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ENE 311
Lecture 8
Effect of applied voltage to p-n junction
Forward Bias
• Under thermal equilibrium, numbers of
electrons crossing in opposite direction are
equal.
• Therefore, there is no net transfer of charge
that leads to no current flow.
• This is the same for the case of holes.
Effect of applied voltage to p-n junction
• If a positive voltage VF is
applied to the p-side with
respect to the n-side, the
p-n junction becomes
forward-biased.
• The flow of electrons
from left to right is not
affected, but the flow of
them from right to left is
certainly affected.
Effect of applied voltage to p-n junction
For conclusion, forward bias causes
• The depletion region decreases.
• Barrier height decreases.
eVbi  e(Vbi - VF)
• Net transfer of charges occurs
as free electrons transported
from right to left producing
electron current flow and free
holes transported from left to
right producing hole current.
VF  I (Ie + Ih) 
electrons: R  L Ie: L  R
holes: L  R Ih: L  R
Effect of applied voltage to p-n junction
Reverse Bias
• If we apply positive VR to
the n-side with respect to
the p-side, the p-n junction
is now reverse-biased. This
reverse bias causes
• Depletion width increases
• Barrier height increases
eVbi  e(Vbi + VR)
Effect of applied voltage to p-n junction
Reverse Bias
• Net transfer of charge occurs
but just few free electrons
are able to be transported
from left to right causing
electron current flow. Alos,
few holes are transported
from right to left causing hole
current flow.
I = Ie + Ih  small current
electrons: L  R
Ie: R  L
holes: R  L
Ih: R  L
Depletion Capacitance
• When p-n junction is
reverse biased, the
presence of 2 layers of
space-charge in the
depletion region makes it
look like a capacitor.
• The solid line indicating
charge and electric field
distribution in the figure
corresponds to an applied
voltage V.
Depletion Capacitance
• The dashed line
represents the charge and
electric field distribution
when the applied voltage
is increased by dV.
• The incremental space
charges on both n- and psides in the space-charge
region are equal but with
opposite charge polarity.
Depletion Capacitance
• The incremental charge
dQ causes an increase in
the electric field by dE =
dQ/s.
• While the corresponding
change in the applied
voltage dV is
approximately equal to
WdE.
Depletion Capacitance
The depletion capacitance per unit area is given by
Or
dQ
dQ
Cj 

dV W dQ
d s
Cj 
s
W
 F/cm 2 
(1)
Equation (1) is a good assumption for the reversebias condition.
Depletion Capacitance
• For forward bias, there is an additional term
called diffusion capacitance due to a large
number of mobile carriers moving across the
junction.
Depletion Capacitance
Therefore, for a one-sided abrupt junction, we
have
Cj 
s
W

q s N B
2 Vbi  V 
1 2 Vbi  V 

2
Cj
e s N B
(2)
Consider a plot of 1/Cj2 versus V of (2). The
slope gives the impurity concentration NB of the
semiconductor and the intercept at 1/Cj2 = 0
yields Vbi.
Depletion Capacitance
• Whereas, the depletion layer capacitance in a
case of linearly graded junction can be
expressed by
1/ 3
 ea

Cj 


W 12 Vbi  V  
s
2
s
 F/cm 2 
(3)
Depletion Capacitance
Ex. For a silicon one-sided abrupt junction with NA
= 2 x 1019 cm-3 and ND = 8 x 1015 cm-3, calculate
the junction capacitance at zero bias and
reversed bias of 4 V.
Depletion Capacitance
Soln
Vbi  0.0259ln
2  1019  8  1015
 9.65 10 
9 2
 0.906 V
2 sVbi
2  11.9  8.85  1014  0.906

eN D
1.6  1019  8  1015
W V 0
 0.386 μm
Cj

V 0
s
W V 0
 2.728  108 F/cm 2
2 s Vbi  V 
2  11.9  8.85  1014   0.906  4 

eN D
1.6  1019  8  1015
W V 4
 0.899 μm
Cj
V 4

s
W V 4
 1.172  108 F/cm 2
Current-Voltage Characteristics
We now consider an ideal case of current-voltage
characteristics based on these assumptions
• the depletion region has abrupt boundaries
• the low-injection condition
• neither generation nor recombination current
exists in the depletion region
• the electron and hole currents are constant
throughout the depletion region
Current-Voltage Characteristics
• At thermal equilibrium, the majority carrier
density (nn0 or pp0) in the neutral regions is
equal to the doping concentration.
• The built-in potential can be written as
eVbi  kT ln
p p 0 nn 0
2
i
n
[eV]
(4)
Current-Voltage Characteristics
By using the mass action law pp0.np0 = ni2, equation (4) can be
rewritten as
(5)
nn 0
eVbi  kT ln
np0
[eV]
From equation (5), we have
nn0  np 0 exp  eVbi / kT 
(6)
p p 0  pn0 exp  eVbi / kT 
(7)
Current-Voltage Characteristics
• We clearly see from (6) and (7) that the charge densities
at the boundaries of the depletion region are related to
the potential difference Vbi at thermal equilibrium.
• If the voltage V is applied to the junction, the potential
Vbi is changed and (6) becomes
nn  n p e
e (Vbi V ) / kT
(8)
where nn and np are the nonequilibrium densities at the
boundaries of the depletion region in the n- and p-sides,
respectively, with V positive for forward bias and negative
for reverse bias.
Current-Voltage Characteristics
• In case of low-injection condition, the injected
minority carrier is much smaller than the
majority carrier density, so that nn  nn0.
Substituting this condition with (6) into (8)
yields
n p  n p 0eeV / kT
n p  n p 0  n p 0  eeV / kT  1
(9)
(10)
where np = the electron density at the boundary of
the depletion region on the p-side at x = -xp.
Current-Voltage Characteristics
• Similarly, we have
pn  pn0eeV / kT
pn  pn 0  pn 0  eeV / kT  1
(11)
(12)
where pn = the hole density at the boundary of
the depletion region on the n-side at x = xn.
Current-Voltage Characteristics
Depletion region, energy
band diagram and carrier
distribution.
• (a) Forward bias.
• (b) Reverse bias.
Current-Voltage Characteristics
• Since we assume that no current is generated
within the depletion region, all currents come
from the neutral regions.
• In the neutral n-region, the steady-state
continuity equation can be expressed as
d 2 pn pn  pn 0

0
2
dx
D p p
(13)
Current-Voltage Characteristics
• The solution of (13) with the boundary
conditions of (12) and pn (x = )= pn0 gives
pn  pn 0  pn 0  e
eV / kT
 1 e
 x  xn  / L p
(14)
where Lp is the diffusion length of holes in the nregion = D .
p p
Current-Voltage Characteristics
dp
J p  xn   eD p n
dx

eD p pn 0
Lp
xn
 eeV / kT  1
(15)
Similarly, for the neutral p-region
np  np0  np0 e
J n   x p   eDn
eV / kT
dn p
dx

 xp
 1 e
eDn n p 0
Ln
 x  x p  / Ln
e
eV / kT
where Lp is the diffusion length of electrons in
the p-region =
Dn. n
 1
(16)
(17)
Current-Voltage Characteristics
• The figure shows that the
injected minority carriers
recombine with the
majority carriers as they
move away from the
boundaries.
• The hole and electron
diffusion current will
decay exponentially in
the n-region with
diffusion length Lp and pregion with diffusion
length Ln, respectively.
Current-Voltage Characteristics
• The figure shows that the
injected minority carriers
recombine with the
majority carriers as they
move away from the
boundaries.
• The hole and electron
diffusion current will
decay exponentially in
the n-region with
diffusion length Lp and pregion with diffusion
length Ln, respectively.
Current-Voltage Characteristics
Injected minority carrier
distribution and electron and
hole currents.
(a) Forward bias.
(b) Reverse bias.
The figure illustrates
idealized currents. For
practical devices, the
currents are not constant
across the space charge
layer.
Current-Voltage Characteristics
• The total current is constant throughout the
device and can be written as
J  J p ( xn )  J n ( x p )  J s  eV / kT  1
(18)
This is called an ideal diode equation.
Js 
eD p pn 0
Lp

eDn n p 0
Ln
where Js is the saturation current density.
(19)
Current-Voltage Characteristics
Ideal current-voltage
characteristics.
• (a) Cartesian plot.
• (b) Semilog plot.
Current-Voltage Characteristics
Ex. Calculate the ideal reverse saturation current
in a Si p-n junction diode with a cross-section
area of 2 x 10-4 cm2. The parameters of the
diode are
• NA = 5 x 106 cm-3, ND = 1016 cm-3, ni = 9.65 x 109 cm3, D = 21 cm2/s, D = 10 cm2/s, and τ = τ = 5 x 10n
p
p
n
7 s.
Current-Voltage Characteristics
Soln From (19) and L p  Dn n
Js 
eD p pn 0
Lp

 1
 en 
 ND

eDn n p 0
Ln
Dn 

p
 n 
 1
19
9 2
 1.6  10   9.65  10   16
 10
 8.58  1012 A/cm 2
2
i
Dp
1

NA
10
1

5  107 5  1016
From the cross-section area A = 2  10-4 cm 2 , we have
I s  A  J s  2  10-4  8.58  1012  1.72  1015 A
21 

5  10 7 