HW1_Soln - January 23, 2008 ECE315 Homework # 1 Solutions...

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January 23, 2008 ECE315 Homework # 1 Solutions (Rev. 0) Spring 2008 1. Textbook Chapter 2, Problem 1. The intrinsic carrier concentration of Ge was given as: n i = 1 . 66(10 15 ) T 3 / 2 exp ± - E g 2 kT ² [ cm - 3 ] , where the bandgap E g is 0.66 eV. (a) Calculate n i at T=300 and T=600 K. Note that the thermal voltage in electron volts is kT/q , where k = 1 . 38066(10 - 23 ) [J/K] and q = 1 . 602(10 - 19 ) [J/eV]. So at 300 K, kT/q is 1 . 38066(10 - 23 )(300) / 1 . 602(10 - 19 ) = 25 . 86 meV or mV. At 600 K, kT/q is double this value, 51.17 meV. So. .. n 300 i = 1 . 66(10 15 )300 3 / 2 exp ³ - 0 . 66 (2)25 . 86(10 - 3 ) ´ = 2 . 48(10 13 ) [ cm - 3 ] , and n 600 i = 1 . 66(10 15 )600 3 / 2 exp ³ - 0 . 66 (2)51 . 17(10 - 3 ) ´ = 3 . 86(10 16 ) [ cm - 3 ] , (b) Law of Mass Action states that pn = n 2 i . So for the Ge sample doped with phosphorous (a donor impurity) then charge neutrality states that n = N d = 5(10 16 ) cm - 3 at any temperature where Ge is still extrinsic (temperatures low enough such that doping far exceeds intrinsic concentration). The hole concentration is temperature dependent from law of mass action. p = n 2 i N d -→ p 300 = [2 . 48(10 13 )] 2 5(10 16 ) = 1 . 23(10 10 ) cm - 3 . -→ p 600 = [3 . 86(10 16 )] 2 5(10 16 ) = 3 . 00(10 16 ) cm - 3 . At room temperature (300 K) the intrinsic concentrations are similar, but as the temperature is raised the intrinsic concentration for Ge increases faster than that of Si. 2. Textbook Chapter 2, Problem 2. A homogeneous electric field of 0.1 V/ μ m (1000 V/cm) is placed on a uniformly doped n-type Si sample. (a) Assuming the velocity has not saturated then v = μ n E . Assuming the doping is light then we take the mobility to be 1350 cm 2 /volt . sec (section 2.1.3 in text). Hence: v = μ n E = 1350(10 3 ) = 1 . 35(10 6 ) [ cm/sec ] . (b) Considering the transport of electrons (by drift) for small electric fields and a drift current of 1 mA/ μm 2 (10 5 amp/cm 2 ): J = σ n E = n nE -→ n = J/ ( n E ) = 10 5 1 . 602(10 - 19 )1350(10 3 ) = 4 . 6(10 17 ) [ cm - 3 ] . Note that this doping level would result in a much lower mobility meaning the actual
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This note was uploaded on 09/28/2008 for the course ECE 3150 taught by Professor Spencer during the Spring '07 term at Cornell University (Engineering School).

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HW1_Soln - January 23, 2008 ECE315 Homework # 1 Solutions...

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