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A comprehensive introduction to semiconductor theory, covering fundamental concepts such as atomic structure, valence electrons, and types of semiconductors. It delves into the characteristics of intrinsic and extrinsic semiconductors, explaining the behavior of diodes under different bias conditions. The document also includes practical examples and exercises to reinforce understanding.
Typology: Summaries
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Semiconductor Materials ( used most frequently )
1. Si (Silicon) 2. Ge (Germanium) 3. GaAs (Gallium Arsenide) “ Semiconductor Materials have a negative temperature coefficient. ” Facts:
Energy Level Insulator Energy Level – Eg > 5 eV Conductor Energy Level – The bands overlap Semiconductor Energy Level Material Diode Energy Gap Si 0.7 V 1.1 eV Ge 0.3 V 6.7 eV GaAs 1.2 V 1.41 eV “The farther an electron is from the nucleus, the higher is the energy state, and any electron that has left its parent atom has a higher energy state than any electron in the atomic structure.”
Semiconductor Crystal Structure & Covalent Bonding Crystalline Structure – atoms that are bound together. Crystal – atoms combined to form a solid material; arranged in at fixed pattern.
Lattice Structure – neutrons and protons form the nucleus, and electrons appear in fixed orbits around the nucleus. Lattice ( geometric grid of points ) – periodic arrangements of the atoms. Valence Electrons – electrons in the outermost shell of an atom (last orbit) No. of Electron in a Shell = 2n^2 Trivalent – 3 valence electrons (at Ga) Tetravalent – 4 valence electrons Pentavalent – 5 valence electrons (at As) Valence – used to indicate that the potential required to remove… Valence Bond – bond in valence electrons located Conduction Bond – bond in which valence electrons can freely move Covalent Bonding – sharing of electrons; bonding of atoms Types of Semiconductors
1. Intrinsic Semiconductor - Carefully refined to reduce impurities to a very low level essentially as pure as can be made. - Pure semi-material Hole – vacancy left in the valence bond Electron-Hole Pair – breaking of covalent bond Recombination – when conduction bond electrons lose energy and fall back into the hole in the valence bond Lifetime – the amount of time between the creation and disappearance of a free electron Electron Flow – the movement of free electrons in one direction Free Electron – the flow of an electric circuit in a wire Hole Flow - the movement of the hole constitutes a current flow 2. Extrinsic Semiconductor
Solution: a. VDQ and IDQ ID=
0.5 kΩ = 20 mA VD= E = 10 V The intersection between the load line and the characteristic curve defines the Q - point as VDQ ≅ 0.78 V IDQ ≅ 18.5 mA b. VR VR= E-VDQ = 10 V - 0.78 V = **9.22 V
Series Diode Configuration Examples:**
1. For the series diode configuration of Fig. 2.13, determine VD, VR, and ID Solution: VD= 0.7 V VR= E-VD = 8 V - 0.7 V = 7.3 V ID= IR =
2.2 kΩ ≅ 3.32 mA
2. For the series diode configuration of Fig. 2.13, determine VD, VR, and ID with the diode reversed. Solution: “Since the diode has been reversed the connection of the diode with the power source will be Reverse Bias which will result in an open circuit” So, VR= 0 V and ID= 0 A E-VD-VR = 0 → VD= E-VR VD= 8 V - 0 V = 8 V
3. For the series diode configuration of Fig. 2.16 , determine VD, VR, and ID Solution: “Since the diode has been reversed the connection of the diode with the power source will be Reverse Bias which will result in an open circuit” So, VR= 0 V and ID= 0 A E-VD-VR = 0 → VD= E-VR VD= 0.5 V - 0 V = 0.5 V 4. Determine VO and ID for the series circuit of Fig.
0.680 kΩ ≅ 13.97 mA
4. Determine ID, VD 2 , and VO for the circuit of Fig. 2.21. Solution: “The combination of a short circuit in series with an open circuit always results in an open circuit.” So,
Solution: a. VO= 10 kΩ^ (12 2 k^ Ω + 10V^ -^ 0.7 V kΩ^ - 0.3 V)≅ 9.17 V b. VO= 10 V
9. Determine VO and ID for the networks of Fig. 2.158. Solution: a. VTH= IR = (10 mA)(2.2 kΩ) = 22 V RTH= 2.2 kΩ ID=
2.2 kΩ + 2.2 kΩ ≅ 4.84 mA VO= (4.84 mA)(2.2 kΩ) = 10.65 V b. ID= 20 V^ + 20 V 6.8 kΩ^ - 0.7 V≅ 5.78 mA VO - 0.7 V + 20 V = 0 VO = - 19.3 V
10. Determine VO 1 and VO 2 for the networks of Fig. 2.159. Solution: a. VO 1 = 12 V - 0.7 V = 11.3 V VO 2 = 1.2 V b. VO 1 = 0 V VO 2 = **0 V
Parallel and Series-Parallel Configurations Examples:** 1. Determine VO, I 1 , ID 1 , and ID 2 for the parallel diode configuration of Fig. 2.28. Solution: “The voltage across parallel elements is always the same.” VO= 0.7 V I 1 =
0.33 kΩ ≅ 28.18 mA ID 1 = ID 2 =
28.18 mA 2 ≅ 14.09 Ma
2. Determine the voltage VO for the network below. Solution: VO= 12 V - 0. 3 V = 11. 7 V 3. Determine the currents I 1 , I 2 , and ID 2 for the network of Fig. 2.
Solution: I 1 =
3.3 kΩ ≅ 0.212 mA E - VK 1 - VK 2 - V 2 = 0 → V 2 = E - VK 1 - VK 2 V 2 = 20 V - 0.7 V - 0.7 V = 18.6 V I 2 =
5.6 kΩ ≅ 3.32 mA ID 2 + I 1 = I 2 → ID 2 = I 2 - I 1 ID 2 = 3.32 mA - 0.212 mA ≅ 3.11 Ma
4. Determine VO and ID for the networks of Fig. 2.160. Solution: a. ID= IR = 12 V - 0.7 V 4.7 kΩ ≅^ 2.4 mA VO 1 = 12 V - 0.7 V = 11.3 V b. ID= IR = 12 V + 4 V 2.2 kΩ^ -^ 0.7 V≅ 10.59 mA VO 1 = 20 V - 0.7 V = 19.3 V 5. Determine VO and I for the networks of Fig. 2. Solution: a. I = 1 V 1 k^ -^ 0.7 VΩ ≅ 0.3 mA VO 1 = 1 V - 0.7 V = 0.3 V b. I = 16 V^ -^ 0.7 V 4.7 k^ -^ 0.7 V + 4 VΩ ≅ 3.96 mA VO 1 =16 V - 0.7 V - 0.7 V = 14.6 V 6. Determine VO 1 , VO 2 , and I for the network of Fig. 2.
1 kΩ = 9.3 mA
2. Determine the output level for the positive logic AND gate of Fig. 2. Solution: VO= 0.7 V I=
1 kΩ = 9.3 mA