Analog Electronics Homework #3 Chris Lee Last worked on: 10/22/2017 Homework Due Date: 10/26/2017 7th Edition of the textbook: Microelectronic Circuits, by Sedra/Smith 1|Page Table of Contents FORMULA SHEET & Constants (Rough Draft) .................................................................................................... 4 Intrinsic Semiconductors............................................................................................................................ 4 Doped Semiconductors: ............................................................................................................................. 4 Current Flow Semiconductors.................................................................................................................... 5 The pn Junction .......................................................................................................................................... 7 The pn Junction with Applied Voltage ....................................................................................................... 9 Capacitive Effects in the pn Junction ......................................................................................................... 9 Intrinsic Semiconductors: ................................................................................................................................ 11 Problem 3.1 .................................................................................................................................................. 11 Part A: At −55°C........................................................................................................................................ 11 Part B: At 0°C ............................................................................................................................................ 12 Part C: At 20°C .......................................................................................................................................... 12 Part D: At 75°C ......................................................................................................................................... 13 Part E: At 125°C ........................................................................................................................................ 13 Problem 3.2 .................................................................................................................................................. 14 Doped Semiconductors: ................................................................................................................................... 15 Problem 3.3 .................................................................................................................................................. 15 Problem 3.4 .................................................................................................................................................. 16 Problem 3.5 .................................................................................................................................................. 17 Current Flow in Semiconductors:..................................................................................................................... 19 Problem 3.6 .................................................................................................................................................. 19 Part A........................................................................................................................................................ 19 Part B ........................................................................................................................................................ 20 Part C ........................................................................................................................................................ 21 Part D ....................................................................................................................................................... 22 Problem 3.7 .................................................................................................................................................. 23 Problem 3.8 .................................................................................................................................................. 24 Problem 3.9 .................................................................................................................................................. 26 Problem 3.10 ................................................................................................................................................ 27 The pn Junction: ............................................................................................................................................... 29 Problem 3.12 ................................................................................................................................................ 29 2|Page Problem 3.14 ................................................................................................................................................ 32 Problem 3.15 ................................................................................................................................................ 33 Problem 3.16 ................................................................................................................................................ 34 The pn Junction with Applied Voltage: ............................................................................................................ 35 Problem 3.17 ................................................................................................................................................ 35 Problem 3.20 ................................................................................................................................................ 37 Problem 3.21 ................................................................................................................................................ 38 Problem 3.23 ................................................................................................................................................ 39 Capacitive Effects in the pn Junction: .............................................................................................................. 41 Problem 3.24 ................................................................................................................................................ 41 Problem 3.25 ................................................................................................................................................ 42 3|Page FORMULA SHEET & Constants (Rough Draft) Intrinsic Semiconductors ππ = π΅π 3/2 π −πΈπ ⁄ 2ππ Where ππ = ππππππππ ⁄ππ3 B material dependent Parameter π΅ = 7.3 ∗ 1015 ππ−3 πΎ −3/2 (silicon) π΅ = 3.56 ∗ 1014 ππ−3 πΎ −3/2 (Gallium arsenide) T is the temperature of the material, π = 273 + π‘(β)πΎ. πΈπ is the bandgap energy is, πΈπ = 1.12 ππ. k is the Boltzmann’s constant π = 8.62 ∗ 10−5 ππ/πΎ π π Fraction of atom ionized: πΉ = ππ‘ππππ πππππππ‘πππ‘πππ @ πΊππ£ππ π‘πππππππ‘π’ππβ Atomic concentration of Silicon: 5 ∗ 1022 ππ‘πππ /ππ3 Doped Semiconductors: p-type semiconductors Intrinsic Carrier Density for Silicon at 300k ni= 1.5 ∗ 1010 ππππππππ ⁄ππ3 Acceptor Doping Concentration or doping concentration of the P side is: NA = 5*1018 carrier/cm3 If the acceptor doping concentration NA >> ni (given >> 1.5*1010), then the majority hole concentration becomes ππ ≅ ππ΄ . Therefore the hole concentration is 5*1018 carrier/cm3 Concentration of minority electrons: π2 ππ = ππ where np is minority electrons, pp is hole concentration, ni is the intrinsic carrier density π n-type semiconductor – concentrations of electrons (nn) ≅ donor concentration (ND) ππ ππ = ππ2 ππ· ππ = ππ2 ππ· = 4|Page ππ2 ππ Phosphorous – Donor Impurity Since phosphorus is a donor impurity, ππ· > ππ (Where ππ· is the donor concentration and ππ is the intrinsic concentration) Since it’s a donor impurity, it’s an n-type semiconductor n-type semiconductor = ππ ≅ ππ· where ππ is the concentration of free electrons in the n-type silicon and ππ· is the donor concentration. ππ = 1.5 ∗ 1010 Current Flow Semiconductors Intrinsic Silicon – Under thermal equilibrium, the concentration of free electrons n is equal to the number of holes p. π = π = ππ Measured in m3 Resistivity π can be calculated as follows: π = 1 π = π π(π π π +ππ ) ππΏ π΄ = β¦ β ππ , π = 1.6 ∗ 10−19 , ππ = πππ‘ππππ ππ πππππππ‘πππ‘πππ ππ π ππππππ, ππ = 1.5 ∗ 1010 ππ2 /π. π ππ = πππππππ‘π¦ ππ πππππ‘ππππ , ππ = 1200 ππ2 /π. π ππ = πππππππ‘π¦ ππ βππππ , ππ = 400 ππ2 /π. π For N semiconductors Doping concentration π = ππ΄ ππ2 π= = ππ3 ππ· N- Semiconductors resistivity π π= 5|Page 1 = ππ2 πππ· ππ P-doped Semiconductors π = ππ· ππ2 ππ· = ππ΄ π= 1 πππ΄ ππ + πππ· ππ Where n is free electron, ND is the doping concentration. Drift Velocity ππ = ππ πΈ Where E is πΈ = π β Formula for conductivity: (bar) π = π(πππ + πππ ) Formula for drift current density π½π· = ππΈ, π½π· = π(πππ + πππ )πΈ π = 1.6 ∗ 10−19 Where π½π· is drift current density π is conductivity E is the electric field n is the electron density p is hole density ππ is electron mobility π½π· = πΌπ· is the drift current A is the cross-sectional area 6|Page πΌπ· π΄ Current Density for a doped silicon π½ = ππππ πΈ Change in hole concentration with respect to x distance (Graphical Representation) πππ 108 ππ0 − ππ0 = ππ₯ 0−π π2 ππ0 = ππ = ππ3 is the concentration of holes π· Hole-current density that flow in the x-direction π½π = −ππ·π πππ = π΄/ππ2 ππ₯ The pn Junction Built-in voltage: ππ΄ ππ· π0 = ππ ln ( 2 ) ππ If the n and p regions are equally doped, then ππ΄ = ππ· At room temperature ππ = 0.026π = 300πΎ ππ 26ππ ππ‘ 300πΎ If not, set up proportion Permeability for a depleted region ππ = 11.7π0 Where π0 = 8.85 ∗ 10−14 ππ ≈ 1.04 ∗ 10−12 Width of the depletion region: 2ππ 1 1 √ [ + ] π0 = ππ π ππ΄ ππ· 7|Page The width of a depletion region is given by: π = π₯π + π₯π π₯π ππ΄ = π₯π ππ· π₯π = π₯π ( ππ΄ ) ππ· Since π₯π and π₯π represents the width of a depletion region that extends into the p and n regions. π = π₯π ( = π₯π ( π₯π = ππ΄ ) + π₯π ππ· ππ΄ + 1) ππ· π π (ππ΄ + 1) π· Charge stored on either side of the junction ππ = π ( ππ΄ ππ· ) π΄π ππ΄ +ππ· Total charge stored for a depletion layer ππ = π΄√2πππ ππ· π0 Total charge stored on one side of the junction Qj is: ππ = π΄πππ· π If NA >> ND 2ππ 1 1 √ [ + ] π0 = ππ π ππ΄ ππ· π0 = 8|Page π 2 πππ· =π 2ππ The pn Junction with Applied Voltage Width of depletion for intrinsic silicon is: π=√ 2π(ππ + ππ ) 1 1 ( + ) π ππ΄ ππ· Where NA is the acceptor concentration ππ΄ = 1017 ππ3 Where ND is the donor concentration ππ· = 1016 ππ3 Saturation Current: π·π π·π πΌπ = π΄πππ2 ( + ) πΏπ ππ· πΏπ ππ΄ A is cross-sectional area Lp is distance traveled by holes during their lifetime Ln distance traveled by electrons during their lifetime DP is the diffusion constant for holes DN is the diffusion constant for electrons NA represents the acceptor concentration ND represents the donor concentration Forward current: πΌ = πΌπ (π π ( )−1 ππ ) Power dissipation at break-down region: ππ§ = π 2 Average breakdown Current πΌπ§(ππ£π) = ( πππ‘πππ£ππ πππππ‘β )πΌ πππππ‘β ππ ππππππππ€π π§ Capacitive Effects in the pn Junction ππ π ππ΄ ππ· πΆπ0 = π΄√ ( )=πΉ 2π0 ππ΄ +ππ· 9|Page πΆπ = πΆπ0 π √1 + π π0 =πΉ Particular junction Depletion Capacitance πΆπ = 10 | P a g e πΆπ0 =πΉ ππ π (1 + π ) 0 Intrinsic Semiconductors: Problem 3.1 Find values of the intrinsic carrier concentration ni for silicon at −55°C, 0°C, 20°C, 75°C, and 125°C. At each temperature, what fraction of the atoms is ionized? Recall that a silicon crystal has approximately 5 × 1022 atoms/cm3. ππ = π΅π 3/2 π −πΈπ ⁄ 2ππ Where ππ = ππππππππ ⁄ππ3 B material dependent Parameter π΅ = 7.3 ∗ 1015 ππ−3 πΎ −3/2 T is the temperature of the material, π = 273 + π‘(β)πΎ. πΈπ is the bandgap energy, πΈπ = 1.12 ππ. k is the Boltzmann’s constant π = 8.62 ∗ 10−5 ππ/πΎ π π Fraction of atom ionized: πΉ = ππ‘ππππ πππππππ‘πππ‘πππ Atomic concentration of Silicon: 5 ∗ 1022 ππ‘πππ /ππ3 Part A: At −55°C π = 273 − 55 = 218β ππ = π΅π 3/2 π = (7.3 ∗ −πΈπ ⁄ 2ππ 1.12 ) 15 )(218)3/2 −(2(8.62∗10−5 )(218) 10 π = 2.68406 ∗ 106 ππππππππ ⁄ππ3 π π Fraction ionized = πΉ = ππ‘ππππ πππππππ‘πππ‘πππ πΉ= 11 | P a g e 2.68406 ∗ 106 = 5.36813623 ∗ 10−17 5 ∗ 1022 Part B: At 0°C π = 273 − 0 = 273β ππ = π΅π 3/2 π −πΈπ ⁄ 2ππ = (7.3 ∗ 1015 )(273)3/2 π 1.12 ) −( 2(8.62∗10−5 )(273) = 1.523205601 ∗ 109 ππππππππ ⁄ππ3 π π Fraction ionized = πΉ = ππ‘ππππ πππππππ‘πππ‘πππ 1.523205601 ∗ 109 πΉ= = 3.0464112 ∗ 10−14 @ 0β 22 5 ∗ 10 Part C: At 20°C π = 273 + 20 = 293β ππ = π΅π 3/2 π −πΈπ ⁄ 2ππ = (7.3 ∗ 1015 )(253)3/2 π 1.12 ) −( 2(8.62∗10−5 )(253) = 8.595351739 ∗ 109 ππππππππ ⁄ππ3 π π Fraction ionized = πΉ = ππ‘ππππ πππππππ‘πππ‘πππ 8.595351739 ∗ 109 πΉ= = 1.71907035 ∗ 10−13 @ 20β 5 ∗ 1022 12 | P a g e Part D: At 75°C π = 273 + 75 = 348β ππ = π΅π 3/2 π = (7.3 ∗ 1015 )(348)3/2 π −πΈπ ⁄ 2ππ 1.12 ) −( 2(8.62∗10−5 )(348∗102 ) = 3.700094354 ∗ 1011 ππππππππ ⁄ππ3 π π Fraction ionized = πΉ = ππ‘ππππ πππππππ‘πππ‘πππ 3.700094354 ∗ 1011 πΉ= = 7.40018871 ∗ 10−12 @ 75β 22 5 ∗ 10 Part E: At 125°C π = 273 + 125 = 398β ππ = π΅π 3/2 π = (7.3 ∗ 1015 )(398)3/2 π −πΈπ ⁄ 2ππ 1.12 ) −( 2(8.62∗10−5 )(398∗102 ) = 4.722761829 ∗ 1012 ππππππππ ⁄ππ3 π π Fraction ionized = πΉ = ππ‘ππππ πππππππ‘πππ‘πππ 4.722761829 ∗ 1012 πΉ= = 9.44552366 ∗ 10−11 @ 125β 5 ∗ 1022 π = 273 + π‘(β)πΎ 218 K 273 K 293 K 348 K 398 K 13 | P a g e ππ = π΅π 3/2 π −πΈπ ⁄ 2ππ 2.68406 ∗ 106 ππππππππ ⁄ππ3 1.52320 ∗ 109 ππππππππ ⁄ππ3 8.59535 ∗ 109 ππππππππ ⁄ππ3 3.7000 ∗ 1011 ππππππππ ⁄ππ3 . 72276 ∗ 1012 ππππππππ ⁄ππ3 πΉ= ππ 5 ∗ 1022 5.36813623 ∗ 10−17 3.0464112 ∗ 10−14 1.71907035 ∗ 10−13 7.40018871 ∗ 10−12 9.44552366 ∗ 10−11 Problem 3.2 Calculate the value of ni for gallium arsenide (GaAs) at T = 300 K. The constant B = 3.56×1014 cm−3K−3/2 and the bandgap voltage Eg= 1.42 eV. ππ = π΅π 3/2 π 3 −πΈπ ⁄ 2ππ ππ = (3.56 ∗ 1014 )(3002 )π −1.42⁄ 2(8.62∗10−5 )(300) = 2.204686 ∗ 106 ππππππππ ⁄ππ3 14 | P a g e Doped Semiconductors: Problem 3.3 For a p-type silicon in which the dopant concentration NA = 5×1018/cm3, find the hole and electron concentrations at T = 300 K. p-type semiconductors 10 Intrinsic Carrier Density for Silicon at 300k ni= 1.5 ∗ 10 ππππππππ ⁄ππ3 Acceptor Doping Concentration or doping concentration of the P side is: NA = 5*1018 carrier/cm3 If the acceptor doping concentration NA >> ni (5*1018 >> 1.5*1010), then the majority hole concentration becomes ππ ≅ ππ΄ . Therefore the hole concentration is 5*1018 carrier/cm3 The concentration of the minority electrons is: ππ ππ = ππ2 ππ2 ππ = ππ ππ = 1.5 ∗ 1010 ππππππππ ⁄ππ3 5 ∗ 1018 ππππππππ ⁄ππ3 45 ππππππππ ⁄ππ3 15 | P a g e Problem 3.4 For a silicon crystal doped with phosphorus, what must ND be if at T =300 K the hole concentration drops below the intrinsic level by a factor of 108? ππ = π΅π 3/2 π 15 ππ = (7.3 ∗ 10 )(300) 3/2 −πΈπ ⁄ 2ππ −(1.12 ) ⁄ 2(8.62∗10−5 )(300) π ππ = 1.493926 ∗ 1010 ππ3 Concentration of minority electrons: π2 ππ = ππ where np is minority electrons, pp is hole concentration, ni is the intrinsic carrier density π n-type semiconductor – concentrations of electrons (nn) = donor concentration (ND) ππ = ππ2 ππ· ππ = ππ2 ππ· = ππ2 ππ ππ = ππ(πππ€) ππ(πππ) ππ = ππ(πππ€) 108 (1.5 ∗ 1010 )2 ππ· = ππ(πππ€) 108 ππ· = (1.5 ∗ 1010 )2 1.493926 ∗ 1010 108 ππ· = 1.5 ∗ 1018 16 | P a g e Problem 3.5 In a phosphorus-doped silicon layer with impurity concentration of 1017/cm3, find the hole and electron concentrations at 27°C and 125°C. Solve: Electron Concentration: Convert from Celsius to Kelvin: π = 273 + π‘(β)πΎ. π = 273 + 27(β)πΎ. π = 300°πΎ π = 273 + 125(β)πΎ π = 398°πΎ Since phosphorus is a donor impurity, ππ· > ππ (Where ππ· is the donor concentration and ππ is the intrinsic concentration) Since it’s a donor impurity, it’s an n-type semiconductor n-type semiconductor = ππ ≅ ππ· where ππ is the concentration of free electrons in the n-type silicon and ππ· is the donor concentration. ππ· = 1017 ππ3 Thus, the electron concentration at 27°C and 125°C is 1017 ππ3 Hole concentration is given as: concentrations of electrons (nn) ≅ donor concentration (ND) ππ is hole concentration ππ ππ = ππ2 ππ· ππ = ππ2 ππ· = 17 | P a g e ππ2 ππ ππ2 ππ = ππ· First ππ needs to be calculated: ππ = π΅π 3/2 π −πΈπ ⁄ 2ππ At 27β (300K) ππ = (7.3 ∗ 1015 ππ−3 )(300)3/2 π −1.12⁄ 2(8.62∗10−5 )(300) ππ = 1.493926 ∗ 1010 /ππ3 Substituting ππ = ππ2 ππ· (1.493926 ∗ 1010 )2 ππ = 1017 = 2.23 ∗ 103 /ππ3 At 125β (398K) ππ = (7.3 ∗ 1015 ππ−3 )(398)3/2 π −1.12⁄ 2(8.62∗10−5 )(398) ππ = 4.7227618/ππ3 ππ = ππ = ππ2 ππ· (4.7227618)2 1017 = 2.23 ∗ 108 /ππ3 Recall (from Formula Sheet) B material dependent Parameter π΅ = 7.3 ∗ 1015 ππ−3 πΎ −3/2 (silicon) π΅ = 3.56 ∗ 1014 ππ−3 πΎ −3/2 (Gallium arsenide) T is the temperature of the material, π = 273 + π‘(β)πΎ. πΈπ is the bandgap energy is, πΈπ = 1.12 ππ. k is the Boltzmann’s constant π = 8.62 ∗ 10−5 ππ/πΎ 18 | P a g e Current Flow in Semiconductors: Problem 3.6 A young designer, aiming to develop intuition concerning conducting paths within an integrated circuit, examines the end-to-end resistance of a connecting bar 10-μm long, 3-μm wide, and 1 μm thick, made of various materials. The designer considers: (a) intrinsic silicon (b) n-doped silicon with ND = 5×1016/cm3 (c) n-doped silicon with ND = 5×1018/cm3 (d) p-doped silicon with NA= 5×1016/cm3 (e) aluminum with resistivity of 2.8 μβ¦·cm Find the resistance in each case. For intrinsic silicon, use the data in Table 3.1. For doped silicon, assume μn = 3μp = 1200 cm2/V · s. (π πππππ π‘βππ‘ π = ππΏ/π΄. ) Solve: Part A Intrinsic Silicon – Under thermal equilibrium, the concentration of free electrons n is equal to the number of holes p. π = π = ππ Resistivity π can be calculated as follows: π = 1 π = π π(π π π +ππ ) ππΏ π΄ = β¦ β ππ , π = 1.6 ∗ 10−19 , ππ = πππ‘ππππ ππ πππππππ‘πππ‘πππ ππ π ππππππ, ππ = 1.5 ∗ 1010 ππ3 /π. π ππ = πππππππ‘π¦ ππ πππππ‘ππππ , ππ = 1200 ππ2 /π. π ππ = πππππππ‘π¦ ππ βππππ , ππ = 400 ππ2 /π. π π= π= 1 ππ π(ππ + ππ ) 1 1.5∗1010 (1.6∗10−19 )(1200+400) π = 260416.667β¦ β ππ 19 | P a g e The length is given as: 10-μm long The area is given as: 3-μm wide * 1 μm thick = 3 μm2 π = π = ππΏ π΄ (260416.667 ∗ 104 ππ)(10μm) 3 μm2 π = 8.68055 ∗ 109 β¦μm Part B n-doped silicon with ND = 5×1016/cm3 n-type semiconductor = ππ ≅ ππ· ππ ππ = ππ2 ππ· = ππ2 ππ ππ = 1.5 ∗ 1010 5 × 1016 = ππ = (1.5 ∗ 1010 )2 ππ (1.5 ∗ 1010 )2 5 × 1016 ππ = 4500ππ3 N- Semiconductors resistivity π π= π= (1.6 ∗ 10 −19 1 πππ· ππ 1 )(5 × 1016 )(1200 ππ2 /π. π ) π = .1042β¦ππ 20 | P a g e The resistance is calculated as: ππΏ π΄ π = 4 π = (. 1042 ∗ 10 )10μm 3 μm2 π = 3473.33 πβππ Part C (c) n-doped silicon with ND = 5×1018/cm3 For N semiconductors Doping concentration π = ππ΄ ππ2 π= ππ· N- Semiconductors resistivity π π= π= 1 πππ· ππ 1 (1.6 ∗ 10 −19 18 )(5 × 10 /cm3 )(1200 ππ2 /π. π ) π = 1.0416 ∗ 10−3 π = π = (1.0416 ∗ 10 ππΏ π΄ −3 ∗ 104 ) 10μm 3 μm2 π = 34.72222223 πβππ 21 | P a g e Part D (d) p-doped silicon with NA= 5×1016/cm3 P-doped Semiconductors π = ππ· ππ· = π= ππ2 ππ΄ 1 πππ΄ ππ + πππ· ππ Where n is free electron, ND is the doping concentration. ππ· = (1.5 ∗ 1010 )2 5 ∗ 1016 ππ· = 4500 ππ = πππππππ‘π¦ ππ πππππ‘ππππ , ππ = 1200 ππ2 /π. π ππ = πππππππ‘π¦ ππ βππππ , ππ = 400 ππ2 /π. π π= (1.6 ∗ 10 −19 )(5 ∗ 1 + (1.6 ∗ 10−19 )(4500)(1200) 1016 )(400) π = .3125 4 π = (. 3125 ∗ 10 ) ∗ 10−6 3 μm2 π = 1041.667 πβππ 22 | P a g e Problem 3.7 Contrast the electron and hole drift velocities through a 10-μm layer of intrinsic silicon across which a voltage of 3 V is imposed. Let μn = 1350 cm2/V γ» s and μp = 480 cm2/V γ» sγ» Solve ππ = ππ πΈ The electric field is given as πΈ= πΈ= π π 3π 10ππ = .3π/ππ ππ = (1350 ∗ ππ2 /π. π )(. 3π/ππ) ππ2 ππ = 405 ( ) ππ. π (10−2 )2 ππ = 405 ( ) 10−6 10−4 ππ = 405 ( −6 ) 10 ππ = 405(100) = 40500π/π 23 | P a g e Problem 3.8 Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hole densities of 104/cm3 and 1016/cm3, respectively, when a 1 V is applied end-to-end. Use μn = 1200 cm2/V γ» s and μp = 500 cm2/V γ»s. Formula for conductivity: π = π(πππ + πππ ) Formula for drift current density π½π· = ππΈ, π½π· = π(πππ + πππ )πΈ Where π½π· is drift current density π is conductivity E is the electric field Where E is πΈ = π β n is the electron density ππ is electron mobility n is the electron density p is hole density πΈ= πΈ= π β 1π 10ππ π½π· = π(πππ + πππ )πΈ π½π· = (1.6 ∗ 10−19 ) ((1016 )500 + (104 )(1200)) ∗ 1000 = 800π΄/ππ2 24 | P a g e The cross-sectional area is given by: (5μm × 4μm ) ∗ 104 = 2 ∗ 10−7 ππ2 Calculating the drift current: π½π· = πΌπ· π΄ πΌπ· is the drift current A is the cross-sectional area πΌπ· = (800π΄/ππ2 )(2 ∗ 10−7 ππ2 ) = 160ππ΄ 25 | P a g e Problem 3.9 In a 10-μm-long bar of donor-doped silicon, what donor concentration is needed to realize a current density of 2 mA/μm2 in response to an applied voltage of 1 V? (Note: Although the carrier mobilities change with doping concentration, as a first approximation you may assume μn to be constant and use 1350 cm2/V γ» s, the value for intrinsic silicon.) Current Density for a doped silicon π½ = ππππ πΈ n-type semiconductor – concentrations of electrons (nn) ≅ donor concentration (ND) π= π½ πππ πΈ Solving for E πΈ= πΈ= π β 1π 10μm πΈ = .1π/μm π= 2 ∗ 10−3 /(10−6 ) (1.6 ∗ 10−19 )(1350 ∗ (10−2 )2 ). 1 π = 9.25925 ∗ 1023 26 | P a g e Problem 3.10 Holes are being steadily injected into a region of n-type silicon (connected to other devices, the details of which are not important for this question). In the steady state, the excess-hole concentration profile shown in Fig. P3.10 is established in the n-type silicon region. Here “excess” means over and above the thermalequilibrium concentration (in the absence of hole injection), denoted pn0. If ND = 1016/cm3, ni=1.5×1010/cm3, Dp=12 cm2/s, and W =50 nm, find the density of the current that will flow in the x direction. n-type semiconductor – concentrations of electrons (nn) ≅ donor concentration (ND) ππ ππ = ππ2 ππ· ππ = ππ2 ππ· = ππ2 ππ Change in hole concentration with respect to x distance πππ 108 ππ0 − ππ0 = ππ₯ 0−π π2 ππ0 = ππ is the concentration of holes π· Solving for the concentration of holes ππ0 = 27 | P a g e (1.5 × 1010 )2 = 22500ππ3 1016 πππ 108 (22500ππ3 ) = ππ₯ 0−π πππ 108 (22500ππ3 ) =− ππ₯ π πππ 108 (22500ππ3 ) =− ππ₯ 50 ∗ 10−9 ∗ 102 ππ2 πππ = 4.5 ∗ 1017 ππ−4 ππ₯ Hole-current density that flow in the x-direction π½π = −ππ·π πππ = π΄/ππ2 ππ₯ π½π = −(1.6 ∗ 10−19 )(12 ∗ 10)(4.5 ∗ 1017 ) = 0.864 π΄/ππ2 28 | P a g e The pn Junction: Problem 3.12 Calculate the built-in voltage of a junction in which the p and n regions are doped equally with 5×1016 atoms/cm3. Assume ni= 1.5 × 1010/cm3. With the terminals left open, what is the width of the depletion region, and how far does it extend into the p and n regions? If the cross-sectional area of the junction is 20 μm2, find the magnitude of the charge stored on either side of the junction. Built-in voltage: ππ΄ ππ· π0 = ππ ln ( 2 ) ππ If the n and p regions are equally doped, then ππ΄ = ππ· This means ππ΄ = 5 ∗ 1016 ππ· = 5 ∗ 1016 ππ΄ ππ· π0 = ππ ln ( 2 ) ππ π0 = 0.026π ln ( (5 ∗ 1016 )(5 ∗ 1016 ) ) 1.5 ∗ 1010 ππ = 0.026π = 300πΎ ππ 26ππ ππ‘ 300πΎ π0 = 0.7810 π Permeability for a depleted region π = 11.7π0 Where π0 = 8.85 ∗ 10−14 Width of the depletion region: 2π 1 1 √ [ + ] π ππ΄ ππ· 29 | P a g e =√ 2(11.7(8.85 ∗ 10−14 )) 1 1 [ + ] −19 5 ∗ 1016 5 ∗ 1016 (1.6 ∗ 10 ) = 2.275 ∗ 10−5 ππ = 2.275 ∗ (10−5 )(10−2 )π = 2.275 ∗ 10−7 π = 0.2275 ∗ 10−6 ππ The width of a depletion region is given by: π = π₯π + π₯π π₯π ππ΄ = π₯π ππ· π₯π = π₯π ( ππ΄ ) ππ· Since π₯π and π₯π represents the width of a depletion region that extends into the p and n regions. π = π₯π ( = π₯π ( π₯π = ππ΄ ) + π₯π ππ· ππ΄ + 1) ππ· π π (ππ΄ + 1) π· Solving for the extension in the P region π₯π = 0.2 5 ∗ 1016 ( + 1) 5 ∗ 1016 0.2ππ = 0.1ππ 2 30 | P a g e Solving for the extension in the n region π − π₯π = π₯π π − π₯π = π₯π 0.2ππ − 0.1ππ = 0.1ππ 31 | P a g e Problem 3.14 Estimate the total charge stored in a 0.1-μm depletion layer on one side of a 10-μm×10-μm junction. The doping concentration on that side of the junction is 1018/cm3. Total charge stored on one side of the junction Qj is: ππ ≈ π΄πππ· π Solve: Calculate the area: 10ππ × 10ππ = 100ππ2 ππ = (100ππ2 )(1.6 ∗ 10−19 πΆ)(1018 /cm3 )(0.1μm) ππ = 1.6 ∗ 10−12 32 | P a g e Problem 3.15 In a pn junction for which NA >>ND, and the depletion layer exists mostly on the shallowly doped side with W =0.2 μm, find V0 if ND= 1016/cm3. Also calculate QJ for the case A = 10 μm2. Given: W If NA >> ND 2ππ 1 1 √ [ + ] π0 = ππ π ππ΄ ππ· π0 = π0 = π0 = π 2 πππ· =π 2ππ (0.2 ∗ 10−4 )2 (1.6 ∗ 10−19 )1016 2(11.7(8.85 ∗ 10−14 )) (0.2 ∗ 10−4 )2 (1.6 ∗ 10−19 )1016 2(11.7(8.85 ∗ 10−14 )) =π =π π0 = .309π Given: For a depletion layer ππ = π΄√2πππ ππ· π0 = (10 ∗ 10−8 )√2(1.6 ∗ 10−19 )(1.04 ∗ 10−12 )(1016 )(. 309π) = 3.1997 ∗ 10−15 33 | P a g e Problem 3.16 By how much does V0 change if NA or ND is increased by a factor of 10? ππ΄ ππ· π0 = ππ ln ( 2 ) ππ At room temperature ππ = 0.026π = 300πΎ ππ 26ππ ππ‘ 300πΎ Original: π0 = 26ππ ln ( 1(1) ) = −1218.42ππ (1.5 ∗ 1010 )2 π0 = 26ππ ln ( 1(10) ) = −1158.56ππ (1.5 ∗ 1010 )2 New Value: −1218.42ππ − −1158.56ππ = 59.86mV Thus, the value increases by 59.86mV when NA or ND is increased by a factor of 10. 34 | P a g e The pn Junction with Applied Voltage: Problem 3.17 If a 3V reverse-bias voltage is applied across the junction specified in Problem 3.13, find W and QJ . Width of depletion for intrinsic silicon is: π=√ 2π(ππ + ππ ) 1 1 ( + ) π ππ΄ ππ· Where NA is the acceptor concentration ππ΄ = 1017 ππ3 Where ND is the donor concentration ππ· = 1016 ππ3 ππ΄ ππ· π0 = ππ ln ( 2 ) ππ ππ = 0.026π = 300πΎ ππ 26ππ ππ‘ 300πΎ π2 = 1.5 ∗ 1010 ππππππππ ⁄ππ3 Solve: (1017 ππ3 )(1016 ππ3 ) π0 = 26ππ ln ( 2) (1.5 ∗ 1010 ππππππππ ⁄ππ3 ) π0 = 757.18 ππ π0 = .75718 π 2(1.04 ∗ 10−12 )(. 75718 + 3) 1 1 π=√ ( 17 + 16 ) −19 10 10 (1.6 ∗ 10 ) = 7.3299 ∗ 10−5 ππ = .73299 πππ π = 1.6 ∗ 10−19 Charge stored on either side of the junction 35 | P a g e ππ = π ( ππ΄ ππ· ) π΄π ππ΄ +ππ· (1017 ππ3 )(1016 ππ3 ) ππ = (1.6 ∗ 10−19 ) ( 17 3 ) (100ππ2 )(. 73299 πππ) 10 ππ + 1016 ππ3 ππ = 10.63 ∗ 1014 πΆ 36 | P a g e Problem 3.20 Calculate IS and the current I for V = 750 mV for a pn junction for which NA = 1017/cm3, ND = 1016/cm3, A = 100 μm2, ni= 1.5×1010/cm3, Lp = 5 μm, Ln= 10 μm, Dp = 10 cm2/s, and Dn= 18 cm2/s. Saturation Current: π·π π·π πΌπ = π΄πππ2 ( + ) πΏπ ππ· πΏπ ππ΄ A is cross-sectional area Lp is distance traveled by holes during their lifetime Ln distance traveled by electrons during their lifetime DP is the diffusion constant for holes DN is the diffusion constant for electrons NA represents the acceptor concentration ND represents the donor concentration π = 1.6 ∗ 10−19 10ππ2 /π 18ππ2 /π −19 2( 10 3 )2 (100ππ) )(1.5 πΌπ = 1.6 ∗ 10 πΆ ∗ 10 /ππ ( + ) (5ππ)(1016 /cm3 ) (10 μm)( 1017 /cm3 ) ππ2 10ππ2 cm3 18ππ2 cm3 πΌπ = (3600) ( 6 ) ( + ) (5ππ)(1016 )π (10 μm)( 1017 )π ππ ππ2 10ππ5 18ππ5 πΌπ = (3600) ( 6 ) ( + ) (5 ∗ 1016 ππ)π (1 ∗ 1018 ππ)π ππ ππ2 ππ5 (2 ∗ 10−16 ) ππ5 (1.8 ∗ 10−17 ) πΌπ = (3600) ( 6 ) ( + ) (ππ)π (ππ)π ππ 2.18 ∗ 10−16 1π πΌπ = (3600)(ππ)π ( )( ) ππ 100ππ πΌπ = 0.7848 ∗ 10−16 π΄ Forward current: 750 πΌ = (0.7848 ∗ 10−16 ) (π ( 26 ) − 1) πΌ = 2.645 ∗ 10−2 37 | P a g e Problem 3.21 Assuming that the temperature dependence of IS arises mostly because IS is proportional to n2i, use the expression for ni in Eq. (3.2) to determine the factor by which n2i changes as T changes from 300 K to 305 K. This will be approximately the same factor by which IS changes for a 5°C rise in temperature. What is the factor? Utilizing Intrinsic Semiconductor Formula ππ = π΅π 3/2 π −πΈπ ⁄ 2ππ B material dependent Parameter π΅ = 7.3 ∗ 1015 ππ−3 πΎ −3/2 (silicon) π΅ = 3.56 ∗ 1014 ππ−3 πΎ −3/2 (Gallium arsenide) T is the temperature of the material, π = 273 + π‘(β)πΎ. πΈπ is the bandgap energy is, πΈπ = 1.12 ππ. k is the Boltzmann’s constant π = 8.62 ∗ 10−5 ππ/πΎ ππ = (7.3 ∗ 1015 ππ−3 )(300)3/2 π (−1.12⁄ ) 2(8.62∗10−5 )300 ππ = 1.493926 ∗ 1010 ππ 2 = 2.2318 ∗ 1020 ππ = (7.3 ∗ 1015 ππ−3 )(305)3/2 π (−1.12⁄ ) 2(8.62∗10−5 )305 ππ = 2.18409 ∗ 1010 ππ 2 = 4.770283 ∗ 1020 ππ2 (305πΎ) ππ2 (300πΎ) = 4.770283 ∗ 1020 2.2318 ∗ 1020 = 2.1 38 | P a g e Problem 3.23 A pn junction for which the breakdown voltage is 12 V has a rated (i.e., maximum allowable) power dissipation of 0.25 W. What continuous current in the breakdown region will raise the dissipation to half the rated value? If breakdown occurs for only 10ms in every 20 ms, what average breakdown current is allowed? Power dissipation at break-down region: ππ§ = ππ§ = π 2 0.25 2 ππ§ = .125 Continuous current in the breakdown region ππ§ = πΌπ ππ (. 125) = πΌπ (12) (. 125) = πΌπ (12) πΌπ = 0.01041666π΄ Breakdown Current at .25W . 25π = πΌπ (12) πΌπ = .0208π΄ Average breakdown Current πΌπ§(ππ£π) = ( πππ‘πππ£ππ πππππ‘β )πΌ πππππ‘β ππ ππππππππ€π π§ 20 ∗ 10−3 πΌπ§(ππ£π) = ( ) . 0208π΄ 10 ∗ 10−3 πΌπ§(ππ£π) = .0416 π΄ 39 | P a g e 40 | P a g e Capacitive Effects in the pn Junction: Problem 3.24 For the pn junction specified in Problem 3.13, find Cj0 and Cj at VR =3 V. Built-in voltage: ππ΄ ππ· π0 = ππ ln ( 2 ) ππ NA = 1017/cm3 ND = 1016/cm3 ππ2 = 1.5 ∗ 1010 ππππππππ ⁄ππ3 ππ = 0.026π = 300πΎ ππ 26ππ ππ‘ 300πΎ ππ ≈ 1.04 ∗ 10−12 π = 1.6 ∗ 10−19 A = 100µm2 ππ π ππ΄ ππ· πΆπ0 = π΄√ ( ) 2π0 ππ΄ +ππ· πΆπ0 πΆπ = √1 + ππ π0 Solve (1017 )(1016 ) π0 = (26) ln ( ) = 757.189ππ (1.5 ∗ 1010 )2 −12 (1.04 ∗ 10 ) (1.6 ∗ 10−19 ) (1017 )(1016 ) √ πΆπ0 = (100ππ ) ( ) (1017 ) + (1016 ) 2(. 757189) 2 πΆπ0 = 3.16 ∗ 10−14 πΆπ = 3.16 ∗ 10−14 √1 + 3 . 757189 πΆπ = 1.418 ∗ 10−14 41 | P a g e Problem 3.25 For a particular junction for which Cj0 = 0.4 pF, V0 = 0.75 V, and m = 1/3, find Cj at reverse-bias voltages of 1 V and 10 V. Particular junction Depletion Capacitance πΆπ = πΆπ = 0.4 1 1/3 (1 + ) 0.75 πΆπ = 42 | P a g e πΆπ0 π π (1 + ππ ) 0 0.4 10 1/3 (1 + ) 0.75 = .30157 ππΉ = .1646 ππΉ
