43 mW 2.15 (a) 40 − 12 = 0.233 A 120 P = ( 0.233)(12 ) = 2.8 W IZ = (b) IR = 0.233 A, IL = (0.9)(0.233) = 0.21 A 12 So 0.21 = ⇒ RL = 57.1Ω RL P = ( 0.1)( 0.233)(12 ) ⇒ P = 0.28 W (c) 2.16 Ri VI II IZ ϩ VZ Ϫ V0 RL IL VI = 6.3 V, Ri = 12Ω, VZ = 4.8
  39. 39. a. 6.3 − 4.8 ⇒ 125 mA 12 I L = I I − I Z = 125 − I Z II = 25 ≤ I L ≤ 120 mA ⇒ 40 ≤ RL ≤ 192Ω b. PZ = I Z VZ = (100 )( 4.8 ) ⇒ PZ = 480 mW PL = I LV0 = (120 )( 4.8 ) ⇒ PL = 576 mW 2.17 a. 20 − 10 ⇒ I I = 45.0 mA 222 10 IL = ⇒ I L = 26.3 mA 380 I Z = I I − I L ⇒ I Z = 18.7 mA II = b. PZ ( max ) = 400 mW ⇒ I Z ( max ) = ⇒ I L ( min ) = I I − I Z ( max ) = 45 − 40 ⇒ I L ( min ) = 5 mA = 400 = 40 mA 10 10 RL ⇒ RL = 2 kΩ (c) For Ri = 175Ω I I = 57.1 mA I L = 26.3 mA I Z = 30.8 mA I Z ( max ) = 40 mA ⇒ I L ( min ) = 57.1 − 40 = 17.1 mA RL = 10 ⇒ RL = 585Ω 17.1 2.18 a. From Eq. (2-31) 500 [ 20 − 10] − 50 [15 − 10] I Z ( max ) = 15 − ( 0.9 )(10 ) − ( 0.1)( 20 ) 5000 − 250 4 I Z ( max ) = 1.1875 A I Z ( min ) = 0.11875 A = From Eq. (2-29(b)) Ri = 20 − 10 ⇒ Ri = 8.08Ω 1187.5 + 50 b. PZ = (1.1875 )(10 ) ⇒ PZ = 11.9 W PL = I L ( max ) V0 = ( 0.5 )(10 ) ⇒ PL = 5 W 2.19 (a) As approximation, assume I Z ( max ) and I Z ( min ) are the same as in problem 2.18. V0 ( max ) = V0 ( nom ) + I Z ( max ) rZ = 10 + (1.1875)(2) = 12.375 V V0 ( min ) = V0 ( nom ) + I Z ( min ) rZ = 10 + (0.11875)(2) = 10.2375 V
  40. 40. % Reg = b. 2.20 % Reg = = 12.375 − 10.2375 × 100% ⇒ % Reg = 21.4% 10 VL ( max ) − VL ( min ) VL ( nom ) × 100% VL ( nom ) + I Z ( max ) rz − (VL ( nom ) + I Z ( min ) rz ) VL ( nom ) ⎡ I Z ( max ) − I Z ( min ) ⎤ ( 3) ⎦ =⎣ = 0.05 6 So I Z ( max ) − I Z ( min ) = 0.1 A 6 6 = 0.012 A, I L ( min ) = = 0.006 A 500 1000 VPS ( min ) − VZ Now I L ( max ) = Now Ri = or 280 = I Z ( min ) + I L ( max ) 15 − 6 ⇒ I Z ( min ) = 0.020 A I Z ( min ) + 0.012 Then I Z ( max ) = 0.1 + 0.02 = 0.12 A and Ri = or 280 = VPS ( max ) − 6 0.12 + 0.006 VPS ( max ) − VZ I Z ( max ) + I L ( min ) ⇒ VPS ( max ) = 41.3 V 2.21 Using Figure 2.21 a. VPS = 20 ± 25% ⇒ 15 ≤ VPS ≤ 25 V For VPS ( min ) : I I = I Z ( min ) + I L ( max ) = 5 + 20 = 25 mA Ri = b. VPS ( min ) − VZ II = 15 − 10 ⇒ Ri = 200Ω 25 For VPS ( max ) ⇒ I I ( max ) = 25 − 10 ⇒ I I ( max ) = 75 mA Ri For I L ( min ) = 0 ⇒ I Z ( max ) = 75 mA VZ 0 = VZ − I Z rZ = 10 − ( 0.025 )( 5 ) = 9.875 V V0 ( max ) = 9.875 + ( 0.075 )( 5 ) = 10.25 V0 ( min ) = 9.875 + ( 0.005 )( 5 ) = 9.90 ΔV0 = 0.35 V c. % Reg = ΔV0 × 100% ⇒ % Reg = 3.5% V0 ( nom ) 2.22 From Equation (2.29(a)) VPS ( min ) − VZ 24 − 16 or Ri = 18.2Ω Ri = = I Z ( min ) + I L ( max ) 40 + 400 Also Vr = VM VM ⇒C = 2 fRC 2 fRVr R ≅ Ri + rz = 18.2 + 2 = 20.2Ω Then
  41. 41. C= 24 ⇒ C = 9901 μ F 2 ( 60 )(1)( 20.2 ) 2.23 VZ = VZ 0 + I Z rZ VZ ( nom ) = 8 V 8 = VZ 0 + ( 0.1)( 0.5 ) ⇒ VZ 0 = 7.95 V Ii = VS ( max ) − VZ ( nom ) Ri = 12 − 8 = 1.333 A 3 For I L = 0.2 A ⇒ I Z = 1.133 A For I L = 1 A ⇒ I Z = 0.333 A VL ( max ) = VZ 0 + I Z ( max ) rZ = 7.95 + (1.133)( 0.5 ) = 8.5165 VL ( min ) = VZ 0 + I Z ( min ) rZ = 7.95 + ( 0.333)( 0.5 ) = 8.1165 ΔVL = 0.4 V ΔVL 0.4 = ⇒ % Reg = 5.0% % Reg = V0 ( nom ) 8 Vr = VM VM ⇒C = 2 fRC 2 fRVr R = Ri + rz = 3 + 0.5 = 3.5Ω Then C = 12 ⇒ C = 0.0357 F 2 ( 60 )( 3.5 )( 0.8 ) 2.24 (a) For −10 ≤ vI ≤ 0, both diodes are conducting ⇒ vO = 0 For 0 ≤ vI ≤ 3, Zener not in breakdown, so i1 = 0, vO = 0 vI − 3 mA 20 1 ⎛ v −3⎞ vo = ⎜ I ⎟ (10 ) = vI − 1.5 20 ⎠ 2 ⎝ At vI = 10 V, vo = 3.5 V, i1 = 0.35 mA For vI > 3 i1 = ␯O(V) 4 3.5 Ϫ10 (b) 3.0 For vI < 0, both diodes forward biased 0 − vI . At vI = −10 V , i1 = −1 mA 10 v −3 For vI > 3, i1 = I . At vI = 10 V , i1 = 0.35 mA 20 −i1 = 10 ␯I(V)
  42. 42. i1(mA) 0.35 Ϫ10 10 ␯I(V) 3 Ϫ1 2.25 (a) 1K ␯I ␯0 1K 2K ϩ15 V1 1 V1 = × 15 = 5 V ⇒ for vI ≤ 5.7, v0 = vI 3 For vI > 5.7 V vI − (V1 + 0.7 ) 15 − V1 V1 + = , v0 = V1 + 0.7 1 2 1 15 − ( v0 − 0.7 ) v0 − 0.7 vI − v0 + = 1 2 1 vI 15.7 0.7 ⎛ 1 1 1⎞ + + = v0 ⎜ + + ⎟ = v0 ( 2.5 ) 1 2 1 ⎝ 1 2 1⎠ 1 vI + 8.55 = v0 ( 2.5 ) ⇒ v0 = vI + 3.42 2.5 vI = 5.7 ⇒ v0 = 5.7 vI = 15 ⇒ v0 = 9.42 ␯0(V) 9.42 5.7 0 5.7 15 ␯I (V) (b) iD = 0 for 0 ≤ vI ≤ 5.7 Then for vI > 5.7 V ⎛ v ⎞ vI − ⎜ I + 3.42 ⎟ vI − vO ⎝ 2.5 ⎠ or i = 0.6vI − 3.42 For v = 15, i = 5.58 mA = iD = I D D 1 1 1
  43. 43. iD(mA) 5.58 5.7 15 ␯I (V) 2.26 ⎛ 20 ⎞ For D off, vo = ⎜ ⎟ (20) − 10 = 3.33 V ⎝ 30 ⎠ Then for vI ≤ 3.33 + 0.7 = 4.03 V ⇒ vo = 3.33 V For vI > 4.03, vo = vl − 0.7; For vI = 10, vo = 9.3 (a) ␯O(V) 9.3 3.33 0 (b) 4.03 10 ␯I (V) For vI ≤ 4.03 V , iD = 0 10 − vo vo − ( −10 ) = 10 20 3 Which yields iD = vI − 0.605 20 For vI = 10, iD = 0.895 mA For vI > 4.03, iD + iD(mA) 0.895 0 2.27 Ϫ30 4.03 ␯O 12.5 10.7 10.7 30 ␯I Ϫ30 30 − 10.7 = 0.175 A 100 + 10 v0 = i(10) + 10.7 = 12.5 V For vI = 30 V, i = 10 ␯I(V)
  44. 44. b. ␯O 12.5 10.7 0 Ϫ30 2.28 ϩ 5Ϫ ␯I ␯O R ϭ 6.8 K Vγ = 0.6 V vI = 15sin ω t ␯O Ϫ4.4 Ϫ19.4 2.29 a. Vγ = 0 0 Ϫ3 V Vγ = 0.6 0 Ϫ2.4 b. Vγ = 0 20 5 Vγ = 0.6 19.4 5 2.30 ␯O
  45. 45. 10 6.7 0 Ϫ4.7 Ϫ10 2.31 One possible example is shown. Ri Ii ϩ Ϫ DZ Vign L D ϩ VZ ϭ 14 V Ϫ ϩ Ϫ RADIO VRADIO L will tend to block the transient signals Dz will limit the voltage to +14 V and −0.7 V. Power ratings depends on number of pulses per second and duration of pulse. 2.32 ␯O(V) 40 (a) 0 ␯O(V) 35 (b) 0 Ϫ5 2.33 C ␯I ␯O ϩ Vx Ϫ a. For Vγ = 0 ⇒ Vx = 2.7 V b. For Vγ = 0.7 V ⇒ Vx = 2.0 V 2.34 C ϩ ␯I Ϫ 2.35 ϩ Ϫ 10 V ϩ ␯O Ϫ
  46. 46. 20 ␯O 10 VB ϭ 0 0 ␯I Ϫ10 20 ␯O 13 10 3 0 VB ϭ ϩ3 V ␯I ϩ VB Ϫ7 20 ␯O 10 7 VB ϭ Ϫ3 V 0 Ϫ3 ␯ I ϩ VB Ϫ13 2.36 For Figure P2.32(a) 10 ␯I 0 Ϫ10 ␯O Ϫ20 2.37 a. 10 − 0.6 ⇒ I D1 = 0.94 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) ⇒ V0 = 8.93 V I D1 = ID2 = 0 b. 5 − 0.6 ⇒ I D1 = 0.44 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) ⇒ V0 = 4.18 V I D1 = c. d. 10 = ID2 = 0 Same as (a) (I ) 2 ( 0.5 ) + 0.6 + I ( 9.5 ) ⇒ I = 0.964 mA V0 = I ( 9.5 ) ⇒ V0 = 9.16 V I D1 = I D 2 = 2.38 a. b. I ⇒ I D1 = I D 2 = 0.482 mA 2 I = I D1 = I D 2 = 0 V0 = 10
  47. 47. 10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) ⇒ I = I D 2 = 0.94 mA I D1 = 0 V0 = 10 − I ( 9.5 ) ⇒ V0 = 1.07 V c. 10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) + 5 ⇒ I = I D 2 = 0.44 mA I D1 = 0 V0 = 10 − I ( 9.5 ) ⇒ V0 = 5.82 V d. 10 = I ( 9.5 ) + 0.6 + I ( 0.5) ⇒ I = 0.964 mA 2 I ⇒ I D1 = I D 2 = 0.482 mA 2 V0 = 10 − I ( 9.5 ) ⇒ V0 = 0.842 V I D1 = I D 2 = 2.39 a. V1 = V2 = 0 ⇒ D1 , D2 , D3 , on V0 = 4.4 V 10 − 4.4 ⇒ I = 0.589 mA 9.5 4.4 − 0.6 = ID2 = ⇒ I D1 = I D 2 = 7.6 mA 0.5 = I D1 + I D 2 − I = 2 ( 7.6 ) − 0.589 ⇒ I D 3 = 14.6 mA I= I D1 I D3 b. V1 = V2 = 5 V D1 and D2 on, D3 off I 10 = I ( 9.5 ) + 0.6 + ( 0.5 ) + 5 ⇒ I = 0.451 mA 2 I I D1 = I D 2 = ⇒ I D1 = I D 2 = 0.226 mA 2 I D3 = 0 V0 = 10 − I ( 9.5 ) = 10 − ( 0.451)( 9.5 ) ⇒ V0 = 5.72 V V1 = 5 V, V2 = 0 D1 off, D2, D3 on V0 = 4.4 V c. 10 − 4.4 ⇒ 9.5 4.4 − 0.6 = ⇒ 0.5 I= I D2 I = 0.589 mA I D 2 = 7.6 mA I D1 = 0 I D 3 = I D 2 − I = 7.6 − 0.589 ⇒ I D 3 = 7.01 mA V1 = 5 V, V2 = 2 V D1 off, D2, D3 on V0 = 4.4 V d. 10 − 4.4 ⇒ 9.5 4.4 − 0.6 − 2 = ⇒ 0.5 I= I D2 I = 0.589 mA I D 2 = 3.6 mA I D1 = 0 I D 3 = I D 2 − I = 3.6 − 0.589 ⇒ I D 3 = 3.01 mA 2.40 (a) D1 on, D2 off, D3 on So I D 2 = 0 Now V2 = −0.6V , I D1 = 10 − 0.6 − ( −0.6 ) R1 + R2 = 10 ⇒ I D1 = 1.25 mA 2+6
  48. 48. V1 = 10 − 0.6 − (1.25 )( 2 ) ⇒ V1 = 6.9 V I R3 = I D3 −0.6 − ( −5 ) = 2.2 mA 2 = I R 3 − I D1 = 2.2 − 1.25 ⇒ I D 3 = 0.95 mA (b) D1 on, D2 on, D3 off So I D 3 = 0 V1 = 4.4 V , I D1 = 10 − 0.6 − 4.4 5 = 6 R1 or I D1 = 0.833 mA I R2 = 4.4 − ( −5 ) R2 + R3 = 9.4 = 0.94 mA 10 I D 2 = I R 2 − I D1 = 0.94 − 0.833 ⇒ I D 2 = 0.107 mA V2 = I R 2 R3 − 5 = ( 0.94 )( 5 ) − 5 ⇒ V2 = −0.3 V All diodes are on V1 = 4.4V , V2 = −0.6 V (c) I D1 = 0.5 mA = 10 − 0.6 − 4.4 ⇒ R1 = 10 k Ω R1 I R 2 = 0.5 + 0.5 = 1 mA = I R 3 = 1.5 mA = 4.4 − ( −0.6 ) −0.6 − ( −5 ) R3 R2 ⇒ R2 = 5 k Ω ⇒ R3 = 2.93 k Ω 2.41 ⎛ 0.5 ⎞ For vI small, both diodes off vO = ⎜ ⎟ vI = 0.0909vI ⎝ 0.5 + 5 ⎠ When vI − vO = 0.6, D1 turns on. So we have vI − 0.0909vI = 0.6 ⇒ vI = 0.66, vO = 0.06 vI − 0.6 − vO vI − vO vO 2v − 0.6 + = which yields vO = I 5 5 0.5 12 2vI − 0.6 When vO = 0.6, D2 turns on. Then 0.6 = ⇒ vI = 3.9 V 12 v − 0.6 − vO vI − vO vO vO − 0.6 + = + Now for vI > 3.9 I 5 5 0.5 0.5 2vI + 5.4 ; For vI = 10 ⇒ vO = 1.15 V Which yields vO = 22 For D1 on 2.42 ϩ10 V 10 K D2 D1 ␯I ␯0 D3 D4 10 K Ϫ10 V 10 K
  49. 49. For vI > 0. when D1 and D4 turn off 10 − 0.7 = 0.465 mA 20 v0 = I (10 kΩ ) = 4.65 V I= ␯0 4.65 Ϫ10 Ϫ4.65 4.65 10 ␯I Ϫ4.65 v0 = vI for − 4.65 ≤ vI ≤ 4.65 2.43 a. R1 D2 ϩ10 V V0 D1 ID1 R1 = 5 kΩ, R2 = 10 kΩ D1 and D2 on ⇒ V0 = 0 R2 Ϫ10 V 10 − 0.7 0 − ( −10 ) − = 1.86 − 1.0 5 10 = 0.86 mA I D1 = I D1 b. R1 = 10 kΩ, R2 = 5 kΩ, D1 off, D2 on I D1 = 0 I= 10 − 0.7 − ( −10 ) = 1.287 15 V0 = IR2 − 10 ⇒ V0 = −3.57 V 2.44 If both diodes on (a) VA = −0.7 V, VO = −1.4 V I R1 = IR2 I R1 + I D1 10 − ( −0.7 ) = 1.07 mA 10 −1.4 − ( −15 ) = = 2.72 mA 5 = I R 2 ⇒ I D1 = 2.72 − 1.07 I D1 = 1.65 mA D1 off, D2 on 10 − 0.7 − ( −15 ) I R1 = I R 2 = = 1.62 mA 5 + 10 VO = I R 2 R2 − 15 = (1.62 )(10 ) − 15 ⇒ VO = 1.2 V (b) VA = 1.2 + 0.7 = 1.9 V ⇒ D1 off , I D1 = 0 2.45
  50. 50. (a) D1 on, D2 off 10 − 0.7 I D1 = = 0.93 mA 10 VO = −15 V (b) D1 on, D2 off 10 − 0.7 I D1 = = 1.86 mA 5 VO = −15 V 2.46 15 − (V0 + 0.7 ) V0 + 0.7 V0 + 10 20 20 15 0.7 0.7 1 1 1 ⎞ ⎛ ⎛ 4.0 ⎞ − − = V0 ⎜ + + ⎟ = V0 ⎜ ⎟ 10 10 20 ⎝ 10 20 20 ⎠ ⎝ 20 ⎠ V0 = 6.975 V ID = = V0 ⇒ I D = 0.349 mA 20 2.47 10 K V1 Va Ϫ VD ϩ 10 K Vb V2 ID 10 K 10 K a. V1 = 15 V, V2 = 10 V Diode off Va = 7.5 V, Vb = 5 V ⇒ VD = −2.5 V ID = 0 b. V1 = 10 V, V2 = 15 V Diode on V2 − Vb Vb Va Va − V1 = + + ⇒ Va = Vb − 0.6 10 10 10 10 15 10 ⎛1 1⎞ ⎛1 1⎞ ⎛ 1 1⎞ + = Vb ⎜ + ⎟ + Vb ⎜ + ⎟ − 0.6 ⎜ + ⎟ 10 10 10 10 ⎠ 10 10 ⎠ ⎝ ⎝ ⎝ 10 10 ⎠ ⎛ 4⎞ 2.62 = Vb ⎜ ⎟ ⇒ Vb = 6.55 V ⎝ 10 ⎠ 15 − 6.55 6.55 ID = − ⇒ I D = 0.19 mA 10 10 VD = 0.6 V 2.48 vI = 0, D1 off, D2 on 10 − 2.5 = 0.5 mA 15 vo = 10 − ( 0.5 )( 5 ) ⇒ vo = 7.5 V for 0 ≤ vI ≤ 7.5 V I= For vI > 7.5 V , Both D1 and D2 on vI − vo vo − 2.5 vo − 10 = + or vI = vo ( 5.5 ) − 33.75 15 10 5 When vo = 10 V, D2 turns off vI = (10 )( 5.5 ) − 33.75 = 21.25 V For vI > 21.25 V, vo = 10 V
  51. 51. 2.49 a. V01 = V02 = 0 b. V01 = 4.4 V, V02 = 3.8 V c. V01 = 4.4 V, V02 = 3.8 V Logic “1” level degrades as it goes through additional logic gates. 2.50 a. V01 = V02 = 5 V b. V01 = 0.6 V, V02 = 1.2 V c. V01 = 0.6 V, V02 = 1.2 V Logic “0” signal degrades as it goes through additional logic gates. 2.51 (V1 AND V2 ) OR (V3 AND V4 ) 2.52 10 − 1.5 − 0.2 I= = 12 mA = 0.012 R + 10 8.3 R + 10 = = 691.7Ω 0.012 R = 681.7Ω 2.53 10 − 1.7 − VI =8 0.75 VI = 10 − 1.7 − 8 ( 0.75 ) ⇒ VI = 2.3 V I= 2.54 ϩ VR Ϫ ϩ VPS Ϫ Rϭ 2 K VR = 1 V, I = 0.8 mA VPS = 1 + ( 0.8 )( 2 ) VPS = 2.6 V 2.55 I Ph = η eΦA 0.6 × 10−3 = (1) (1.6 × 10−19 )(1017 ) A A = 3.75 × 10−2 cm 2
  52. 52. Chapter 3 Exercise Solutions EX3.1 VTN = 1 V , VGS = 3 V , VDS = 4.5 V VDS = 4.5 > VDS ( sat ) = VGS − VTN = 3 − 1 = 2 V Transistor biased in the saturation region I D = K n (VGS − VTN ) ⇒ 0.8 = K n ( 3 − 1) ⇒ K n = 0.2 mA / V 2 2 2 (a) VGS = 2 V, VDS = 4.5 V Saturation region: I D = ( 0.2 )( 2 − 1) ⇒ I D = 0.2 mA 2 (b) VGS = 3 V, VDS = 1 V Nonsaturation region: 2 I D = ( 0.2 ) ⎡ 2 ( 3 − 1)(1) − (1) ⎤ ⇒ I D = 0.6 mA ⎣ ⎦ EX3.2 VTP = −2 V , VSG = 3 V VSD ( sat ) = VSG + VTP = 3 − 2 = 1 V (a) (b) (c) VSD = 0.5 V ⇒ Nonsaturation VSD = 2 V ⇒ Saturation VSD = 5 V ⇒ Saturation EX3.3 ⎛ R2 ⎞ ⎛ 160 ⎞ VG = ⎜ ⎟ (VDD ) = ⎜ ⎟ (10 ) = 3.636 V = VGS ⎝ 160 + 280 ⎠ ⎝ R1 + R2 ⎠ I D = 0.25 ( 3.636 − 2 ) = 0.669 mA 2 VDS = 10 − ( 0.669 )(10 ) = 3.31 V P = I DVDS = ( 0.669 )( 3.31) = 2.21 mW EX3.4 I DQ = K P (VSG + VTP ) 2 1.2 = 0.4 (VSG − 1.2 ) ⇒ VSG = 2.932 V 2 ⎛ R1 ⎞ 1 VSG = ⎜ ⋅ VTTN − VDD ⎟ VDD = R2 ⎝ R1 + R2 ⎠ Note K = kΩ 1 2.932 = ( 200 )(10 ) ⇒ R2 = 682 K R2 682 R1 = 200 ⇒ R1 = 283 K 682 + R1 RD = 10 − 4 =5K 1.2 EX3.5 ⎛ R2 ⎞ ⎛ 40 ⎞ VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 = −1 V R1 + R2 ⎠ ⎝ 40 + 60 ⎠ ⎝ V − ( −5 ) 2 ID = S = K n (VGS − VTN ) RS (a) VS = VG − VGS
  53. 53. 2 ( 5 − 1) − VGS = ( 0.5 )(1) (VGS − 2VGS + 1) 2 2 0.5VGS − 3.5 = 0 VGS = 7 VGS = 2.646 V I D = ( 0.5 )( 2.646 − 1) ⇒ I D = 1.354 mA VDS = 10 − (1.354 )( 3) = 5.937 V 2 4 − VGS = K n (1)(VGS − VTN ) (b) 2 (1) K n = (1.05 )( 0.5 ) = 0.525 (2) K n = ( 0.95)( 0.5 ) = 0.475 (3) VTN = (1.05 )(1) = 1.05 V (4) VTN = ( 0.95 )(1) = 0.95 V (1)-(3) 2 4 − VGS = 0.525 (VGS − 2.1VGS + 1.1025 ) 2 0.525VGS − 0.1025VGS − 3.421 = 0 VGS = 0.1025 ± 0.010506 + 7.1841 = 2.652 V 2 ( 0.525 ) I D = 0.525 ( 2.652 − 1.05 ) = 1.348 mA VDS = 10 − (1.348 )( 3) = 5.957 V (2)-(4) 2 4 − VGS = 0.475 (VGS − 1.9VGS + 0.9025 ) 2 2 0.475VGS + 0.0975VGS − 3.5713 = 0 VGS = −0.0975 ± 0.00950625 + 6.78547 2 ( 0.475 ) VGS = 2.641 V I D = 0.475 ( 2.641 − 0.95 ) = 1.359 mA VDS = 10 − (1.359 )( 3) = 5.924 V (1)-(4) 2 4 −VGS = ( 0.525) (VGS −1.9VGS + 0.9025) 2 2 0.525 VGS + 0.0025VGS − 3.5262 = 0 VGS = −0.0025 ± 0.00000625 + 7.40502 2 ( 0.525) = 2.5893 V I D = ( 0.525)( 2.5893 − 0.95) = 1.411 2 VDS = 10 − I D ( 3) = 5.7678 V (2)-(3) 2 4 − VGS = 0.475 (VGS − 2.1VGS +1.1025 ) 2 0.475VGS + 0.0025VGS − 3.4763 = 0 −0.0025 ± 0.00000625 + 6.60499 2(0.475) = 2.7027 VGS = VGS I D = (0.475)(2.7027 − 1.05) 2 = 1.2973 mA VDS = 10 − I D (3) = 6.108 V 1.297 ≤ I DQ ≤ 1.411 mA 5.768 ≤ VDS ≤ 6.108 V EX3.6
  54. 54. ⎛ R2 ⎞ VG = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠ ⎛ 200 ⎞ =⎜ ⎟ (10 ) − 5 = 0.714 V ⎝ 350 ⎠ VS = 5 − I D RS = 5 − (1.2 ) I D So VSG = VS − VG = 5 − (1.2 ) I D − 0.714 = 4.286 − (1.2 ) I D ID = 4.286 − VSG 1.2 I D = K p (VSG + VTP ) 2 ( 2 4.286 − VSG = (1.2 )( 0.25 ) × VSG − 2VSG ( −1) + ( −1) 2 4.286 − VSG = ( 0.3) V − 0.6VSG + 0.3 ) 2 SG 2 0.3VSG + 0.4VSG − 3.986 = 0 VSG = ( 0.4 ) −0.4 ± + 4 ( 0.3)( 3.986 ) 2 2 ( 0.3) Must use + sign ⇒ VSG = 3.04 V I D = ( 0.25 )( 3.04 − 1) ⇒ I D = 1.04 mA 2 VSD = 10 − I D ( RS + RD ) = 10 − (1.04 )(1.2 + 4 ) ⇒ VSD = 4.59 V VSD > VSD ( sat ) , Yes EX3.7 VSD = 10 − I DQ ( RS + RP ) VSD = 10 − K P (VSG + VTP ) ( RS + RP ) 2 Set VSD = VSG + VTP VSG + VTP = 10 − ( 0.25 )(VSG + VTP ) ( 5.2 ) 2 1.3 (VSG + VTP ) + (VSG + VTP ) − 10 = 0 2 (VSG + VTP ) = −1 ± 1 + 4 (1.3)(10 ) 2 (1.3) = 2.415 V ( 3.42 V ) VSD = 2.415 V ( 2.42 V ) 2 I D = ( 0.25 )( 2.415 ) = 1.46 mA VSG = 3.415 V EX3.8 ⎛ R2 ⎞ ⎛ 240 ⎞ VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 R1 + R2 ⎠ ⎝ 240 + 270 ⎠ ⎝ VG = −0.294 V ID = VS − ( −5 ) RS = VG − VGS + 5 2 = K n (VGS − VTN ) RS ⎛ 0.08 ⎞ 2 4.706 − VGS = ⎜ ⎟ ( 4 )( 3.9 ) (VGS − 2.4VGS + 1.44 ) ⎝ 2 ⎠ 2 0.624VGS − 0.4976VGS − 3.80744 = 0
  55. 55. VGS = 0.4976 ± 0.2476 + 9.50337 2 ( 0.624 ) VGS = 2.90 V 2 ⎛ 0.08 ⎞ ID = ⎜ ⎟ ( 4 )( 2.90 − 1.2 ) ⇒ I D = 0.463 mA 2 ⎠ ⎝ VDS = 10 − I D ( 3.9 + 10 ) ⇒ VDS = 3.57 V EX3.9 10 − VSG 2 ID = and I D = K p (VSG + VTP ) RS 0.12 = ( 0.050 )(VSG − 0.8 ) VSG = 2.35 V 2 10 − 2.35 ⇒ RS = 63.75 kΩ 0.12 VSD = 8 = 20 − I D ( RS + RD ) 8 = 20 − ( 0.12 )( 63.75 ) − ( 0.12 ) RD RS = RD = (1) (2) (3) (4) 20 − ( 0.12 )( 63.75 ) − 8 KP KP VTP VTP ID ⇒ RD = 36.25 kΩ 0.12 = ( 0.05 )(1.05 ) = 0.0525 = ( 0.05 )( 0.95 ) = 0.0475 = −0.8 (1.05 ) = −0.84 V = −0.8 ( 0.95 ) = −0.76 V 10 − VSG 2 = = K P (VSG + VTP ) RS (1)-(3) 2 10 − VSG = ( 63.75 )( 0.0525 ) ⎡VSG − 1.68VSG + 0.7056 ⎤ ⎣ ⎦ 2 3.347VSG − 4.623VSG − 7.6384 = 0 VSG = 4.623 ± 21.372 + 102.263 2 ( 3.347 ) VSG = 2.352 V ⇒ I D ≅ 0.120 mA VSD ≈ 8.0 V (2)-(4) 2 10 − VSG = ( 63.75 )( 0.0475 ) ⎡VSG − 1.52VSG + 0.5776 ⎤ ⎣ ⎦ 2 3.028VSG − 3.603VSG − 8.251 = 0 VSG = 3.603 ± 12.9816 + 99.936 2 ( 3.028 ) VSG = 2.35 V I D ≈ 0.120 VSD ≈ 8.0 (1)-(4) 2 10 − VSG = ( 63.75 )( 0.0525 ) ⎡VSG − 1.52VSG + 0.5776 ⎤ ⎣ ⎦ 2 3.347VSG − 4.087VSG − 8.06685 = 0
  56. 56. VSG = 4.087 ± 16.7036 + 107.999 2 ( 3.347 ) VSG = 2.279 V I D = 0.121 mA VSD = 7.89 V (2)-(3) 2 10 − VSG = ( 63.75 )( 0.0475 ) ⎡VSG − 1.68VSG + 0.7056 ⎤ ⎣ ⎦ 2 3.028VSG − 4.0873VSG − 7.8634 = 0 VSG = 4.0873 ± 16.706 + 95.242 2 ( 3.028 ) VSG = 2.422 V I D = 0.119 mA VSD = 8.11 V Summary 0.119 ≤ I D ≤ 0.121 mA 7.89 ≤ VSD ≤ 8.11 V EX3.10 V − VGS , I D = DD RS I D = K n (VGS − VTN ) 2 2 2 10 − VGS = (10 )( 0.2 ) (VGS − 2VGSVTN + VTN ) 2 10 − VGS = 2VGS − 8VGS + 8 2 2VGS − 7VGS − 2 = 0 VGS = 7± (7) + 4 ( 2) 2 2 ( 2) 2 Use + sign: VGS = VDS = 3.77 V 10 − 3.77 ⇒ I D = 0.623 mA 10 Power = I DVDS = ( 0.623)( 3.77 ) ⇒ Power = 2.35 mW ID = EX3.11 (a) VI = 4 V, Driver in Non ⋅ Sat. K nD ⎡ 2 (VI − VTND ) VO − VO2 ⎤ = K nL [VDD − VO − VTNL ] ⎣ ⎦ 2 2 5 ⎡ 2 ( 4 − 1) VD − VD ⎤ = ( 5 − VD − 1) = ( 4 − VO ) = 16 − 8VO + VO2 ⎣ ⎦ 2 2 2 6VD − 38VO + 16 = 0 VD = 38 ± 1444 − 384 2 ( 6) VD = 0.454 V (b) VI = 2 V Driver: Sat K nD [VI − VTND ] = K nL [VDD − VO − VTNL ] 2 2 5 [ 2 − 1] = [5 − VO − 1] 2 2 5 = 4 − VO ⇒ VO = 1.76 V EX3.12 If the transistor is biased in the saturation region
  57. 57. I D = K n (VGS − VTN ) = K n ( −VTN ) 2 2 I D = ( 0.25 )( 2.5 ) ⇒ I D = 1.56 mA 2 VDS = VDD − I D RS = 10 − (1.56 )( 4 ) ⇒ VDS = 3.75 VDS > VGS − VTN = −VTN 3.75 > − ( −2.5 ) Yes — biased in the saturation region Power = I DVDS = (1.56 )( 3.75 ) ⇒ Power = 5.85 mW EX3.13 (a) For VI = 5 V, Load in saturation and driver in nonsaturation. I DD = I DL K nD ⎡ 2 (VI − VTND ) VO − VO2 ⎤ = K nL ( −VTNL ) ⎣ ⎦ 2 K nD ⎡ K 2 2 5 − 1)( 0.25 ) − ( 0.25 ) ⎤ = 4 ⇒ nD = 2.06 ⎣ ( ⎦ K nL K nL (b) I DL = K nL ( −VTNL ) ⇒ 0.2 = K nL ⎡ − ( −2 ) ⎤ ⎣ ⎦ 2 2 K nL = 50 μ A / V 2 and K nD = 103 μ A / V 2 EX3.14 For M N I DN = I DP K n (VGSN − VTN ) = K p (Vscop + VTP ) 2 2 VGSN = 1 + ( 5 − 3.25 − 1) = 1.75 V = VI Vo = VDSN ( sat ) = 1.75 − 1 ⇒ Vo = 0.75 V For M P : VI = 1.75 V VDD − VO = VSD ( sat ) = VSGP + VTP = ( 5 − 3.25 ) − 1 = 0.75 V So Vot = 5 − 0.75 ⇒ Vot = 4.25 V EX3.15 For RD = 10 k Ω, VDD = 5 V, and Vo = 1 V 5 −1 = 0.4 mA 10 2 I D = K n ⎡ 2 (VGS − VTN ) VDS − VDS ⎤ ⎣ ⎦ ID = 2 I D = 0.4 = K n ⎡ 2 ( 5 − 1)(1) − (1) ⎤ ⇒ K n = 0.057 mA / V 2 ⎣ ⎦ P = I D ⋅ VDS = ( 0.4 )(1) ⇒ P = 0.4 mW EX3.16 a. V1 = 5 V, V2 = 0, M 2 cutoff ⇒ I D 2 = 0 I D = K n ⎡ 2 (VI − VTN ) VO − VO2 ⎤ = ⎣ ⎦ 5 − VO RD ( 0.05 )( 30 ) ⎡ 2 ( 5 − 1)V0 − V02 ⎤ = 5 − V0 ⎣ ⎦
  58. 58. 1.5V02 − 13V0 + 5 = 0 V0 = 13 ± (13) − 4 (1.5 )( 5) ⇒ V0 = 0.40 V 2 (1.5 ) 2 5 − 0.40 ⇒ I R = I D1 = 0.153 mA 30 V1 = V2 = 5 V I R = I D1 = b. 5 − VO = 2 K n ⎡ 2 (VI − VTN ) VO − VO2 ⎤ ⎣ ⎦ RD { } 5 − V0 = 2 ( 0.05 )( 30 ) ⎡ 2 ( 5 − 1)V0 − V02 ⎤ ⎣ ⎦ 3V02 − 25V0 + 5 = 0 V0 = 25 ± ( 25) − 4 ( 3)( 5 ) ⇒ V0 = 0.205 V 2 ( 3) 2 5 − 0.205 ⇒ I R = 0.160 mA 30 = I D 2 = 0.080 mA IR = I D1 EX3.17 M 2 & M 3 watched ⇒ I Q1 = I REF 1 = 0.4 mA 0.4 = 0.3 (VGS 3 − 1) ⇒ VGS 3 = VGS 2 = 2.15 V 2 0.4 = 0.6 (VGS 1 − 1) ⇒ VGS1 = 1.82 V 2 EX3.18 2 ⎛ 0.04 ⎞ 0.1 = ⎜ ⎟ (15 )(VSGC − 0.6 ) ⎝ 2 ⎠ VSGC = 1.177 V = VSGB 2 ⎛ 0.04 ⎞ ⎛ W ⎞ 0.2 = ⎜ ⎟ ⎜ ⎟ (1.177 − 0.6 ) ⎝ 2 ⎠ ⎝ L ⎠B ⎛W ⎞ ⎜ ⎟ = 30 ⎝ L ⎠B 2 ⎛ 0.04 ⎞ 0.2 = ⎜ ⎟ ( 25 )(VSGA − 0.6 ) ⎝ 2 ⎠ VSGA = 1.23 V EX3.19 (a) I REF = K n 3 (VGS 3 − VTN ) = K n 4 (VGS 4 − VTN ) VGS 3 = 2 V ⇒ VGS 4 = 3 V K K 1 2 2 ( 2 − 1) = n 4 ( 3 − 1) ⇒ n 4 = K n3 K n3 4 (b) I Q = K n 2 (VGS 2 − VTN ) But VGS 2 = VGS 3 = 2 V 2 0.1 = K n 2 ( 2 − 1) ⇒ 2 (c) 2 K n 2 = 0.1 mA / V 2 0.2 = K n 3 ( 2 − 1) ⇒ K n 3 = 0.2 mA / V 2 2 0.2 = K n 4 ( 3 − 1) ⇒ K n 4 = 0.05 mA / V 2 2 EX3.20 2
  59. 59. VS 2 = 5 − 5 = 0 RS 2 = I D 2 = K n 2 (VGS 2 − VTN 2 ) 5 = 16.7 K 0.3 2 0.3 = 0.2 (VGS 2 − 1.2 ) ⇒ VGS 2 = 2.425 V ⇒ VG 2 = VGS 2 + VS = 2.425 V 2 5 − 2.425 = 25.8 K 0.1 VS 1 = VG 2 − VDSQ1 = 2.425 − 5 = −2.575 V RD1 = RS 1 = −2.575 − ( −5 ) 0.1 ⇒ RS 1 = 24.3 K I D1 = K n1 (VGS 1 − VTN 1 ) 2 0.1 = 0.5 (VGS1 − 1.2 ) ⇒ VGS 1 = 1.647 V VG1 = VGS 1 + VS 1 = 1.647 + ( −2.575 ) ⇒ VG1 = −0.928 V 2 ⎛ R2 ⎞ 1 VG1 = ⎜ ⎟ (10 ) − 5 = ⋅ RTN ⋅ (10 ) − 5 R1 ⎝ R1 + R2 ⎠ 1 −0.928 = ( 200 )(10 ) − 5 ⇒ R1 = 491 K R1 491 R2 = 200 ⇒ R2 = 337 K 491 + R2 EX3.21 VS1 = I D RS − 5 = (0.25)(16) − 5 = −1 V I DQ = K n (VGS1 − VTN ) 2 ⇒ 0.25 = 0.5(VGS 1 − 0.8) 2 ⇒ VGS 1 = 1.507 V VG1 = VGS 1 + VS 1 = 1.507 − 1 = 0.507 V ⎛ ⎞ R3 R3 (5) ⇒ R3 = 50.7 K VG1 = ⎜ ⎟ (5) ⇒ 0.507 = 500 ⎝ R1 + R2 + R3 ⎠ VS 2 = VS 1 + VDS1 = −1 + 2.5 = 1.5 V VG 2 = VS 2 + VGS = 1.5 + 1.507 = 3.007 V ⎛ R2 + R3 ⎞ ⎛ R2 + R3 ⎞ VG 2 = ⎜ ⎟ (5) ⇒ 3.007 = ⎜ ⎟ (5) R1 + R2 + R3 ⎠ ⎝ 500 ⎠ ⎝ R2 + R3 = 300.7 R2 = 300.7 − 50.7 ⇒ R2 = 250 K R1 = 500 − 250 − 50.7 ⇒ R1 = 199.3 K VD 2 = VS 2 + VDS 2 = 1.5 + 2.5 = 4 V 5−4 ⇒ RD = 4 K RD = 0.25 EX3.22 VDS ( sat ) = VGS − VP = −1.2 − ( −4.5 ) ⇒ VDS ( sat ) = 3.3 V ⎛ ( −1.2 ) ⎞ ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ = 12 ⎜ 1 − ⇒ I D = 6.45 mA ⎜ ( −4.5 ) ⎟ ⎟ VP ⎠ ⎝ ⎝ ⎠ 2 2 EX3.23 Assume the transistor is biased in the saturation region.
  60. 60. ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ ⎞ V 8 = 18 ⎜ 1 − GS ⎟ ⇒ VGS = −1.17 V ⇒ VS = −VGS = 1.17 ⎜ ( −3.5 ) ⎟ ⎝ ⎠ VD = 15 − ( 8 )( 0.8 ) = 8.6 VDS = 8.6 − (1.17 ) = 7.43 V VDS = 7.43 > VGS − VP = −1.17 − ( −3.5 ) = 2.33 Yes, the transistor is biased in the saturation region. EX3.24 I D = 2.5 mA ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ V ⎞ 2.5 = 6 ⎜1 − GS ⎟ ⇒ VGS = −1.42 V ⎜ ( −4 ) ⎟ ⎝ ⎠ VS = I D RS − 5 = ( 2.5 )( 0.25 ) − 5 VS = −4.375 VDS = 6 ⇒ VD = 6 − 4.375 = 1.625 5 − 1625 RD = ⇒ RD = 1.35 kΩ 2.5 ( 20 ) 2 R1 + R2 = 2 ⇒ R1 + R2 = 200 kΩ VG = VGS + VS = −1.42 − 4.375 = −5.795 ⎛ R2 ⎞ VG = ⎜ ⎟ ( 20 ) − 10 ⎝ R1 + R2 ⎠ ⎛ R ⎞ −5.795 = ⎜ 2 ⎟ ( 20 ) − 10 ⇒ R2 = 42.05 kΩ → 42 kΩ ⎝ 200 ⎠ R1 = 157.95 kΩ → 158 kΩ EX3.25 VS = −VGS . I D = 0 − VS VGS = RS RS ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 ⎛ V VGS V2 ⎞ ⎛ V ⎞ = 6 ⎜ 1 − GS ⎟ = 6 ⎜ 1 − GS + GS ⎟ 1 4 ⎠ 2 16 ⎠ ⎝ ⎝ 2 0.375VGS − 4VGS + 6 = 0 2 VGS = 4 ± 16 − 4 ( 0.375 )( 6 ) 2 ( 0.375 ) VGS = 8.86 or VGS = 1.806 V impossible ID = VGS = 1.806 mA RS
  61. 61. VD = I D RD − 5 = (1.81)( 0.4 ) − 5 = −4.278 VSD = VS − V0 = −1.81 − ( −4.276 ) ⇒ VSD = 2.47 V VSD ( sat ) = VP − VGS = 4 − 1.81 = 2.19 So VSD > VSD ( sat ) EX3.26 Rin = R1 R2 = R1 R2 = 100 kΩ R1 + R2 I DQ = 5 mA, VS = − I DQ RS = − ( 5 )(1.2 ) = −6 V VSDQ = 12 V ⇒ VD = VS − VSDQ = −6 − 12 = −18 V RD = −18 − ( −20 ) 5 ⇒ RD = 0.4 kΩ 2 ⎛ V ⎞ ⎛ V ⎞ I DQ = I DSS ⎜ 1 − GS ⎟ ⇒ 5 = 8 ⎜ 1 − GS ⎟ VP ⎠ 4 ⎠ ⎝ ⎝ VGS = 0.838 V 2 VG = VGS + VS = 0.838 − 6 = −5.162 ⎛ R2 ⎞ VG = ⎜ ⎟ ( −20 ) ⎝ R1 + R2 ⎠ 1 −5.162 = (100 )( −20 ) ⇒ R1 = 387 kΩ R1 R1 R2 = 100 ⇒ ( 387 ) R2 = 100 ( 387 ) + 100 R2 R1 + R2 ( 387 − 100 ) R2 = (100 )( 387 ) ⇒ R2 = 135 kΩ TYU3.1 VTN = 1.2 V , VGS = 2 V (a) V DS ( sat ) = VGS − VTN = 2 − 1.2 = 0.8 V (i) VDS = 0.4 ⇒ Nonsaturation (ii) VDS = 1 ⇒ Saturation (iii) VDS = 5 ⇒ Saturation (b) VTN = −1.2 V , VGS = 2 V V DS ( sat ) = VGS − VTN = 2 − ( −1.2 ) = 3.2 V (i) VDS = 0.4 ⇒ Nonsaturation (ii) VDS = 1 ⇒ Nonsaturation (iii) VDS = 5 ⇒ Saturation TYU3.2 (a) W μ n Cox 2L −14 ∈ox ( 3.9 ) ( 8.85 × 10 ) Cox = = = 7.67 × 10−8 F / cm −8 tox 450 × 10 Kn = Kn = (100 )( 500 ) ( 7.67 ×10−8 ) ⇒ K n = 0.274 mA / V 2 2 (7) (b) VTN = 1.2 V, VGS = 2 V
  62. 62. (i) VDS = 0.4 V ⇒ Nonsaturation (ii) 2 I D = ( 0.274 ) ⎡ 2 ( 2 − 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ I D = 0.132 mA ⎣ ⎦ VDS = 1 V ⇒ Saturation I D = ( 0.274 )( 2 − 1.2 ) ⇒ I D = 0.175 mA 2 (iii) VDS = 5 V ⇒ Saturation I D = ( 0.274 )( 2 − 1.2 ) ⇒ I D = 0.175 mA 2 VTN = −1.2 V , VGS = 2 V (i) VDS = 0.4 V ⇒ Nonsaturation (ii) 2 I D = ( 0.274 ) ⎡ 2 ( 2 + 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ I D = 0.658 mA ⎣ ⎦ VDS = 1 V ⇒ Nonsaturation 2 I D = ( 0.274 ) ⎡ 2 ( 2 + 1.2 )(1) − (1) ⎤ ⇒ I D = 1.48 mA ⎣ ⎦ (iii) VDS = 5 V ⇒ Saturation I D = ( 0.274 )( 2 + 1.2 ) ⇒ I D = 2.81 mA 2 TYU3.3 (a) VSD (sat) = VSG + VTP = 2 − 1.2 = 0.8 V (i) Non Sat (ii) Sat (iii) Sat (b) VSD (sat) = 2 + 1.2 = 3.2 V (i) Non Sat (ii) Non Sat (iii) Sat TYU3.4 (a) (3.9)(8.85 × 10−14 ) ⎛ W ⎞ ⎛ μ p Cox ⎞ KP = ⎜ ⎟ ⎜ ⎟ Cox = 350 × 10−8 ⎝ L ⎠⎝ Z ⎠ = 9.861× 10−8 KP = (40) ⎛ ( 300 ) ( 9.861× 10 ⎜ (2) ⎜ 2 ⎝ K P = 0.296 mA / V (b) (i) −8 )⎞ ⎟ ⎟ ⎠ 2 I D = (0.296) ⎡ 2(2 − 1.2)(0.4) − (0.4) 2 ⎤ ⎣ ⎦ = 0.142 mA (ii) I D = (0.296) [ 2 − 1.2] ⇒ I D = 0.189 mA 2 (iii) ID = 0.189 mA 2 (i) I D = (0.296) ⎡ 2 ( 2 + 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⎣ ⎦ = 0.710 mA (ii) 2 I D = (0.296) ⎡ 2 ( 2 + 1.2 )(1) − (1) ⎤ ⎣ ⎦ =1.60 mA (iii) I D = ( 0.296 )( 2 + 1.2 ) = 3.03 mA TYU3.5 2
  63. 63. (a) λ = 0, VDS ( sat ) = 2.5 − 0.8 = 1.7 V For VDS = 2 V , VDS = 10 V ⇒ Saturation Region I D = ( 0.1)( 2.5 − 0.8 ) ⇒ I D = 0.289 mA 2 (b) λ = 0.02 V −1 I D = K n (VGS − VTN ) (1 + λVDS ) For VDS = 2 V 2 I D = ( 0.1)( 2.5 − 0.8 ) ⎡1 + ( 0.02 )( 2 ) ⎤ ⇒ I D = 0.300 mA ⎣ ⎦ VDS = 10 V 2 (c) 2 I D = ( 0.1) ⎡( 2.5 − 0.8 ) (1 + ( 0.02 )(10 ) ) ⎤ ⇒ I D = 0.347 mA ⎣ ⎦ For part (a), λ = 0 ⇒ ro = ∞ For part (b), λ = 0.02 V −1 , −1 2 2 ro = ⎡λ K n (VGS − VTN ) ⎤ = ⎡( 0.02 )( 0.1)( 2.5 − 0.8 ) ⎤ ⎣ ⎦ ⎣ ⎦ −1 or ro = 173 k Ω TYU3.6 VTN = VTNO + γ ⎡ 2φ f + VSB − 2φ f ⎤ ⎣ ⎦ 2φ f = 0.70 V , VTNO = 1 V (a) VSB = 0 ⇒, VTN = 1 V (b) ⎡ ⎤ VSB = 1 V , VTN = 1 + ( 0.35 ) ⎣ 0.7 + 1 − 0.7 ⎦ ⇒ VTN = 1.16 V VSB = 4 V , VTN = 1 + ( 0.35 ) ⎡ 0.7 + 4 − 0.7 ⎤ ⇒ VTN = 1.47 V ⎣ ⎦ (c) TYU3.7 I D = K n (VGS − VTN ) 2 0.4 = 0.25 (VGS − 0.8 ) ⇒ VGS = 2.06 V 2 ⎛ R2 ⎞ VGS = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 2.06 = ⎜ 2 ⎟ ( 7.5 ) ⇒ R2 = 68.8 kΩ ⎝ 250 ⎠ R1 = 181.2 kΩ VDS = 4 = VDD − I D RD 7.5 − 4 RD = ⇒ RD = 8.75 kΩ 0.4 VDS > VDS ( sat ) , Yes TYU3.8
  64. 64. VS − ( −5 ) ID = and VS = −VGS RS So RS = 5 − VGS 0.1 I D = K n (VGS − VTN ) 2 0.1 = ( 0.080 )(VGS − 1.2 ) ⇒ VGS = 2.32 V 5 − 2.32 So RS = ⇒ RS = 26.8 kΩ 0.1 VDS = VD − VS ⇒ VD = VDS + VS = 4.5 − 2.32 2 VD = 2.18 5 − VD 5 − 2.18 RD = = ⇒ RD = 28.2 kΩ ID 0.1 VDS > VDS ( sat ) , Yes TYU3.9 For VDS = 2.2 V 5 − 2.2 ⇒ I D = 0.56 mA ID = 5 I D = K n (VGS − VTN ) 0.56 = K n ( 2.2 − 1) 2 2 K n = 0.389 mA / V = W μ n Cox ⋅ 2 L W ( 389 )( 2 ) W = ⇒ = 19.4 L L ( 40 ) TYU3.10 (a) The transition point is ( VDD − VTNL + VTND 1 + K nD / K nL VIt = = ) 1 + K nD /K nL ( 5 − 1 + 1 1 + 0.05/ 0.01 ) 1 + 0.05/ 0.01 7.236 = ⇒ VIt = 2.236 V 3.236 VOt = VIt − VTND = 2.24 − 1 ⇒ VOt = 1.24 V (b) We may write I D = K n D (VGSD − VTND ) = ( 0.05 )( 2.236 − 1) ⇒ I D = 76.4 μ A 2 TYU3.11 2 ( VDD − VTNL + VTND 1 + K nD /K nL VIt = ) 1 + K nD /K nL 2.5 = ( 5 − 1 + 1 1 + K nD /K nL ) 1 + K nD /K nL 2.5 + 2.5 K nD /K nL = 5 + K nD /K nL ⇒ K nD /K nL = b. For VI = 5, driver in nonsaturated region. 5 − 2.5 = 1.67 ⇒ K nD /K nL = 2.78 1.5
  65. 65. I DD = I DL K nD ⎡ 2 (VI − VTND ) VO − VO2 ⎤ = K nL (VGSL − VTNL ) ⎣ ⎦ 2 K nD 2 ⎡ 2 (VI − VTND ) VO − VO2 ⎤ = [VDD − VO − VTNL ] ⎦ K nL ⎣ 2.78 ⎡ 2 ( 5 − 1) V0 − V02 ⎤ = [5 − V0 − 1] ⎣ ⎦ 2 22.24V0 − 2.78V02 = ( 4 − V0 ) 2 = 16 − 8V0 + V02 3.78V02 − 30.24V0 + 16 = 0 V0 = 30.24 ± ( 30.24 ) − 4 ( 3.78 )(16 ) ⇒ V0 = 0.57 V 2 ( 3.78 ) 2 TYU3.12 We have VDS = 1.2 V < VGS − VTN = −VTN = 1.8 V Transistor is biased in the nonsaturation region. V − VDS 5 − 1.2 2 ⎡ ⎤ = ⇒ I D = 0.475 mA I D = K n ⎣ 2 (VGS − VTN ) VDS − VDS ⎦ and I D = DD 8 RS 0.475 = K n ⎡ 2 ( 0 − ( −1.8 ) ) (1.2 ) − (1.2 ) ⎤ ⎣ ⎦ 0.475 = K n ( 2.88 ) ⇒ K n = 0.165 mA/V 2 2 W μ n Cox ⋅ 2 L 165 )( 2 ) W ( W = ⇒ = 9.43 35 L L Kn = TYU3.13 (a) Transition point for the load transistor – Driver is in the saturation region. I DD = I DL K nD (VGSD − VTND ) = K nL (VGSL − VTNL ) 2 2 VDSL ( sat ) = VGSL − VTNL = −VTNL ⇒ VDSL = VDD − VOt = 2 V Then VOt = 5 − 2 = 3 V , VOt = 3 V K nD (VIt − 1) = ( −VTNL ) K nL 0.08 (VIt − 1) = 2 ⇒ VIt = 1.89 V 0.01 (b) For the driver: VOt = VIt − VTND VIt = 1.89 V , VOt = 0.89 V TYU3.14 2 I D = K n ⎡ 2 (VGS − VTN ) VDS − VDS ⎤ ⎣ ⎦ 2 = ( 0.050 ) ⎡ 2 (10 − 0.7 )( 0.35 ) − ( 0.35 ) ⎤ ⎣ ⎦ I D = 0.319 mA RD = VDD − Vo 10 − 0.35 = ⇒ RD = 30.3 kΩ ID 0.319 TYU3.15 (a) Transistor biased in the nonsaturation region
  66. 66. 5 − 1.5 − VDS = 12 R 2 I D = K n ⎡ 2 (VGS − VTN ) VDS − VDS ⎤ ⎣ ⎦ ID = 2 12 = 4 ⎡ 2 ( 5 − 0.8 ) VDS − VDS ⎤ ⎣ ⎦ 2 4VDS − 33.6VDS + 12 = 0 ⇒ VDS = 0.374 V Then R = 5 − 1.5 − 0.374 ⇒ R = 261 Ω 12 TYU3.16 a. ID = 5 − VO = K n ⎡ 2 (V2 − VTN ) VO − VO2 ⎤ ⎣ ⎦ RD 5 − ( 0.10 ) b. 2 = K n ⎡ 2 ( 5 − 1)( 0.10 ) − ( 0.10 ) ⎤ ⇒ K n = 0.248 mA / V 2 ⎣ ⎦ 25 5 − V0 = 2 ( 0.248 ) ⎡ 2 ( 5 − 1) V0 − V02 ⎤ ⎣ ⎦ 25 2 5 − V0 = 12.4 ⎡8V0 − V0 ⎤ ⎣ ⎦ 12.4V02 − 100.2V0 + 5 = 0 V0 = 100.2 ± (100.2 ) − 4 (12.4 )( 5 ) ⇒ V0 = 0.0502 V 2 (12.4 ) 2 TYU3.17 2 2 I DQ = K (VGS − VTN ) ⇒ 5 = 50 (VGS − 0.15 ) ⇒ VGS = 0.466 V VS = ( 0.005 )(10 ) = 0.050 V ⇒ VGG = VGS + VS = 0.466 + 0.050 ⇒ VGG = 0.516 V VD = 5 − ( 0.005 )(100 ) ⇒ VD = 4.5 V VDS = VD − VS = 4.5 − 0.050 ⇒ VDS = 4.45 V TYU3.18 2 I D = K ⎡ 2 (VGS − VTN ) VDS − VDS ⎤ ⎣ ⎦ 2 = 100 ⎡ 2 ( 0.7 − 0.2 )( 0.1) − ( 0.1) ⎤ ⎣ ⎦ ID = 9 μA RD = 2.5 − 0.1 ⇒ RD = 267 kΩ 0.009
  67. 67. Chapter 3 Problem Solutions 3.1 ⎛ W ⎞ ⎛ k ′ ⎞ ⎛ 10 ⎞ ⎛ 0.08 ⎞ 2 Kn = ⎜ ⎟ ⎜ n ⎟ = ⎜ ⎟⎜ ⎟ = 0.333 mA/V L ⎠ ⎝ 2 ⎠ ⎝ 1.2 ⎠ ⎝ 2 ⎠ ⎝ For VDS = 0.1 V ⇒ Non Sat Bias Region (a) VGS = 0 ⇒ I D = 0 (b) 2 VGS = 1 V I D = 0.333 ⎡ 2 (1 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.01 mA ⎣ ⎦ (c) VGS = 2 V 2 I D = 0.333 ⎡ 2 ( 2 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.0767 mA ⎣ ⎦ (d) VGS = 3 V 2 I D = 0.333 ⎡ 2 ( 3 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.143 mA ⎣ ⎦ 3.2 All in Sat region ⎛ 10 ⎞⎛ 0.08 ⎞ 2 Kn = ⎜ ⎟⎜ ⎟ = 0.333 mA/V ⎝ 1.2 ⎠⎝ 2 ⎠ (a) ID = 0 (b) I D = 0.333[1 − 0.8] = 0.0133 mA (c) I D = 0.333[ 2 − 0.8] = 0.480 mA (d) I D = 0.333[3 − 0.8] = 1.61 mA 2 2 2 3.3 (a) Enhancement-mode (b) From Graph VT = 1.5 V Now 2 0.03 = K n ( 2 − 1.5 ) = 0.25 K n ⇒ K n = 0.12 0.15 = K n ( 3 − 1.5 ) = 2.25 K n K n = 0.0666 0.39 = K n ( 4 − 1.5 ) = 6.25 K n K n = 0.0624 0.77 = K n ( 5 − 1.5 ) = 12.25 K n K n = 0.0629 2 2 2 From last three, K n (Avg) = 0.0640 mA/V 2 (c) 3.4 a. iD (sat) = 0.0640(3.5 − 1.5) 2 ⇒ iD (sat) = 0.256 mA for VGS = 3.5 V iD (sat) = 0.0640(4.5 − 1.5) 2 ⇒ iD (sat) = 0.576 mA for VGS = 4.5 V VGS = 0 VDS ( sat ) = VGS − VTN = 0 − ( −2.5 ) = 2.5 V i. VDS = 0.5 V ⇒ Biased in nonsaturation ii. 2 I D = (1.1) ⎡ 2 ( 0 − (−2.5) )( 0.5 ) − ( 0.5 ) ⎤ ⇒ I D = 2.48 mA ⎣ ⎦ VDS = 2.5 V ⇒ Biased in saturation I D = (1.1) ( 0 − ( −2.5 ) ) ⇒ I D = 6.88 mA 2 iii. VDS = 5 V Same as (ii) ⇒ I D = 6.88 mA b. VGS = 2 V VDS ( sat ) = 2 − ( −2.5 ) = 4.5 V i. VDS = 0.5 V ⇒ Nonsaturation I D = (1.1) ⎡ 2(2 − (−2.5))(0.5) − (0.5) 2 ⎤ ⇒ I D = 4.68 mA ⎣ ⎦
  68. 68. VDS = 2.5 V ⇒ Nonsaturation ii. I D = (1.1) ⎡ 2(2 − (−2.5))(2.5) − (2.5) 2 ⎤ ⇒ I D = 17.9 mA ⎣ ⎦ VDS = 5 V ⇒ Saturation iii. I D = (1.1) ( 2 − ( −2.5 ) ) ⇒ I D = 22.3 mA 2 3.5 VDS > VGS − VTN = 0 − ( −2 ) = 2 V Biased in the saturation region k′ W 2 I D = n ⋅ (VGS − VTN ) 2 L 2 W ⎛ 0.080 ⎞ ⎛ W ⎞ = 9.375 1.5 = ⎜ ⎟ ⎜ ⎟ ⎡ 0 − ( −2 ) ⎤ ⇒ ⎣ ⎦ L ⎝ 2 ⎠⎝ L ⎠ 3.6 ′ kn = μ n Cox = μ n ∈ox tox = ( 600 )( 3.9 ) (8.85 ×10−14 ) tox (a) 500 A 250 ′ kn = 82.8 μ A/V 2 (c) 100 ′ kn = 207 μ A/V 2 (d) 50 ′ kn = 414 μ A/V 2 (e) 25 2.071× 10−10 tox ′ kn = 41.4 μ A/V 2 (b) = ′ kn = 828 μ A/V 2 3.7 a. Cox = Kn = ∈ox ( 3.9 ) ( 8.85 × 10 = t0 x 450 × 10−8 −14 )⇒∈ ox t0 x = 7.67 ×10−8 F/cm 2 μ n Cox W 2 ⋅ L 1 64 ( 650 ) ( 7.67 ×10−8 ) ⎛ ⎞ ⎜ ⎟ 2 ⎝ 4 ⎠ K n = 0.399 mA / V 2 = b. VGS = VDS = 3 V ⇒ Saturation I D = K n (VGS − VTN ) = ( 0.399 )( 3 − 0.8 ) ⇒ I D = 1.93 mA 2 3.8 2 ⎛ ω ⎞⎛ k′ ⎞ I D = ⎜ ⎟ ⎜ n ⎟ (VGS − VTN ) 2 ⎠⎝ 2 ⎠ ⎝ 2 ⎛ ω ⎞ ⎛ 0.08 ⎞ 1.25 ⎜ ⎟⎜ ⎟ ( 2.5 − 1.2 ) ⇒ ω = 23.1 μ m ⎝ 1.25 ⎠ ⎝ 2 ⎠ 3.9 ∈ Cox = ox t0 x = ( 3.9 ) (8.85 ×10−14 ) 400 × 10−8 = 8.63 × 10−8 F/cm 2 2
  69. 69. Kn = μ n Cox W ⋅ L 1 ⎛W ⎞ = ( 600 ) ( 8.63 × 10−8 ) ⎜ ⎟ 2 ⎝ 2.5 ⎠ 2 K n = (1.036 × 10−5 ) W I D = K n (VGS − VTN ) 2 1.2 × 10 −3 = (1.036 × 10 −5 ) W ( 5 − 1) ⇒ W = 7.24 μ m 2 3.10 Biased in the saturation region in both cases. ′ kp W 2 I D = ⋅ (VSG + VTP ) 2 L 2 ⎛ 0.040 ⎞⎛ W ⎞ (1) 0.225 = ⎜ ⎟⎜ ⎟ ( 3 + VTP ) 2 ⎠⎝ L ⎠ ⎝ 2 ⎛ 0.040 ⎞ ⎛ W ⎞ 1.40 = ⎜ (2) ⎟ ⎜ ⎟ ( 4 + VTP ) ⎝ 2 ⎠⎝ L ⎠ Take ratio of (2) to (1): (4 + VTP ) 2 1.40 = 6.222 = 0.225 (3 + VTP ) 2 6.222 = 2.49 = 4 + VTP ⇒ VTP = −2.33 V 3 + VTP W 2 ⎛ 0.040 ⎞ ⎛ W ⎞ Then 0.225 = ⎜ = 25.1 ⎟ ⎜ ⎟ ( 3 − 2.33) ⇒ L ⎝ 2 ⎠⎝ L ⎠ 3.11 VS = 5 V, VG = 0 ⇒ VSG = 5 V VTP = −0.5 V ⇒ VSD ( sat ) = VSG + VTP = 5 − 0.5 = 4.5 V a. VD = 0 ⇒ VSD = 5 V ⇒ Biased in saturation I D = 2 ( 5 − 0.5 ) ⇒ I D = 40.5 mA 2 b. VD = 2 V ⇒ VSD = 3 V ⇒ Nonsaturation c. 2 I D = 2 ⎡ 2 ( 5 − 0.5 )( 3) − ( 3) ⎤ ⇒ I D = 36 mA ⎣ ⎦ VD = 4 V ⇒ VSD = 1 V ⇒ Nonsaturation d. 2 I D = 2 ⎡ 2 ( 5 − 0.5 )(1) − (1) ⎤ ⇒ I D = 16 mA ⎣ ⎦ VD = 5 V ⇒ VSD = 0 ⇒ I D = 0 3.12 (a) (b) Enhancement-mode From Graph VTP = + 0.5 V 0.45 = k p ( 2 − 0.5 ) = 2.25 K p ⇒ K p = 0.20 1.25 = k p ( 3 − 0.5 ) = 6.25 K p 0.20 2.45 = k p ( 4 − 0.5 ) = 12.25 K p 0.20 2 2 2 4.10 = k p ( 5 − 0.5 ) = 20.25 K p 2 (c) 0.202 Avg K p = 0.20 mA/V 2 iD (sat) = 0.20 (3.5 − 0.5) 2 = 1.8 mA iD (sat) = 0.20 (4.5 − 0.5) 2 = 3.2 mA
  70. 70. 3.13 VSD ( sat ) = VSG + VTP (a) VSD ( sat ) = −1 + 2 ⇒ VSD ( sat ) = 1 V (b) VSD ( sat ) = 0 + 2 ⇒ VSD ( sat ) = 2 V (c) VSD ( sat ) = 1 + 2 ⇒ VSD ( sat ) = 3 V ID = (a) (b) (c) ′ kp W k′ W 2 2 p ⋅ (VSG + VTP ) = ⋅ ⋅ ⎡VSD ( sat ) ⎤ ⎣ ⎦ 2 L 2 L 2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )(1) ⇒ I D = 0.12 mA ⎝ 2 ⎠ 2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )( 2 ) ⇒ I D = 0.48 mA ⎝ 2 ⎠ 2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )( 3) ⇒ I D = 1.08 mA ⎝ 2 ⎠ 3.14 VSD (sat) = VSG + VTP = 3 − 0.8 = 2.2 V ⎛ 15 ⎞⎛ 0.04 ⎞ 2 KP = ⎜ ⎟⎜ ⎟ = 0.25 mA/V ⎝ 1.2 ⎠⎝ 2 ⎠ a) b) c) d) e) 2 VSD = 0.2 Non Sat I D = 0.25 ⎡ 2 ( 3 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = 0.21 mA ⎣ ⎦ 2 VSD = 1.2 V Non Sat I D = 0.25 ⎡ 2 ( 3 − 0.8 )(1.2 ) − (1.2 ) ⎤ = 0.96 mA ⎣ ⎦ 2 VSD = 2.2 V Sat I D = 0.25(3 − 0.8) = 1.21 mA VSD = 3.2 V Sat ID = 1.21 mA VSD = 4.2 V Sat ID = 1.21 mA 3.15 ′ k p = μ p Cox = μ p ∈ox t0 x = ( 250 )( 3.9 ) (8.85 ×10−14 ) t0 x (a) tox = 500Å ⇒ k ′ = 17.3 μ A/V 2 p (b) 250Å ⇒ k ′ = 34.5 μ A/V 2 p (c) 100Å ⇒ k ′ = 86.3 μ A/V 2 p (d) ′ 50Å ⇒ k p = 173 μ A/V 2 (e) = 25Å ⇒ k ′ = 345 μ A/V 2 p 3.16 −14 ∈ox ( 3.9 ) ( 8.85 × 10 ) = = 6.90 × 10−8 F/cm 2 Cox = −8 t0 x 500 × 10 ′ kn = ( μ n Cox ) = ( 675 ) ( 6.90 × 10−8 ) ⇒ 46.6 μ A/V 2 k ′ = ( μ p Cox ) = ( 375 ) ( 6.90 × 10−8 ) ⇒ 25.9 μ A/V 2 p PMOS: 8.629 × 10−11 t0 x
  71. 71. ID = k′ ⎛ W ⎞ 2 p ⎜ ⎟ (VSG + VTP ) 2 ⎝ L ⎠p 2 ⎛ 0.0259 ⎞⎛ W ⎞ ⎛W ⎞ 0.8 = ⎜ ⎟⎜ ⎟ ( 5 − 0.6 ) ⇒ ⎜ ⎟ = 3.19 ⎝ 2 ⎠⎝ L ⎠ p ⎝ L ⎠p L = 4 μ m ⇒ W p = 12.8 μ m ⎛ 0.0259 ⎞ 2 Kp = ⎜ ⎟ ( 3.19 ) ⇒ K p = 41.3 μ A/V = K n ⎝ 2 ⎠ Want Kn = Kp k′ ⎛ W ⎞ ′ kn ⎛ W ⎞ p ⎜ ⎟ = ⎜ ⎟ = 41.3 2 ⎝ L ⎠N 2 ⎝ L ⎠p ⎛ 46.6 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎜ ⎟ ⎜ ⎟ = 41.3 ⇒ ⎜ ⎟ = 1.77 2 ⎠ ⎝ L ⎠N ⎝ ⎝ L ⎠N L = 4 μ m ⇒ WN = 7.09 μ m 3.17 VGS = 2 V, I D = ( 0.2 )( 2 − 1.2 ) = 0.128 mA 1 1 r0 = = ⇒ r0 = 781 kΩ λ I D ( 0.01)( 0.128 ) 2 VGS = 4 V, I D = ( 0.2 )( 4 − 1.2 ) = 1.57 mA 1 r0 = ⇒ r = 63.7 kΩ ( 0.01)(1.57 ) 0 2 VA = 1 λ = 1 ⇒ VA = 100 V ( 0.01) 3.18 2 2 ⎛ 0.080 ⎞ ID = ⎜ ⎟ ( 4 )( 3 − 0.8 ) = ( 0.16 )( 3 − 0.8 ) ⇒ I D = 0.774 mA 2 ⎠ ⎝ 1 1 1 ⇒λ = = ⇒ λ (max) = 0.00646 V −1 r0 = λ ID r0 I D ( 200 )( 0.774 ) VA ( min ) = 1 λ ( max ) = 1 ⇒ VA ( min ) = 155 V 0.00646 3.19 VTN = VTNO + γ ⎡ 2φ f + VSB − 2φ f ⎤ ⎣ ⎦ ΔVTN = 2 = ( 0.8 ) ⎡ 2φ f + VSB − 2 ( 0.35 ) ⎤ ⎣ ⎦ 2.5 + 0.837 = 2 ( 0.35 ) + VSB ⇒ VSB = 10.4 V 3.20 VTN = VTNo + r ⎡ 2φ f + VSB − 2φ f ⎤ ⎣ ⎦ ⎡ 2 ( 0.37 ) + 3 − 2 ( 0.37 ) ⎤ = 0.75 + 0.6 ⎣ ⎦ = 0.75 + 0.6 [1.934 − 0.860] VTN = 1.39 V VDS (sat) = 2.5 − 1.39 = 1.11 V
  72. 72. 2 ⎛ 0.08 ⎞ Sat Region I D = (15 ) ⎜ ⎟ ( 2.5 − 1.39 ) ⎝ 2 ⎠ I D = 0.739 mA (a) 2 ⎛ 0.08 ⎞ ⎡ Non-Sat I D = (15 ) ⎜ ⎟ 2 ( 2.5 − 1.39 )( 0.25 ) − ( 0.25 ) ⎤ ⎦ 2 ⎠⎣ ⎝ I D = 0.296 mA (b) 3.21 a. VG = %ox t0 x = ( 6 × 106 )( 275 × 10−8 ) VG = 16.5 V VG = b. 16.5 ⇒ VG = 5.5 V 3 3.22 Want VG = ( 3)( 24 ) = %ox t0 x = ( 6 × 106 ) t0 x t0 x = 1.2 ×10−5 cm = 1200 Angstroms 3.23 ⎛ R2 ⎞ ⎛ 18 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) = 3.6 V ⎝ 18 + 32 ⎠ ⎝ R1 + R2 ⎠ Assume transistor biased in saturation region V V − VGS 2 = K n (VGS − VTN ) ID = S = G RS RS 3.6 − VGS = ( 0.5 )( 2 )(VGS − 0.8 ) 2 = VGS − 1.6VGS + 0.64 2 2 VGS − 0.6VGS − 2.96 = 0 VGS = ID = 0.6 ± VG − VGS RS ( 0.6 ) 2 + 4 ( 2.96 ) ⇒ VGS = 2.046 V 2 3.6 − 2.046 = ⇒ I D = 0.777 mA 2 VDS = VDD − I D ( RD + RS ) = 10 − ( 0.777 )( 4 + 2 ) ⇒ VDS = 5.34 V VDS > VDS ( sat ) 3.24
  73. 73. ID(mA) 4 (a) Q-pt Q-pt 1.67 (b) 4 5 V (V) DS VGS = 4 V VDS (sat) = 4 − 0.8 = 3.2 V (a) If Sat I D = 0.25 ( 4 − 0.8 ) = 2.56 2 VDS = 1.44 × Non-Sat 2 4 = I D RD + VDS = K n RD ⎡ 2 (VGS − VT ) VDS − VDS ⎤ + VDS ⎣ ⎦ 2 4 = ( 0.25 )(1) ⎡ 2 ( 4 − 0.8 ) VDS − VDS ⎤ + VDS ⎣ ⎦ 2 4 = 2.6VDS − 0.25VDS 2 0.25VDS − 2.6VDS + 4 = 0 2.6 ± 6.76 − 4 = 1.88 V 2 ( 0.25 ) VDS = 4 − 1.88 = 2.12 mA 1 (b) Non-Sat region 2 5 = I D RD + VDS = K n RD ⎡ 2 (VGS − VT )VDS − VDS ⎤ + VDS ⎣ ⎦ ID = 2 5 = ( 0.25 )( 3) ⎡ 2 ( 5 − 0.8 ) VDS − VDS ⎤ + VDS ⎣ ⎦ 2 5 = 7.3VDS − 0.75VDS 2 0.75 VDS − 7.3VDS + 5 = 0 VDS = 7.3 ± 53.29 − 15 2 ( 0.75 ) VDS = 0.741 V 5 − 0.741 ID = = 1.42 mA 3 3.25 ID(mA) 2.92 (a) Q-pt 1.25 (b) 3.5 5 V (V) SD
  74. 74. VSG = VDD = 3.5 (a) VSD ( sat ) = 3.5 − 0.8 = 2.7 V If biased in Sat region, I D = ( 0.2 )( 3.5 − 0.8 ) = 1.46 mA VSD = 3.5 − (1.46 )(1.2 ) = 1.75 V 2 × Biased in Non-Sat Region. 2 3.5 = VSD + I D RD = VSD + K p RD ⎡ 2 (VSG + VTP ) VSD − VSD ⎤ ⎣ ⎦ 2 3.5 = VSD + ( 0.2 )(1.2 ) ⎡ 2 ( 3.5 − 0.8 ) VSD − VSD ⎤ ⎣ ⎦ 2 3.5 = VSD + 1.296 VSD − 0.24 VSD 2 0.24 VSD − 2.296 VSD + 3.5 = 0 VSD = +2.296 ± 5.272 − 3.36 use − sign VSD = 1.90 V 2 ( 0.24 ) 2 I D = ( 0.2 ) ⎡ 2 ( 3.5 − 0.8 )(1.9 ) − (1.9 ) ⎤ = 0.2 [10.26 − 3.61] ⎣ ⎦ 3.5 − 1.90 ID = = 1.33 mA 1.2 I D = 1.33 mA VSG = VDD = 5 V VSD ( sat ) = 5 − 0.8 = 4.2 V (b) If Sat Region I D = ( 0.2 )( 5 − 0.8 ) = 3.53 mA, VSD < 0 2 Non-Sat Region. 2 5 = VSD + K p RD ⎡ 2 (VSG + VTP ) VSD − VSD ⎤ ⎣ ⎦ 2 5 = VSD + ( 0.2 )( 4 ) ⎡ 2 ( 5 − 0.8 ) VSD − VSD ⎤ ⎣ ⎦ 2 5 = VSD + 6.72 VSD − 0.8 VSD 2 0.8 VSD − 7.72 VSD + 5 = 0 VSD = ID = 7.72 ± 59.598 − 16 use − sign VSD = 0.698 V 2 ( 0.8 ) 5 − 0.698 ⇒ I D = 1.08 mA 4 3.26 10 − VS 2 = K p (VSG + VTP ) RS Assume transistor biased in saturation region ⎛ R2 ⎞ VG = ⎜ ⎟ ( 20 ) − 10 ⎝ R1 + R2 ⎠ ID = ⎛ 22 ⎞ =⎜ ⎟ ( 20 ) − 10 ⇒ VG = 4.67 V ⎝ 8 + 22 ⎠ VS = VG + VSG 10 − ( 4.67 + VSG ) = (1)( 0.5 )(VSG − 2 ) 2 2 5.33 − VSG = 0.5 (VSG − 4VSG + 4 ) 2 0.5VSG − VSG − 3.33 = 0 VSG = 1± (1) 2 + 4 ( 0.5 )( 3.33) 2 ( 0.5 ) ⇒ VSG = 3.77 V
  75. 75. VSD 10 − ( 4.67 + 3.77 ) ⇒ I D = 3.12 mA 0.5 = 20 − I D ( RS + RD ) = 20 − ( 3.12 )( 0.5 + 2 ) ⇒ VSD = 12.2 V ID = VSD > VSD ( sat ) 3.27 VG = 0, VSG = VS Assume saturation region 2 I D = 0.4 = K p (VSG + VTP ) 0.4 = ( 0.2 )(VS − 0.8 ) 2 0.4 + 0.8 ⇒ VS = 2.21 V 0.2 VD = I D RD − 5 = ( 0.4 )( 5 ) − 5 = −3 V VSD = VS − VD = 2.21 − ( −3) ⇒ VSD = 5.21 V VS = VSD > VSD ( sat ) 3.28 VDD = I DQ RD + VDSQ + I DQ RS 2 ⎛ k ′ ⎞⎛ W ⎞ (1) 10 = I DQ ( 5 ) + 5 + VGS and I DQ = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN ) ⎝ 2 ⎠⎝ L ⎠ 2 ⎛ 0.060 ⎞ ⎛ W ⎞ or (2) I DQ = ⎜ ⎟ ⎜ ⎟ (VGS − 1.2 ) ⎝ 2 ⎠⎝ L ⎠ Let VGS = 2.5 V Then from (1), 10 = I DQ ( 5 ) + 5 + 2.5 ⇒ I D = 0.5 mA W 2 ⎛ 0.060 ⎞⎛ W ⎞ Then from (2), 0.5 = ⎜ = 9.86 ⎟⎜ ⎟ ( 2.5 − 1.2 ) ⇒ 2 ⎠⎝ L ⎠ L ⎝ V 2.5 I DQ RS = VGS ⇒ RS = GS = ⇒ RS = 5 k Ω I DQ 0.5 IR = 10 = ( 0.5 )( 0.05 ) = 0.025 mA R1 + R2 Then R1 + R2 = 10 = 400 k Ω 0.025 ⎛ R2 ⎞ ⎛ R2 ⎞ ⎜ ⎟ (VDD ) = 2VGS ⇒ ⎜ ⎟ (10 ) = 2 ( 2.5 ) ⇒ R1 = R2 = 200 k Ω R1 + R2 ⎠ ⎝ 400 ⎠ ⎝ 3.29 ⎛ 75 ⎞ K n = ( 25 ) ⎜ ⎟ ⇒ 0.9375 mA/V 2 ⎝ 2⎠ ⎛ 6 ⎞ VG = ⎜ ⎟ (10 ) − 5 = −2 V ⎝ 6 + 14 ⎠ (VG − VGS ) − ( −5) 2 = I D = K n (VGS − VTN ) RS −2 − VGS + 5 = ( 0.9375 )( 0.5 )(VGS − 1) 2 3 − VGS = 0.469 (VGS − 2VGS + 1) 2
  76. 76. 2 0.469 VGS + 0.0625 VGS − 2.53 = 0 VGS = −0.0625 ± 0.003906 + 4.746 ⇒ VGS = 2.26 V 2 ( 0.469 ) I D = 0.9375 ( 2.26 − 1) ⇒ I D = 1.49 mA 2 VDS = 10 − (1.49 )(1.7 ) ⇒ VDS = 7.47 V 3.30 20 = I DQ RS + VSDQ + I DQ RD (1) 20 = VSG + 10 + I DQ RD ⎛ k′ ⎞⎛ W ⎞ 2 p I DQ = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) 2 ⎠⎝ L ⎠ ⎝ 2 ⎛ 0.040 ⎞ ⎛ W ⎞ (2) I DQ = ⎜ ⎟ ⎜ ⎟ (VSG − 2 ) ⎝ 2 ⎠⎝ L ⎠ For example, let I DQ = 0.8 mA and VSG = 4 V ⎛ 0.040 ⎞ ⎛ W Then 0.8 = ⎜ ⎟⎜ ⎝ 2 ⎠⎝ L I DQ RS = VSG ⇒ ( 0.8 ) RS W 2 ⎞ = 10 ⎟ ( 4 − 2) ⇒ L ⎠ = 4 ⇒ RS = 5 k Ω From (1) 20 = 4 + 10 + ( 0.8 ) RD ⇒ RD = 7.5 k Ω IR = 20 = ( 0.8 )( 0.1) ⇒ R1 + R2 = 250 k Ω R1 + R2 ⎛ R1 ⎞ ⎜ ⎟ ( 20 ) = 2VSG = ( 2 )( 4 ) ⎝ R1 + R2 ⎠ R1 ( 20 ) = 8 ⇒ R1 = 100 k Ω, R2 = 150 k Ω 250 3.31 (a) (i) I Q = 50 = 500 (VGS − 1.2 ) ⇒ VGS = 1.516 V 2 VDS = 5 − ( −1.516 ) =⇒ VDS = 6.516 V (ii) I Q = 1 = ( 0.5 )(VGS − 1.2 ) ⇒ VGS = 2.61 V 2 VDS = 5 − ( −2.61) ⇒ VDS = 7.61 V (b) (i) Same as (a) VGS = VDS = 1.516 V (ii) VGS = VDS = 2.61 V 3.32 I D = K n (VGS − VTN ) 2 0.25 = ( 0.2 )(VGS − 0.6 ) 2 0.25 + 0.6 ⇒ VGS = 1.72 V ⇒ VS = −1.72 V 0.2 VD = 9 − ( 0.25 )( 24 ) ⇒ VD = 3 V VGS = 3.33 (a)
  77. 77. ID(mA) 1.0 0.808 Q-pt 0.5 3.81 RD = 10 V (V) DS 5 −1 ⇒ RD = 8 K 0.5 I DQ = 0.5 = 0.25 (VGS − 1.4 ) ⇒ VGS = 2.81 V 2 RS = −2.81 − ( −5 ) ⇒ RS = 4.38 K 0.5 Let RD = 8.2 K, RS = 4.3 K (b) Now −VGS − ( −5 ) 5 − VGS = I D = 0.25 (VGS − 1.4 ) 4.3 2 = 1.075 (VGS − 2.8 VGS + 1.96 ) 2 2 1.075 VGS − 2.01 VGS − 2.89 = 0 VGS = 2.01 ± 4.04 + 12.427 ⇒ VGS = 2.82 V 2 (1.075 ) I D = 0.25 ( 2.82 − 1.4 ) ⇒ I D = 0.504 mA 2 VDS = 10 − ( 0.504 )( 8.2 + 4.3) ⇒ VDS = 3.70 V (c) If RS = 4.3 + 10% = 4.73 K 2 5 − VGS = 1.18 (VGS − 2.8VGS + 1.96 ) 2 1.18 VGS − 2.31 VGS − 2.68 = 0 VGS = 2.31 ± 5.336 + 12.65 = 2.78 V 2 (1.18 ) I D = ( 0.25 )( 2.78 − 1.4 ) ⇒ I D = 0.476 mA 2 If Rs = 4.3 − 10% = 3.87 K 2 5 − VGS = ( 0.9675 ) (VGS − 2.8VGS + 1.96 ) 2 0.9675VGS − 1.71VGS − 3.10 = 0 VGS = 1.71 ± 2.924 + 12.0 = 2.88 V 2 ( 0.9675 ) I D = ( 0.25 )( 2.88 − 1.4 ) = 0.548 mA 2 3.34 VDD = VSD + I DQ R 9 = 2.5 + ( 0.1) R ⇒ R = 65 k Ω ⎛ k′ ⎞⎛ W ⎞ 2 p I DQ = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) ⎝ 2 ⎠⎝ L ⎠ W 2 ⎛ 0.025 ⎞ ⎛ W ⎞ =8 ( 0.1) = ⎜ ⎟ ⎜ ⎟ ( 2.5 − 1.5 ) ⇒ L⎠ L ⎝ 2 ⎠⎝ Then for L = 4 μ m, W = 32 μ m
  78. 78. 3.35 5 = I DQ RS + VSDQ = I DQ ( 2 ) + 2.5 I DQ = 1.25 mA IR = 10 = (1.25 )( 0.1) ⇒ R1 + R2 = 80 k Ω R1 + R2 I DQ = K p (VSG + VTP ) 2 1.25 = 0.5 (VSG + 1.5 ) ⇒ 2 1.25 − 1.5 = VSG 0.5 VSG = 0.0811 V VG = VS − VSG = 2.5 − 0.0811 = 2.42 V ⎛ R2 ⎞ VG = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠ ⎛R ⎞ 2.42 = ⎜ 2 ⎟ (10 ) − 5 ⇒ R2 = 59.4 k Ω, R1 = 20.6 k Ω ⎝ 80 ⎠ 3.36 (a) ID(mA) 0.429 Q-pt RD = VD − ( −5 ) I DQ 5 V (V) SD = 5−2 ⇒ RD = 12 K 0.25 2 ⎞⎛ k′ ⎞ p ⎟ ⎜ ⎟ (VSG + VTP ) ⎠⎝ 2 ⎠ 2 ⎛ 0.035 ⎞ 0.25 = (15 ) ⎜ ⎟ (VSG − 1.2 ) ⇒ VSG = 2.18 V ⎝ 2 ⎠ 5 − 2.18 RS = ⇒ RS = 11.3 K 0.25 VSD = 2.18 − ( −2 ) = 4.18 V (b) k ′ = 35 + 5% = 36.75 μ A/V 2 p ⎛W ID = ⎜ ⎝L 5 − VSG 2 ⎛ 0.03675 ⎞ I D = (15 ) ⎜ ⎟ (VSG − 1.2 ) = 2 11.3 ⎝ ⎠ 2 3.11(VSG − 2.4VSG + 1.44 ) = 5 − VSG 2 3.11VSG − 6.46VSG − 0.522 = 0
  79. 79. VSG = 6.46 ± 41.73 + 6.49 = 2.155 V 2 ( 3.11) 5 − 2.155 = 0.252 mA 11.3 = 10 − ( 0.252 )(12 + 11.3) = 4.13 V ID = VSD k ′ = 35 − 5% = 33.25 μ A/V 2 p 5 − VSG 2 ⎛ 0.03325 ⎞ I D = (15 ) ⎜ ⎟ (VSG − 1.2 ) = 2 11.3 ⎝ ⎠ 2 2.82 (VSG − 2.4VSG + 1.44 ) = 5 − VSG 2 2.82VSG − 5.77VSG − 0.939 = 0 VSG = 5.77 ± 33.29 + 10.59 = 2.198 V 2 ( 2.82 ) 5 − 2.198 = 0.248 mA 11.3 = 10 − ( 0.248 )(12 + 11.3) = 4.22 V ID = VSD 3.37 ID = −VSD − ( −10 ) RD ⇒5= −6 + 10 ⇒ RD = 0.8 kΩ RD I D = K P (VSG + VTP ) ⇒ 5 = 3 (VSG − 1.75 ) 2 VSG = 2 5 + 1.75 = 3.04 V ⇒ VG = −3.04 3 ⎛ R2 ⎞ VG = ⎜ ⎟ (10 ) − 5 = −3.04 ⎝ R1 + R2 ⎠ Rin = R1 || R2 = 80 kΩ 1 ⋅ ( 80 )(10 ) = 5 − 3.04 ⇒ R1 = 408 kΩ R1 408 R2 = 80 ⇒ R2 = 99.5 kΩ 408 + R2 3.38 ⎛ 60 ⎞ K n1 = ⎜ ⎟ ( 4 ) = 120 μ A/V 2 ⎝ 2 ⎠ ⎛ 60 ⎞ K n 2 = ⎜ ⎟ (1) = 30 μ A/V 2 ⎝ 2 ⎠ For vI = 1 V , M1 Sat. region, M2 Non-sat region. (a) I D 2 = I D1 30 ⎡ 2 ( −VTNL )( 5 − vO ) − ( 5 − vO ) ⎤ = 120 (1 − 0.8 ) ⎣ ⎦ 2 We find vO − 6.4vO + 7.16 = 0 ⇒ vO = 4.955 V 2 (b) 2 For vI = 3 V , M1 Non-sat region, M2 Sat. region. I D 2 = I D1 2 30 ⎡ − ( −1.8 ) ⎤ = 120 ⎡ 2 ( 3 − 0.8 ) vO − vO ⎤ ⎣ ⎦ ⎣ ⎦ 2 2 We find 4vO − 17.6vO + 3.24 = 0 ⇒ vO = 0.193 V (c) For vI = 5 V , biasing same as (b) 2 30 ⎡ − ( −1.8 ) ⎤ = 120 ⎡ 2 ( 5 − 0.8 ) vO − vO ⎤ ⎣ ⎦ ⎣ ⎦ 2
  80. 80. 2 We find 4vO − 33.6vO + 3.24 = 0 ⇒ vO = 0.0976 V 3.39 For vI = 5 V , M1 Non-sat region, M2 Sat. region. I D1 = I D 2 ′ ′ 2 ⎛ kn ⎞ ⎛ W ⎞ ⎛ kn ⎞ ⎛ W ⎞ 2 ⎜ 2 ⎟ ⎜ L ⎟ ⎡ 2 (VGS1 − VTN 1 ) VDS 1 − VDS 1 ⎤ = ⎜ 2 ⎟ ⎜ L ⎟ (VGS 2 − VTN 2 ) ⎣ ⎦ ⎝ ⎠ ⎝ ⎠1 ⎝ ⎠ ⎝ ⎠2 2 W⎞ ⎡ 2 ⎛ ⎜ ⎟ ⎣ 2 ( 5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎤ = (1) ⎡0 − ( −2 ) ⎤ ⎣ ⎦ ⎦ ⎝ L ⎠1 ⎛W ⎞ which yields ⎜ ⎟ = 3.23 ⎝ L ⎠1 3.40 a. M1 and M2 in saturation K n1 (VGS 1 − VTN 1 ) = K n 2 (VGS 2 − VTN 2 ) K n1 = K n 2 , VTN 1 = VTN 2 ⇒ VGS1 = VGS 2 = 2.5 V, V0 = 2.5 V 2 2 I D = (15 )( 40 )( 2.5 − 0.8 ) ⇒ I D = 1.73 mA 2 b. ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ > ⎜ ⎟ ⇒ VGS1 < VGS 2 ⎝ L ⎠1 ⎝ L ⎠ 2 40 (VGS1 − 0.8 ) = (15 )(VGS 2 − 0.8 ) 2 2 VGS 2 = 5 − VGS 1 1.633 (VGS 1 − 0.8 ) = ( 5 − VGS1 − 0.8 ) 2.633VGS 1 = 5.506 ⇒ VGS 1 = 2.09 V VGS 2 = 2.91 V, V0 = VGS1 = 2.91 V I D = (15 )(15 )( 2.91 − 0.8 ) ⇒ I D = 1.0 mA 2 3.41 (a) V1 = VGS 3 = 2.5 V 2 ⎛ W ⎞ ⎛ 0.06 ⎞ I D = 0.5 = ⎜ ⎟ ⎜ ⎟ ( 2.5 − 1.2 ) ⎝ L ⎠3 ⎝ 2 ⎠ ⎛W ⎞ ⎜ ⎟ = 9.86 ⎝ L ⎠3 V2 = 6 V ⇒ VGS 2 = V2 − V1 = 6 − 2.5 = 3.5 V 2 ⎛ W ⎞ ⎛ 0.06 ⎞ ⎛W ⎞ 0.5 = ⎜ ⎟ ⎜ ⎟ ( 3.5 − 1.2 ) ⇒ ⎜ ⎟ = 3.15 ⎝ L ⎠2 ⎝ 2 ⎠ ⎝ L ⎠2 VGS1 = 10 − V2 = 10 − 6 = 4 V 2 ⎛ W ⎞ ⎛ 0.06 ⎞ ⎛W ⎞ 0.5 = ⎜ ⎟ ⎜ ⎟ ( 4 − 1.2 ) ⇒ ⎜ ⎟ = 2.13 ⎝ L ⎠1 ⎝ 2 ⎠ ⎝ L ⎠1 (b) ′ kn1 = 0.06 + 5% = 0.063 mA/V 2 ′ ′ kn 2 = k n3 = 0.6 − 5% = 0.057 mA/V 2 2 ⎛ 0.057 ⎞ For M3: I D = ( 9.86 ) ⎜ ⎟ (V1 − 1.2 ) ⎝ 2 ⎠ 2 ⎛ 0.057 ⎞ For M2: I D = ( 3.15 ) ⎜ ⎟ (V2 − V1 − 1.2 ) 2 ⎠ ⎝
  81. 81. 2 ⎛ 0.063 ⎞ For M1: I D = ( 2.13) ⎜ ⎟ (10 − V2 − 1.2 ) ⎝ 2 ⎠ 0.281(V1 − 1.2 ) = 0.0898 (V2 − V1 − 1.2 ) = 0.0671( 8.8 − V2 ) 2 2 Take square root. 0.530 (V1 − 1.2 ) = 0.300 (V2 − V1 − 1.2 ) = 0.259 ( 8.8 − V2 ) (1) 0.830V1 = 0.300V2 + 0.276 (2) 0.559V2 = 0.300V1 + 2.64 From (2) ⇒ V2 = 0.537V1 + 4.72 Substitute into (1) 0.830V1 = 0.300 [ 0.537V1 + 4.72] + 0.276 = 0.161V1 + 1.69 V1 = 2.53 V Then V2 = 0.537 ( 2.53) + 4.72 V2 = 6.08 V 3.42 ML in saturation MD in nonsaturation 2 ⎛W ⎞ ⎛W ⎞ 2 ⎡ ⎤ ⎜ ⎟ (VGSL − VTNL ) = ⎜ ⎟ ⎣ 2 (VGSD − VTND )VDSD − VDSD ⎦ ⎝ L ⎠L ⎝ L ⎠D W 2 2 ⎤ (1)( 5 − 0.1 − 0.8) = ⎛ ⎞ ⎣ 2 ( 5 − 0.8)( 0.1) − ( 0.1) ⎦ ⎜ ⎟ ⎡ L ⎠D ⎝ ⎛W ⎞ 16.81 = ⎜ ⎟ [ 0.83] ⎝ L ⎠D ⎛W ⎞ ⎜ ⎟ = 20.3 ⎝ L ⎠D 3.43 ML in saturation MD in nonsaturation 2 ⎛W ⎞ ⎛W ⎞ 2 ⎡ ⎤ ⎜ ⎟ (VGSL − VTNL ) = ⎜ ⎟ ⎣ 2 (VGSD − VTND )VDSD − VDSD ⎦ ⎝ L ⎠L ⎝ L ⎠D W 2 2 (1)(1.8 ) = ⎛ ⎞ ⎡ 2 ( 5 − 0.8 )( 0.05) − ( 0.05) ⎤ ⎜ ⎟ ⎣ ⎦ ⎝ L ⎠D ⎛W ⎞ 3.24 = ⎜ ⎟ [ 0.4175] ⎝ L ⎠D ⎛W ⎞ ⎜ ⎟ = 7.76 ⎝ L ⎠D 3.44 VDD − V0 5 − 0.1 = = 0.49 mA 10 RD Transistor biased in nonsaturation I D = 0.49 ID = 2 ⎛W ⎞ = ( 0.015 ) ⎜ ⎟ ⎡ 2 ( 4.2 − 0.8 )( 0.1) − ( 0.1) ⎤ ⎣ ⎦ ⎝L⎠ W ⎛W ⎞ 0.49 = ⎜ ⎟ 0.01005 ⇒ = 48.8 L ⎝L⎠ 3.45 2
  82. 82. 5 = I D RD + Vγ + VDS 5 = (12 ) RD + 1.6 + 0.2 ⇒ RD = 267 Ω 2 ⎛ k′ ⎞⎛ W ⎞ I D = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN ) ⎝ 2 ⎠⎝ L ⎠ W 2 ⎛ 0.040 ⎞ ⎛ W ⎞ = 34 12 = ⎜ ⎟ ⎜ ⎟ ( 5 − 0.8 ) ⇒ 2 ⎠⎝ L ⎠ L ⎝ 3.46 5 = VSD + I D RD + Vγ 5 = 0.15 + (15 ) RD + 1.6 ⇒ RD = 217 Ω ⎛ k′ ⎞⎛ W ⎞ 2 p I D = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) ⎝ 2 ⎠⎝ L ⎠ W 2 ⎛ 0.020 ⎞⎛ W ⎞ = 85 15 = ⎜ ⎟⎜ ⎟ ( 5 − 0.8 ) ⇒ L ⎝ 2 ⎠⎝ L ⎠ 3.47 (a) VDD − VO ⎛W = 2⎜ RD ⎝L 5 − 0.2 ⎛W = 2⎜ 20 ⎝L ⎞⎛ 0.060 ⎞ 2 ⎟⎜ ⎟ ⎡( 2 )(VGS − VTN ) VO − VO ⎤ ⎣ ⎦ ⎠⎝ 2 ⎠ 2 ⎞ ⎟ ( 0.030 ) ⎡ 2 ( 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ ⎣ ⎦ ⎠ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ 0.24 = 0.0984 ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 2.44 ⎝ L ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2 (b) 5 − VO ⎛ 0.06 ⎞ 2 = ( 2.44 ) ⎜ ⎟ ⎡ 2 ( 5 − 0.8 ) VO − VO ⎤ ⎦ 20 2 ⎠⎣ ⎝ 5 − VO = 12.30VO − 1.464VO2 1.464VO2 − 13.30VO + 5 = 0 VO = 13.30 ± 176.89 − 29.28 2 (1.464 ) VO = 0.393 V 3.48 ′ 2 ⎞ ⎛ kn ⎞ ⎟ ⎜ ⎟ (VGS 1 − VTN ) 2⎠ ⎠1 ⎝ ′ 2 ⎞ ⎛ kn ⎞ ⎟ ⎜ ⎟ (VDS 2 ( sat ) ) ⎠2 ⎝ 2 ⎠ 2 ⎛ W ⎞ ⎛ 0.08 ⎞ ⎛W ⎞ ⎛W ⎞ 0.1 = ⎜ ⎟ ⎜ ⎟ ( 0.5 ) ⇒ ⎜ ⎟ = 10 = ⎜ ⎟ L ⎠2 ⎝ 2 ⎠ L ⎠2 ⎝ ⎝ ⎝ L ⎠1 ⎛ W ⎞ ⎛ 200 ⎞ ⎛ W ⎞ ⎜ ⎟ =⎜ ⎟ ⎜ ⎟ = 20 ⎝ L ⎠3 ⎝ 100 ⎠ ⎝ L ⎠ 2 M1 & M2 matched. 2 ⎛ 0.08 ⎞ Then 0.1 = (10 ) ⎜ ⎟ (VGS 1 − 0.25 ) 2 ⎠ ⎝ VGS1 = 0.75 V ⎛W I Q1 = ⎜ ⎝L ⎛W I Q1 = ⎜ ⎝L VD1 = −0.75 + 2 = 1.25 V RD = 2.5 − 1.25 ⇒ RD = 12.5 K 0.1
  83. 83. 3.49 (a) 2 ⎛ W ⎞ ⎛ k′ ⎞ p I Q 2 = ⎜ ⎟ ⎜ ⎟ (VSDB ( sat ) ) ⎝ L ⎠B ⎝ 2 ⎠ 2 ⎛ W ⎞ ⎛ 0.04 ⎞ 0.25 = ⎜ ⎟ ⎜ ⎟ ( 0.8 ) ⇒ ⎝ L ⎠B ⎝ 2 ⎠ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = 19.5 = ⎜ ⎟ ⎝ L ⎠B ⎝ L ⎠A I KQ 2 ⎛ W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ IQ 2 ⎝ L ⎠B ⎝ L ⎠C ⎛ 100 ⎞ =⎜ ⎟ (19.5 ) = 7.81 ⎝ 250 ⎠ ′ 2 ⎛W ⎞ ⎛ kp ⎞ I Q 2 = ⎜ ⎟ ⎜ ⎟ (VSGA + VTP ) ⎝ L ⎠A ⎝ 2 ⎠ 2 ⎛ 0.04 ⎞ 0.25 = (19.5 ) ⎜ ⎟ (VSGA − 0.5 ) ⎝ 2 ⎠ VSGA = 1.30 V (b) VDA = 1.3 − 4 = −2.7 V RD = −2.7 − ( −5 ) 0.25 ⇒ RD = 9.2 K 3.50 ′ 2 ⎞ ⎛ kn ⎞ ⎟ ⎜ ⎟ (VDS 2 ( sat ) ) 2⎠ ⎠2 ⎝ 2 ⎞ ⎛ 0.06 ⎞ ⎛W ⎞ ⎛W ⎞ ⎟ ⎜ ⎟ ( 0.5 ) ⇒ ⎜ ⎟ = 53.3 = ⎜ ⎟ ⎠2 ⎝ 2 ⎠ ⎝ L ⎠2 ⎝ L ⎠1 2 ⎛ W ⎞ ⎛ k′ ⎞ I Q = ⎜ ⎟ ⎜ n ⎟ (VGS 1 − VTN ) L ⎠1 ⎝ 2 ⎠ ⎝ 2 ⎛ 0.06 ⎞ 0.4 = ( 53.3) ⎜ ⎟ (VGS 1 − 0.75 ) ⎝ 2 ⎠ VGS1 = 1.25 V VD1 = −1.25 + 4 = 2.75 V 5 − 2.75 RD = ⇒ RD = 5.625 K 0.4 ⎛W IQ = ⎜ ⎝L ⎛W 0.4 = ⎜ ⎝L 3.51 VDS ( sat ) = VGS − VP So VDS > VDS ( sat ) = −VP , I D = I DSS 3.52 VDS ( sat ) = VGS − VP = VGS + 3 = VDS ( sat ) a. VGS = 0 ⇒ I D = I DSS = 6 mA b. ⎛ V ⎞ ⎛ −1 ⎞ I D = I DSS ⎜ 1 − GS ⎟ = 6 ⎜ 1 − ⎟ ⇒ I D = 2.67 mA VP ⎠ ⎝ −3 ⎠ ⎝ 2 2 c. d. 2 ⎛ −2 ⎞ I D = 6 ⎜ 1 − ⎟ ⇒ I D = 0.667 mA ⎝ −3 ⎠ ID = 0
  84. 84. 3.53 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ ⎛ 1 ⎞ 2.8 = I DSS ⎜1 − ⎟ ⎝ VP ⎠ 2 2 ⎛ 3 ⎞ 0.30 = I DSS ⎜1 − ⎟ ⎝ VP ⎠ ⎛ 1 ⎜1 − 2.8 ⎝ VP = 0.30 ⎛ 3 ⎜1 − VP ⎝ 2 2 ⎞ ⎟ ⎠ = 9.33 2 ⎞ ⎟ ⎠ ⎛ 1 ⎜1 − VP ⎝ ⎛ 3 ⎜1 − ⎝ VP ⎞ ⎟ ⎠ = 3.055 ⎞ ⎟ ⎠ 1 9.165 = 3.055 − 1− VP VP 8.165 = 2.055 ⇒ VP = 3.97 V VP 2 1 ⎞ ⎛ 2.8 = I DSS ⎜1 − ⎟ = I DSS ( 0.560 ) ⇒ I DSS = 5.0 mA ⎝ 3.97 ⎠ 3.54 VS = −VGS , VSD = VS − VDD Want VSD ≥ VSD ( sat ) = VP − VGS VS − VDD ≥ VP − VGS − VGS − VDD ≥ VP − VGS ⇒ VDD ≤ −VP So VDD ≤ −2.5 V ⎛ V ⎞ I D = 2 = I DSS ⎜1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ V ⎞ 2 = 6 ⎜ 1 − GS ⎟ ⇒ VGS = 1.06 V ⇒ VS = −1.06 V ⎝ 2.5 ⎠ 3.55 I D = K n (VGS − VTN ) 2 18.5 = K n ( 0.35 − VTN ) 86.2 = K n ( 0.5 − VTN ) 2 2 Then ( 0.35 − VTN ) 18.5 = 0.2146 = ⇒ VTN = 0.221 V 2 86.2 ( 0.50 − VTN ) 2 18.5 = K n ( 0.35 − 0.221) ⇒ K n = 1.11 mA / V 2 2 3.56 I D = K (VGS − VTN ) 2 250 = K ( 0.75 − 0.24 ) ⇒ K = 0.961 mA / V 2 2
  85. 85. 3.57 2 ⎛ V ⎞ V V I D = I DSS ⎜ 1 − GS ⎟ = S = − GS VP ⎠ RS RS ⎝ 2 V ⎛ V ⎞ 10 ⎜ 1 − GS ⎟ = − GS 0.2 −5 ⎠ ⎝ 2 ⎛ 2V V ⎞ 2 ⎜1 + GS + GS ⎟ = −VGS 5 25 ⎠ ⎝ 2 2 9 VGS + VGS + 2 = 0 25 5 2 2VGS + 45VGS + 50 = 0 VGS = −45 ± ID = − ( 45 ) − 4 ( 2 )( 50 ) ⇒ VGS 2 ( 2) 2 = −1.17 V VGS 1.17 = ⇒ I D = 5.85 mA RS 0.2 VD = 20 − ( 5.85 )( 2 ) = 8.3 V VDS = VD − VS = 8.3 − 1.17 ⇒ VDS = 7.13 V 3.58 VDS = VDD − VS 8 = 10 − VS ⇒ VS = 2 V = I D RS = ( 5 ) RS ⇒ RS = 0.4 kΩ ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ ⎝ VP ⎠ 2 2 ⎛ −1 ⎞ 5 = I DSS ⎜1 − ⎟ Let I DSS = 10 mA ⎝ VP ⎠ 2 ⎛ −1 ⎞ 5 = 10 ⎜ 1 − ⎟ ⇒ VP = −3.41 V ⎝ VP ⎠ VG = VGS + VS = −1 + 2 = 1 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 1 = ( 500 )(10 ) ⇒ R1 = 5 MΩ R1 5R2 = 0.5 ⇒ R2 = 0.556 MΩ 5 + R2 3.59 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ ⎝ VP ⎠ 2 2 ⎛ V ⎞ 5 = 8 ⎜ 1 − GS ⎟ ⇒ VGS = 0.838 V 4 ⎠ ⎝ VSD = VDD − I D ( RS + RD ) = 20 − ( 5 )( 0.5 + 2 ) ⇒ VSD = 7.5 V
  86. 86. VS = 20 − ( 5 )( 0.5 ) = 17.5 V VG = VS + VGS = 17.5 + 0.838 = 18.3 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 18.3 = (100 ) ( 20 ) ⇒ R1 = 109 kΩ R1 109 R2 = 100 ⇒ R2 = 1.21 MΩ 109 + R2 3.60 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ V ⎞ 5 = 7 ⎜ 1 − GS ⎟ ⇒ VGS = 0.465 V 3 ⎠ ⎝ VSD = VDD − I D ( RS + RD ) 6 = 12 − ( 5 )( 0.3 + RD ) ⇒ RD = 0.9 kΩ VS = 12 − ( 5 )( 0.3) = 10.5 V VG = VS + VGS = 10.5 + 0.465 = 10.965 V ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 10.965 = ⎜ 2 ⎟ (12 ) ⇒ R2 = 91.4 kΩ ⇒ R1 = 8.6 kΩ ⎝ 100 ⎠ 3.61 ⎛ R2 ⎞ ⎛ 60 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ ( 20 ) ⇒ VG = 6 V ⎝ 140 + 60 ⎠ ⎝ R1 + R2 ⎠ 2 ⎛ V ⎞ V V − VGS I D = I DSS ⎜ 1 − GS ⎟ = S = G VP ⎠ RS RS ⎝ ⎛ (8 )( 2 ) ⎜1 − ⎜ ⎝ 2 VGS ⎞ ⎟ = 6 − VGS ( −4 ) ⎟ ⎠ ⎛ V V2 ⎞ 16 ⎜ 1 + GS + GS ⎟ = 6 − VGS 2 16 ⎠ ⎝ 2 VGS + 9VGS + 10 = 0 VGS = −9 ± (9) 2 − 4 (10 ) 2 ⇒ VGS = −1.30 ⎛ ( −1.30 ) ⎞ I D = 8 ⎜1 − ⎟ ⇒ I D = 3.65 mA ⎜ ( −4 ) ⎟ ⎝ ⎠ VDS = VDD − I D ( RS + RD ) 2 = 20 − ( 3.65 )( 2 + 2.7 ) VDS = 2.85 V VDS > VDS ( sat ) = VGS − VP = −1.30 − ( −4 ) = 2.7 V (Yes) 3.62
  87. 87. VDS = VDD − I D ( RS + RD ) 5 = 12 − I D ( 0.5 + 1) ⇒ I D = 4.67 mA VS = I D RS = ( 4.67 ) ( 0.5 ) ⇒ VS = 2.33 V ⎛ R2 ⎞ ⎛ 20 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (12 ) ⇒ VG = 0.511 V R1 + R2 ⎠ ⎝ 450 + 20 ⎠ ⎝ VGS = VG − VS = 0.511 − 2.33 ⇒ VGS = −1.82 V ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 ⎛ ( −1.82 ) ⎞ 4.67 = 10 ⎜ 1 − ⎟ ⇒ VP = −5.75 V ⎜ VP ⎟ ⎝ ⎠ 2 3.63 2 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ , VGS = 0 VP ⎠ ⎝ I D = I DSS = 4 mA RD = VDD − VDS 10 − 3 ⇒ RD = 1.75 kΩ = 4 ID 3.64 VSD = VDD − I D RS 10 = 20 − (1) RS ⇒ RS = 10 kΩ R1 + R2 = VDD 20 = = 200 kΩ 0.1 I ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ 2 2 ⎛ V ⎞ 1 = 2 ⎜1 − GS ⎟ ⇒ VGS = 0.586 V 2 ⎠ ⎝ VG = VS + VGS = 10 + 0.586 = 10.586 ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 10.586 = ⎜ 2 ⎟ ( 20 ) ⇒ R2 = 106 kΩ ⎝ 200 ⎠ R1 = 94 kΩ 3.65
  88. 88. VDS = VDD − I D ( RS + RD ) 2 = 3 − ( 0.040 )(10 + RD ) ⇒ RD = 15 kΩ I D = K (VGS − VTN ) 2 40 = 250 (VGS − 0.20 ) ⇒ VGS = 0.60 V 2 VG = VGS + VS = 0.60 + ( 0.040 )(10 ) = 1.0 V ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 1 = ⎜ 2 ⎟ ( 3) ⇒ R2 = 50 kΩ ⎝ 150 ⎠ R1 = 100 kΩ 3.66 For VO = 0.70 V ⇒ VDS = 0.70 > VDS ( sat ) = VGS − VTN 0.75 − 0.15 = 0.6 Biased in the saturation region V − VDS 3 − 0.7 ⇒ I D = 46 μ A I D = DD = 50 RD I D = K (VGS − VTN ) ⇒ 46 = K ( 0.75 − 0.15 ) ⇒ K = 128 μ A / V 2 2 2
  89. 89. Chapter 4 Exercise Solutions EX4.1 g m = 2 K n (VGS − VTN ) and I D = K n (VGS − VTN ) 2 0.75 = 0.5 (VGS − 0.8 ) ⇒ VGS = 2.025 V g m = 2 ( 0.5 )( 2.025 − 0.8 ) ⇒ g m = 1.22 mA / V 2 ro = 1 λ I DQ 1 (0.01)(0.75) = 133 k Ω ro = 133 k Ω = EX4.2 Av = − g m RD g m = 2 K n I DQ = 2 ( 0.5)( 0.4 ) = 0.8944 mA/V Av = − ( 0.8944 )(10 ) = −8.94 EX4.3 (a) ⎛ R2 ⎞ ⎛ 320 ⎞ VGS = ⎜ ⎟ VDD = ⎜ ⎟ ( 5 ) = 1.905 V R1 + R2 ⎠ ⎝ 520 + 320 ⎠ ⎝ I DQ = 0.20 (1.905 − 0.8 ) = 0.244 mA 2 g m = 2 K n I DQ = 2 ( 0.2 )( 0.244 ) = 0.442 mA/V ro = ∞ (b) Av = − g m RD = − ( 0.422 )(10 ) = −4.22 (c) (d) Ri = R1 R2 = 520 320 = 198 K RO = RD = 10 K EX4.4 At transition point, I D = 1 mA I D = K n (VGSt − VTN ) = K n (VDS ( sat ) ) 2 2 1 = 0.2 (VDS ( sat ) ) ⇒ VDS ( sat ) = 2.236 V 2 5 − 2.236 + 2.236 = 3.62 V 2 5 − 3.62 RD = = 2.76 K 0.5 Want VDSQ = 0.5 = 0.2 (VGSQ − 0.8 ) ⇒ VGSQ = 2.38 V 2 ⎛ R2 ⎞ 1 VGSQ = ⎜ ⎟ VDD = ( R1 R2 ) VDD R1 ⎝ R1 + R2 ⎠ 1 So 2.38 = ( 200 )( 5 ) ⇒ R1 = 420 K and R2 = 382 K R1

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