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

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: C. C. Hu, Modern Semiconductor Devices for ICs, Figure 4-1 A pn junction has a rectifying current-vs.-voltage characteristic: C. C. Hu, Modern Semiconductor Devices for ICs, Figure 4-2 EE130/230A Fall 2013 Lecture 9, Slide 2

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

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

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

“Game Plan” to obtain r(x), E(x), V(x) Find the built-in potential Vbi Use the depletion approximation  r (x) (depletion widths xp, xn unknown) Integrate r (x) to find E(x) Apply boundary conditions E(-xp)=0, E(xn)=0 Integrate E(x) to obtain V(x) Apply boundary conditions V(-xp)=0, V(xn)=Vbi For E(x) to be continuous at x=0, NAxp = NDxn Solve for xp, xn EE130/230A Fall 2013 Lecture 9, Slide 6

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

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

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

Electric Field Distribution E(x) -xp xn x The electric field is continuous at x = 0  NAxp = NDxn EE130/230A Fall 2013 Lecture 9, Slide 10

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

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

Depletion Width Eliminating xp, we have: Eliminating xn, we have: Summing, we have: EE130/230A Fall 2013 Lecture 9, Slide 13

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

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

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

Example: One-Sided pn Junction A p+n junction has NA=1020 cm-3 and ND =1017cm-3. Find (a) Vbi (b) W (c) xn and (d) xp . EE130/230A Fall 2013 Lecture 9, Slide 17

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

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

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

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