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Unformatted text preview: ECE 315 Homework 3 Solution Spring 2009 1. (PN junction, continued from Prob. 7 of HW 2. Please follow the numbers there. ) For a pn junction well described by the depletion region approximation as shown below, assume the electron mobility μ n =1000cm 2 /Vs, the hole mobility μ p =400cm 2 /Vs, and the minority lifetime τ n = τ p =107 s (a) In equilibrium, which region(s) satisfy the charge neutrality condition? Explain briefly. (4 pts) Regions 1 and 4. In these regions, there is negligible voltage drop and is therefore characteristically similar to bulk silicon. In regions 2 and 3, we have majority carrier diffused across the junction, which leaves behind ionized immobile dopant charges whose electric charge is no longer neutralized by their corresponding free charge carriers. Regions 2 and 3 are thus approximated with the “depletion regions”, which means the majority carriers are depleted (noted that the minority carriers are actually increased to keep the np product constant.) (b) In equilibrium, which region(s) satisfy the relation of np=n i 2 ? Explain briefly. (4 pts) All Regions of 14. In equilibrium, np=n i 2 is always true everywhere. Rigorous proof has to come from statistical mechanics and the definition of the Fermi levels, but you can think of the situation as the chemical equation of e + h + ↔ nil. (c) At V D =0, Calculate the depletion capacitance in F/cm 2 . (4 pts) We can use the calculation in Homework 2 and have x p = x n = 216 nm. Hence the depletion capacitance, or the capacitance across the depletion region, is: C jo = ε si ε o /W = ε si ε o /(x p +x n = 24.0 nF/cm 2 . (d) Estimate the ideal saturation current I S at 300K and 400K. You can use the n i values from Problem 2 of Homework 2. Assume A = 1cm 2 . (4 pts) We will ignore the temperature dependence of mobility and recombination lifetime here. They do have some temperature dependence, but in view of the exponential dependence of n i , are not very important to include. I S = J S × A J S = qD p p no /L p + qD n n po /L n p no = n i 2 (T)/ n no ≈ n i 2 /N D n po = n i 2 (T)/ p po ≈ n i 2 /N A L p = (D p τ p ) 1/2 L n = (D n τ n ) 1/2 J S = qD p n i 2 / N D L p + qD n n i 2 /N A L n Since we have μ n =1,000 cm 2 /Vs and μ p =400 cm 2 /Vs, as well as τ n = τ p =0.1 μ s Then, D n = kT/q × μ n , D p = kT/q × μ p Therefore, I S = qA (D p n i 2 / N D L p + qD n n i 2 /N A L n ) 1 at T = 300 K, n i = 1.5 × 10 10 cm3 , I S = 94.5 pA (×1012 A) at T = 400 K, n i = 3.06 × 10 12 cm3 , I S = 4.54 μA (×106 A) You can see that the direct dependence of n i 2 has caused the current to change over 4 orders of magnitude!!! Your computer “can” behave quite differently in Fargo and in Phoenix, though there are often other feedback/stabilizing techniques applied to minimize this effect....
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 Spring '07
 SPENCER
 Microelectronics, Pn junction, depletion region, forward bias, Charge carriers in semiconductors, Vnode Vdiode Vnode VD

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