Download - Lecture 9
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:
EE130/230M Spring 2013 Lecture 9, Slide 2
• A pn junction has a rectifying current-vs.-voltage characteristic:
Idealized pn Junctions
EE130/230M Spring 2013 Lecture 9, Slide 4
• 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
Electrostatics (Step Junction)
Band diagram:
Electrostatic potential:
Electric field:
Charge density:
EE130/230M 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 (x)(depletion widths xp, xn unknown)
3. Integrate (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 EE130/230M Spring 2013 Lecture 9, Slide 6
Built-In Potential Vbi
i
A
isidepFi
n
NkT
n
pkTEE
ln
ln)(
sideniFsidepFisiden Ssidep Sbi )()( EEEEqV
i
D
isideniF
n
NkT
n
nkTEE
ln
ln)(
For non-degenerately doped material:
EE130/230M Spring 2013 Lecture 9, Slide 7
What if one side is degenerately doped?
sideniFsidepFibi EEEEqV )()(
p+n junction n+p junction
EE130/230M Spring 2013 Lecture 9, Slide 8
The Depletion ApproximationIn the depletion region on the p side, = –qNA
EE130/230M Spring 2013 Lecture 9, Slide 9
ps
A
s
A xxqN
CxqN
x
1)(
In the depletion region on the n side, = qND
xxqN
CxqN
x ns
A
s
D
1)(
Electric Field Distribution
The electric field is continuous at x = 0
NAxp = NDxn
EE130/230M Spring 2013 Lecture 9, Slide 10
xxn-xp
E(x)
On the p side:
Choose V(-xp) to be 0
On the n side:
12)(
2)( Dxx
qNxV p
s
A
22
2 )(2
)(2
)( xxqN
VDxxqN
xV ns
Dbin
s
D
EE130/230M Spring 2013 Lecture 9, Slide 11
Electrostatic Potential Distribution
V(xn) = Vbi
• At x = 0, expressions for p side and n side must be equal:
• We also know that NAxp = NDxn
EE130/230M Spring 2013 Lecture 9, Slide 12
Derivation of Depletion Width
Depletion Width• Eliminating xp, we have:
• Eliminating xn, we have:
• Summing, we have:
)(
2
DAD
Abisn NNN
N
q
Vx
)(
2
DAA
Dbisp NNN
N
q
Vx
DA
bispn NNq
VWxx
112
EE130/230M 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 density dopantlighter NNN AD
1111
nD
bis xqN
VW
2
0 ADnp NNxx
qNVW bis2
EE130/230M Spring 2013 Lecture 9, Slide 14
Peak E-Field in a One-Sided Junction
biVWdx )0(2
1
s
bibi qNV
W
V
22
)0(
bis V
qNW
2
EE130/230M Spring 2013 Lecture 10, Slide 15
V(x) in a One-Sided Junction
EE130/230M Spring 2013 Lecture 9, Slide 16
biDA
D VNN
NV
)0(
p side n side2)(
2)( p
s
A xxqN
xV
2)(2
)( xxqN
VxV ns
Dbi
Example: One-Sided pn JunctionA p+n junction has NA=1020 cm-3 and ND =1017cm-3. Find (a) Vbi (b) W (c) xn and (d) xp .
i
DGbi n
N
q
kT
q
EV ln
2
D
bis
qN
VW
2
Wxn
ADnp NNxx
EE130/230M 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.
EE130/230M Spring 2013 Lecture 9, Slide 18
Summary• For a non-degenerately-doped pn junction:
Built-in potential
Depletion width
• For a one-sided junction:
Built-in potential
Depletion width
2ln
i
ADbi n
NN
q
kTV
DA
Abispn NNq
VVxxW
112
i
Gbi n
N
q
kTEV ln
2
qN
VVW Abis
2
EE130/230M Spring 2013 Lecture 9, Slide 20
WNN
Nx
DA
Dp W
NN
Nx
DA
An