Q & A

A Collection of Questions & Answers Asked by Current and Past ECE 3040 Students

A: Basically, if you take the y-axis as energy and x-axis as the distance, the energy states split and form multiple energy states. As the # of atoms (,say, N atoms) increases, the split states may overlap for certain energy levels and start to fill up the energy space. As N approaches infinity, the spacing between levels becomes indistinguishable. Energy “bands”, instead of discrete “energy states”, are formed. For simplicity, we can say a semiconductor forms 2 groups of “bands” (i,e, C.B. and V.B.) and in between these bands exists the “forbidden gap”, or the “bandgap” |

A:Introductory semiconductor books usually do not cover this topic. You may check other advanced solid-state physics books on this topic. You may read the following book chapters if you wish: 1. Chapter 1, “Fundamentals of semiconductor devices,” Anderson&Anderson, McGraw Hill, 2005 2. Chapter 15-17, “solid state physics,” Ashcroft & Mermin,1976 |

A:You may assume constant mobility for electrons and holes, respectively, at low doping concentration (<10^14 cm^-3) from Fig. 3.5 and read the intrinsic ni values form Fig. 2.20. The mobility at low impurity level is dominated by lattice scattering. |

A:Please read the exam-related section in the |

A:You may assume constant mobility for electrons and holes, respectively, at low doping concentration (<10^14 cm^-3) from Fig. 3.5 and read the intrinsic ni values form Fig. 2.20. The mobility at low impurity level is dominated by lattice scattering. |

A: 0 |

A: At steady state, the recombination and diffusion process for minority carrier in a semiconductor reach balanced status so the spatial distribution of minority carriers are “frozen” in the time domain. Therefore the “time derivative” term is set to zero. |

A: Good analogy. |

A: This is exactly what happens in a |

A: 1. The fermi level can be below the valence band. Please note that the fermi level represent the energy level where one can find 50% of electron occupancy (by definition) 2. There must be a unique solution for electric field at any given points in a PN junction diode The elecrostatics law states that the electric flux density (D) is continuous through out the material. Unless the p-side has different material than that for the n-side, the electric field is continuous across x=0. 3. rephrase 6.1 (e): Why minority carrier diffusion equations are not applicable in the depletion region? |

A: Please note the units of Boltzmann constant “k”. k = 1.38 E-23 (J/K) or 8.617 E-5 (eV/K). Also note that 1 eV = 1.6 E-19 J. If you apply 1 volt on an electron, the electron gain energy with the amount of 1.6E-19 Joule (E=q*V), which is equal to 1 eV. |

A: You should use Poisson’s equation to conceptualize this question. In the |

A: Good Question! The Vbi-VA is the “potential difference” between the p-side QNR and n-side QNR in the band diagram, or the potential barrier that electrons and holes need to overcome if they wish to reach the other side of the junction. In chapter 6, VA – Vohmic is the case where the ohmic loss in the QNR is taken into account. If we refer to the “voltage applied across the DR”, we are not referring to the POTENTIAL DIFFERENCE in the DR, instead, we are referring to the voltage difference applied across the DR. The electron still see a potential across the DR with a value of Vbi-(VA-Vohmic). Think: Electrons see a potential barrier across the DR with a value of Vbi at zero voltage.) |

A: I guess you are talking about the PNP and NPN under “active” biasing mode. I will not give you a straight answer and I would like you to follow the thoughts below: 1. The EB junction is forward biased and the BC junction is reverse biased for Active mode 2. In the base, the minority carriers (injected from the emitter) diffuse across the quasi-neutral base and are swept into the collector (due to the electric field inside of the DR for the reverse-biased BC junction), contributing to one of the collector current components 3. The electron flux density and the electron current are directing in the opposite directions because electrons are negatively charged. Once you understand these facts in the BJT, it should not be a problem for you to find the answer. |

A:The answer is “False” because the built-in voltage is a fixed value that is only dependent on the doping concentration, temperature, and intrinsic carrier concentration. |

A: The electron are “injected” from n-type into the p-type region due to the reduction of potential barrier across the space-charge region. Once the electrons are injected into the p-side, they become the minority carriers and start to diffuse further into the p-side QNR. As these electrons diffuse in the p-side QNR, the electrons recombine with the hole simultaneously. — That is how you see the “electron diffusion current” in the p-side under forward bias. Similar analogy can be obtained on the holes at the n-side under forward bias condition.. |

A:In the quasi-neutral region, you will see the minority carrier concentration is depleted as it approaches the depletion edge. This is due to the existence of electric field at the DR with reverse bias. Under steady-state, the minority carrier concentration is “quenched” near the depletion edge and is restored back to equilibrium at a position far away from the DR. The minority carrier will tend to “diffuse and generate” in a region where there is a lack of minority carriers. The resulting “minority carrier diffusion current” components in the depletion edge are used to characterize the total reverse bias saturation current. As you see in the lecture slides, the diffusion current components reach their maximum at the depletion edge. It implies that the corresponding the minority carrier flux will be moving into the DR — note that this is for rev. biased case only. Inside of the DR, the minority carriers experience an electric field and hence the (minority) carrier transport mechanism inside the DR is due the “Drift.” |

A:The difference between phiM and phisS gives you the bandgap alignment guideline at 0 V. Whether it is ohmic or rectifying depends on the bandgap prfile. (please see textbook 14.1, 14.2) It is nothing to do with the applied voltage. Once you apply positive voltage on the metal side, the Fermi level of metal moves downward in the band diagram. |

A:The energy band bending under strong inversion results in the hole accumulation at the O-S interface. Mathematically, one can calculate the hole concentration at the O-S interface from chapter 2 of Pierret. One will find that the hole concentration equals the background donor concentration at the “onset of strong inversion”, i.e., p=ND. Beyond the onset of strong inversion, hole concentration increases exponentially and p>>n. This is why we define such biasing mode as “strong inversion” |

A:In the Pierret book, it assumes Phi/sub/MS =0 (for an ideal MOS), which is not true in reality. If Phi/sub/MS is a nonzero value, one would need to apply an extra voltage V/sub/FB to achieve the “flat band” condition. From MOS band diagram, if one assumes the perfect oxide layer and interface (which is not true either as there is no “perfect oxide” in the world), the flat band voltage will be equal to Phi/sub/MS. The resulting threshold voltage will be shifted by an amount of Phi/sub/MS from that for the “ideal” MOS. |