Lecture #10
ANNOUNCEMENT
– Quiz #2 next Thursday (2/27) covering
• carrier action (drift, diffusion, R-G)
• continuity & minority-carrier diffusion equations
• metal-semiconductor contacts
OUTLINE
• pn junction electrostatics
Reading: Chapter 5
Spring 2003
EE130 Lecture 10, Slide 1
pn Junctions
Donor s
N- type
P- type
– V +
I
I
V
Reverse bias
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Forward bias
N
P
diode
symbol
EE130 Lecture 10, Slide 2
1
Terminology
Doping Profile:
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EE130 Lecture 10, Slide 3
Idealized Junctions
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EE130 Lecture 10, Slide 4
2
Energy Band Diagram
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EE130 Lecture 10, Slide 5
Qualitative Electrostatics
Band diagram
Electrostatic potential
Electric field
Charge density
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EE130 Lecture 10, Slide 6
3
“Game Plan” for Obtaining ρ(x), (x), V(x)
• Find the built-in potential Vbi
• Use the depletion approximation → ρ (x)
(depletion-layer widths xp, xn unknown)
• Integrate ρ (x) to find (x)
– boundary conditions (-xp)=0, (xn)=0
• Integrate (x) to obtain V(x)
– boundary conditions V(-xp)=0, V(xn)=Vbi
• For (x) to be continuous at x=0, NAxp = NDxn
→ solve for xp, xn
Spring 2003
EE130 Lecture 10, Slide 7
Built-In Potential Vbi
qVbi = Φ S p −side − Φ S n −side = ( Ei − EF ) p −side + ( EF − Ei ) n −side
For non-degenerately doped material:
 p
( Ei − E F ) p − side = kT ln 
 ni 
n
( E F − Ei ) n − side = kT ln 
 ni 
N 
= kT ln A 
 ni 
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N 
= kT ln D 
 ni 
EE130 Lecture 10, Slide 8
4
Vbi for “One-Sided” pn Junctions
qVbi = ( Ei − E F ) p − side + ( E F − Ei ) n − side
p+n junction
Spring 2003
n+p junction
EE130 Lecture 10, Slide 9
The Depletion Approximation
On the p-side, ρ = –qNA
qN
d
=− A
εs
dx
( x) = −
qN A
εs
x + C1 =
-qNA
εs
(x + x p )
On the n-side, ρ = qND
( x) =
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-qN D
εs
( xn − x )
EE130 Lecture 10, Slide 10
5
Electric Field in the Depletion Layer
The electric field is continuous at x = 0
⇒
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NAxp = NDxn
EE130 Lecture 10, Slide 11
Electrostatic Potential in the Depletion Layer
On the p-side:
V ( x) =
qN A
( x + x p ) 2 + D1
2ε s
(arbitrarily choose the voltage at x = xp to be 0)
On the n-side:
V ( x) = −
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qN D
qN D
( xn − x ) 2 + D2 = Vbi −
( xn − x ) 2
2ε s
2ε s
EE130 Lecture 10, Slide 12
6
• At x = 0, expressions for p-side and n-side must be equal:
• We also know that NAxp = NDxn
Spring 2003
EE130 Lecture 10, Slide 13
Depletion Layer Width
• Eliminating xp, we have:
xn =
2ε sVbi
q


NA


 ND (N A + ND ) 
• Eliminating xn, we have:
xp =
2ε sVbi
q


ND


 N A(N A + ND ) 
• Summing, we have:
xn + x p = W =
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2ε sVbi
q
 1
1 


+
 N A ND 
EE130 Lecture 10, Slide 14
7
One-Sided Junctions
If NA >> ND as in a p+n junction:
W=
2ε sVbi
≈ xn
qN D
x p = xn N D N A ≅ 0
What about a n+p junction?
W = 2ε s Vbi qN
where
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1
1
1
1
=
+
≈
N N D N A lighter dopant density
EE130 Lecture 10, Slide 15
Peak Electric Field
1
∫ ε dx = 2 ε (0) W = Vbi − VA
2ε s
(Vbi − VA )
qN
• For a one-sided junction, W ≅
so
ε (0) = 2(V W− V ) ≅
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bi
A
2qN (Vbi − VA )
εs
EE130 Lecture 10, Slide 16
8
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
−14
b) W ≈ 2ε sVbi =  2 × 12 × 8.85 × 10 × 1 
 1.6 × 10−19 × 1017 
qN D


= 0.12 µm
c) xn ≈ W = 0.12 µm
d) x p = xn N D N A = 1.2 × 10−4 µm = 1.2 Å ≈ 0
Spring 2003
EE130 Lecture 10, Slide 17
Linearly Graded Junction
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EE130 Lecture 10, Slide 18
9
Biased PN Junctions
REQUIREMENT: VA < Vbi, otherwise, we cannot assume low-level injection
Spring 2003
EE130 Lecture 10, Slide 19
Current Flow - Qualitative
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EE130 Lecture 10, Slide 20
10
Effect of Bias on Electrostatics
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EE130 Lecture 10, Slide 21
11