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Brief Article
The Author
September 15, 2003
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Name on Answer sheet (1 point). You can keep the actual exam (suitable for framing).
Final Exam
July 29, 2003
EE 3329 Summer 2003
Dr. Lush
A semi-infinite semiconductor wafer (NA = 1015 cm−3 and ND = 0) has been illuminated for a
long time such that the generation rate is uniform throughout. The minority carrier lifetime is
τp = τn = 10−9 seconds. Go = 1023 cm−3 /s. The bare front surface is damaged so that all excess
minority carriers recombine if they are there.
1. Which is the minority carrier?
(a) Holes
(b) Electrons
(c) Cannot tell with the information given
2. What is the general solution?
t
p,n
−τ
(a) ∆p, n(t) = Ae
+ Go τp,n
−Lx
(b) ∆p, n(x) = Ae
x
p,n
+ Be Lp,n + Go τp,n
t
p,n
−τ
(c) ∆p, n(t) = Ae
−Lx
(d) ∆p, n(x) = Ae
x
p,n
+ Be Lp,n
(e) None of the above
3. What are the boundaries and boundary conditions?
(a) ∆p, n(t = 0) = 0
(b) ∆p, n(t = 0) = Go τp,n
(c) ∆p, n(x = 0) = 0; ∆p, n(x = ∞) = Go τp,n
(d) ∆p, n(x = 0) = −p, no ; ∆p, n(x = ∞) = Go τp,n
(e) ∆p, n(x = 0) = 1015 ; ∆p, n(x = ∞) = 0
4. What is the exact solution?
t
p,n
−τ
(a) ∆p, n(t) = Go τp,n (1 − e
)
−Lx
p,n
(b) ∆p, n(x) = 1015 e
−Lx
(c) ∆p, n(x) = Go τp,n (1 − e
(d) ∆p, n(t) =
p,n
)
− t
1015 e τp,n
−Lx
(e) ∆p, n(x) = Go τp,n e
p,n
5. The semiconductor is in low injection.
(a) True
(b) False
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6. What is the minority carrier current at x=0?
(a) Zero.
(b) Jp,n = −qGo Lp,n
(c) Jp,n = qGo Lp,n
(d) Jp,n = qDp,n Go /Lp,n
(e) Jp,n = q1015 /Lp,n
(f) None of the above
7. What is the majority carrier current at x=0?
(a) Zero.
(b) Jp,n = −qGo Lp,n
(c) Jp,n = qGo Lp,n
(d) Jp,n = qDp,n Go /Lp,n
(e) Jp,n = q1015 /Lp,n
(f) None of the above or we cannot know
8. A silicon wafer has po = 2x1016 cm−3 and no = 8x1010 cm−3 . What is ni ?
(a) 1010 cm−3
(b) 2x1016 cm−3
(c) 4x1016 cm−3
(d) 4x1013 cm−3
(e) None of the above.
9. What is the temperature?
(a) 200 Kelvin
(b) 300 Kelvin
(c) 400 Kelvin
(d) 450 Kelvin
(e) None of the above.
10. Where is the Fermi level for this situation?
(a) Above ED but below Ec
(b) Above Ei but below ED
(c) Below Ei but above EA
(d) Below EA but above Ev
(e) None of the above.
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11. The temperature is changed so that no = 5x103 cm−3 . What is the temperature?
(a) 100 Kelvin
(b) 200 Kelvin
(c) 300 Kelvin
(d) 400 Kelvin
(e) None of the above.
12. Where is the Fermi level for this situation compared to where it was previously?
(a) Higher but still below ED
(b) Higher and above ED
(c) Lower but still above EA
(d) Lower and below EA
(e) None of the above
13. Now one thing is changed, resulting in po = 1016 cm−3 . What was done?
(a) The donor concentration was increased by 1016 cm−3
(b) The acceptor concentration was increased by 1016 cm−3
(c) The acceptor concentration is now 4x1016 cm−3
(d) The donor concentration was increased by 2x1016 cm−3
(e) The acceptor concentration was increased by 1016 cm−3
14. What is the electron concentration?
(a) 1010 cm−3
(b) 1016 cm−3
(c) 104 cm−3
(d) 10−6 cm−3
(e) None of the above.
15. What would have to be changed from this situation for the Fermi level to be right at ED ?
(a) Change the temperature to between 0 and 100 Kelvin
(b) Change the temperature to 150 Kelvin
(c) Change the temperature to 0 Kelvin
(d) Change the temperature to 300 Kelvin
(e) None of the above.
16. Semiconductor A is at room temperature and Semiconductor B is at 350 Kelvin. Compare the
mobilities. Both have the same doping concentration.
(a) µA > µB
(b) µA < µB
(c) µA ≈ µB
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17. Semiconductor A is at room temperature and Semiconductor B is at 350 Kelvin. Compare the
conductivities. Both have the same doping concentration.
(a) σA > σB
(b) σA < σB
(c) σA ≈ σB
18. Dopants are supposed to be scattering mechanisms yet even if at room temperature Semiconductor
A has 100 times the doping that Semiconductor B has, and Semiconductor B is doped with ND =
1014 cm−3 , the two hole mobilities are very similar. Why? Use the figure on the cheat sheet.
(a) The donors are un-ionized.
(b) Phonons are the dominant scattering mechanism.
(c) Because holes are minority carriers
(d) None of the above
19. Nearing zero Kelvin a semiconductor becomes an insulator while a conductor becomes a super
conductor. Why?
(a) The conductor has a wider bandgap.
(b) In the semiconductor the carrier concentration decreases while a conductor has more carriers
at zero Kelvin than the conductor had at room temperature.
(c) The semiconductor has no carriers and the conductor mobility has become huge.
(d) The semiconductor carrier have become too cold to move.
(e) None of the above
20. Which is the dominant recombination mechanism in silicon?
(a) Auger recombination
(b) Shockley-Read-Hall or thermal recombination
(c) Radiative recombination
(d) Genetic recombination
21. All donors in GaAs (a III-V semiconductor) come from...
(a) ...the column in periodic table to the right of the element it replaces.
(b) ...Column III.
(c) ...Column V.
(d) ...Column IV.
(e) None of the above satisfy the needed description for GaAs.
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22. Which of the following is true in an p-type semiconductor with both donors and acceptors?
(a) At all temperatures all donors are always ionized.
(b) At all temperatures only some donors are always ionized.
(c) Only some of the donors are ionized at very low temperatures (T<50 K).
(d) Only some of the acceptors are ionized at room temperature.
(e) All of the acceptors are always ionized.
23. Which of the following is true in an n-type semiconductor with bND /NA = 2?
(a) At all temperatures all donors are always ionized.
(b) At all temperatures half of the acceptors are always ionized.
(c) Half of the donors are ionized at very, very low temperatures (T<10 K).
(d) Only some of the donors are ionized at room temperature.
(e) None of the above.
24. For the figures shown here, the mechanisms are from left to right...
(a) Radiative, Shockley-Read-Hall, Auger
(b) Radiative, Auger, Shockley-Read-Hall
(c) Shockley-Read-Hall, Radiative, Auger
(d) Auger, Shockley-Read-Hall, Radiative
(e) All of the above are required
25. Which sketch from Figures B looks like the potential? Refer to band diagram in Figures B.
(a) (a)
(b) (b)
(c) (c)
(d) None of the above.
26. Which sketch from Figures B looks like the electric field? Refer to band diagram in Figures B.
(a) (a)
(b) (b)
(c) (c)
(d) None of the above.
27. Which sketch from Figures B looks like the hole concentration? Refer to band diagram in Figures
B.
(a) (a)
(b) (b)
(c) (c)
(d) None of the above.
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28. Which sketch from Figures B looks like the electron diffusion current? Refer to band diagram in
Figures B.
(a) (a)
(b) (b)
(c) (c)
(d) None of the above.
29. Is there a region where po no 6= n2i ? Refer to band diagram in Figures B.
(a) Yes.
(b) No.
(c) Cannot tell.
30. Is this semiconductor in equilibrium? Refer to band diagram in Figures B.
(a) Yes.
(b) No.
(c) Cannot tell.
31. Is there a region where this semiconductor is intrinsic? Refer to band diagram in Figures B.
(a) Yes.
(b) No.
(c) Cannot tell.
32. Looking at the ideal diode equation for an n+ p diode, what happens to the reverse bias current for
a given voltage if we were to increase significantly the doping concentration on the n-type side?
(a) The current magnitude would increase significantly.
(b) The current magnitude would decrease significantly.
(c) The current magnitude would increase a little.
(d) The current magnitude would decrease a little.
(e) The current magnitude would stay essentially the same.
33. Which of the following contributes the most to the total current in an n+ p diode under low reverse
bias conditions if all the non-ideal current mechanisms could be important for this device?
(a) Electrons diffusing over the barrier and into the opposite neutral region.
(b) Impact ionization in the depletion region.
(c) Electrons and holes generated in the space charge region.
(d) Electrons and holes recombining in the space charge region.
(e) None of the above contribute at all
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34. Which of the following contributes the most to the total current in an ideal p+ n diode under low
forward bias conditions?
(a) Electrons and holes recombining in the space charge region.
(b) Electrons diffusing over the barrier and into the opposite neutral region.
(c) Holes diffusing over the barrier and into the opposite neutral region.
(d) Holes drifting down the barrier and into the opposite neutral region.
(e) None of the above contribute at all
35. Which of the following can we learn with the greatest accuracy about a n+ p diode by making
measurements of capacitance versus voltage–and, of course, analyzing the results?
(a) Doping on the p-type side
(b) Doping on the n-type side
(c) Built-in potential (Vbi )
(d) None of the above
36. The electric field at the metallurgical junction of a p+ p junction is positive.
(a) True
(b) False
An infinite p-type semiconductor (NA = 1015 cm−3 andτp,n = 10−6 s) has been illuminated for a
long time with a generation rate equal to Go . At t=0 the light intensity is increased to 10Go .
Go = 1019 cm−3 /s.
37. Which carriers do we need to model?
(a) Holes
(b) Electrons
38. What is the general solution for the minority carriers?
t
p,n
−τ
(a) ∆p, n(t) = Ae
+ Go τp,n
−τ t
p,n
(b) ∆p, n(t) = Ae
−x
x
(c) ∆p, n(x) = Ae Lp,n + Be Lp,n + Go τp,n
−x
x
(d) ∆p, n(x) = Ae Lp,n + Be Lp,n
t
p,n
−τ
(e) ∆p, n(t) = Ae
+ Go τp,n /3
39. What are the boundary conditions?
(a) ∆p, n(t = 0) = 0
(b) ∆p, n(t = 0) = 10Go τp,n
(c) ∆p, n(t = 0) = Go τp,n
(d) ∆p, n(x = 0) = 10Go τp,n ; ∆p, n(x = ∞) = 0
(e) None of the above
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40. What is the exact solution for the minority carrier concentration?
(a) ∆p, n(t) = Go τp,n
(b) ∆p, n(t) = Go τp,n (10 − 9e−t/τp,n )
−t
(c) ∆p, n(x) = 10Go τp,n (1 − e τp,n )
−t
(d) ∆p, n(x) = Go τp,n (1 + 9e τp,n )
(e) None of the above
41. At what time does the device leave low level injection?
(a) A long time
(b) t = τp,n ln(9/5)
(c) t = τp,n ln(10/9)
(d) 10τp,n
(e) None of the above or it is always in low injection.
42. What is Vbi for the diode below (Figure C)? Assume the bandgap of the semiconductor is 2.0 eV.
(a) 0.0 volt
(b) 0.5 volt
(c) 1.0 volt
(d) 1.5 volt
(e) None of the above
43. Which of the following (Figure C) accurately describes the device?
(a) A p+ n junction
(b) A pn junction
(c) An n+ p junction
(d) An np junction
(e) None of the above
44. Which of the following statements represents Figure D? Assume the device is same as in Figure C.
(a) A diode with 1.0 volt forward bias.
(b) A diode with 1.0 volt reverse bias.
(c) A diode with 2.0 volt forward bias.
(d) A diode with 2.0 volt reverse bias.
(e) None of the above
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45. Which of the following statements represents the Figure E?
(a) An np junction in forward bias
(b) An np junction in reverse bias
(c) A pn junction in forward bias
(d) A pn junction in reverse bias
(e) None of the above
46. Which BJT mode of operation is depicted in Figure F?
(a) Forward active (pnp)
(b) Inverse active (npn)
(c) Forward Active (npn)
(d) Cut-off (pnp)
(e) None of the above or not a possible plot
47. The biases needed for a npn transistor in inverse active mode are:
(a) Veb < 0; Vcb < 0;
(b) Veb < 0; Vcb > 0;
(c) Veb > 0; Vcb < 0;
(d) Veb > 0; Vcb > 0;
48. Which of the following does NOT impact gain in an ideal BJT?
(a) Base transport factor
(b) Base width modulation
(c) Emitter injection efficiency
(d) Emitter doping
(e) All have some impact on the gain
49. Which mode of operation is depicted in Figure G?
(a) Forward active
(b) Inverse active
(c) Cut-off
(d) Saturation
(e) None of the above or not a possible plot
50. For the problem just before this one, the plot shown is of excess carriers.
(a) True
(b) False
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51. The base transport factor in a pnp transistor is
(a) The fraction of holes injected from the collector that make it to the emitter.
(b) The ratio of electrons back-injected into emitter and holes injected into base.
(c) The ratio of electrons back-injected into emitter and total emitter current.
(d) The ratio of holes back-injected into emitter and total emitter current.
(e) The fraction of holes injected from the emitter that make it to the collector.
52. The biases needed for an npn transistor in cut-off mode are:
(a) Veb < 0; Vcb < 0;
(b) Veb < 0; Vcb > 0;
(c) Veb > 0; Vcb < 0;
(d) Veb > 0; Vcb > 0;
53. Which of the following impacts gain in an non-ideal BJT?
(a) Collector series resistance
(b) Emitter-base space charge recombination
(c) Base width modulation
(d) Emitter injection efficiency
(e) All have some impact on the gain in an non-ideal BJT
54. What type of semiconductor was used to make the MOS-C depicted in Figure H?
(a) p-type
(b) n-type
55. The MOS-C depicted in Figure H is non-ideal.
(a) True
(b) False
56. The regions left to right in Figure I should be labeled:
(a) Accumulation, depletion, and inversion
(b) Depletion, accumulation, and inversion
(c) Accumulation, inversion, and depletion
(d) Inversion, depletion, and accumulation
(e) None of the above
57. The regions left to right in Figure I should be labeled:
(a) Accumulation, depletion, and inversion
(b) Depletion, accumulation, and inversion
(c) Accumulation, inversion, and depletion
(d) Inversion, depletion, and accumulation
(e) None of the above
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58. The device in Figure I is ideal.
(a) True
(b) False
(c) Cannot be sure
59. The MOSFET depicted in Figure J is PMOS and it is correctly labeled.
(a) True
(b) False
60. The MOSFET depicted in Figure J is in which mode of operation?
(a) Saturation
(b) Right at pinch-off
(c) Threshold
(d) Linear region
61. The area the arrow points to under the gate of the MOSFET in Figure J is in which mode of
operation? Think of it as a MOS-C.
(a) Accumulation
(b) Depletion
(c) Inversion
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