hwk10_sol

# hwk10_sol - EE40 Homework #10 Solutions Problem 1 a) Boron,...

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Unformatted text preview: EE40 Homework #10 Solutions Problem 1 a) Boron, p‐type Phosphorous, n‐type Carbon, since it is in the same column as Silicon, so it is neither donor nor acceptor, so neither n‐type nor p‐type. b) Ga is in III‐column, and As is in VI‐column, P is in VI‐column, Si is in IV‐column. If replace Ga with P, then it is n‐type doping; if replace As with P, then effectively no change, neither n‐type nor p‐type. Problem 2 Because the depletion region is depleted of carriers, and the quasi‐neutral region has a lot of carriers. Note that carriers are the current‐conducting component in semiconductors. If temperature goes up, the intrinsic carrier concentration, ni, will go up, which means resistance will go down. Silicon is a semiconductor, it is resistivity is in between conductors and insulators. if do nothing to it, resistance is high If add Boron / Phosphorous to it, the hole / electron carrier concentration will increase, the resistance will be lower If add equal amount of Boron and Phosphorous, it is called compensated doping, and effectively the electron and hole carrier concentration will remain the same, so the resistance will be roughly equal to an intrinsic silicon, so resistance will be high. Note that in ee130, you will learn that in this case, actually the resistance will be higher than the resistance of an intrinsic silicon, which, of course, will not be on your test; just point out for the sake of accuracy. If add carbon, the resistance will be roughly the same as intrinsic silicon, so resistance will be high. If replace Ga with Si, then n‐type if replace As with Si, then p‐type Problem 3 a) b) c) p = NA = 10^18. n = ni^2 / p = 10^2. n = ND = 10^17. p = ni^2 / n = 10^3. According to the charge neutrality condition, p + Nd = n + Na, where Nd = 0, Na = 10^10. In addition, we have np = ni^2. Therefore, (n+10^10)*n = 10^20. Solving this second‐order equation, we have n = 0.62*10^10, p = 1.62*10^10. What is interesting here, is that you cannot simply assume p = Na. Why? If we carefully examine this equation: p + Nd = n + Na, or p = n + Na (if Nd = 0). Since Na is not large enough, Na becomes comparable with n, therefore p > Na, and we cannot assume p = Na. What if Na = 10ni = 10^11? If Na = 10ni, we have p = n + Na > Na, therefore p > 10ni. Since np = ni^2, n < 0.1ni, then, n + Na is the sum of a quantity which is less than 0.1ni and another quantity which is equal to 10ni, since n < 0.01Na, we can say p = n + Na is roughly equal to Na (within 1% error). Problem 4 a) The depletion approximation basically says that, in the depletion region of a PN junction, the electron/hole carrier concentration is effectively zero. As a result of the depletion approximation, the charge density in the depletion region is equal to the doping concentration times the 1.6*10^‐19. b) c) According to the charge neutrality constraint, NA * xp = ND * xn, where xp and xn are the depletion region width within p and n region, respectively. The built in potential is equal to Vbi = kT NAND ln( 2 ) = 26mV × ln(1015 ) = 26mV × 15 × 2.3 = 0.9V q ni Note: interestingly, this built‐in potential, is in fact the VDON for the piecewise model of a diode, which you will learn from ee130. Problem 5 a) obeys charge neutrality. The total charge on the n‐side is proportional to 10^17 * 15 = 1.5 *10^18 The total charge on the p‐side is proportional to (basically integral of charge density over x) 10^18 * 1 / 2 + 10^18 * 1 = 1.5 * 10^18. b) ε is the dielectric constant for the material. Regarding the electric field, since the charge density is never infinite, so the electric field is a CONTINUOUS FUNCTION across the entire region. E = 0, for x ≥ xn0 , or , x ≤ − x p1 dE + qND + qND = , for 0 ≤ x ≤ xn0 , so E (x) = ( x − xn 0 ) dx ε ε dE −qNA −qNA = , for − x p 0 ≤ x ≤ 0, so E (x) = (x + x AA ) ε ε dx where x AA = ND xn 0 NA dE −qNA (x + x p1 ) −qNA x 2 = , for − x p1 ≤ x ≤ − x p 0 , so E (x) = ( + x p1 x + xBB 2 ) dx ε ε 2 where 2 x2 −qNA x p1 ( − x p12 + xBB 2 ) = 0, so xBB12 = p1 2 2 ε Regarding the electrostatic potential, since the electric field is never infinite, so the electrostatic potential is a CONTINUOUS function over the entire region. Since dφ = −E dx for 0 ≤ x ≤ xn0 , φ (x) = +qND 1 2 ( x − xn0 x + C1) ε2 −qNA x 2 for − x p 0 ≤ x ≤ 0, φ (x) = ( + x AA x + C 2) ε 2 for − x p1 ≤ x ≤ − x p 0 , φ (x) = −qNA x 3 x p1 2 (+ x + xBB 2 x + C 3) ε 6 2 where C1, C2 and C3 are three arbitrary constants which make the electrostatic potential function CONTINUOUS over the entire region. Problem 6 Problem 7 K = Kp /2 * W / L = 250 uA/V^2 Vto = 1V (a) Vgs = 5V id = K (Vgs‐Vto)^2 = 4mA (b) Vgs = 3V, Vds = 1V id = K[ 2(Vgs‐Vto)Vds – Vds2] = 0.75mA (c) Vgs = 3V id = K (Vgs‐Vto)^2 = 1mA (d) id = 0 ...
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## This note was uploaded on 10/24/2009 for the course EE 40 taught by Professor Chang-hasnain during the Fall '07 term at Berkeley.

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