ch5

# ch5 - SOLUTIONS CHAPTER 5 5.1 A silicon pn junction is...

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SOLUTIONS CHAPTER 5 5.1. A silicon pn junction is formed between n-type silicon doped with N D =10 17 cm -3 and p-type silicon doped with N A =10 16 cm -3 . 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 0 = N D = 10 17 cm - 3 p n 0 = n i 2 n n 0 = 1.08 × 10 10 cm - 3 ( 29 2 10 17 cm - 3 = 1.17 × 10 3 cm - 3 p p 0 = N A = 10 16 cm - 3 n p 0 = n i 2 p p 0 = 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 built-in voltage? q V b i = E g - δ n - δ p = 1 . 1 2 - 0 . 1 4 7 - 0 . 2 0 9 = 0 . 7 6 4 e V or V bi =0.764V Anderson& Anderson 1 November 19, 2007 Solutions Chapter 5

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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 p-side is the anode and the n-side 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 cm -3 . 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 built-in voltage? V bi = E g - δ n - δ p = 1.12 - 0.165 - 0 = 0.955 eV 5.4. Fill in the missing steps to derive Equation 5.35 for the junction width on the n side of the junction. Anderson& Anderson 2 November 19, 2007 Solutions Chapter 5
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 n-side having ' D N =2 × 10 16 cm -3 and on the p-side the net doping is ' A N =5 × 10 15 cm -3 . a) Draw, to scale, the energy band diagram of the junction at equilibrium. We begin by finding the locations of the Fermi levels on each side of the junction. On the n side, E C - E f = - kT ln n 0 N C = - kT ln N D ' N C = - 0.026 eV ln 2 × 10 16 2.89 × 10 19 = 0.19 eV This assumes complete ionization and nondegeneracy. On the p side, we have ' 15 0 19 5 10 ln

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