Homework 5. (b) Given n i = cm 3, in the quasi-neutral p-region, ECE344 Fall

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1 Homework 5 1. Problem: A silicon p-n junction is formed between n-type Si doped with N D = cm 3 and p-type Si doped with N A = cm 3. (a) Sketch the energy band diagram. Label all axes and all important energy levels. (b) Find n n0, n p0, p p0, and p n0. Sketch the carrier concentration (of both electrons and holes) as a function of position. (c) Calculate the built-in potential V bi in ev. Solution: (a) The energy band diagram with labeled important energy levels and axes is (b) Given n i = cm 3, in the uasi-neutral p-region, ECE344 Fall

2 In the uasi-neutral n-region, In the depletion region, p p0 = N A = cm 3. (1) n n0 = N D = cm 3. (2) for p-side and n p0 = n2 i N A = cm 3 (3) p n0 = n2 i = cm 3 (4) N D for n-side. The diagram for carrier concentration is: ECE344 Fall

3 (c) Using E. (123) or (124) on lecture notes Part 2, the build-in potential is V bi = k BT ln(n AN D n 2 ) = k BT ln( p p ) = eV. (5) p i n 2. Problem: A Si p-n junction has dopant concentrations N D = cm 3 and N A = cm 3. Calculate the built-in potential V bi in ev and the total width of the depletion region W = x n0 + x p0 at zero bias (that is, V a = 0) and under a reverse bias V a = 8V. Solution: (a) Given n i = cm 3, the build-in potential is: V bi = k BT ln(n AN D n 2 ) = k BT ln( p p ) = eV. (6) p i n (b) According to E. (140) & (141) in lecture notes Part 2, the depletion width with a bias V a is give by: W = 2ɛ s (V bi V a ) ( ). (7) N A N D For V a = 0V, we have W = 0.691µm and for V a = 8V, we have W = 2.475µm. 3. Problem: A Si p-n junction is reverse-biased with V a = 10V. Determine the percent change in junction (depletion) capacitance and built-in potential if the doping in the p region is increased by a factor of 2. Solution: Let us assume the N A and N D values from the Prob. 2, then the build-in potential is calculated by E. 6 and we have: and V bi = k BT V bi = k BT ln(n AN D n 2 ) = eV, (8) i ln(2n AN D n 2 ) = eV. (9) i ECE344 Fall

4 Therefore, V bi is increased by 2.67%. From E. (184) on lecture note Part 2, the depletion capacitance can be calculated by C j = ɛa W, (10) where A is the cross-sectional area of the junction and W is the depletion width calculated by: W = 2ɛ s (V bi V a ) ( ) = µm, (11) N A N D W 2ɛ s (V bi = V a) 1 ( + 1 ) = µm. (12) 2N A N D Therefore, the percent change in junction depletion capacitance is given by: C j C j C j = W W W = 2.27%. (13) 4. Problem: Consider a p + -n Si junction at T = 300 K with N A = cm 3 and N D = cm 3. The minority carrier hole diffusion constant is D p = 12 cm 2 /s and the minority carrier hole lifetime is τ p = 100 ns. The cross-sectional area of the junction is A = 10 4 cm 2. Calculate the reverse saturation current I s = AJ s. Calculate also the current at a forward bias V a = 0.5 V. Solution: Since N A N D, it is an asymmetric junction and the total current is dominated by the most heavily-doped side of the junction. According to E. (151) in lecture notes Part 2, the saturation current density is given by J s = D pp n0 L p (14) ECE344 Fall

5 where L p = (D p τ p ) 1/2 = cm and p n0 = n2 i N D = cm 3. So For a forward bias V a = 0.5 V, the current is: J s = A/cm 2 (15) I s = AJ s = A. (16) I = I s [e V a/k B T 1] = 8.93µA. (17) 5. Which are the most important breakdown mechanisms in a reverse-biased p-n junction? Describe succintly (no more than one paragraph for each process) how they work and how they trigger the breakdown process. Solution: The important breakdown mechanisms are Zener and Avalanche. Refer to pages in lecture notes Part Problem: (a) Calculate the maximum width of the depletion layer w max (at the onset of inversion) and the maximum depletion charge Q d,max in p-type Si, GaAs, and Ge semiconductors of an MOS structure with N A = cm 3 and at T = 300 K. (b) Repeat the calculations for T = 77 K. Solution: and (a) At T = 300K, For Si, W max = 4ɛ rɛ 0 k B T ln N A ni 2 (18) N A Q d = N A W max. (19) n i = cm 3 (20) ɛ r = (21) ECE344 Fall

6 From the above euations we find out W max and Q d,max W max = µm (22) Q d,max = C cm2. (23) For Ge, n i = cm 3 (24) ɛ r = 16. (25) From the above euations we find out W max and Q d,max W max = 0.234µm (26) Q d,max = C cm2. (27) For GaAs n i = cm 3 (28) ɛ r = (29) From the above euations we find out W max and Q d,max W max = µm (30) Q d,max = C cm2. (31) ECE344 Fall

7 (b) At T = 77 K, note that For Si, n i (T ) T 3/2 e E g/2k B T. (32) n i = cm 3 (33) ɛ r = (34) From the above euations we find out W max and Q d,max W max = µm (35) Q d,max = C cm2. (36) For Ge, n i = cm 3 (37) ɛ r = 16. (38) From the above euations we find out W max and Q d,max W max = µm (39) Q d,max = C cm2. (40) For GaAs, n i = cm 3 (41) ɛ r = (42) ECE344 Fall

8 From the above euations we find out W max and Q d,max W max = 0.45µm (43) Q d,max = C cm2. (44) 7. Problem:Determine the metal-semiconductor work function difference Φ MS (in ev) in an MOS structure with p-type Si for the case where the gate is Al (χ M = 3.2 ev), n + polysilicon (polycrystalline Si, assume it is identical to normal Si), and p + polysilicon. Assume N A = cm 3 and T = 300 K. Solution:Given, where χ S = 4.05eV and E g = 1.11eV. For Al, For n + polysilicon gate, N A = cm 3 (45) T = 300K (46) Φ S = χ S + E g 2 + Φ F (47) Φ S = χ S + E g 2 + k BT ln N A n i (48) Φ M = χ M = 3.2eV (49) Φ MS = Φ M Φ S = 1.73eV. (50) Φ M = χ S = 4.05eV (51) Φ MS = Φ M Φ S = 0.889eV. (52) ECE344 Fall

9 For p + polysilicon gate, Φ M = χ M = 5.16eV (53) Φ MS = Φ M Φ S = 0.221eV. (54) ECE344 Fall