Lecture 9 OUTLINE pn Junction Diodes Electrostatics step
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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: • A pn junction has a rectifying current-vs. -voltage characteristic: EE 130/230 M Spring 2013 Lecture 9, Slide 2
Terminology Doping Profile: EE 130/230 M Spring 2013 Lecture 9, Slide 3
Idealized pn Junctions • 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 EE 130/230 M Spring 2013 Lecture 9, Slide 4
Electrostatics (Step Junction) Band diagram: Electrostatic potential: Electric field: Charge density: EE 130/230 M Spring 2013 Lecture 9, Slide 5
“Game Plan” to obtain (x), E(x), V(x) 1. Find the built-in potential Vbi 2. Use the depletion approximation r (x) (depletion widths xp, xn unknown) 3. Integrate r (x) to find E(x) Apply boundary conditions E(-xp)=0, E(xn)=0 4. Integrate E(x) to obtain V(x) Apply boundary conditions V(-xp)=0, V(xn)=Vbi 5. For E(x) to be continuous at x=0, NAxp = NDxn Solve for xp, xn EE 130/230 M Spring 2013 Lecture 9, Slide 6
Built-In Potential Vbi For non-degenerately doped material: EE 130/230 M Spring 2013 Lecture 9, Slide 7
What if one side is degenerately doped? p+n junction EE 130/230 M Spring 2013 n+p junction Lecture 9, Slide 8
The Depletion Approximation In the depletion region on the p side, = –q. NA In the depletion region on the n side, = q. ND EE 130/230 M Spring 2013 Lecture 9, Slide 9
Electric Field Distribution E(x) -xp xn x The electric field is continuous at x = 0 N A xp = N D xn EE 130/230 M Spring 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: EE 130/230 M Spring 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 EE 130/230 M Spring 2013 Lecture 9, Slide 12
Depletion Width • Eliminating xp, we have: • Eliminating xn, we have: • Summing, we have: EE 130/230 M Spring 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 EE 130/230 M Spring 2013 Lecture 9, Slide 14
Peak E-Field in a One-Sided Junction EE 130/230 M Spring 2013 Lecture 10, Slide 15
V(x) in a One-Sided Junction p side EE 130/230 M Spring 2013 n side Lecture 9, Slide 16
Example: One-Sided pn Junction A p+n junction has NA=1020 cm-3 and ND =1017 cm-3. Find (a) Vbi (b) W (c) xn and (d) xp. EE 130/230 M Spring 2013 Lecture 9, Slide 17
Voltage Drop across a pn Junction Note that VA should be significantly smaller than Vbi in order for low -level injection conditions to prevail in the quasi-neutral regions. EE 130/230 M Spring 2013 Lecture 9, Slide 18
Effect of Applied Voltage EE 130/230 M Spring 2013 Lecture 9, Slide 19
Summary • For a non-degenerately-doped pn junction: Built-in potential Depletion width • For a one-sided junction: Built-in potential Depletion width EE 130/230 M Spring 2013 Lecture 9, Slide 20
Linearly Graded pn Junction EE 130/230 M Spring 2013 Lecture 9, Slide 21