1. Lecture 9 OUTLINE pn Junction Diodes – Electrostatics (step junction) Reading: Pierret 5; Hu 4.1-4.2

2. pn Junctions A pn junction is typically fabricated by implanting or diffusing donor atoms into a p-type substrate to form an n-type layer: EE130/230A Fall 2013Lecture 9, Slide 2 A pn junction has a rectifying current-vs.-voltage characteristic: C. C. Hu, Modern Semiconductor Devices for ICs, Figure 4-2 C. C. Hu, Modern Semiconductor Devices for ICs, Figure 4-1

3. Terminology Doping Profile: Lecture 9, Slide 3EE130/230A Fall 2013 Net R.F. Pierret, Semiconductor Fundamentals, Figure 5.1

4. Idealized pn Junctions Lecture 9, Slide 4 In the analysis going forward, we will consider only the net dopant concentration on each side of the pn junction: N A  net acceptor doping on the p side: (N A -N D ) p-side N D  net donor doping on the n side: (N D -N A ) n-side EE130/230A Fall 2013 R.F. Pierret, Semiconductor Fundamentals, Figure 5.2

5. Electrostatics (Step Junction) Band diagram: Electrostatic potential: Electric field: Charge density: Lecture 9, Slide 5EE130/230A Fall 2013 R.F. Pierret, Semiconductor Fundamentals, Figure 5.4

6. “Game Plan” to obtain  (x), E (x), V(x) 1.Find the built-in potential V bi 2.Use the depletion approximation   (x) (depletion widths x p, x n unknown) 3.Integrate  (x) to find E (x) Apply boundary conditions E (-x p )=0, E (x n )=0 4.Integrate E (x) to obtain V(x) Apply boundary conditions V(-x p )=0, V(x n )=V bi 5.For E (x) to be continuous at x=0, N A x p = N D x n Solve for x p, x n Lecture 9, Slide 6EE130/230A Fall 2013

7. Built-In Potential V bi For non-degenerately doped material: Lecture 9, Slide 7EE130/230A Fall 2013 R.F. Pierret, Semiconductor Fundamentals, Figure 5.4a

8. What if one side is degenerately doped? p + n junctionn + p junction Lecture 9, Slide 8EE130/230A Fall 2013

9. The Depletion Approximation In the depletion region on the p side,  = –qN A Lecture 9, Slide 9 In the depletion region on the n side,  = qN D EE130/230A Fall 2013 R.F. Pierret, Semiconductor Fundamentals, Figure 5.6

10. Electric Field Distribution The electric field is continuous at x = 0  N A x p = N D x n Lecture 9, Slide 10 x xnxn -xp-xp E(x)E(x) EE130/230A Fall 2013

11. On the p side: Choose V(-x p ) to be 0 On the n side: Lecture 9, Slide 11 Electrostatic Potential Distribution  V(x n ) = V bi EE130/230A Fall 2013

12. At x = 0, expressions for p side and n side must be equal: We also know that N A x p = N D x n Lecture 9, Slide 12 Derivation of Depletion Width EE130/230A Fall 2013

13. Depletion Width Eliminating x p, we have: Eliminating x n, we have: Summing, we have: Lecture 9, Slide 13EE130/230A Fall 2013

14. Depletion Width in a One-Sided Junction If N A >> N D as in a p + n junction: What about a n + p junction? where Lecture 9, Slide 14EE130/230A Fall 2013

15. Peak E -Field in a One-Sided Junction Lecture 9, Slide 15EE130/230A Fall 2013

16. V(x) in a One-Sided Junction Lecture 9, Slide 16 p siden side EE130/230A Fall 2013

17. Example: One-Sided pn Junction A p + n junction has N A =10 20 cm -3 and N D =10 17 cm -3. Find (a) V bi (b) W (c) x n and (d) x p. Lecture 9, Slide 17EE130/230A Fall 2013

18. Voltage Drop across a pn Junction Note that V A should be significantly smaller than V bi in order for low-level injection conditions to prevail in the quasi-neutral regions. Lecture 9, Slide 18EE130/230A Fall 2013 R.F. Pierret, Semiconductor Fundamentals, Figure 5.10

19. Effect of Applied Voltage Lecture 9, Slide 19EE130/230A Fall 2013 R.F. Pierret, Semiconductor Fundamentals, Figure 5.11

20. Summary For a non-degenerately-doped pn junction: Built-in potential Depletion width For a one-sided junction: Built-in potential Depletion width Lecture 9, Slide 20EE130/230A Fall 2013

21. Linearly Graded pn Junction Lecture 9, Slide 21EE130/230A Fall 2013