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

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

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

4. 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

5. 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

6. “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

7. 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

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

9. 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

10. 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

11. 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

12. 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

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

14. 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

15. Peak E-Field in a One-Sided Junction
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Lecture 9, Slide 15

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

17. 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

18. 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

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

20. 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

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