Reverse biased PN junction- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
p n
bias RV Vtotal bi RV V V
WW
- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
p n
0
( )
0
ap p
s
dn n
s
eNx x x x
E xeN
x x x x
a dmax p n
s s
eN eNE x x
2 1s bi R an
d d a
V V Nx
e N N N
2 s bi R d a
d a
V V N N
e N N
a p d nN x N xn pW x x
1 dn
a
NW x
N
d an
a
N Nx
N
2 1s bi R a d a
d d a a
V V N N NW
e N N N N
Reverse biased PN junctionenergy diagram
cE
cEvE
vE
FiE
FiE
FnE
FpE
ReV
total bi ReV e V V
Voltage-dependent capacitor
px
nx
with RV
with R RV dV
Voltage-dependent capacitor
px
nx
with RV
with R RV dV
dQ
dQ
differential R
dQC'
dV
nd
R R
dxdQC' eN
dV dV
deN
aeN
Example 5.5 Consider a gallium arsenide PN junction at room temperature. Determine the junction capacitance with a reverse bias voltage of 5 V.
& 15 3 16 3a dN 1 10 cm N 2 10 cm
2cross-sectional area A 410 cm
a dBbi 2
i
N NTV ln
e n
15 16
bi 26
1 10 2 10V 0.0259 ln 1.12V
1.8 10
Example 5.5 Consider a gallium arsenide PN junction at room temperature. Determine the junction capacitance with a reverse bias voltage of 5 V.
& 15 3 16 3a dN 1 10 cm N 2 10 cm
2cross-sectional area A 410 cm
s a d
bi R a d
e N NC A
2 V V N N
19 9 15 16
415 16
13.11.6 10 10 10 2 10
36C 102 1.12 5 10 2 10
0.380 pF
Forward biased PN junction
p n
DV
appliedE
space charg eE
- +
Basic model – thermal equilibrium
FE
- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
- - - - - -
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
+ + + + + +
p n+-
cE
cE
vE
vE
FiEbi biE eV
FiEbi biE eV
bi biE eV
FneVFpeV
Reverse biased PN junctionenergy diagram
cE
cEvE
vE
FiE
FiE
FnE
FpE
ReV
totaleV
Forward biased PN junctioncE
cE
vE
vE
FiE
FiE
FnEFpE
DeV
bi DeV eV
Schottky barrier junction
metal surface
Schottky barrier junctionEnergy levels before joining
metal surface
cE
vE
FiE
FE
FE
Schottky barrier junctionThermal equilibrium
cE
vE
FiE
FE
nx
W
bi biE eVSchottky barriermetal surface
FE -
electrons depart leaving a region of ions – Fermi energies lineup
Schottky barrier junctionReverse bias conditions
cE
vE
FE
nx
W
bi ReV eV
metal surface
FE
Schottky barrier
Schottky barrier junctionForward bias conditions
cE
vE
FE
DeV
metal surface
FE
Schottky barrierDV
electrons can easily move creating a current
Characteristics of a Schottky diode and a PN junction diode
VDVT
D sI I e 1
reverse
saturation current
thermal potential
BT
TV
e
Example 5.7 A silicon PN junction is forward biased at room temperature. Determine the voltage that is required to induce a particular diode current.
IV DD lnV IT s
VDVT
D sI I e 1
saturation current 14sI 5 10 A
desired current 3DI 4.25 10 A
VDVT
sI e
IDV V lnD T Is
3
14
4.25 10V 0.0259 lnD 5 10
0.652V
02
20
1 expB e
eVn e
Td V
dx
2
2
dVU
dx
d V dU dV dUU
dx dV dx dV
0
0
1 expB e
eVn e
TUdU dV
0
0
1 expB e
eVn e
TUdU dV
20
10
exp2
B e
B e
n e TU eVV C
e T
01
0
2expB e
B e
n e TdV eVU V C
dx e T
00 1
0
2expx B e
appliedB e
n e T eVV V V C dx
e T
00 1
0
2expx B e
appliedB e
n e T eVV V V C dx
e T
1applied
B e
eV
T
0
0
2n eVdVU
dx
00
00
2
applied
L
V
n edVdx
V0
0
2 appliedV
Ln e
You think that you had a bad day.
Simple three-dimensional unit cell
a
b
c
Top Related