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Unformatted text preview: SOLUTIONS CHAPTER 5 5.1. A silicon pn junction is formed between ntype silicon doped with N D =10 17 cm3 and ptype silicon doped with N A =10 16 cm3 . a) Sketch the energy band diagram. Label both axes and all important energy levels. Neither side is degenerately doped. We find δ n =  kT ln 10 17 2.86 × 10 19 = 0.147 eV and δ p =  kT ln 10 16 3.1 × 10 19 = 0.209 eV . The resulting energy band diagram is E C E i E f E V E C E i E f E V distance (x) qV bi =0.764 eV δ n =0.147 eV δ p =0.209 eV b) Find n n0 , p n0 , n p0 , and p p0 . Sketch the carrier concentrations. n n = N D = 10 17 cm 3 p n = n i 2 n n = 1.08 × 10 10 cm 3 ( 29 2 10 17 cm 3 = 1.17 × 10 3 cm 3 p p = N A = 10 16 cm 3 n p = n i 2 p p = 1.08 × 10 10 cm 3 ( 29 2 10 16 cm 3 = 1.17 × 10 4 cm 3 See above for carrier concentration sketches. c) What is the builtin voltage? qV bi = E g δ n δ p = 1.12 0.147 0.209 = 0.764 eV or V bi =0.764V Anderson& Anderson 1 May 7, 2009 Solutions Chapter 5 5.2 Recall that the circuit symbol for a diode is as shown in Figure II.3. Which is the anode (the end labeled “+”) of a pn junction diode, the p side or the n side? Explain your reasoning. In a diode, current can flow from the anode to the cathode but not from the cathode to the anode. In a pn junction, current can flow from p to n but only a small leakage current can flow from n to p. Thus, the pside is the anode and the nside is the cathode. 5.3. A p + n junction is formed in silicon. On the p side, the Fermi level is at the (intrinsic) valence band edge E f = E V0 . The n side is doped with N D =5 × 10 16 cm3 . a) Sketch the energy band diagram. First, we have to find δ n δ n =  kT ln N D N C =  0.026 eV ln 5 × 10 16 cm 3 2.86 × 10 19 cm 3 = 0.165 eV The resulting energy band diagram is E C E i E f E V E C E i E f E V distance (x) qV bi =0.955 eV δ n =0.165 eV δ p =0 eV = b) Sketch the carrier concentrations. see above c) What is the builtin voltage? V bi = E g δ n δ p = 1.12 0.165 = 0.955 eV Anderson& Anderson 2 May 7, 2009 Solutions Chapter 5 5.4. Fill in the missing steps to derive Equation 5.35 for the junction width on the n side of the junction. We want to show that ' ' 2 ' 1 j n D D A V w N qN N ε = + using ' 2 n j n D V w qN ε = and ' ' n j A p j D V N V N = . From this last equation, and V j = Vj n + V j p we can write ' ' 1 1 p j n p n n D j j j j j n j A V N V V V V V V N = + = + = + Rearranging, ' ' 1 j n j D A V V N N = + and substituting this into ' 2 n j n D V w qN ε = to get ' ' ' 2 1 j n D D A V w N qN N ε = + 5.5. A step pn junction diode is made in silicon with the nside having ' D N =2 × 10 16 cm3 and on the pside the net doping is...
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 Spring '06
 McCann
 Pn junction, Anderson Solutions Chapter, Anderson& Anderson Solutions

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