NAME: PUID: : SOLUIONS: ECE 66 Exam : February 14, 13 Mark Lundstrom Purdue University his is a closed book exam. You may use a calculator and the formula sheet at the end of this exam. here are four equally weighted questions. o receive full credit, you must show your work (scratch paper is attached). he exam is designed to be taken in 6 minutes, but you may use the full, 75 minute class period. Be sure to fill in your name and Purdue student ID at the top of the page. DO NO open the exam until told to do so, and stop working immediately when time is called. 4 points possible, 1 per question 1) 1 point each part no partial credit a) points b) 4 points c) 4 points 3a) points 3b) 4 points 3c) 4 points 4) points for each part ECE- 66 1
Exam : ECE 66 1) Consider the E(k) plot shown below, and then answer the questions. 1a) Is this a direct or indirect bandgap semiconductor? Direct bandgap semiconductor (conduction band minimum and valence band maximum are at same location in k- space. 1b) What is the bandgap of this semiconductor?.7 ev 1c) Where is the effective mass for electrons the largest, point A or C? C (effective mass goes as 1/curvature. Curvature is lower at C than at A) 1d) Where is the effective mass for electrons negative, point A, B, or C? B (curvature is negative there) 1e) What is the velocity of an electron at point D positive, zero, or negative? Negative. (Velocity is proportional to the slope of E(k).) 1f) If we apply an electric field in the - x direction, which direction in k- space does the electron at point D move? (in the +kx direction or in the kx direction). +kx direction (Force = - q time electric field is positive. Force = d(hbar k)/dt), so kx increases with time. 1g) If we apply an electric field in the - x direction, which direction in real- space does the electron at point D move? (in the +x direction or in the x direction). x direction (Velocity is negative, so the electron moves in the x direction (later on, the electric field will turn it around and eventually it will move in the +x direction.) ECE- 66
Exam Solutions: ECE 66 1h) Compare the density- of- states in energy, D(E), at points A and C. Which one is larger? DOS is larger at point C (Because D(E) is proportional to mass to some power (depending on 1D, D, or 3D), so bigger mass means bigger DOS.) 1i) For common, cubic semiconductors, what is the shape of the constant energy surface near point A? (spherical or ellipsoidal) spherical 1j) For common, cubic semiconductors, what is the shape of the constant energy surface near point C? (spherical or ellipsoidal) ellipsoidal - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ) Graphene is a D semiconductor of great interest these days. For E >, it has a density of states per J- m that is ( ) = E D E π υ F. Answer the following questions. a) Write down an expression for the total energy of all the electrons in the graphene conduction band (i.e. for E > ). HIN: his should be an integral. It should begin at E = and include the graphene density- of- states. Solution: W = ( ) ED E E 1 f ( E)dE = E F ( 1+ e E ) de k B b) Work out the integral in a) assuming = K. W = ED( E) de = υ F E de = 3 υ F 3 ECE- 66 3
Exam Solutions: ECE 66 c) Work out the integral in a) but DO NO assume = K. W = F E de ( 1+ e E ) k B η = E k B de = k B dη ( ) η k B dη k W = B where η F 1+ e η η F F = k B W = F k B ( ) 3 η dη 1+ e η η F F j ( ) = η F 1 Γ( j +1) η j dη 1+ e η η F η dη = F 1+ e η η F ( η F )Γ 3 ( ) Γ( 3) =!= W = 4 ( k B ) 3 F υ ( η F ) F 3) InSb is a semiconductor with a very small bandgap of.18 ev. You may assume that the conduction and valence bands are spherical with m n * =.116m and m p * =.4m. Using these numbers, we find that the intrinsic carrier concentration is about 1.4 1 16 cm - 3 at room temperature. Answer the following questions about InSb. 3a) For intrinsic InSb, do you expect the intrinsic Fermi level to lie at E C + E V ( E C + E V ), or below ( E C + E V )? Explain your answer. ( ), above Conduction band effective mass is smaller than the valence band effective mass, so conduction band DOS is smaller than valence band DOS. o get the same density of electrons and holes n = p = n i, the intrinsic Fermi level must lie ( ). closer to the conduction band. Above E C + E V ECE- 66 4
Exam Solutions: ECE 66 3b) Derive an expression for the location of the intrinsic level. Assume non- degenerate carrier statistics. Solution: E n = N C F F E C 1/ k B = N F E V V 1/ k B For non- degenerate carrier statistics: N C exp E C k B = N exp V E V k B N C = exp E + E E C V F N V k B E C + E V = k B ln N C N V = E C + E V k B ln N C N V = E C + E V k B ln m * n * m p 3/ = E C + E V 3k B 4 ln m * n * m p Since m * n < m * p, EF lies above E + E C V. Putting in numbers and assuming = 3K, we find: =.9 +.7 ev = E C. So EF is too close to EC to use nondegenerate carrier statistics! But solving this problem with FD statistics produces a similar answer. 3c) Assume that InSb is doped n- type at 1 16 cm - 3 with shallow dopants that are fully ionized at room temperature. Compute the electron concentration. Begin with space charge neutrality: p n + N D + N A = his becomes: n i n n + N = which can be solved for: n = N D D + N D + n i 1/ n = 1 16 + 1 16 n =.89 1 16 cm -3 ( ) + ( 1.4 1 16 ) 1/ = 1 16 +.96 1 3 =.89 1 16 ECE- 66 5
4) Polycrystalline materials consist of crystalline grains that are randomly oriented within the material. he interfaces between grains (so- called grain boundaries) are often charged. Consider the energy band diagram below, which shows a region near a grain boundary. 4a) Sketch the electric field vs. position. HIN: It is easiest to do this for x > + and for x < - and not worry about what happens exactly at x =. 4b) Sketch the electron density vs. position. ECE- 66 6
4) continued Exam : ECE 66 4c) Sketch the electrostatic potential vs. position. 4d) Sketch the space charge density vs. position. 4e) Is this grain boundary charged positively or negatively? Explain your answer. Negatively charged. Since the charge is positive for x < and for x > the positive charges must be imaged on negative charge at x =. Electric fields point from + charge to charge, so from the electric field plot, we also deduce that the charge at x = must be negative. ECE- 66 7