Lecture #14
OUTLINE
• pn junction electrostatics
Reading: Chapter 5
Spring 2007
EE130 Lecture 14, Slide 1
Qualitative Electrostatics
Band diagram
Electrostatic potential
Electric field
Charge density
Spring 2007
EE130 Lecture 14, Slide 2
“Game Plan” for Obtaining r(x), E(x), V(x)
• Find the built-in potential Vbi
• Use the depletion approximation  r (x)
(depletion-layer widths xp, xn unknown)
• Integrate r (x) to find E(x)
– boundary conditions E(-xp)=0, E(xn)=0
• Integrate E(x) to obtain V(x)
– boundary conditions V(-xp)=0, V(xn)=Vbi
• For E(x) to be continuous at x=0, NAxp = NDxn
 solve for xp, xn
Spring 2007
EE130 Lecture 14, Slide 3
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
( E F  Ei ) n  side  kT ln  
 ni 
 NA 

 kT ln 
 ni 
Spring 2007
EE130 Lecture 14, Slide 4
 ND 

 kT ln 
 ni 
Vbi for “One-Sided” pn Junctions
qVbi  ( Ei  EF ) p  side  ( EF  Ei )n  side
p+n junction
Spring 2007
n+p junction
EE130 Lecture 14, Slide 5
The Depletion Approximation
On the p-side, r = –qNA
d E   qN A
es
dx
-qNA
qN A
p
E( x)   e s x  C  e s (x  x )
1
On the n-side, r = qND
-qN D xn 
x)
(
E( x) 
s
e
Spring 2007
EE130 Lecture 14, Slide 6
Electric Field in the Depletion Layer
The electric field is continuous at x = 0

Spring 2007
NAxp = NDxn
EE130 Lecture 14, Slide 7
Electrostatic Potential in the Depletion Layer
On the p-side:
qN A
V ( x) 
( x  x p ) 2  D1
2e s
(arbitrarily choose the voltage at x = xp to be 0)
On the n-side:
V ( x)  
Spring 2007
qN D
qN D
( xn  x ) 2  D2  Vbi 
( xn  x )2
2e s
2e s
EE130 Lecture 14, Slide 8
• At x = 0, expressions for p-side and n-side must be equal:
• We also know that NAxp = NDxn
Spring 2007
EE130 Lecture 14, Slide 9
Depletion Layer Width
• Eliminating xp, we have:
xn 

2e sVbi 
NA


q  ND (N A  ND ) 
• Eliminating xn, we have:
xp 

2e sVbi 
ND


q  N A(N A  ND ) 
• Summing, we have:
2e sVbi  1
1 


xn  x p  W 

q  N A ND 
Spring 2007
EE130 Lecture 14, Slide 10
One-Sided Junctions
If NA >> ND as in a p+n junction:
2e sVbi
W
 xn
qN D
x p  xn N D N A  0
What about a n+p junction?
W  2e s Vbi qN
Spring 2007
where
1
1
1
1



N N D N A lighter dopant density
EE130 Lecture 14, Slide 11
Example
A p+n junction has NA=1020 cm-3 and ND =1017cm-3. What
is a) its built in potential, b)W , c)xn , and d) xp ?
Solution:
a)
Vbi 
EG kT N D

ln
1V
2q
q
ni
1/ 2
b) W  2e sVbi   2  12  8.85  10  1 
 1.6  1019  1017 
qN D


14
c) xn  W  0.12 μm
d) x p  xn N D N A  1.2 104 μm  1.2 Å  0
Spring 2007
EE130 Lecture 14, Slide 12
 0.12 μm
Summary
• For a non-degenerately-doped pn junction at equilibrium:
– Built-in potential Vbi 
kT N D N A
ln
q
ni2
– Depletion-layer width W  xn  x p 
N A x p  N D xn
2e sVbi  1
1 



q  N A ND 
• For a one-sided (p+n or pn+) junction at equilibrium:
EG kT N

ln
– Built-in potential Vbi 
2
q
ni
– Depletion-layer width W 
Spring 2007
2e sVbi
qN
EE130 Lecture 14, Slide 13