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:
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Lecture 9, Slide 2
Terminology
Doping Profile:
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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
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Lecture 9, Slide 4
Electrostatics (Step Junction)
Band diagram:
Electrostatic potential:
Electric field:
Charge density:
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Lecture 9, Slide 5
“Game Plan” to obtain r(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
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Lecture 9, Slide 6
Built-In Potential Vbi
qVbi  S pside  S nside  (Ei  EF )pside  (EF  Ei )nside
For non-degenerately doped material:
 p
( Ei  EF ) p  side  kT ln 
 ni 
n
( EF  Ei )n  side  kT ln 
 ni 
 NA 

 kT ln
 ni 
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 ND 

 kT ln
 ni 
Lecture 9, Slide 7
What if one side is degenerately doped?
qVbi  ( Ei  EF ) pside  ( EF  Ei )nside
p+n junction
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n+p junction
Lecture 9, Slide 8
The Depletion Approximation
In the depletion region
on the p side, r = –qNA
 ( x)   qN

A
x  C1  
s
qNA
s
x  x 
p
In the depletion region
on the n side, r = qND
 ( x) 
EE130/230M Spring 2013
Lecture 9, Slide 9
qND
s
x  C1 
qN A
s
xn  x 
Electric Field Distribution
E(x)
-xp
xn
x
The electric field is continuous at x = 0
 NAxp = NDxn
EE130/230M Spring 2013
Lecture 9, Slide 10
Electrostatic Potential Distribution
On the p side:
qNA
V ( x) 
( x  x p ) 2  D1
2 s
Choose V(-xp) to be 0
 V(xn) = Vbi
On the n side:
qND
qND
2
V ( x)  
( xn  x )  D2  Vbi 
( xn  x )2
2 s
2 s
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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
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Lecture 9, Slide 12
Depletion Width
• Eliminating xp, we have:
xn 
2 sVbi
q


NA


 ND (N A  ND ) 
• Eliminating xn, we have:
xp 
2 sVbi
q


ND


 N A(N A  ND ) 
• Summing, we have:
2 sVbi
xn  x p  W 
q
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 1
1 



 N A ND 
Lecture 9, Slide 13
Depletion Width in a One-Sided Junction
If NA >> ND as in a p+n junction:
W 
2 sVbi
 xn
qN D
x p  xn N D N A  0
What about a n+p junction?
W  2 s Vbi qN where
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1
1
1
1



N N D N A lighter dopantdensity
Lecture 9, Slide 14
Peak E-Field in a One-Sided Junction

1
dx   (0) W  Vbi
2
2 s
W
Vbi
qN

2Vbi
2qNVbi
(0) 

W
s
EE130/230M Spring 2013
Lecture 10, Slide 15
V(x) in a One-Sided Junction
p side
n side
qN A
V ( x) 
( x  x p )2
2 s
qND
V ( x)  Vbi 
( xn  x) 2
2 s
ND
V (0) 
Vbi
N A  ND
EE130/230M Spring 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 .
EG kT N D
Vbi 

ln
2q q
ni
W
2 sVbi
qN D
xn  W
x p  xn N D N A
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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.
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Lecture 9, Slide 18
Effect of Applied Voltage
W
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Lecture 9, Slide 19
 1
2 s
1 

(Vbi  VA )

q
 N A ND 
Summary
• For a non-degenerately-doped pn junction:
kT N D N A
ln
q
ni2
Built-in potential
Vbi 
Depletion width
W  xn  x p 
2 s Vbi  VA   1
1 



q
 N A ND 
NA
W
N A  ND
xn 
• For a one-sided junction:
Built-in potential Vbi 
Depletion width
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EG kT N

ln
2
q
ni
2 s Vbi  VA 
W
qN
Lecture 9, Slide 20
xp 
ND
W
N A  ND
Linearly Graded pn Junction
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Lecture 9, Slide 21