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Unformatted text preview: ECE 315 Homework 2 Solution Fall 2007 1. (Compensational doping) For a piece of homogeneous semiconductor under equilibrium at room temperature with n i =1.5 × 10 10 cm3 , find the electron and hole concentrations for the semiconductor contains both N A =10 18 cm3 and N D =10 15 cm3 : (a) Use the charge neutrality condition and np=n i 2 to obtain the exact solution of n and p (4 pts) Charge neutrality: n – p +10 18 – 10 15 = 0 = n – p + 9.99 × 10 17 Equilibrium carrier statistics: np = n i 2 = 2.25 × 10 20 . Substituting the first equation into the second and obtain the quadratic equation, p 2 – 9.99 × 10 17 p – 2.25 × 10 20 =0, or ( 29 17 20 2 17 17 10 99 . 9 2 10 25 . 2 4 10 99 . 9 10 99 . 9 × = × × × ± × = p cm3 . We can notice that the majority carrier p = N A – N D as long as N A – N D >> 2n i . The minority carrier concentration n = n i 2 /p = 2.25 × 10 2 cm3 . (b) Explain why p = N A – N D is a good approximation for the majority carrier here (4 pts) See the equation above, and the majority carrier p = N A – N D as long as N A – N D >> 2n i . (c) Repeat (a) and (b) for N A =10 18 cm3 and N D =10 17 cm3 . (4 pts) The difference of 9 × 10 17 cm3 is much larger than 2n i here still, and therefore, p = N A – N D = 9 × 10 17 cm3 and n = 2.50 × 10 2 cm3 . (d) If you are asked to calculate the electrostatic potential for the case in (a), will you use p or N A ? What is the potential in relation to the intrinsic level? (4 pts) The electrostatic potential in equilibrium is defined on the carrier concentration, instead of the doping. Therefore, we should use p . Since = kT ψ q n p i exp , we obtain ψ = kT/q × ln( p/n i ) = 26mV × ln(9.99 × 10 17 /1.5 × 10 10 ) = 0.47V in reference of the intrinsic level. I will introduce a calculation by inspection here. Notice that kT/q × ln10 = 60mV at room temperature, we can estimate ψ = 60mV × log 10 ( p/n i ) 2245 60mV × log 10 (10 18 /10 10 ) = 0.48V. We can see this approximation is pretty good. (e) If we build a resistor with this semiconductor, how will you compare the mobility for the compensational doping with both N A =10 18 cm3 and N D =10 17 cm3 and the mobility for the case with only N A =9 × 10 17 cm3 ? Briefly explain. (4 pts) The mobility in the compensational doping case will be lower, since any impurity, regardless of type, is a defect to the original crystalline lattice, and hence will cause scattering. Although for the concentration sake, the effect is mostly subtractive, but for mobility, the two dopants will have an 1 additive effect. The more scattering, the lower the mobility. This is still a good analogy to anions and cations for acidity in water....
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This note was uploaded on 03/26/2008 for the course ECE 3150 taught by Professor Spencer during the Fall '07 term at Cornell.
 Fall '07
 SPENCER
 Microelectronics

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