ECE
hw02sol

# hw02sol - ECE 315 Homework 2 Solution Spring 2009...

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ECE 315 Homework 2 Solution Spring 2009 1. (Compensational doping) For a piece of homogeneous silicon under equilibrium at room temperature with n i =1.5 × 10 10 cm -3 , find the electron and hole concentrations where both types of exist: N A =10 18 cm -3 and N D =10 15 cm -3 : (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 cm -3 . 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 cm -3 . (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 cm -3 and N D =10 17 cm -3 . (4 pts) The difference of 9 × 10 17 cm -3 is much larger than 2n i here still, and therefore, p = N A – N D = 9 × 10 17 cm -3 and n = 2.50 × 10 2 cm -3 . (d) If you are asked to calculate the electrostatic potential for the case in (c), will you use p or N A ? What is the potential using the intrinsic level as reference? (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 × 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 cm -3 and N D =10 17 cm -3 and the mobility for the case with only N A =9 × 10 17 cm -3 ? 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

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additive effect. The more scattering, the lower the mobility. This is still a good analogy to anions and cations for acidity in water for carrier concentration concerns. 2. (Carrier concentration in different temperatures) The intrinsic silicon has E gap = 1.1eV and n i = 1.5 × 10 10 cm -3 at 300K. If we assume E gap and the dopant (impurity that is either donor or acceptor) ionization are temperature insensitive, given that n i exp( -E gap /2kT ), (a) What will be the intrinsic concentration at 250K and 400K? (4 pts) k = Boltzmann constant = 8.62×10 -5 eV/K n i (T) = exp {-E gap /2kT} n i (T=300 K) = 1.5×10 10 cm -3 = c× exp {-1.1 eV/2×k×300 K}, i.e., c = 2.59×10 19 cm -3 n i (T=250 K) = c× exp {-1.1eV/2×k×250 K} = 2.13×10 8 cm -3 n i (T=400 K) = c× exp {-1.1eV/2×k×400 K} = 3.06×10
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