EE 130 Intro to MS Junctions Week 6 Notes. What is the work function? Energy to excite electron from Fermi level to the vacuum level

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1 EE 13 Intro to S Junctions eek 6 Notes Problem 1 hat is the work function? Energy to ecite electron from Fermi level to the vacuum level Electron affinity of 4.5eV Electron affinity of Ge 4.eV orkfunction of Al 4.1eV n i of Ge cm -3 Bandgap of Ge.7437eV ielectric constant raw the band diagram for this material. 2. How do you deal with the electric field? 3. hat is flatband between Al-?

2 Problem 2 A number of S contacts are formed on substrates maintained at T 3K. For each of the contacts listed below, draw the energy-band diagram (indicating the Schottky-barrier height B, the depletion-layer width, E c - E F in the neutral region, and the built-in potential V ), and indicate whether the contact is ohmic or rectifying: a) an ideal contact between Al ( 4.1eV) and n-type silicon with N 1 17 cm -3, V A V. b) an ideal contact between Al ( 4.1eV) and p-type silicon with N A 1 17 cm -3, V A V. c) an ideal contact between Al ( 4.1eV) and p-type silicon with N A 1 17 cm -3, V A -.5 V. d) a contact between Ni and n-type silicon with N 1 17 cm -3, with V A V. e) a contact between Ni and p-type silicon with N A 1 17 cm -3, with V A V. a) This is an ohmic contact, because the barrier to majority-carrier (electron) flow from the metal into the semiconductor is very small. Al n-type E o 4.1 ev χ 4.3 ev.7 evqv Bn.7 ev.14 ev E c E F E v Accumulation Region idth Bn χ ev qv Bn c F 7 ( E E ) ev Accumulation Region idth (etra information, not covered in class): qv ε kt qv 6 2L ep 1 2 ep cm 52. 1nm 2 2kT q N 2kT where ε kt is the ebye length, which is the charge-screening distance. q N L 2

3 b) This is a rectifying contact, because the barrier to majority-carrier (hole) flow from the metal into the semiconductor is large. (Under forward as, holes can readily flow from the semiconductor into the metal, but under reverse as very few holes have enough thermal energy to surmount the Schottky barrier to flow into the semiconductor.) Al p-type χ 4.3 ev E o 4.1 ev E c.14 ev E F E 1.5 ev qv (E F E v ) v.18 μm χ + Eg ev qv ( E E ) ev F v ( F / cm)(.91v ) 2ε siv cm qn C 1 cm A.18µ m

4 c) This is a rectifying contact (same as in part (b), under forward as. (The barrier to hole flow from the semiconductor into the metal is reduced, so that current readily flows.) Al p-type E o χ 4.3 ev 4.1 ev E c E F qv A.5 ev 1.5 ev.14 ev q(v -V A ).41 ev E FS E v.728 μm χ + Eg ev qv ( E E ) ev F v ( F / cm)(.91v.5v ) 2ε si ( V + VA ) cm qn C 1 cm A 72.8nm

5 d) From Lecture 1 Slide 11, Bn.65eV. This is a rectifying contact, because the barrier to majority-carrier (electron) flow from the metal into the semiconductor is large. (Under forward as, electrons can readily flow from the semiconductor into the metal, but under reverse as very few electrons have enough thermal energy to surmount the Schottky barrier to flow into the semiconductor.) Ni n-type E o 4.68 ev χ 4.3 ev.65 ev Bn qv.51 ev.14 ev E c E F.813 μm E v χ + Bn ev qv ( E E ) ev Bn c F ( F / cm)(.51v ) 2ε si ( V ) cm qn C 1 cm 81.3nm

6 e) From Lecture 1 Slide 11,.47eV. This is also a rectifying contact, because the barrier to majoritycarrier (hole) flow from the metal into the semiconductor is large. (Under forward as, holes can readily flow from the semiconductor into the metal, but under reverse as very few holes have enough thermal energy to surmount the Schottky barrier to flow into the semiconductor.). Ni p-type E o χ 4.3 ev 4.68 ev E c.47 ev qv.33 ev.14 ev E F E v.654 μm χ + Eg eV qv ( E E ) ev F v ( F / cm)(.33v ) 2ε si ( V ) cm qn C 1 cm A 65.4nm Problem 3 Consider the charge density distribution, ρ(), the electric field distribution, ε(), and potential distribution, V(), for a S (n-type) diode under equilibrium conditions, shown in Figure 14.4 of Pierret. a) Illustrate with simple sketches how ρ(), ε(), and V() change if i) the Schottky barrier B decreases, ii) the semiconductor doping concentration N increases, iii) a reverse as is applied. b) Construct plots of the equilibrium (V A V) depletion width and peak electric field versus the doping concentration N in silicon Schottky diodes maintained at T 3K. Vary N over the range 1 15 cm -3 to 1 2 cm -3 and include curves corresponding to Bn.3 V and Bn.8 V.

7 a) i) ii) iii) ρ ρ ρ qn qn qn ε ε ε V V V -V -V -V i) If the Schottky barrier B decreases, the built-in voltage V B (E c E F ) decreases, i.e. there will be less band bending in the semiconductor. This means that decreases, since V. The slope of the electric field distribution ε() in the semiconductor remains the same, since N is unchanged, and therefore the peak electric field ε ma is smaller. ii) If the semiconductor doping concentration N increases, then 1/ N decreases. The slope of the electric field distribution ε() in the semiconductor increases, since it is proportional to N, and ε ma increases. V B (E c E F ) increases because (E c E F ) decreases. iii) If a reverse as is applied, then the depletion region widens (i.e. increases), and ε ma and V both increase since they are proportional to and 2, respectively. N

8 b) ith an increase in semiconductor doping, the charge density (ρqn ) in the depletion region increases. Therefore the built-in voltage (V B (E c E F ), which changes relatively slowly with increasing doping because of the logarithmic dependence of (E c E F ) on N ), which is proportional to the second integral of charge density, can be dropped across a shorter distance, i.e. the depletion width decreases. ith a decrease in Schottky barrier height, the voltage drop across the semiconductor (V ) decreases and is therefore dropped across a shorter distance, i.e. the depletion width decreases. Note that for doping concentrations above ~1E19/cm 3, the width of the depletion region is so narrow that significant quantum-mechanical tunneling of carriers through the Schottky barrier can occur, i.e. a practical ohmic contact is achieved. Equilibrium epletion idth versus oping Phi.3 V.6 Phi.8 V epletion idth (um) E+15 1.E+16 1.E+17 1.E+18 1.E+19 1.E+2 oping (cm^-3) Peak Electric Field versus oping 6.E+6 5.E+6 4.E+6 Phi.3 V Electric Field (V/cm) 3.E+6 2.E+6 Phi.8 V 1.E+6.E+ 1.E+15 1.E+16 1.E+17 1.E+18 1.E+19 1.E+2 oping (cm^-3)