Semiconductor Physics And Devices [SOLUTIONS] (Donald Neamen)

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 1 Exercise Solutions TYU 1.2 (a) Number of atoms per (100) lattice plane Ex 1.1 (a) Number of atoms per unit cell Surface Density (b) Volume Density = cm (b) Number of atoms per (110) lattice plane cm _______________________________________ Ex 1.2 Intercepts of plane; p=1, q=2, s=2 Inverse; Multiply by lowest common denominator, plane _______________________________________ Ex 1.3 (a) Number of atoms per (100) plane Surface Density cm (c) Number of atoms per (111) lattice plane Lattice plane area where Surface Density cm (b) Number of atoms per (110) plane Surface Density Then lattice plane area Surface Density cm cm _______________________________________ Test Your Understanding Solutions _______________________________________ TYU 1.1 TYU 1.3 Number of atoms per unit cell (a) For (100) planes, distance (b) For (110) planes, distance Volume Density cm Radius _______________________________________ _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU 1.4 (a) 8 corner atoms (b) 6 face-centered atoms (c) 4 atoms totally enclosed _______________________________________ TYU 1.5 Number of atoms in the unit cell Volume Density cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 1 Problem Solutions 1.1 (a) fcc: 8 corner atoms atom 6 face atoms atoms Total of 4 atoms per unit cell (b) bcc: 8 corner atoms atom 1 enclosed atom =1 atom Total of 2 atoms per unit cell (c) Diamond: 8 corner atoms atom 6 face atoms atoms 4 enclosed atoms = 4 atoms Total of 8 atoms per unit cell _______________________________________ Then Ratio (d) Diamond lattice Body diagonal Unit cell vol 8 atoms per cell, so atom vol 1.2 (a) Simple cubic lattice: Unit cell vol Then 1 atom per cell, so atom vol Ratio Then _______________________________________ Ratio (b) Face-centered cubic lattice 1.3 (a) ; From Problem 1.2d, Unit cell vol Then 4 atoms per cell, so atom vol Then Center of one silicon atom to center of nearest neighbor (b) Number density Ratio (c) Body-centered cubic lattice Unit cell vol 2 atoms per cell, so atom vol cm (c) Mass density grams/cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) 1.4 (a) 4 Ga atoms per unit cell (b) Number density Density of Ga atoms 4 As atoms per unit cell Density of As atoms (b) 8 Ge atoms per unit cell (c) A-atoms: # of atoms cm Density cm cm B-atoms: # of atoms Number density Density of Ge atoms cm _______________________________________ 1.5 From Figure 1.15 (a) Density cm _______________________________________ 1.9 (a) # of atoms (b) Number density cm _______________________________________ Mass density 1.6 gm/cm _______________________________________ (b) # of atoms 1.7 (a) Simple cubic: Number density (b) fcc: cm (c) bcc: (d) diamond: _______________________________________ 1.8 Mass density gm/cm _______________________________________ 1.10 From Problem 1.2, percent volume of fcc Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ atoms is 74%; Therefore after coffee is Surface density cm ground, Volume = 0.74 cm For 1.12(a) and (b), Same material _______________________________________ 1.11 (b) (b) For 1.12(a), A-atoms; Surface density (c) Na: Density cm Cl: Density (d) Na: At. Wt. = 22.99 Cl: At. Wt. = 35.45 So, mass per unit cell cm cm B-atoms; Surface density cm For 1.12(b), A-atoms; Surface density Then mass density cm grams/cm B-atoms; Surface density _______________________________________ cm 1.12 (a) For 1.12(a) and (b), Same material _______________________________________ Then 1.14 Density of A: cm (a) Vol. Density Surface Density Density of B: cm (b) Same as (a) (c) Same material _______________________________________ (b) Same as (a) _______________________________________ 1.15 (i) (110) plane (see Figure 1.10(b)) (ii) (111) plane (see Figure 1.10(c)) 1.13 (iii) (220) plane (a) For 1.12(a), A-atoms Same as (110) plane and [110] direction Surface density (iv) (321) plane cm For 1.12(b), B-atoms: Intercepts of plane at Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ [321] direction is perpendicular to (321) plane _______________________________________ 1.16 (a) So Area of plane (b) cm _______________________________________ Surface density cm 1.17 (b) bcc (i) (100) plane: Intercepts: 2, 4, 3 (634) plane _______________________________________ Surface density 1.18 Surface density cm (ii) (110) plane: (a) cm (b) (iii) (111) plane: (c) Surface density _______________________________________ cm (c) fcc (i) (100) plane: 1.19 (a) Simple cubic (i) (100) plane: Surface density cm (ii) (110) plane: Surface density Surface density cm cm (ii) (110) plane: (iii) (111) plane: Surface density cm (iii) (111) plane: Area of plane where Now Surface density cm _______________________________________ 1.20 (a) (100) plane: - similar to a fcc: Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ = Surface density cm cm (b) (110) plane: _______________________________________ Surface density 1.22 Density of silicon atoms cm and 4 valence electrons per atom, so Density of valence electrons cm _______________________________________ cm (c) (111) plane: 1.23 Density of GaAs atoms Surface density cm cm _______________________________________ 1.21 An average of 4 valence electrons per atom, So Density of valence electrons cm _______________________________________ 1.24 (a) #/cm (a) cm (b) _______________________________________ (b) #/cm cm 1.25 (a) Fraction by weight (c) (b) Fraction by weight (d) # of atoms Area of plane: (see Problem 1.19) _______________________________________ 1.26 Volume density Area So cm cm cm We have Then #/cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.27 Volume density So cm cm We have Then _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 2 Exercise Solutions or Ex. 2.1 eV Then (a) eV, eV, eV J (b) or eV (b) J J or or eV eV _______________________________________ Then eV eV Ex 2.2 (a) eV _______________________________________ kg-m/s m Ex 2.4 or (c) J kg-m/s Now Set = Then J or eV _______________________________________ Ex 2.3 or m (a) (a) J m or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ % kg-m/s (b) m (b) or % _______________________________________ J or eV _______________________________________ Ex 2.5 (a) m Then TYU 2.2 (a) eV s (b) (b) Same as part (a), s _______________________________________ _______________________________________ TYU 2.3 (a) Ex 2.6 From Example 2.6, we have eV = meV, meV, meV _______________________________________ Test Your Understanding TYU 2.1 (a) (b) m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ m _______________________________________ TYU 2.4 so that m Then m or _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 2 2.1 2.6 Sketch _______________________________________ (a) kg-m/s 2.2 Sketch _______________________________________ m/s or 2.3 Sketch _______________________________________ cm/s (b) kg-m/s 2.4 m/s From Problem 2.2, phase = constant Then or cm/s (c) Yes _______________________________________ 2.7 (a) (i) From Problem 2.3, phase = constant kg-m/s Then m _______________________________________ 2.5 or (ii) kg-m/s Gold: So, eV m J cm or (iii) or kg-m/s m Cesium: So, eV m J cm or m _______________________________________ or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) 2.10 (a) kg-m/s kg-m/s m or _______________________________________ m/s or cm/s 2.8 eV J Now or eV (b) or kg-m/s J Now m or eV or kg-m/s _______________________________________ 2.9 m or Now _______________________________________ and Set and Then 2.11 (a) J which yields Now V (b) J keV _______________________________________ kg-m/s kV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then m J or or eV _______________________________________ _______________________________________ 2.14 kg-m/s m/s _______________________________________ 2.12 2.15 (a) kg-m/s _______________________________________ 2.13 (a) (i) s (b) kg-m/s (ii) kg-m/s _______________________________________ 2.16 (a) If and are solutions to Schrodinger's wave equation, then Now and kg-m/s so Adding the two equations, we obtain J or (b) (i) eV kg-m/s (ii) kg-m/s which is Schrodinger's wave equation. So is also a solution. Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) If were a solution to Schrodinger's wave equation, then we could write so or which can be written as _______________________________________ 2.18 Dividing by , we find or _______________________________________ Since is a solution, then 2.19 Subtracting these last two equations, we have Note that Function has been normalized. (a) Now Since is also a solution, we have Subtracting these last two equations, we obtain This equation is not necessarily valid, which means that is, in general, not a solution to Schrodinger's wave equation. _______________________________________ 2.17 or which yields (b) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or which yields (c) or (c) which yields _______________________________________ or _______________________________________ 2.20 2.21 (a) (a) or or (b) (b) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or cm/s m or (ii) kg-m/s eV _______________________________________ or 2.23 (a) (c) (b) so m/s For electron traveling in cm/s direction, cm/s kg-m/s or m _______________________________________ m or 2.22 (a) (i) or rad/s _______________________________________ m/s cm/s m 2.24 (a) kg-m/s m or (ii) kg-m/s m J or (b) (i) rad/s eV m/s (b) kg-m/s Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ m m rad/s _______________________________________ 2.25 m or nm _______________________________________ J or 2.27 or Then (a) eV eV eV eV _______________________________________ or (b) 2.26 mJ (c) No (a) _______________________________________ J or eV 2.28 For a neutron and : Then eV J or eV eV eV For an electron in the same potential well: (b) J J or eV _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.29 Schrodinger's time-independent wave equation Use separation of variables, so let We know that for and We have Substituting into the wave equation, we obtain for so in this region The solution is of the form where Dividing by and letting , we find Boundary conditions: at (1) We may set First mode solution: where Solution is of the form Boundary conditions: Second mode solution: where and where Similarly, let and Third mode solution: where Applying the boundary conditions, we find , , Fourth mode solution: where From Equation (1) above, we have or _______________________________________ 2.30 The 3-D time-independent wave equation in cartesian coordinates for is: so that Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ 2.31 J (a) Solution is of the form: or eV We find (ii) For Substituting into the original equation, we find: (1) From the boundary conditions, , where So J or eV _______________________________________ 2.33 (a) For region II, , Also So cm , where , General form of the solution is where Substituting into Eq. (1) above (b)Energy is quantized - similar to 1-D result. There can be more than one quantum state per given energy - different than 1-D result. _______________________________________ 2.32 (a) Derivation of energy levels exactly the same as in the text Term with term with Region I, represents incident wave and represents reflected wave. General form of the solution is where (b) For Then (i) For Term involving represents the transmitted wave and the term involving represents reflected wave: but if a particle is transmitted into region I, it will not be reflected so that . Then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) Boundary conditions: (1) _______________________________________ 2.35 (2) Applying the boundary conditions to the solutions, we find Combining these two equations, we find where or m (a) For The reflection coefficient is (b) For m m The transmission coefficient is (c) , where is the density of transmitted electrons. eV J _______________________________________ 2.34 m/s cm/s electrons/cm Density of incident electrons, where cm _______________________________________ m (a) For 2.36 m (a) For (b) For m (c) For m or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ m Then or m _______________________________________ 2.38 Region I Region II Region III (a) Region I: or (b) For , ; , , (incident) = (reflected) where Region II: or m Then where Region III: or _______________________________________ 2.37 (b) In Region III, the term represents a reflected wave. However, once a particle is transmitted into Region III, there will not be a reflected wave so that . (c) Boundary conditions: At : where m At : (a) (b) The transmission coefficient is defined as so from the boundary conditions, we want to solve for in terms of . Solving Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ for in terms of , we find We then find Finally, _____________________________________ 2.39 Region I: We have If we assume that be large so that , then will reflected transmitted reflected Region II: We can then write which becomes Substituting the expressions for , we find incident where where and Region III: and transmitted where There is no reflected wave in Region III. The transmission coefficient is defined as: Then From the boundary conditions, solve for Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ in terms of . The boundary conditions are: At : At : or At (c) : and since , then From , we can write or But Then, eliminating , , from the boundary condition equations, we find This equation can be written as or _______________________________________ 2.40 (a) Region I: Since Region II: , we can write , so This last equation is valid only for specific values of the total energy . The energy levels are quantized. _______________________________________ 2.41 (J) Region III: The general solutions can be written, keeping in mind that must remain finite for , as (eV) or where (eV) and (b) Boundary conditions At : eV eV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ eV eV _______________________________________ so 2.42 We have We then obtain and Substituting into the wave equation, we have or To find the maximum probability where which gives Then the above equation becomes or is the radius that gives the greatest probability. _______________________________________ 2.43 is independent of and , so the wave equation in spherical coordinates reduces to or where For Then which gives 0 = 0 and shows that is indeed a solution to the wave equation. _______________________________________ 2.44 All elements are from the Group I column of the periodic table. All have one valence Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ electron in the outer shell. _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 3 Ex 3.3 Exercise Solutions (a) Ex 3.1 So or m/s or cm/s _______________________________________ Ex 3.2 At m or cm , we have (b) From Example 3.2, we have or At , J. , we see that so m or or cm _______________________________________ J Then Ex 3.4 J Or eV _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ m or cm _______________________________________ Ex 3.5 _______________________________________ Ex 3.6 (a) Then (b) _______________________________________ Test Your Understanding Solutions _______________________________________ TYU 3.1 At , we see that , so Ex 3.7 or J At the other point, so K _______________________________________ Ex 3.8 is in the range . Then from we find, by trial and error, . Then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ J Then or J or _______________________________________ eV _______________________________________ TYU 3.4 We have so that TYU 3.2 From Example 3.2, for and For , We have J. , we have and By trial and error, Then rad. or _______________________________________ TYU3.5 J Then J (a) or eV _______________________________________ TYU 3.3 We have (b) _______________________________________ TYU 3.6 We find so that We have (a) eV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ (b) _______________________________________ TYU 3.7 We find (a) (b) _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 3 3.1 If were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If were to decrease, the bandgap energy would increase and the material would begin to behave more like an insulator. _______________________________________ Setting becomes: for region I, the equation where Q.E.D. In Region II, form of the solution: . Assume the same 3.2 Schrodinger's wave equation is: Substituting into Schrodinger's wave equation, we find: Assume the solution is of the form: Region I: . Substituting the assumed solution into the wave equation, we obtain: This equation can be written as: which becomes Setting equation becomes for region II, this where again Q.E.D. This equation may be written as _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ which yields 3.3 We have The second boundary condition is Assume the solution is of the form: which yields The first derivative is The third boundary condition is which yields and the second derivative becomes and can be written as Substituting these equations into the differential equation, we find The fourth boundary condition is which yields Combining terms, we obtain and can be written as We find that Q.E.D. For the differential equation in and the proposed solution, the procedure is exactly the same as above. _______________________________________ 3.4 We have the solutions for and for . The first boundary condition is _______________________________________ 3.5 (b) (i) First point: Second point: By trial and error, (ii) First point: Second point: By trial and error, _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.6 (b) (i) First point: Second point: By trial and error, 3.8 (a) (ii) First point: Second point: By trial and error, _______________________________________ J 3.7 From Problem 3.5 Let Then Consider , of this function. J We find J or eV (b) Then For , So that, in general, J From Problem 3.5, And So J This implies that for J _______________________________________ or eV _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.9 (a) At At 3.10 (a) , J , By trial and error, J From Problem 3.6, J J J or (b) At eV , J or eV (b) At J . From Problem 3.5, J From Problem 3.6, J J J or eV _______________________________________ J Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or J eV _____________________________________ J or eV _______________________________________ 3.11 (a) At 3.12 For , K, eV At J , By trial and error, K, eV K, eV K, eV K, eV K, eV _______________________________________ 3.13 The effective mass is given by J We have J or (b) At eV , so that _______________________________________ 3.14 The effective mass for a hole is given by At J , From Problem 3.6, We have that so that _______________________________________ 3.15 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Points A,B: velocity in -x direction Points C,D: velocity in +x kg direction or Points A,D: negative effective mass For B: Points B,C: positive effective mass _______________________________________ 3.16 For A: At m , eV kg Or J So or _______________________________________ Now 3.18 (a) (i) kg or or Hz (ii) For B: At m , Or eV cm J nm (b) (i) So Hz (ii) Now kg or _______________________________________ 3.17 For A: cm nm _______________________________________ 3.19 (c) Curve A: Effective mass is a constant Curve B: Effective mass is positive around , and is negative around . _______________________________________ 3.20 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then and 3.23 For the 3-dimensional infinite potential well, when , , and . In this region, the wave equation is: Use separation of variables technique, so let Then Substituting into the wave equation, we have or _______________________________________ Dividing by , we obtain 3.21 (a) Let The solution is of the form: (b) Since so that Also, at , then at , so that where . . Then _______________________________________ Similarly, we have 3.22 (a) and From the boundary conditions, we find and where and From the wave equation, we can write (b) The energy can be written as _______________________________________ _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.24 The total number of quantum states in the 3-dimensional potential well is given Then (in k-space) by where Divide by the "volume" a, so We can then write Taking the differential, we obtain So Substituting these expressions into the density of states function, we have m J _______________________________________ Noting that 3.26 (a) Silicon, this density of states function can be simplified and written as Dividing by states so that will yield the density of _______________________________________ 3.25 For a one-dimensional infinite potential well, Distance between quantum states (i) At K, eV J Now Then m cm or Now (ii) At K, Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ eV (ii)At K, J J Then m cm or (b) GaAs, m cm or (b) GaAs, (i) At K, m cm or (ii) At (i)At J K, J (ii)At K, J m or cm _______________________________________ 3.28 3.27 (a) Silicon, (a) For or J m cm or m cm _______________________________________ (i)At K, K, J m cm (b) ; eV; m J eV; m J eV; m J eV; m J Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) For , ; eV; m J eV; m J eV; m J eV; m J _______________________________________ (b) , (c) , _______________________________________ 3.33 3.29 (a) or (b) _______________________________________ (a) , (b) 3.30 Plot _______________________________________ , (c) , _______________________________________ 3.34 3.31 (a) (a) ; (b) (i) ; (ii) ; _______________________________________ ; 3.32 ; Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) Then Or ; _______________________________________ ; 3.36 ; For ; , Filled state J or For ; eV , Empty state J or _______________________________________ eV Therefore eV _______________________________________ 3.37 (a) For a 3-D infinite potential well 3.35 For 5 electrons, the 5th electron occupies the quantum state ; so and J or So eV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For the next quantum state, which is empty, the quantum state is . This quantum state is at the same energy, so eV (b) For 13 electrons, the 13th electron occupies the quantum state ; so This expression can be written as J or eV or The 14th electron would occupy the quantum state . This state is at the same energy, so eV _______________________________________ 3.38 The probability of a state at being occupied is The probability of a state at being empty is Then or (b) At , which yields _______________________________________ 3.40 (a) (b) or so Q.E.D. _______________________________________ 3.39 (a) At energy (c) , we want or eV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or (b) For eV, eV which yields K _______________________________________ 3.41 (a) At , or At or 0.304% (b) At Then K, , eV or _______________________________________ or 14.96% 3.43 (a) At (c) or 99.7% (d) or At , for all temperatures _______________________________________ At So , eV 3.42 (a) For or Then (b) For , eV For Then , eV At , or or At , Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or _______________________________________ For , For , At , 3.44 (eV) _______________________________________ so 3.45 (a) At , or Si: (a) At K, For eV, or Ge: eV At (b) At K, eV For , or For , GaAs: At eV , (eV) or (b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a). _______________________________________ (c) At K, eV 3.46 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) or eV so K (b) or K _______________________________________ 3.47 (a) At K, eV eV By symmetry, for , eV Then (b) For eV K, eV , from part (a), eV Then eV _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 4 Exercise Solutions Ex 4.1 or cm For cm _______________________________________ K, or cm (b) Ex 4.2 (a) _______________________________________ cm Ex 4.4 (a) GaAs eV meV (b) Ge cm meV _______________________________________ (b) Ratio _______________________________________ Ex 4.3 (a) For Ex 4.5 K, cm We find eV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ (a) K cm eV cm _______________________________________ Ex 4.6 (b) So K cm From Figure 4.10, eV _______________________________________ Ex 4.7 (a) eV K cm _______________________________________ eV Ex 4.9 From Example 4.3, cm cm Then (a) (b) K K cm eV cm cm (b) (c) Fraction increases as temperature decreases. _______________________________________ Ex 4.8 K Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Ex 4.11 (a) cm cm _______________________________________ Ex 4.10 For or For K, cm cm K, cm (b) cm _______________________________________ or (a) cm K Ex 4.12 cm cm cm (b) eV _______________________________________ K Ex 4.13 We have cm cm _______________________________________ eV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU 4.3 So (a) Or Then cm _______________________________________ cm (b) cm Test Your Understanding Solutions TYU 4.1 cm Now eV (c) _______________________________________ cm _______________________________________ TYU 4.4 (a) TYU 4.2 cm Now cm eV (b) cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ cm (b) meV _______________________________________ (c) _______________________________________ TYU 4.7 (a) GaAs: TYU 4.5 (a) J or Also meV cm (b) Ge: Conductivity effective mass (b) cm J (c) _______________________________________ or Also meV _______________________________________ TYU 4.8 (a) For , and TYU 4.6 cm (a) (b) meV For , and Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ cm _______________________________________ Now, percentage ionized atoms % (b) For K, TYU 4.9 or _______________________________________ % (c) For TYU 4.10 We have For Now, percentage ionized atoms K, cm K, cm K, K, cm K, cm Now, percentage ionized atoms % (d) For K, For K, eV K, eV K, eV K, eV Fraction of ionized impurity atoms Now, percentage ionized atoms % _______________________________________ (a) For K, TYU 4.11 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ cm Then or cm _______________________________________ TYU 4.12 (b) Then which yields Now By trial and error, K _______________________________________ TYU 4.13 At or Set Then K, cm cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 4 4.1 where and are the values at 300 K. (K) (a) Silicon (cm ) (eV) (b) By trial and error, (K) (b) Germanium (cm ) (c) GaAs (cm ) K _______________________________________ 4.4 _______________________________________ At K, eV 4.2 Plot _______________________________________ At K, eV 4.3 (a) or By trial and error, K or Now eV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ which yields so cm _______________________________________ The maximum value occurs at 4.5 For For For K, K, K, eV eV eV (a) For K, (b) For K, (c) For K, (b) Let _______________________________________ Then 4.6 To find the maximum value (a) Same as part (a). Maximum occurs at or Let _______________________________________ Then To find the maximum value: 4.7 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (K) (eV) ( )(eV) where and _______________________________________ Then 4.12 (a) meV or _______________________________________ (b) meV 4.8 Plot _______________________________________ _______________________________________ 4.9 Plot _______________________________________ 4.10 Silicon: , eV Germanium: , eV Gallium Arsenide: , 4.13 Let Then eV _______________________________________ 4.11 Let constant Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ so that We can write So so that _______________________________________ The integral can then be written as 4.15 We have which becomes For germanium, Then , _______________________________________ or 4.14 Let Then for The ionization energy can be written as eV eV _______________________________________ 4.16 We have Let so that We can write Then For gallium arsenide, , Then The ionization energy is or or eV _______________________________________ 4.17 (a) We find that Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ eV (b) eV (c) (d) cm eV (d) Holes _______________________________________ (e) 4.19 (a) eV _______________________________________ eV 4.18 (a) eV (b) eV cm (b) eV (c) p-type _______________________________________ (c) cm 4.20 (a) eV cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ eV eV cm cm _______________________________________ (b) eV 4.22 (a) p-type eV (b) eV cm _______________________________________ cm eV 4.21 (a) eV cm _______________________________________ cm 4.23 eV cm (a) cm (b) eV cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) 4.25 eV cm cm cm cm _______________________________________ 4.24 cm (a) (a) eV eV (b) (b) eV eV (c) cm (c) (d) Holes (e) cm (d) Holes eV _______________________________________ (e) 4.26 eV _______________________________________ (a) cm eV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ cm cm (b) eV cm (b) cm eV cm eV eV cm eV eV cm cm _______________________________________ _____________________________________ 4.28 (a) 4.27 For (a) , cm eV Then cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) cm _______________________________________ 4.29 Define Then To find maximum , set So We find eV _______________________________________ or 4.30 which yields (a) Then For the hole concentration Using the Boltzmann approximation cm or (b) cm _______________________________________ 4.31 For the electron concentration Define The Boltzmann approximation applies, so Then To find maximum value of or , set Using the results from above, we find the maximum at Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ cm _______________________________________ 4.33 Plot _______________________________________ 4.32 (a) Silicon: We have 4.34 (a) We can write cm cm For eV and eV we can write (b) cm cm (c) or cm (d) cm We also have cm Again, we can write cm For and eV cm Then (e) or cm cm (b) GaAs: assume Then eV cm or cm Assume Then or cm eV _______________________________________ 4.35 (a) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ cm cm (b) (b) GaAs: cm (i) cm (c) cm cm cm cm cm (ii) (d) cm cm (c) The result implies that there is only one minority carrier in a volume of cm . _______________________________________ cm cm 4.37 (a) For the donor level cm (e) cm or cm cm (b) We have _______________________________________ Now 4.36 (a) Ge: cm or (i) Then or cm or cm _______________________________________ (ii) cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.38 (a) p-type cm (b) Silicon: cm or _______________________________________ cm Then 4.40 cm cm n-type _______________________________________ Germanium: 4.41 or cm Then cm cm Gallium Arsenide: cm and So cm Then cm , cm _______________________________________ 4.39 (a) n-type (b) cm cm so that cm _______________________________________ (c) 4.42 Plot _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.43 4.47 Plot (a) n-type _______________________________________ 4.44 Plot _______________________________________ (b) cm electrons are majority carriers 4.45 cm holes are minority carriers (c) so cm _______________________________________ 4.48 so cm cm _______________________________________ 4.46 (a) For Germanium (K) (eV) ) p-type Majority carriers are holes cm Minority carriers are electrons and cm cm (K) (b) Boron atoms must be added So (cm (cm ) (eV) cm cm _______________________________________ _______________________________________ 4.49 (a) For cm , eV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ cm , eV cm , eV cm , eV eV At K, eV (b) For cm , eV cm , eV cm , eV cm cm , eV _______________________________________ 4.50 (a) cm eV then so eV (c) Closer to the intrinsic energy level. Now _______________________________________ 4.51 By trial and error, (b) At K At K, K, K, At K, eV eV eV K, cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ At K, _______________________________________ 4.53 (a) cm At K, or eV (b) Impurity atoms to be added so cm At K and At K, eV K, cm (i) p-type, so add acceptor atoms (ii) eV Then cm Then, K, eV K, eV K, eV _______________________________________ 4.52 (a) For or cm _______________________________________ cm , eV cm , eV cm , eV cm , eV (b) 4.54 For cm , eV cm , eV cm , eV cm , eV so Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ eV or cm _______________________________________ (c) For part (a); cm 4.55 (a) Silicon (i) cm eV (ii) For part (b): eV cm cm cm cm Additional donor atoms _______________________________________ (b) GaAs (i) 4.57 eV (ii) eV cm cm Additional donor atoms _______________________________________ 4.56 cm Add additional acceptor impurities cm _______________________________________ (a) 4.58 eV (a) eV (b) (b) eV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) eV (e) eV _______________________________________ (d) eV (e) eV _______________________________________ 4.59 (a) eV 4.60 n-type (b) eV ______________________________________ (c) eV (d) eV 4.61 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (d) eV _______________________________________ eV cm cm Now So eV _______________________________________ 4.62 (a) Replace Ga atoms Replace As atoms (b) Silicon acts as a donor cm Silicon acts as an acceptor cm p-type (c) cm cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 5 Exercise Solutions (d) cm /V-s Ex 5.1 _______________________________________ which yields cm _______________________________________ Ex 5.5 Ex 5.2 Using Figure 5.2; (a) C, (i) cm , (ii) (b) (i) cm , cm , C, (ii) cm /V-s cm /V-s cm /V-s C, cm /V-s _______________________________________ Ex 5.3 (a) For (a) For cm , cm /V-s (b) For (b) , A/cm cm, A/cm ( -cm) (c) (c) For -cm cm, A/cm _______________________________________ _______________________________________ Ex 5.4 Ex 5.6 (a) (b) -cm (c) ( So -cm) Then Using Figure 5.3 and trial and error, cm or V/cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ (b) Ex 5.7 or ( cm /V-s _______________________________________ -cm) We find ( -cm) _______________________________________ Ex 5.8 From Equation (5.59), TYU 5.3 So A or mA From Equation (5.53), Using Figure 5.3 and trial and error, cm and V or mV _______________________________________ cm /V-s _______________________________________ TYU 5.4 We have cm /s Test Your Understanding Solutions cm m Then TYU 5.1 A/cm cm Also (a) cm A/cm (b) (c) Now , m, A/cm , _______________________________________ TYU 5.5 or A/cm _______________________________________ TYU 5.2 (a) For cm cm /V-s; So , cm /V-s And Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 5 5.1 From Figure 5.3, for cm we (a) find -cm (b) ( cm /V-s which gives -cm) _______________________________________ 5.2 -cm or (b) cm _______________________________________ 5.3 (a) From Figure 5.3, for then From Figure 5.3, for find cm we cm , cm /V-s which gives ( -cm) cm /V-s which gives _______________________________________ ( -cm) (b) 5.5 From Figure 5.3, for find cm we or cm /V-s which gives -cm _______________________________________ cm /V-s _______________________________________ 5.4 5.6 (a) (a) cm and cm (b) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ cm For GaAs doped at cm , So cm /V-s cm Then Note: For the doping concentrations obtained, the assumed mobility values are or valid. _______________________________________ A/cm (b) (i) cm 5.8 cm (ii) For GaAs doped at (a) cm , For cm , then cm /V-s cm /V-s A or A/cm _______________________________________ or mA 5.7 (a) or (b) (b) A or ( (c) or or -cm) mA (c) For (a), Then cm A/cm Then (d) cm/s or For (b), Then A/cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or cm/s cm/s _______________________________________ 5.9 (a) For cm , then cm /V-s _______________________________________ 5.11 (a) Silicon: For kV/cm, cm/s Then s For GaAs: Then or cm/s s cm (b) Silicon: For cm/s Then (b) kV/cm, or s For GaAs: Then cm/s cm/s s (c) _______________________________________ A or mA _______________________________________ 5.12 (a) cm cm /V-s cm /V-s 5.10 (a) (b) or cm /V-s (b) -cm V/cm cm cm /V-s cm /V-s Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or (300 K) Now -cm (c) cm cm cm /V-s cm /V-s or -cm _______________________________________ 5.13 (a) GaAs: which gives eV From Figure 5.3, and using trial and error, we find cm and cm /V-s Then cm (b) Silicon: Now (500K) or (500 K) Then cm which gives (500 K) ( -cm) _______________________________________ or 5.15 (a) (i) Silicon: which gives cm or and cm Note: For the doping concentrations obtained in part (b), the assumed mobility values are valid. _______________________________________ ( -cm) ( -cm) ( -cm) (ii) Ge: or (iii) GaAs: or 5.14 (b) Then (i) Si: Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.18 (a) V/cm (ii) Ge: (b) (iii) GaAs: _______________________________________ 5.16 (a) From Figure 5.3, for then cm , ( cm /V-s So -cm) (c) ( -cm) A (b) Using Figure 5.2, (i) For K( or C), cm /V-s ( (ii) For (d) Top surface; -cm) K( A ( -cm) C), A/cm cm /V-s Bottom surface: ( -cm) _______________________________________ ( -cm) A/cm 5.17 _______________________________________ 5.19 Plot _______________________________________ ( -cm) _______________________________________ 5.20 (a) so V/cm cm/s or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ m/s Then which becomes and yields or J (b) eV kV/cm cm/s or or m/s Then By trial and error, we find or J eV _______________________________________ K _______________________________________ 5.21 (a) or cm For cm 5.22 >> cm (a) and Then Then To find the minimum conductivity, set or A/cm which yields (b) A 5% increase is due to a 5% increase in electron concentration, so (Answer to part (b)) Substituting into the conductivity expression Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ( -cm) compensated: ( which simplifies to -cm) (d) The intrinsic conductivity is defined as n-type: V/cm p-type: V/cm The minimum conductivity can then be compensated: written as V/cm _______________________________________ 5.24 _______________________________________ 5.23 (a) n-type: cm or cm p-type: cm cm compensated: Then cm /V-s _______________________________________ 5.25 cm cm (a) At K, cm /V-s (b) At K, cm /V-s _______________________________________ (b) From Figure 5.3, n-type: p-type: cm /V-s cm /V-s compensated: (c) n-type: cm /V-s Then cm /V-s _______________________________________ ( p-type: 5.26 -cm) 5.27 Plot _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.28 cm Plot _______________________________________ _______________________________________ 5.29 5.32 Then (a) For , which yields cm _______________________________________ 5.30 A/cm _______________________________________ 5.31 (b) For A/cm m, (c) For A/cm m, _______________________________________ 5.33 For electrons: (a) At which yields , A/cm cm For holes: For , A/cm (b) A/cm _______________________________________ 5.34 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) (a) (i) A/cm (b) (ii) A/cm A/cm (b) (i) A/cm (ii) A/cm _______________________________________ (c) We have So V/cm _______________________________________ 5.37 (a) We have cm /V-s, so that 5.35 cm /s or Then which yields Then Solution is of the form We find so that or Substituting into the differential equation, we _______________________________________ 5.36 have This equation is valid for all x, so Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or which yields Also Then so which yields cm At so that (b) , which yields Assume Then K, so cm eV and Then cm (b) At Or , or cm At m, (i) At or cm (c) At , A/cm (ii) At m, A/cm _______________________________________ 5.39 m, (a) or A/cm Then where or A/cm _______________________________________ We find 5.38 or (a) , cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ Solving for the electric field, we find 5.41 From Example 5.6 V/cm (b) For A/cm or mV _______________________________________ Then V/cm _______________________________________ 5.40 5.42 For (a) So V/cm Which yields cm _______________________________________ 5.43 (a) We have or V/cm We have (b) V or mV or cm /s Then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or 5.47 A/cm (a) (b) Now or mV or (b) We have or so V/cm (c) which yields V/cm _______________________________________ 5.44 Plot _______________________________________ or m /V-s 5.45 (a) (i) cm /s (ii) _______________________________________ cm /s (b) (i) cm /V-s (ii) cm /V-s _______________________________________ 5.48 (a) cm, cm, cm m (a) or or (b) n-type (b) 5.46 cm (c) V mV V/cm _______________________________________ cm /V-s m /V-s or cm /V-s Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ or 5.49 (a) or m cm (c) mV (b) negative n-type (c) or m /V-s or m cm cm /V-s (d) (d) or ( -cm) _______________________________________ or m /V-s cm /V-s _______________________________________ 5.50 (a) (b) negative n-type Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 6 Yes, low-injection condition is met. Exercise Solutions _______________________________________ Ex 6.1 cm For , s cm s Ex 6.4 s, (a) cm cm s or s, cm m (b) s or s, cm s _______________________________________ Ex 6.2 (a) ns (b) (i) ( ) ( ) cm (ii) cm (iii) cm (iv) cm (v) cm _______________________________________ s or ns _______________________________________ Ex 6.5 (a) Ex 6.3 cm or (i) m m. (ii) m, (iii) m, (a) (i) (ii) cm (iii) cm (iv) cm (b) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ (b) m (a) In thermal equilibrium, (i) s, eV (ii) (b) Quasi-Fermi levels, s, (iii) s, eV _______________________________________ Ex 6.6 (a) For cm Figure 5.3, ( in GaAs, from cm /V-s. eV _______________________________________ Ex 6.8 n-type; -cm) cm , cm Then cm , s s or We have ps (b) For cm Figure 5.3, ( in silicon, from cm /V-s. -cm) or cm s _______________________________________ Then s or ps _______________________________________ Ex 6.7 Ex 6.9 (a) For From Equation (6.109), cm As cm As cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Then (b) (b) (c) Note: is a result of . _______________________________________ or (c) As , cm _______________________________________ Ex 6.10 TYU 6.3 (a) (i) For , cm Then (ii) For cm , _______________________________________ (b) TYU 6.4 n-type; Minority carriers = holes (i) For , (ii) For , , constant _______________________________________ Then hole diffusion current density A/cm We have (electrons) (holes) Then electron diffusion current density A/cm _______________________________________ Test Your Understanding Solutions TYU 6.5 TYU 6.1 (a) p-type; Minority carriers = electrons (b) (a) Then (b) cm _______________________________________ (c) TYU 6.2 (a) p-type; Minority carriers = electrons (d) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Now Then (a) m (b) m (c) m (d) m _______________________________________ or (ii) or TYU 6.6 Using the results from TYU 6.5, we find (c) (i) (a) (i) or (ii) or or (ii) or (b) (i) _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 6 (b) Generation rate = recombination rate Then 6.1 cm cm cm s (c) (a) Minority carrier hole lifetime is a constant. s cm s _______________________________________ cm s 6.4 (a) or (b) J; energy of one photon cm s _______________________________________ Now 6.2 1 W = 1 J/s cm photons/s Volume = (1)(0.1) = 0.1 cm cm Then (a) cm s (b) e-h pairs/cm -s (b) s _______________________________________ or 6.3 (a) Recombination rates are equal cm _______________________________________ cm cm 6.5 We have Then and which yields s The hole particle current density is Now Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ We have the continuity equations (1) We can write and and (2) so Then By charge neutrality, and and We can then write Also , _______________________________________ Then we have (1) 6.6 From Equation (6.18), and (2) For steady-state, Multiply Equation (1) by and Equation (2) by , and add the two equations. Then For a one-dimensional case, We find or cm s _______________________________________ Divide by 6.7 From Equation (6.18), or cm s _______________________________________ 6.8 Define , then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ and For very, very low injection, Then we have Q.E.D. _______________________________________ cm /s and 6.9 p-type material; minority carriers are electrons (a) From Figure 5.3, cm /V-s cm /V-s (b) (b) For holes, (c) cm /s s s For electrons, cm s _______________________________________ cm 6.11 so s _______________________________________ With excess carriers and For an n-type semiconductor, we can write Then 6.10 For Ge: cm or so In steady-state, So that cm cm (a) We have: cm /V-s, cm /V-s, cm /s cm /s _______________________________________ 6.12 (a) cm cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.14 ; Now cm For cm Then, cm /V-s cm /V-s Then where ( (b) (i) ( (ii) cm -cm) -cm) ( -cm) _______________________________________ 6.13 (a) For s, A or mA _______________________________________ cm At s, cm cm Then for 6.15 (a) s, cm (b) cm For s (b) s, cm s (c) ( For -cm) s, ( -cm) _______________________________________ (i) s Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ cm (ii) For s, s cm (ii) cm _______________________________________ (iii) s 6.18 (a) For (iv) s s _______________________________________ cm 6.16 At s, For s cm cm cm (a) cm so s (b) (i) (b) cm (ii) cm (c) (iii) cm _______________________________________ s _______________________________________ 6.17 (a) (i)For cm 6.19 p-type; minority carriers - electrons s cm /s cm At s, (a) cm (b) cm For cm s cm (ii) cm (b) (i) For s cm At s, A/cm Holes diffuse at same rate as minority carrier electrons, so Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ A/cm A/cm _______________________________________ A/cm (b) For 6.20 (a) p-type; cm, cm cm and A/cm A/cm cm (b) Excess minority carrier concentration At , (c) For cm cm A/cm so that A/cm cm _______________________________________ (c) For the one-dimensional case, 6.22 n-type, so we have or Assume the solution is of the form where The general solution is of the form Then , For , remains finite, so Then the solution is . _______________________________________ Substituting into the differential equation or 6.21 cm where Dividing by , we have cm The solution for s is which can be rewritten as A/cm (a) For , cm Define Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ which yields Then In order that as , use the minus sign for and the plus sign for . Then the solution is for for where We may note that if , then and (b) where so _______________________________________ cm /s and cm 6.23 Plot _______________________________________ 6.24 (a) From Equation (6.55) or m V/cm, then For cm and or We have that so we can define cm (c) Force on the electrons due to the electric field is in the negative x-direction. Therefore, the effective diffusion of the electrons is reduced and the concentration drops off faster with the applied electric field. _______________________________________ 6.25 p-type so the minority carriers are electrons and Then we can write The solution is of the form where Then Uniform illumination means that . For left with which gives and Substituting into the differential equation, we find For Then , for or For And , so that , we are Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ (no recombination) _______________________________________ 6.27 6.26 V/cm n-type, so minority carriers are holes and We have , , and cm /V-s (steady-state). Then we have or For , = constant. Then and cm /s We find V For , so we have so that For and , 6.28 (a) so that so that This value is very close to 0.0259 for K. _______________________________________ and Assume that is the solution to the differential equation The boundary conditions are: (1) at (2) at (3) continuous at (4) continuous at (5) continuous at (6) continuous at To prove: we can write and Applying the boundary conditions, we find for Also for for Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Substituting the expressions for and eV into the differential equation, we find 0 = 0. Q.E.D. (b) Consider eV Let , then or (c) eV or Let Now meV _______________________________________ Then _______________________________________ 6.29 Plot _______________________________________ 6.30 6.31 (a) p-type (a) eV or eV (b) (b) cm cm and cm Then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) or or eV and cm (b) or eV eV _______________________________________ (c) (i) 6.32 (a) For n-type, (ii) So eV Which yields cm or meV _______________________________________ (b) eV (c) eV _______________________________________ 6.33 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Quasi-Fermi level for holes: we have 6.34 (a) (i) We have We find eV cm ( (ii) m) and ( . ) (eV) 0 +0.58115 50 +0.58140 _______________________________________ 6.36 (a) We can write eV (b) (i) eV and (ii) eV _______________________________________ 6.35 Quasi-Fermi level for minority carrier electrons: so that or cm Then We have or low injection, so that Then cm (b) We find ( m) 0 1 2 10 20 50 ( -0.581 +0.361 +0.379 +0.420 +0.438 +0.462 ) (eV) or eV _______________________________________ 6.37 Plot _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ where since these are the thermal equilibrium generation and recombination 6.38 rates. If , then and (a) cm , eV so that Thus a negative recombination rate implies a net positive generation rate. _______________________________________ 0.22805 6.40 We have that 0.28768 (b) If cm , and , then eV 0.365274 0.365286 If 0.365402 Then , we can neglect 0.366536 _______________________________________ (a) For n-type; Then , s (b) For intrinsic, Then 6.39 (a) or Let . For s (b) We had defined the net generation rate as (c) For p-type; Then , : also Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ cm s (ii) For cm/s, _______________________________________ 6.41 (a) From Equation (6.56) (iii) For , Solution is of the form At , so that , Then (b) (i) For , cm cm/s, cm , (ii) For We have (iii) For _______________________________________ We can write and 6.42 Then cm (a) At Solving for , cm , we find or cm For The excess concentration is then The solution is of the form where cm Now At At , , Solving these two equations, we find or and (i) For , Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Substituting into the general solution, we find so that cm and The solution is now which can be written as (a) For , cm Then where cm and , we have (b) If m or A/cm (b) For cm/s, so the solution is of the form Applying the boundary conditions, we find cm Also A/cm _______________________________________ _______________________________________ 6.43 For 6.44 For , we have so that So the solution is of the form and At For or , which yields so that At The boundary conditions are (1) at so that , the flux of excess holes is Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (2) at so that cm /V-s, cm /V-s (3) continuous at (4) ( continuous at Let Applying the boundary conditions, we find -cm) m Then cm and So Which yields cm _______________________________________ Then for and for _______________________________________ 6.45 Plot _______________________________________ 6.48 (a) GaAs: and cm For cm , from Figure 5.3, cm /V-s, ( Let cm /V-s -cm) m Then cm So Which yields (b) Silicon: cm , For cm cm , from Figure 5.3, Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 7 Exercise Solutions Ex 7.1 cm (a) (i) V (ii) V (b) cm Now (i) V cm Now (ii) V _______________________________________ V/cm _______________________________________ Ex 7.3 (a) Ex 7.2 V V Then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ cm or cm m or m cm cm or m _______________________________________ or m Ex 7.4 V cm or (b) m Now V cm or m Then V _______________________________________ Ex 7.5 (a) V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU 7.1 (b) (a) V F/cm (c) cm F/cm or m _______________________________________ Ex 7.6 For a one-sided junction cm or So m cm We have Then cm cm _______________________________________ or Test Your Understanding Solutions m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ V V/cm (b) V or cm m or m cm or m cm cm or m cm or m cm or m V/cm _______________________________________ V/cm _______________________________________ TYU 7.2 TYU 7.3 (a) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ V Now for V, or cm or cm or Also cm Now cm V/cm _______________________________________ cm Now cm Also V/cm (b) For V TYU 7.4 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ V or (a) For V, F (b) For pF V, F pF _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 7 7.1 (b) cm , Si: Ge: GaAs: (a) (c) (i) cm V V V cm , cm Si: V Ge: V GaAs: V _______________________________________ V (ii) V 7.3 (a) Silicon ( (iii) K) V (b) For cm (i) V (b) GaAs ( (ii) ; V ; ; ; V V V K) V For (iii) cm ; V ; ; ; V _______________________________________ V V V (c) Silicon (400 K), 7.2 Si: cm cm Ge: For cm GaAs: cm cm and V ; V ; ; ; V V V GaAs(400 K), (a) Then cm Si: Ge: GaAs: , cm V V V ' For cm cm ; V ; V ; V ; V _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or 7.4 (a) n-side cm m We have or eV or p-side V/cm _______________________________________ 7.5 (a) n-side or eV (b) or or eV V p-side (c) or or eV V (b) (d) or V (c) or cm Now m or V (d) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For 300 K; V For 400 K; or cm m cm m By symmetry V _______________________________________ Now 7.8 or V/cm _______________________________________ 7.6 So (a) (b) or cm or which yields cm cm or (c) cm V _______________________________________ 7.7 200 K; ; 300 K; ; 400 K; For 200 K; or or cm m cm ; V cm cm m cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Now V/cm (b) From part (a), we can write which yields cm cm or cm cm m or m Now cm or m or V/cm _______________________________________ cm m (c) 7.9 (a) or V (b) or V/cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 7.10 (a) (b) At V increases as temperature decreases K, we can write At K, eV For At Also V V, K eV K, Then V V _______________________________________ So Then 7.12 (b) For cm , V We find or eV _______________________________________ For cm 7.11 or eV Then or Using the procedure from Problem 7.10, we can write, for K, V _______________________________________ 7.13 (a) At K, Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 7.14 Assume silicon, so or V (b) or (a) or cm (b) cm (c) (c) Now (a) (b) (c) Also , cm cm m , , m m V V V or cm Then (a) (b) (c) Now m m m (a) (b) (d) (c) _______________________________________ 7.15 or V/cm _______________________________________ We find Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) (i) For , (i) For , ; V (ii) (iii) (iv) ; ; ; (i) For , ; (ii) (iii) (iv) (b) (i) For ; ; ; V/cm V/cm V/cm , ; V ; ; ; V V V or m (ii) For V, , (ii) (iii) (iv) (c) cm V/cm (ii) (iii) (iv) (i) For V V V cm ; V/cm ; ; ; V/cm V/cm V/cm increases as the doping increases, or m (c) (i)For , and the electric field extends further into the low-doped side of the pn junction. _______________________________________ V/cm (ii)For 7.16 V, (a) V/cm V _______________________________________ (b) 7.17 (a) V (b) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V/cm (d) cm or m F or pF _______________________________________ 7.18 (a) We find cm or m cm cm cm or Also (c) (b) m m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V cm or m (b) So (c) For a larger doping, the space charge width narrows which results in a larger capacitance. _______________________________________ cm or 7.20 (a) m or (c) V Now V/cm (d) or F/cm or _______________________________________ so that 7.19 (a) V which yields V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ which yields (b) V or _______________________________________ V We have 7.21 (a) so that V or We find V which yields V We find V (c) or or V We have (b) so that V or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) We can then write which yields cm and cm _______________________________________ 7.23 or V _______________________________________ 7.22 (a) We have So or For which yields V _______________________________________ V, we find 7.24 (a) or V (b) Then Now so V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) (i) For , F pF (ii) For V, pF (b) H (b) (i) For V, (ii) For V, mH pF V Hz MHz pF Hz MHz _______________________________________ 7.26 (i) For Let , pF (ii) For V (a) V, cm pF (b) _______________________________________ cm _______________________________________ 7.25 7.27 V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) or cm Also By trial and error, cm , cm , cm , V (b) From part (a), By trial and error, cm , V _______________________________________ or 7.28 (a) cm or (c) For V (b) which becomes m, we have Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ We find V _______________________________________ 7.29 An junction with cm , (a) A one-sided junction and assume . Then (i) For V, F (ii) For V, F (iii) For V, F _______________________________________ or 7.31 (a) which yields V (b) cm so (b) cm m (c) or V/cm cm _______________________________________ (c) 7.30 (a) V (b) V _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 7.32 Plot _______________________________________ so that for , we have We also have at 7.33 Then (a) (c) p-region which gives or Then for We have we have at _______________________________________ Then for 7.34 n-region, (a) For So m, or At So m , n-region, Then or At We have at , , so Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or V/cm 7.36 (c) Magnitude of potential difference is cm _______________________________________ Let at , then 7.37 (a) For cm , from Figure 7.15, cm , V Then we can write (b) For V _______________________________________ At m 7.38 (a) From Equation (7.36), or V Potential difference across the intrinsic region Set and V or V By symmetry, the potential difference across the p-region space-charge region is also 3.863 V. The total reverse-bias voltage is then V _______________________________________ Then 7.35 V So (b) (a) or V V Then cm Then (b) Or cm _______________________________________ So V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ 7.39 For a silicon neglecting junction with cm and we have V, then, Assume (a) For m which yields V (b) For m or cm m _______________________________________ 7.40 We find V which yields V Note: From Figure 7.15, the breakdown voltage is approximately 300 V. So, in each case, breakdown is reached first. _______________________________________ 7.42 Impurity gradien Now cm so From Figure 7.15, V _______________________________________ 7.43 (a) For the linearly graded junction which yields Then cm Now Now Then At So and , Then which yields V or V _______________________________________ 7.41 Assume silicon: For an (b) Set junction at , then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) Then _______________________________________ (i) For 7.44 We have that (ii) For V, , pF pF _______________________________________ Then which yields cm _______________________________________ 7.45 (a) Let cm Then V Now cm << Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 8 Exercise Solutions Ex 8.1 cm cm cm cm cm A/cm cm We have that and so low injection applies. _______________________________________ A/cm Ex 8.2 The total current density is: A/cm _______________________________________ A/cm _______________________________________ Ex 8.3 We find Ex 8.4 In the n-region, for cm , cm /V-s or cm V/cm In the p-region, for cm , cm /V-cm cm or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ V/cm _______________________________________ Ex 8.5 From Example 8.5, we have Let and Then K, K, eV, V. which yields V so or V C increase in temperature. _______________________________________ A/cm mV per Ex 8.6 (c) _______________________________________ (a) Ex 8.7 or A/cm or (b) A V or A Then A or cm (a) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ A cm (b) For , A Now (a) so that (b) or V _______________________________________ TYU 8.2 We find (a) , Then (a) A; A (b) A; A or A mA We find (b) So (a) or F nF or (b) A mA (c) or F nF _______________________________________ A or Test Your Understanding Solutions mA _______________________________________ TYU 8.1 cm TYU 8.3 From TYU 8.2, Now mA Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ We find cm cm Then Then or mA _______________________________________ A/cm (c) _______________________________________ TYU 8.4 TYU 8.5 (a) or A or A A/cm Then A/cm So (b) (a) V We find A A (b) A Now Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ (a) ( -cm) Then (b) We have The total resistance is , _______________________________________ We find (a) A; (b) A; A TYU 8.7 (a) A Now Now mA So (a) So F nF (b) From Appendix G, F nF _______________________________________ So that s TYU 8.6 From Figure 5.3, for cm , cm , cm /V-s For cm /V-s In the n-region, ( Then In the p-region, -cm) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ (b) By trial and error s _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 8 or 8.1 In forward bias cm (b) V, Then cm cm or (c) V (a) For , then or mV _______________________________________ mV (b) For 8.3 , then cm cm or mV mV _______________________________________ (a) V, 8.2 cm cm cm cm (b) V cm (a) V, cm cm (c) _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.4 (a) A/cm cm A or mA cm (b) (i) or V (ii) n-region - lower doped side A/cm A (b) or cm mA (c) cm mA _______________________________________ (i) 8.6 For an silicon diode V (ii) p-region - lower doped side _______________________________________ 8.5 or (a) (a) For or A V, Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) For A V, A (b) or (i) A _______________________________________ A (ii) A 8.7 (iii) A _______________________________________ A/cm (a) 8.9 We have A or mA (b) A _______________________________________ 8.8 or we can write this as so that (a) In reverse bias, is negative, so at , we have A/cm or mV _______________________________________ 8.10 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Case 1: A mA mA/cm Case 2: or mA or mA/cm Case 3: (b) From part (a), So V Then or mA Case 4: _______________________________________ mA cm _______________________________________ 8.12 The cross-sectional area is cm 8.11 We have (a) which yields A/cm We can write Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ We want We have or and so = or which yields (b) Using Einstein's relation, we can write Now We find cm and cm _______________________________________ 8.13 Plot _______________________________________ We have and Also Then 8.14 (a) _______________________________________ 8.15 (a) p-side; Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or or eV (A) Also on the n-side; Then _______________________________________ or 8.16 eV (b) We can find (a) cm /s cm /s Now A (b) or A A/cm Then (c) or V A We find V or A (c) The hole current is A cm (d) A Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ We find cm and cm m Then (e) or cm A A (b) We have Now At A cm, or A/cm Then (c) We have A We can determine that cm and Then _______________________________________ 8.17 (a) The excess hole concentration is given by or A/cm We can also find A/cm m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then at m, Then or A/cm _______________________________________ We find that cm /s and Also 8.18 (a) Problem 8.7 m cm Then or or Then, we find the total number of excess electrons in the p-region to be: V (b) Problem 8.8 or (a) V, (b) V, (c) V, Similarly, the total number of excess holes in the n-region is found to be We find that cm /s and V _______________________________________ 8.19 The excess electron concentration is given by The total number of excess electrons is We may note that m Also cm Then So (a) V, (b) V, (c) V, _______________________________________ 8.20 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then so or For K, For K, , , (i) Germanium: eV We then have or (ii) Silicon: or eV or _______________________________________ Then or eV _______________________________________ 8.22 Plot _______________________________________ 8.23 First case: 8.21 (a) We have or V which can be written in the form Now K or (b) Taking the ratio Second case: Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or Now A By trial and error, K The reverse-bias current is limiting factor. _______________________________________ A or mA (b) (i) 8.24 cm or m; (a) or V (i) (ii) or A V (ii) A A Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ A or mA _______________________________________ 8.25 (a) We can write for the n-region We can also find The general solution is of the form The boundary condition at gives The solution can now be written as and the boundary condition at gives or finally From this equation, we have Then, from the first boundary condition, we obtain (b) = We then obtain Then which can be written as _______________________________________ 8.26 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ temperature. As a first approximation, neglect the variation of and with temperature For the temperature range K, over the range of interest. We can then write neglect the change in and . Then Taking the ratio of currents, but maintaining a constant, we have where is another constant, independent of temperature. We find or _______________________________________ We then have 8.28 (a) We have K, V and eV, V K, eV, V eV, V A K, (b) We find For K, which yields V For K, V which yields V _______________________________________ 8.27 (a) We can write where C is a constant, independent of and Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then cm A or Then A (b) From Problem 8.28 A A A So (c) _______________________________________ 8.29 (a) Set , V _______________________________________ 8.30 so cm cm /s Then cm /s (a) (i) By trial and error, K We have A Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.31 (ii) Using results from Problem 8.30, we find V, A, A, V, A A A A, (iii) V, A A A, A V, (iv) A A A, A V. A A A, A _______________________________________ (b) 8.32 Plot _______________________________________ V 8.33 Plot _______________________________________ cm 8.34 We have that (i)Then Let We can write A (ii) and A We also have (iii) so that A (iv) A _______________________________________ Then and Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Define and Then the recombination rate can be written as or To find the maximum recombination rate, set Q.E.D. _______________________________________ 8.35 We have In this case, cm s and is a constant through the space charge region. Then We find or which simplifies to or V Also The denominator is not zero, so we have or Then the maximum recombination rate becomes or cm Then or or A/cm _______________________________________ which can be written as If , then we can neglect the (-1) term in the numerator and the (+1) term in the denominator, so we finally have Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.36 _______________________________________ or A/cm Now 8.38 (a) , For mA We want C or (b) For mA which can be written as C _______________________________________ We find or V _______________________________________ 8.39 For a diode , 8.37 (a) Now S and F or We have nF (b) where F or F nF We obtain kHz , kHz , MHz , MHz , _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 8.40 Reverse bias 0.60 _______________________________________ 8.41 For a diode, , then V Now F/A Then F or s (V) (pF) 10 5 3 1 0 At 1 mA, 1.555 2.123 2.624 3.818 5.747 6.650 8.179 or F _______________________________________ 8.42 (a) Forward bias For (i) Then or A or mA (ii) A (V) 0.20 0.40 (F) + (F) = (F) V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (iii) (b) (i) or A or We can write mA (ii) (a) (i) For mA, V (iii) _______________________________________ or (ii) For or 8.43 (a) p-region: V mA, V (b) Set (i) For mA, or V so or (ii) For mA, n-region: so or V _______________________________________ 8.45 or The total resistance is (a) or A or (b) which yields mA _______________________________________ 8.44 V (b) A Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ V 8.48 (a) erf _______________________________________ erf = erf erf 8.46 Then (a) or which yields (b) erf (b) which yields By trial and error, _______________________________________ _______________________________________ 8.47 8.49 pF at (a) If pF at Then we have V We have s, mA and mA or So We find or s Also (b) If , then pF The time constant is which yields s Now, the turn-off time is Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 8 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or s or cm _______________________________________ _______________________________________ 8.50 V We find which yields cm _______________________________________ 8.51 Sketch _______________________________________ 8.53 From Figure 7.15, cm Let cm cm Then V A Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 9 Exercise Solutions Ex 9.3 Ex 9.1 V V V (a) V, V cm cm V/cm Then V/cm _______________________________________ Ex 9.2 From Figure 9.3, V V (b) V, cm V/cm Then V _______________________________________ or cm _______________________________________ Ex 9.4 Assume , then A/K -m A/K -cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Ex 9.5 so that or V _______________________________________ For the pn junction: Test Your Understanding Solutions V For the Schottky junction: TYU 9.1 (a) V _______________________________________ V (b) V Ex 9.6 V (c) Then A _______________________________________ or cm Then Ex 9.7 We have or V/cm (d) cm or _______________________________________ Ex 9.8 From Example 9.8, We find F/cm _______________________________________ eV. cm Now or TYU 9.2 (a) (b) V V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ V V (c) (b) For the pn junction diode or cm V Now For the Schottky diode V or V/cm _______________________________________ (d) or F/cm _______________________________________ TYU 9.3 or V or cm _______________________________________ TYU 9.4 Then (a) For the pn junction diode V For the Schottky diode Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 9 9.1 9.2 (a) (a) We have eV (c) =0.2235 V V or V (b) and V or V V increases, Also remains constant (c) V V or decreases, cm remains constant _______________________________________ Then 9.3 (a) V (b) or V/cm (d) Using the figure, So V V V or (c) V We then find cm (i) and cm V/cm _______________________________________ or m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (i) V/cm cm or m (ii) V/cm cm or (ii) m cm or m V/cm _______________________________________ V/cm _______________________________________ 9.4 (a) V (b) 9.6 V (c) V (a) We have V (d) V (i) V cm or m (i) V/cm F (ii) or pF cm or (ii) m V/cm _______________________________________ or 9.5 (b) (c) (d) F V V pF Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) V or V V (d) (i) or V _______________________________________ F or 9.8 From Figure 9.5, pF V (a) (ii) V V (i) F or cm m or pF _______________________________________ V/cm (ii) cm or m 9.7 (a) From the figure, (b) We find V V/cm (b) (i) and We can then write V or cm (c) or cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) V (ii) V cm _______________________________________ V/cm 9.9 Now We have cm or And Now V _______________________________________ Solving for x, we find 9.11 Plot _______________________________________ Substituting this value of into the equation for the potential, we find 9.12 (a) or V which yields (b) We have _______________________________________ 9.10 From Figure 9.5, V which becomes (a) V V cm or m V/cm or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) We find V (c) If A/cm V, then (d) or V From part (b), we have V We then find V With interface states, the barrier height is less sensitive to the metal work function. _______________________________________ 9.13 We have that _______________________________________ 9.15 (a) V A/cm A (i) Let (cm Then we can write eV ) V (ii) V (iii) V We then find cm eV _______________________________________ (b) eV 9.14 (a) V (b) V A (i) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V V (ii) and V V (a) We find for V, (iii) V _______________________________________ 9.16 (a) or cm V m Then (b) A/cm (c) or V/cm V Now (d) V _______________________________________ or 9.17 Plot _______________________________________ 9.18 From the figure, V V Then or A/cm For (b) For cm , we find A V, then or or cm We have Also Now m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or V/cm so that and or V Then We can write The differential volume element is or A/cm Finally, A _______________________________________ The current is due to all x-directed velocities that are greater than and for all y- and z- directed velocities. Then 9.19 We have that The incremental electron concentration is where and assuming the Boltzmann approximation We can write Then Make a change of variables: or If the energy above We can then write and We can also write is kinetic energy, then Taking the differential, we find We may note that when , . We may define other change of variables, Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) V Substituting the new variables, we have (b) V (c) V _______________________________________ _______________________________________ 9.20 For the Schottky diode, 9.22 (a) (i) (ii) mA in each diode V (a) V V (b) Same voltage across each diode Then V (b) cm _______________________________________ 9.21 For the pn junction, A Then (a) V V (b) V mA (c) V A _______________________________________ For the Schottky junction, A 9.23 (a) For mA, we find Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ A/cm We have Then For the pn junction diode, V For the Schottky diode, V (b) For the pn junction diode, or mA _______________________________________ 9.24 Plot _______________________________________ 9.25 Then (a) (b) (c) _______________________________________ or 9.26 Now (a) (i) (ii) or mA For the Schottky diode, (i) mV mV _______________________________________ 9.27 or or mV (b) (ii) Now mV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) Now for V V we have (b) V _______________________________________ 9.28 (b) We need And V or We have or and which yields cm (c) Barrier height = 0.20 V _______________________________________ 9.29 We have that By trial and error, we find cm _______________________________________ 9.30 (b) Then or Let At or at , , so , so V _______________________________________ 9.31 Sketches _______________________________________ 9.32 Sketches _______________________________________ 9.33 Electron affinity rule Also where For GaAs, and for AlAs, . If we assume a linear extrapolation between GaAs and AlAs, then for Al Ga As Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then eV _______________________________________ 9.34 Consider an n-P heterojunction in thermal equilibrium. Poisson's equation is In the n-region, For uniform doping, we have Now or Similarly on the P-side, we find The boundary condition is at , so we obatin We have that Then We can write In the P-region, Substituting and collecting terms, we find which gives Solving for , we have We have the boundary condition that at , so that Similarly on the P-side, we have Then The total space charge width is then Substituting and collecting terms, we obtain Assuming zero surface charge density at , the electric flux density D is continuous, so , which yields We can determine the electric potential as _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 10 Exercise Solutions Ex 10.4 From Figure 10.16, We find Ex 10.1 V V V cm or m _______________________________________ cm Ex 10.2 C/cm Then V V _______________________________________ Ex 10.3 From Figure 10.16, V F/cm V _______________________________________ Ex 10.5 From Figure 10.16, We find V Then V V _______________________________________ cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ We find C/cm _______________________________________ Now Ex 10.7 We find F/cm Then or V _______________________________________ or Now Ex 10.6 A/V mA/V F/cm V cm Now (a) mA (b) mA (c) mA _______________________________________ Ex 10.8 We find F/cm F/cm Then We find Now Then Or F/cm cm /V-s Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Hz We now find GHz _______________________________________ V _______________________________________ Test Your Understanding Solutions Ex 10.9 TYU 10.1 (a) (a) V V F/cm cm or V (b) m (b) V (i) cm V or m (ii) _______________________________________ V TYU 10.2 _______________________________________ or V Ex 10.10 We have or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ V _______________________________________ TYU 10.3 From TYU 10.2, V cm We have or nm _______________________________________ or V _______________________________________ TYU10.4 TYU 10.6 V From Equation (10.17) From Ex 10.7, Then mA/V V _______________________________________ TYU 10.7 F/cm V _______________________________________ F/cm Now TYU 10.5 V A Then V mA V mA V mA _______________________________________ cm C/cm From Figure 10.16, V TYU 10.8 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ _______________________________________ TYU 10.10 _______________________________________ We find F/cm A/V TYU 10.9 Then (a) F/cm _______________________________________ V (b) V (i) V (ii) V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 10 10.1 (a) p-type; inversion (b) p-type; depletion (c) p-type; accumulation (d) n-type; inversion _______________________________________ or cm m (ii) 10.2 V (a) (i) cm or m _______________________________________ V 10.3 (a) cm m or 1st approximation: Let V Then (ii) V cm 2nd approximation: cm m or (b) V V Then cm (b) V V _______________________________________ so cm (i) V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.4 p-type silicon (a) Aluminum gate 10.7 From Problem 10.5, We have V (a) F/cm Then or V V (b) V (b) polysilicon gate F/cm or V (c) V _______________________________________ polysilicon gate 10.8 (a) or V _______________________________________ V V (b) F/cm 10.5 V (i) V (ii) V _______________________________________ 10.6 (a) (c) V F/cm cm (b) Not possible (c) V is always positive. (i) cm V _______________________________________ (ii) V _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ F/cm 10.9 where V (a) n Then poly gate on p-type: V V or (b) p V Now poly gate on p-type: (c) Al gate on p-type: V V V V _______________________________________ or 10.11 We have V or F/cm So now cm C/cm or C/cm cm F/cm _______________________________________ 10.10 V cm (a) n poly gate on n-type: V V C/cm (b) p poly gate on n-type: V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V F/cm (c) Al gate on n-type: V V C/cm _______________________________________ By trial and error, let cm . Now 10.12 V V The surface potential is V We have cm V Now C/cm V We obtain Then or cm Then Then V V _______________________________________ or 10.14 C/cm We also find F/cm or F/cm Then or V _______________________________________ 10.13 C/cm By trial and error, let Now V cm cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) V C/cm V Then F/cm Now Then V, which is within the specified value. _______________________________________ 10.15 We have V (b) F/cm C/cm By trial and error, let Now V cm cm V C/cm cm Now C/cm V Then or V _______________________________________ 10.17 (a) We have n-type material under the gate, so V Then V V which meets the specification. _______________________________________ where 10.16 Then V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or or cm m (b) cm m Now or For an C/cm polysilicon gate, Then or V Now or C/cm or We have C/cm We now find V V _______________________________________ 10.19 Plot _______________________________________ or V _______________________________________ 10.20 Plot _______________________________________ 10.21 Plot _______________________________________ 10.18 (b) 10.22 Plot _______________________________________ where V and V 10.23 (a) For Hz (low freq), Then or V (c) For We find F/cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ F/cm F/cm Now F/cm V Now cm V Then F/cm (b) (inv) MHz (high freq), cm F/cm F/cm (unchanged) F/cm (unchanged) F/cm (unchanged) (inv) Then F/cm F/cm (c) V (inv) (b) F/cm MHz (high freq), Now F/cm (unchanged) F/cm (unchanged) F/cm (unchanged) (inv) C/cm V _______________________________________ 10.24 (a) (c) F/cm V Now Hz (low freq), C/cm Then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ We have V _______________________________________ Now 10.25 The amount of fixed oxide charge at x is C/cm By lever action, the effect of this oxide charge on the flatband voltage is or If we add the effect at each point, we must integrate so that or V _______________________________________ (c) We find 10.26 (a) We have Then or Now or which becomes Then or V (b) or V _______________________________________ 10.27 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Sketch Since , then _______________________________________ The potential is 10.28 Sketch _______________________________________ For zero bias, we can write where 10.29 (b) are the voltage drops across the n-region, the oxide, and the p-region, respectively. For the oxide: For the n-region: or V (c) Apply V, For V Arbitrarily, set at , then so that V, n-side: At , which is the voltage drop across the n-region. Because of symmetry, . Then for zero bias, we have at , then so which can be written as for In the oxide, , so or constant. From the boundary conditions, in the oxide Solving for , we obtain In the p-region, at , then If we apply a voltage , so We find At So that , , then replace by Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ which yields which yields cm F/cm or Now pF _______________________________________ 10.31 (a) Point 1: Inversion 2: Threshold 3: Depletion 4: Flat-band 5: Accumulation _______________________________________ or V We also find 10.32 We have or V _______________________________________ Now let , so 10.30 (a) n-type (b) We have F/cm For a p-type substrate, negative value, so we can write Also is a or cm nm (c) Using the definition of threshold voltage we have or , At saturation which yields C/cm (d) cm which then makes equal to zero at the drain terminal. _______________________________________ 10.33 (a) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.35 mA (a) (b) mA (c) Same as (b), mA (b) mA (d) (c) mA mA _______________________________________ _______________________________________ 10.34 10.36 (a) Assume biased in saturation region (a) mA V Note: V V So the transistor is biased in the saturation region. (b) mA (b) mA (c) or (c) mA _______________________________________ mA (d) Same as (c), mA _______________________________________ 10.37 F/cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ mA V, A/V =1.111 mA/V (a) V , V, V, mA V, V mA V, V, mA V, V mA V, V, (c) mA V, mA V for V V, mA (c) V mA V for V, mA V mA V, mA V mA V, mA _______________________________________ mA V, mA _______________________________________ 10.39 (a) From Problem 10.37, For V, , 10.38 mA/V V mA V, V mA F/cm V, V mA _______________________________________ A/V =0.961 mA/V (a) , V, V 10.40 Sketch _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 10.41 Sketch _______________________________________ 10.42 We have mA/V (b) so that mA Since , the transistor is always biased in the saturation region. Then where, from Problem 10.37, mA/V and V Then (mA) 0 0 1 0.336 2 2.67 3 7.22 4 14.0 5 23.0 _______________________________________ 10.43 From Problem 10.38, mA/V (c) mA _______________________________________ 10.45 We find that Now V where or F/cm We are given . From the graph, for V, we have For For For V, V, , then V, or mA/V _______________________________________ or which yields cm /V-s _______________________________________ 10.46 (a) 10.44 (a) or V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) so _______________________________________ which yields A/V (c) V so 10.48 From Problem 10.37, mA/V (a) or so A (d) mA/V (b) so mA/V _______________________________________ or A _______________________________________ 10.47 10.49 From Problem 10.38, mA/V (a) (a) F/cm (i) or mA/V (b) A/V or or mA/V _______________________________________ A/V 10.50 (ii) (a) Now F/cm (b) (i) A/V or (ii) Then A/V V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) V V V (i) (iv) cm V, V C/cm V _______________________________________ 10.51 V V A/V or or V mA/V For , V Now For V _______________________________________ V (c) (i) For , V 10.52 (a) (ii) V, F/cm V V (iii) V V, (b) V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or V _______________________________________ or 10.53 (a) which yields poly-to-p-type V V V _______________________________________ 10.54 Plot _______________________________________ also 10.55 (a) or cm Now or or C/cm mS Also Now or which yields F/cm We find C/cm or k Then (b) For Then V, mS mS or or V (b) For NMOS, apply positive direction, so for V. So and shifts in a , we want which is a 12% reduction. _______________________________________ 10.56 (a) The ideal cutoff frequency for no overlap Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 10 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ capacitance is, or GHz _______________________________________ or GHz (b) Now 10.57 (a) For the ideal case where or We find GHz (b) With overlap capacitance (using the values from Problem 10.56), We find or F Also or S We have or S or F Then Then or F Now GHz _______________________________________ or F We now find or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 11 Exercise Solutions Ex 11.4 Ex 11.1 V or cm From Example 11.1, we have m Then F/cm m _______________________________________ Ex 11.2 From Figure 11.10, cm /V-s _______________________________________ cm or Ex 11.3 m _______________________________________ F/cm Ex 11.5 V V cm cm m m Also V _______________________________________ or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Ex 11.7 (a) cm cm V Then V _______________________________________ F/cm Ex 11.6 We find Then V V (b) cm cm cm m F/cm Then F/cm The initial threshold voltage is V (c) Threshold voltage shift decreases when the oxide thickness decreases. _______________________________________ V Now Negative Now Test Your Understanding Solutions V implant donor ions or cm _______________________________________ TYU 11.1 or Then V or mV _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU 11.2 (a) A or A (b) A or A _______________________________________ TYU 11.3 m m cm V _______________________________________ TYU 11.4 We have from Example 11.6, V, F/cm (a) cm (b) cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 11 11.1 (a) 11.3 V We find that For V, cm/V A For V, For V, A (a) A Then the total current is: V cm For V, A For V, mA For V, mA (b) Power: Then For V, W For V, mW For V, mW _______________________________________ m (b) V cm m (c) V cm m (d) V 11.2 cm m _______________________________________ 11.4 V We find that (a) V (b) V (c) V _______________________________________ cm/V V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) cm m cm/V Now (a) m (b) (i) cm m cm Now m (ii) m _______________________________________ 11.5 cm m (iii) cm F/cm m (b) m _______________________________________ 11.6 V Now V We find cm/V V (a) Ideal, cm mA (i) V C/cm So cm V V m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 11.7 (a) (i) mA (ii) cm m mA A (ii) A (iii) mA (b) M k (b) (i) k (c) V mA (ii) (i) mA (iii) cm m k _______________________________________ 11.8 Plot _______________________________________ mA 11.9 (a) Assume (ii) cm m V. Then We find ( m) (V/cm) mA (b) Assume k _______________________________________ cm /V-s, we have Then For m, For m, cm/s cm/s For m, cm/s _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) 11.10 A/V mA/V k (b) mA k _______________________________________ 11.11 (a) Now or (mA) and (mA) (b) We find Let Where V/cm cm/V V Let We find V Now cm/V or Then cm cm /V-s and m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ mA -- -- (i) For V, 0.370 cm/s mA (c) The slope of the variable mobility curve is not constant, but is continually decreasing. _______________________________________ 11.12 Plot _______________________________________ (ii) For V, cm/s 11.13 mA (a) (iii) For V, cm/s F/cm mA V, mA (iv) For A/V For mA/V V, V (i) mA (c) For part (a), V For part (b), V _______________________________________ 11.14 Plot _______________________________________ 11.15 (a) Non-saturation region (ii) mA We have (iii) mA and , (iv) also , mA cm/s So (b) A Then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ In the saturation region, V Then (b) cm _______________________________________ 11.16 or _______________________________________ 11.17 (a) V _______________________________________ (i) 11.19 mA F/cm (ii) Scaled device: V V mA/V m m cm Then mA (b) (i) V mW (ii) mW _______________________________________ 11.18 V _______________________________________ 11.20 F/cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ We can then write F/cm V Similarly from (2), we will have cm where The average bulk charge in the trapezoid (per unit area) is or We can write m _______________________________________ which is 11.21 We have Now, and from the geometry (1) replaces in the threshold equation. Then and (2) From (1) or so that which can be written as Then substituting, we obtain or Define Note that if , then and the expression for reduces to that given in the text. _______________________________________ 11.22 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ We have , so Equation (11.27) F/cm becomes V cm or Then Equation (11.28) is Then change in the threshold voltage is or which becomes V _______________________________________ 11.27 F/cm _______________________________________ V 11.23 Plot _______________________________________ 11.24 Plot _______________________________________ cm 11.25 In this case, So m _______________________________________ or _______________________________________ 11.28 Plot _______________________________________ 11.26 11.29 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Assume that is a constant, then Let V and . Now when or _______________________________________ we can write this as 11.30 (a) (i) Now V (ii) V (b) (i) V (ii) _______________________________________ V _______________________________________ 11.33 One Debye length is 11.31 (a) cm or nm or nm cm Six Debye lengths is then cm m From Example 11.5, we have m, which is the zero-biased source-substrate junction width. At near punch-through, we will have (b) cm or _______________________________________ 11.32 Snapback breakdown means and , where where is the reverse-biased drainsubstrate junction width. Now or m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then, at near punch-through we have We have for V, or or cm m Then which yields V or m _______________________________________ From Example 11.5, we have V, so V which is the near punch-through voltage. The ideal punch-through voltage was V _______________________________________ 11.35 With a source-to-substrate voltage of 2 volts, or 11.34 cm V The zero-biased source-substrate junction width is given by We have problem. Now m m from the previous or or cm cm m m Then The Debye length is or m _______________________________________ 11.36 or F/cm cm so that cm Now m Implant acceptor ions for a positive threshold voltage shift. Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) For a positive threshold voltage shift, add acceptor ions. cm _______________________________________ V Then 11.37 F/cm cm _______________________________________ Implant donor ions for a negative threshold voltage shift. 11.39 (a) V F/cm cm _______________________________________ 11.38 (a) V F/cm V V V cm V C/cm cm C/cm V (b) For a negative threshold voltage shift, add donor ions. V V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ cm Then (c) Add acceptor ions. cm V _______________________________________ Then 11.40 (a) cm V _______________________________________ F/cm 11.41 The total space charge width is greater than , so from Chapter 10 cm Now V C/cm and F/cm Then V or (b) For a positive threshold voltage shift, add acceptor ions. V Then Then (V) (V) 1 0.0443 3 0.0987 5 0.1385 _______________________________________ 11.42 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) V and or V _______________________________________ cm poly on n-type We have V 11.44 The areal density of generated holes is cm . Now C/cm Now F/cm Then or V (Enhancement PMOS) (b) For , shift threshold voltage in positive direction, so implant acceptor ions. where x is the fraction of holes that may be trapped. For V we find _______________________________________ so or cm _______________________________________ 11.43 The areal density of generated holes is cm The equivalent surface charge trapped is cm Then 11.45 We have the areal density of generated holes as where is the generation rate and is the radiation dose. The equivalent charge trapped Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 11 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ is where x is the fraction of generated holes trapped. Then or _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 12 Exercise Solutions Ex 12.1 Ex 12.5 From Example 12.5, Then m. _______________________________________ Ex 12.2 m _______________________________________ Ex 12.6 _______________________________________ Ex 12.3 _______________________________________ or Ex 12.4 V _______________________________________ Ex 12.7 (a) cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ cm m Neglecting the B-E space charge width, we find the neutral base width to be: V, m V, m _______________________________________ Then Ex 12.9 Neglecting bandgap narrowing, (b) cm From Figure 12.26, cm eV for , _______________________________________ cm _______________________________________ Ex 12.8 The space charge width extending into the base region is Ex 12.10 (a) V (b) From Figure 7.15, V _______________________________________ Now Ex 12.11 (a) From Figure 7.15, V For (b) V, V V _______________________________________ cm For V, m Ex 12.12 We find Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ _______________________________________ V _______________________________________ Test Your Understanding Solutions Ex 12.13 TYU 12.1 We have Or cm F pF _______________________________________ (a) Ex 12.14 cm (b) Using Equation (12.15a), we find s At ps , Then s ps So s s ps ps or Now cm ps Then (c) For an ideal linear function cm Hz MHz Also Ratio _______________________________________ Hz MHz TYU 12.2 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ cm (a) or cm which yields or cm _______________________________________ (b) We have TYU 12.5 and Then, using Equation (12.21a), cm _______________________________________ TYU 12.3 or Then Now _______________________________________ Then TYU 12.6 or Then so Now _______________________________________ TYU 12.4 We have _______________________________________ cm cm Now TYU 12.7 (a) We find Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ ( -cm) mA Then Then mA or A _______________________________________ (b) TYU 12.9 V or (a) mV For For m, m, So (b) So that Then (c) _______________________________________ _______________________________________ TYU 12.8 or Now mA and Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU 12.10 cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 12 12.1 Sketch _______________________________________ (b) 12.2 Sketch _______________________________________ 12.3 (a) Then A (b) V _______________________________________ (i) A A 12.5 (ii) (a) A mA (b) (i) For A, (iii) A A mA A _______________________________________ (ii) For mA, 12.4 mA (a) A mA (iii) For cm mA, mA Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ mA mA (c) (i) For A A, A (c) A A (ii) For A mA, _______________________________________ mA A mA 12.7 (c) For mA, or mA We have or V _______________________________________ (iii) For mA, mA A 12.8 (a) For (i) V, , mA V mA (ii) For , mA _______________________________________ (b) For (i) V, , mA (ii) For V , mA _______________________________________ 12.6 (a) 12.9 mA (a) cm (b) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) cm cm cm (b) cm cm _______________________________________ 12.11 cm (a) _______________________________________ cm Now 12.10 cm (a) cm Then V cm (b) cm cm + cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ 12.13 In the base of the transistor, we have 12.12 We have or where At The general solution to the differential equation is of the form , From the boundary conditions, we have At , Also From the first boundary condition, we can write Taking the ratio Substituting into the second boundary condition, we find Solving for B, we find (a) For Ratio (b) For Ratio (c) For Ratio We then find Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ 12.14 In the base of the pnp transistor, we have or _______________________________________ 12.15 For the idealized straight line approximation, the total minority carrier concentration is given by where The general solution is of the form The excess carrier concentration is so for the idealized case, we can write From the boundary conditions, we can write At , we have Also For the actual case, we have From the first boundary condition equation, we find Substituting into the second boundary equation, we obtain (a) For and and then we obtain Then Substituting the expressions for A and B into the general solution and collecting terms, we obtain , we have Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ If we assume that , then we find that the ratio is V (b) Using the linear approximation, (b) For , we have Since , Then and Then A/cm (c) Using Equation (12.15a), Again assume that . Then the Now ratio becomes _______________________________________ 12.16 (a) For , , cm Then cm A/cm For , Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ The coefficient D can be written as A/cm (d) For part (b), The excess electron concentration is then given by For part (c), _______________________________________ 12.17 (a) For an npn transistor biased in saturation, the excess minority carrier electron concentration in the base is found from (b) The electron diffusion current density is or or where The general solution is of the form (c) The total excess charge in the base region is If , then also , so that which can be written as which yields The boundary conditions are and _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.18 We find (a) Using the linear approximation, we can write cm Then cm cm _______________________________________ Then 12.19 (b) cm and cm At , V (b) V (c) We have or At cm cm , or cm (c) From the B-C space charge region, (d) In the collector, V Then Now or cm From the B-E space charge region Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (iii) V Then (iv) (v) (b) For or cm Then Now or mA m _______________________________________ 12.20 Low-injection limit is reached when or cm mA We have A cm Also or mA A _______________________________________ 12.22 (a) Using Equation (12.37) or V _______________________________________ 12.21 (a) Now (i) cm (ii) cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ We find cm and cm Then Then or A/cm A mA We also have (b) Now cm and cm Then Then A (c) A _______________________________________ 12.23 (a) We have We find that or A/cm We can find Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.24 (a) We have or or A/cm The recombination current density is (i) Now or A/cm or finally (b) Using the calculated current densities, we find or We also find (ii) We have Then or Also or Then or or Now (b) (i) We find or (ii) _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (ii) We find (d) Device C has the largest . The emitter injection efficiency, base transport, and recombination factors all increase. _______________________________________ 12.25 (a) We have or or finally (i) Then (c) Neglect any change in space charge width. (i) or Now so Then finally (ii) Now Then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or or finally Recombination factor increases _______________________________________ (b) We have 12.26 (b) cm (i) Then (ii) cm Now (c) Neglect any change in space charge width. or (i) A/cm Assuming a long collector where cm and Now cm so Then or or Recombination factor decreases (ii) We have A/cm The collector current is Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.28 We have or mA The emitter current is Now cm or mA _______________________________________ and 12.27 (a) cm Then and or 0.01 0.10 1.0 10.0 (b) For have 0.99995 0.995 0.648 0.0000908 , A/cm 19,999 199 1.84 , Now , we and or (a) 0.01 0.10 1.0 10.0 0.990 0.909 0.50 0.0909 99 9.99 1.0 0.10 (c) For , the value of is unreasonably large, which means that the base transport factor is not the limiting factor. For , the value of is very small, which means that the base transport factor will probably be the limiting factor. If , the emitter injection efficiency is probably not the limiting factor. If, however, , then the current gain is small and the emitter injection efficiency is probably the limiting factor. _______________________________________ and (b) Now 0.20 0.40 0.60 0.7535 0.99316 0.999855 3.06 145 6,902 (c) If V, the recombination factor is likely the limiting factor in the current gain. _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 12.29 For Let K and V. m cm which yields Then or m (b) For Now For cm _______________________________________ K and A/cm , eV, or cm Then 12.30 (a) We have We find A/cm or cm A/cm Finally and cm Then or _______________________________________ or 12.31 Plot _______________________________________ We have 12.32 Plot _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ 12.33 Plot 12.37 _______________________________________ 12.34 Plot _______________________________________ 12.35 (a) (i) k Now (ii) (k (iii) ) mA V (i) For V, m (ii) For V, m (iii) For V, m Neglecting the B-E space charge width, (i) For V, m (ii) For V, m (iii) For V, m (b) (i) k Now (ii) (k ) where (iii) mA cm _______________________________________ so 12.36 A/cm mA A (i)For V, A/cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (ii)For V, (iii)For V, A/cm A/cm or (b) For V _______________________________________ V, For (a) For For m V, m m V, m 12.38 We find Then cm A/cm For and V, m and A/cm We can write or cm We have where A/cm /V Then or A/cm which yields V Neglecting the space charge width at the B-E junction, we have (b) For For Now m V, m Then or V Also A/cm For V, m and A/cm Now Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ A/cm /V We can write V For For Then or which yields V (c) For For V, m V, V, m m m (b) m Then We find A/cm For cm V, m Then and A/cm Now A A/cm /V We can write or which yields V _______________________________________ 12.39 (a) For For V, A mA A mA V, Then mA (c) V (d) k _______________________________________ Now 12.40 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Let , , Then the emitter injection efficiency is which yields where cm (b) Neglecting bandgap narrowing, we would have . For no bandgap narrowing, With bandgap narrowing, . which yields Then cm _______________________________________ 12.42 (a) (a) No bandgap narrowing, so (i) . We find (b) Taking into account bandgap narrowing, we find k (ii) (meV) V _______________________________________ 12.41 (a) We have (iii) For For Then , , cm , we have , we obtain meV Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) Then where (b) and is a constant. In thermal equilibrium (i) so that k (ii) V which becomes (iii) or Then _______________________________________ which is a constant. 12.43 (b) The electric field is in the negative x-direction which will aid the flow of minority carrier electrons across the base. (c) Then V Now Assuming no recombination in the base, will be a constant across the base. Then where The homogeneous solution to the differential equation is found from where cm m _______________________________________ 12.44 The solution is of the form The particular solution is found from Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ where The particular solution is then We find V and V The total solution is then so that and Then or cm m _______________________________________ _______________________________________ 12.45 (a) For cm 12.47 (a) For cm , , V V (b) (b) V (c) For cm , V _______________________________________ 12.46 We want Then V V _______________________________________ 12.48 (a) which yields V For this breakdown voltage, we need cm The depletion width into the collector at this voltage is V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) From Chapter 7, We find mA A (b) For V, we find mA A (c) For V, we find mA A _______________________________________ V/cm 12.51 For an npn transistor biased in the active mode, we have , so that . Now _______________________________________ Then we have 12.49 or cm m _______________________________________ _______________________________________ 12.50 We have 12.52 We can write We can write or (a) For Substituting, we find V, we find From the definition of currents, we have for the case of . Then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ When a C-E voltage is applied, then the B-C becomes reverse biased, so . For V Then Finally, we find A For A V _______________________________________ 12.53 A A (a) For V A For V, (c) For V, For For V A V For A For A For mA V A V A A mA _______________________________________ 12.54 V A (b) For V, A, V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 12 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ A, V A, V A, V _______________________________________ 12.55 (a) (i) k s ps (ii) s ps 12.57 We have (iii) s ps We are given ps and We find (iv) s ps ps s (b) ps or ps (c) Also Hz s GHz or (d) ps Hz MHz _______________________________________ Then ps We obtain 12.56 s Hz or We have GHz _______________________________________ so that s Then or Hz MHz _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 13 Exercise Solutions Ex 13.1 V V V Now mA cm m _______________________________________ Ex 13.2 Or A _______________________________________ Ex 13.4 From Ex 13.3, V, V, mA Then V V V _______________________________________ mA/V _______________________________________ Ex 13.3 Ex 13.5 V A mA V V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ V V _______________________________________ mA Ex 13.6 mA V V V Now Then k m _______________________________________ _______________________________________ Ex 13.9 Ex 13.7 A/V Hz GHz _______________________________________ mA/V From exercise problem Ex 13.5, V Then Test Your Understanding Solutions TYU 13.1 mA _______________________________________ V Ex 13.8 cm m which yields cm m We want cm m m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ So that TYU 13.3 V Hz GHz _______________________________________ Then V TYU 13.4 _______________________________________ TYU 13.2 Hz GHz _______________________________________ or A mA Now or V Also V Now or mA _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 13 13.1 Sketch _______________________________________ cm 13.2 Sketch _______________________________________ m 13.3 (a) (i) V (ii) V (c) (i) V V (b) (ii) (i) V _______________________________________ cm m m 13.4 (a) (ii) (i) V cm m m (iii) (ii) V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V _______________________________________ (b) 13.5 (i) (a) cm m cm m (b) (ii) V cm V m m (c) m (iii) cm m V (d) (c) V _______________________________________ (i) V 13.6 (ii) (a) V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V or Now cm (b) cm V m V (c) (b) m V (c) (i) V (ii) V _______________________________________ V 13.8 (d) (a) V V V _______________________________________ 13.7 (a) cm V (b) (c) V m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ (i) V (ii) 13.11 (a) V _______________________________________ 13.9 (a) Now or mA (b) V We find V or V Also V cm (b) (i) m Now V (ii) V _______________________________________ or 13.10 We have V (a) Then (i) For V (ii) For , V V, V V (iii) For V, V (iv) For cm V m (c) (b) (i) V (ii) V V, Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ S (b) We have (i) For (ii) For , mA V, mA (iii) For V, or V We find mA (iv) For V, mA _______________________________________ or V Then V 13.12 We then obtain For where , V For V, V (c) or S mS Then (mS) 0 0.453 0.523 -0.264 0.590 0.371 -0.528 0.726 0.237 -0.792 0.863 0.114 -1.056 1.0 0 _______________________________________ 13.13 n-channel JFET - GaAs (a) where or mA Then or (mA) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ For , mA mS and For V, _______________________________________ mA 13.16 _______________________________________ n-channel MESFET - GaAs (a) 13.14 We have mA, V, V The maximum transconductance occurs when . Then or V Now where or For mS m, we have or V so that or mS/cm = 1.31 mS/mm _______________________________________ Then or 13.15 The maximum transconductance occurs for , so we have V (b )If for an n-channel device, the device is a depletion mode MESFET. _______________________________________ (a) which can be written as 13.17 n-channel MESFET - GaAs (a) We want V Then We found mS, V V, V Then so or which can be written as mS This is for a channel length of m. (b) If the channel length is reduced to m, then or Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V (c) By trial and error, cm (b) At K (i) Then cm m cm m Also (ii) Then cm m m which becomes V _______________________________________ (iii) 13.18 cm m (a) _______________________________________ cm m 13.19 (a) (b) V We find We find V V Then V V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V or Then By trial and error cm _______________________________________ (b) V V 13.21 n-channel MESFET - silicon (a) For a gold contact, We find V. V or and V With Then and V V, we find so that cm m _______________________________________ or cm Now 13.20 We want Now V, so or We obtain V and (b) or which yields m Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V _______________________________________ A/V 13.22 mA/V (b) (a) (i) V mA (i) A (ii) V mA (ii) V (c) (i) V (ii) V _______________________________________ (iii) 13.24 V (a) mA/V (b) (i) V cm m (ii) (b) V (i) (iii) mA V _______________________________________ A (ii) mA _______________________________________ 13.23 (a) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 13.25 Plot _______________________________________ 13.26 Plot _______________________________________ 13.28 We have that Assuming that we are in the saturation region, then and . We can write 13.27 If V , then We have that V (a) V which can be written as If we write cm then by comparing equations, we have Now The parameter Define cm is not independent of m (b) and consider the function V which is directly proportional to cm Then cm m _______________________________________ 1.5 1.75 2.0 2.25 2.50 2.75 3.0 . Then 0.222 0.245 0.250 0.247 0.240 0.231 0.222 . Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ So that is nearly a constant. _______________________________________ Also 13.29 (a) Saturation occurs when As a first approximation, let V/cm. Then V (b) We have that or V Then and or or For mA _______________________________________ V , we obtain 13.30 (a) If when or cm (c) We then find V We find We have obtain or (d) For m m, then saturation will occur V and for , we mA , we have or cm m Then Now or mA If velocity saturation did not occur, then from the previous problem, we would have mA or mA (b) If velocity saturation occurs, then the relation does not apply. Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ _______________________________________ or 13.31 (a) V cm/s Then Then V Let Now s V or ps (b) Assume Then cm/s s or ps _______________________________________ 13.32 (a) cm/s Then or (a) For (b) For (c) For , V, V, m m m The depletion region volume at the drain is s or ps (b) For cm/s s or or ps _______________________________________ (a) For , (b) For V, cm 13.33 The reverse-bias current is dominated by the generation current. We have (c) For V, cm cm The generation current at the drain is We find or V Also or (a) For (b) For , V, pA pA Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) For V, pA _______________________________________ so 13.34 (a) The ideal transconductance for is or cm m _______________________________________ where 13.35 Considering the capacitance charging time, we have or where mS We find or or We must use V F , so we obtain We have Hz V We can also write so that V Then so s or mS (b) With a source resistance For The channel transit time is s The total time constant is s Taking into account the channel transit time and the capacitance charging time, we find we obtain or (c) Hz GHz _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 13.36 (a) For constant mobility (a) cm m (b) Hz GHz (b) For saturation velocity model Hz GHz _______________________________________ cm m _______________________________________ 13.39 (a) where 13.37 or V Then or V (a) (b) Hz GHz (b) For , we have Hz GHz _______________________________________ 13.38 or or cm _______________________________________ 13.40 (a) We have We find Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 13 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or S/cm (b) At mS/mm , we obtain or A/cm mA/mm _______________________________________ 13.41 We want V, so or V We have or We then obtain cm _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 14 Exercise Solutions Ex 14.1 For silicon, m cm m Let (a) For m Ex 14.3 (a) We find cm cm cm m, cm Now (b) For m, A/cm _______________________________________ Ex 14.2 For Now m in silicon, We find cm eV (a) V (b) W/cm V cm s So (b) W/cm _______________________________________ Ex 14.4 The photocurrent is given by cm s _______________________________________ A A _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Ex 14.5 We find Ex 14.7 (a) For , eV m (b) For , eV m V _______________________________________ Ex 14.8 For GaAs, For GaP, Then for GaAs P , Then cm Then _______________________________________ From Example 14.5, m m Ex 14.9 For GaAs P , (See Exercise Ex 14.8) Then A/cm Now _______________________________________ A/cm Then Test Your Understanding Solutions _______________________________________ Ex 14.6 (a) For TYU 14.1 (a) For (i) cm m, cm m W/cm (ii) A/cm (b) For m mA/cm cm (b) For (i) A/cm mA/cm _______________________________________ W/cm m, m W/cm cm Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ (ii) m Then W/cm _______________________________________ TYU 14.2 From Example 14.3, Now A/cm or So A W _______________________________________ We have TYU 14.5 Use results from Example 14.5; We have V Then which yields or A/cm _______________________________________ TYU 14.3 From Example 14.3, We have or Now m A/cm A/cm A/cm mA Then A/cm Now We have and or V _______________________________________ which yields cm For TYU 14.4 m, s cm eV So or By trial and error, V We have W/cm _______________________________________ A A Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 14 14.1 m (a) Si: m (b) Ge: m (c) GaAs: m (b) m (i) From Figure 14.4, (d) InP: _______________________________________ 14.2 (a) For (ii) nm, Fraction absorbed eV For cm m _______________________________________ nm, eV (b) For eV, 14.4 m For eV, For m For For silicon: Then for eV, eV, m cm W/cm , we obtain m _______________________________________ or 14.3 (a) cm s The excess concentration is m (i) From Figure 14.4, cm or cm _______________________________________ (ii) 14.5 (a) Fraction absorbed cm s Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ GaAs: For eV, For m cm, we have 50% absorbed or 50% transmitted, then m From Figure 14.4, cm We can write or cm This value corresponds to m, eV _______________________________________ W/cm 14.8 The ambipolar transport equation for minority carrier holes in steady state is or (b) where The photon flux in the semiconductor is cm m and the generation rate is _______________________________________ so the differential equation becomes 14.6 The general solution is of the form m From Figure 14.4, cm (a) As cm m At , , we have (b) cm m _______________________________________ 14.7 so we can write so that . Then Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ and Then we have so we can write Solving for A, we find The solution can now be written as Solving for B, we obtain _______________________________________ 14.9 We have The solution is then where B was just given. _______________________________________ or 14.10 cm where The general solution can be written in the form cm Now For at At , means . Then A/cm Now A (a) We find and Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ V A mA mW (d) cm Then _______________________________________ A mA 14.12 From Problem 14.10, (a) (b) A (i) V (c) V _______________________________________ 14.11 From Problem 14.10, A (ii) (a) By trial and error, Now V V (b) V A mA Then (c) mW (b) (i) By trial and error, Now V V Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (ii) _______________________________________ By trial and error, Now V 14.14 (a) A We have or A mA Then mW (c) which becomes _______________________________________ A/cm or A We have or 14.13 We see that when We find where (V) 0 0.1 0.2 0.3 0.4 0.45 0.50 0.55 0.57 0.59 which becomes or Then (cm ) (A/cm ) (V) , V. (mA) 50 50 50 50 49.98 49.84 48.89 42.36 33.46 14.19 (b) The voltage at the maximum power point is found from Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ By trial and error, By trial and error, V At this point, we find mA so the maximum power is V Then A mW _______________________________________ or mW (c) We have 14.16 (a) or V _______________________________________ 14.15 (b) (a) V By trial and error, Then (b) V A By trial and error, mA mW V (c) cells (d) Now V A A Then mA mW (e) Then A So (c) _______________________________________ (d) Now 14.17 Let correspond to the edge of the space charge region in the p-type material. Then in the p-region Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ cm or or m and for Then where eV, cm . cm Then we have or m _______________________________________ The general solution is of the form 14.19 (a) As , so that cm . Then A We also have , mA (b) cm (c) which yields ( -cm) (d) We then obtain where is the incident flux at . _______________________________________ A mA (e) 14.18 For 90% absorption, we have Then _______________________________________ or For Then eV , cm 14.20 n-type, so holes are the minority carrier (a) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or cm (b) which becomes Now or ( -cm) (c) or A _______________________________________ 14.22 or mA (d) V cm or cm _______________________________________ 14.21 The electron-hole generation rate is and the excess carrier concentration is Now Then and (a) cm The photocurrent is now found from A A (b) In n-region, cm In p-region, cm Then (c) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ A mA _______________________________________ 14.23 In the n-region under steady state and for , we have since So and are in opposite directions. or Finally where and where is positive in the negative x direction. The homogenerous solution is found from _______________________________________ 14.24 (a) The general solution is found to be The particular solution is found from Diode A: A/cm Diode B: which yields A/cm Diode C: The total solution is the sum of the homogeneous and particular solutions, so we have A/cm (b) Diode A: One boundary condition is that remains finite as which means that . Then at , , so that . We find that Diode B: The solution is then written as Diode C: A/cm A/cm A/cm _______________________________________ The diffusion current density is found as But 14.25 Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or (a) m _______________________________________ cm s Then 14.28 For At system, a direct bandgap for , we have , eV, so for the direct bandgap (b) eV which yields m _______________________________________ 14.29 (a) From Figure 14.24, A/cm eV mA/cm _______________________________________ m (b) From Figure 14.24, eV m 14.26 (a) _______________________________________ 14.30 A/cm eV (b) From Figure 14.23, _______________________________________ 14.31 eV A/cm _______________________________________ 14.27 The minimum which gives occurs when cm From Figure 14.24, _______________________________________ 14.32 (a) For GaAs, The critical angle is m and for air, . We want which can be written as The fraction of photons that will not experience total internal reflection is Then (b)Fresnel loss: cm . Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ The fraction of photons emitted is then _______________________________________ _______________________________________ 14.33 We can write the external quantum efficiency as 14.34 For an optical cavity, we have where and where is the reflection coefficient (Fresnel loss), and the factor is the fraction of photons that do not experience total internal reflection. We have If changes slightly, then N changes slightly also. We can write so that If we define which reduces to Now consider the solid angle from the source point. The surface area described by the solid angle is . The factor is given by From the geometry, we have Then the area is Now From a trig identity, we have Then The external quantum efficiency is now or Rearranging terms, we find We can approximate , then we have , then Then which yields _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 14 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 14.35 For GaAs: eV m Then cm or m _______________________________________ Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 15 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Chapter 15 Exercise Solutions Ex 15.1 W (a) V _______________________________________ A Test Your Understanding Solutions A TYU 15.1 (a) Collector Region, (b) V Neglecting , A A (c) or m (b) Base Region, V W A or m _______________________________________ _______________________________________ So Ex 15.2 (a) V A mA Maximum power at the center of the load line W _______________________________________ W (b) TYU 15.2 V A Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 15 15.1 See diagrams in Figure 8.29 _______________________________________ (c) Hz GHz _______________________________________ 15.6 15.2 Hz _______________________________________ GHz _______________________________________ 15.3 15.7 (a) cm Hz MHz _______________________________________ 15.4 (a) (i) cm (i) cm m (ii) s V (iii) Hz (ii) Neglecting any recombination in the base GHz (b) (i) cm m (ii) s (iii) Hz A GHz _______________________________________ 15.5 cm (i) (a) (b) (b) V/cm V cm/s (ii) Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ cm m (Minimum base width) Depletion width into the collector A _______________________________________ or cm m (Minimum collector width) _______________________________________ 15.8 (a) From Figure 7.15, 15.10 V (a) (b) (i) (ii) V V (b) V (c) From Figure 7.15, V _______________________________________ (i) V 15.9 From the junction breakdown curve, for V, we need the collector doping (ii) V concentration to be cm Depletion width into the base (neglect _______________________________________ . ). 15.11 (a) We have so or which yields or (b) We have Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or (b) A _______________________________________ so which yields A _______________________________________ 15.14 If V, then A The power 15.12 Now, to find the maximum power point (b) which yields A A So or W So maximum is V _______________________________________ V 15.15 Now (c) Power dissipated in the transistor V We have V so we can write (d) Same as part (b) _______________________________________ For Then C, 15.13 which yields . V (a) The power is A W Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ We then have (C) ( ) (V) (W) 25 2.0 3.92 7.69 50 2.33 4.56 8.91 75 2.67 5.19 10.1 100 3.0 5.83 11.3 _______________________________________ cm m = channel length 15.16 (a) We have, for three devices in parallel, cm = drift region width (b) Assume cm V or V Then, Now, , so that m A A A , so W W W (b) Now cm V m = channel length Then A, W A, W A, W _______________________________________ 15.17 (a) Let the n-drift region doping concentration be cm V For the base region, . cm m = drift region width _______________________________________ 15.18 (b) In the saturation region, For V, A, V W For For V, A, V W V, transistor biased in the nonsaturation region. Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Expanding, we find We obtain V, A W so transistor may For V, be damaged. _____________________________________ 15.19 (a) A which yields _______________________________________ 15.21 The reverse-biased p-well to substrate junction corresponds to the junction in an SCR. The photocurrent generated in this junction will be similar to the avalanche generated current in an SCR, which can trigger the device. _______________________________________ (b) Then Or V _______________________________________ 15.20 We have . Now and so which can be written as or 15.22 Case 1: Terminal 1(+), terminal 2(-), and negative: this triggering was discussed in the text. Case 2: Terminal 1(+), terminal 2(-), and positive: the gate current enters the P2 region directly so that J3 becomes forward biased. Electrons are injected from N2 and diffuse into N1, lowering the potential of N1. The junction J2 becomes more forward biased, and the increased current triggers the SCR so that P2N1P1N4 turns on. Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 15 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Case 3: Terminal 1(-), terminal 2(+), and positive: the gate current enters the P2 region directly so that the J3 junction becomes more forward biased. More electrons are injected from N2 into N1 so that J1 also becomes more forward biased. The increased current triggers the P1N1P2N2 device into its conducting state. Case 4: Terminal 1(-), terminal 2(+), and negative: in this case, the J4 junction becomes forward biased. Electrons are injected from N3 and diffuse into N1. The potential of N1 is lowered which increases the forward biased potential of J1. This increased current then triggers the P1N1P2N2 device into its conducting state. _______________________________________

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