consider the formation of a p-n junction by placing two n- doped and p-doped crystals side-by-side: ...
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Consider the formation of a p-n junction by placing two n-doped and p-doped crystals side-by-side:
(doping density)
Nd
Na
e-
Positive charges occur when donors on the n-side are ionized during the electron transfer to the p-side.
n-typep-type
0 x
Negative charges form when holes are ionized by the capture of electrons. Electrons (e-) flow to the p-side holes (h+) flow to the n-side. Consider a p-n junction in equilibrium: Assume that non-degenerate conditions hold for all x .
“Abrupt junction model” (1)
0,
0,0)(
0,0
0,)(
xN
xxN
x
xNxN
aa
dd
There is a transfer of charge between the n- and p-type regions in order to equalize the Fermi-level (or chemical potential) on both sides.
h+
Before: p-side n-side
EC
EV
After: p-side n-side
(EF)
EC
EV
In order to describe the spatial variation in the band edges, EC and EV, we introduce a potential function, (x), such that each level is shifted by -e(x). That is, EC EC - e(x) and EV EV - e(x). Therefore, the carrier concentrations can be expressed by
Tk
xeETPxp
Tk
xeETNxn
B
VV
B
CC
)(exp)()(
)(exp)()(
Consider limits at x =
Tk
eE
P
N
Tk
eE
N
N
Tk
eETPpN
Tk
eETNnN
B
V
V
a
B
C
C
d
B
VVa
B
CCd
)(ln
)(ln
)(exp)()(
)(exp)()(
(2)
We can solve for the difference of the two potentials at
V
a
C
dBg
V
a
C
dBVC
P
N
N
NTkEe
P
N
N
NTkEEee
ln
ln)()(
The electric field is related to the potential by Gauss’s Law:
)(4
)(4
2
22 x
dx
d
E
rE
for our 1D problem.
We assume that donor and acceptor impurities are fully ionized at all x.
)()()()()( xpxnxNxNex vcad (3)
We can substitute Eqs. (1) and (2) into Eq.(3) to get a difficult nonlinear differential equation.
We can make some important approximations to simplify and solve:
Assume that the potential change occurs in a finite region near the junction that is defined by -dp x dn. This is referred to as the depletion region. Also n << Nd
and p << Na in this region since is closer to the middle of the gap (i.e., midgap).
)()()( xNxNex ad We can solve this problem assuming an abrupt junction model which will give a simple linear second order ordinary differential equation:
xd
dx
xd
dx
eN
eN
dx
d
p
p
n
n
a
d
0
0
0
4
4
,0
2
2
xd
dx
xd
dx
dxeN
dxeN
x
p
p
n
n
pa
nd
0
0
)(
)(2
)(
)(2
)(
),(
)(2
2
Note that we have first used the boundary conditions at the edges of the depletion region (see graph on next slide) to solve for (x):
)()(
)()(
p
n
d
d
0)('|
0)('|
pdx
ndx
dE
dE
p
n
The potential must be its limiting value at the boundaries of the depletion region.
The electric field must be zero outside the depletion region for equilibrium to hold.
Nd
Na
Carrier densitync(x)
pv(x)
n-typep-type
-dp dn 0
Depletion layer
Charge density (x)
eNd
- eNa
dn -dp
x
x
dn
-dp
(x)
()
(-)
x
Graphical illustration of the Abrupt-junction model.
dn
-dp
(x)
()
(-)
x
x
dx
dE
maxE
Comparison of the potential and electric field in the abrupt-junction model:
There are important relations that we get by imposing continuity of (x) and ’(x).
Continuity of the derivative gives:
pand
pa
nd
dNdN
deN
deN
dx
d
dx
d
)(4
)(4)0()0(
Secondly, continuity of (x) gives
)()(
2)0()0( 22
pand dNdNe
Together with the derivative condition: Nddn=Nadp this gives
eNN
NNd
da
dapn
2
1
,
For Na,d ranging from 1014 to 1018 cm-3
dn,p ranges from 104 to 102 Å
The total depletion width is LD = dn +dp
We can also write this in a numerically convenient form:
o
da
eVdapn NN
eNNd
18
1
, 10105
The maximum electric f ield within the depletion layer is of the order of / (dn+adp) and ranges from 105 to 107 volts/m.
Charge density, (x)
- eNa
dn -dp
x
- eNa
dn -dp
x
- eNa
dn -dp
x
(x)
o
o- |V1|
o + |V2|
1(x)
2(x)
Consider the effect of an applied voltage on the depletion width of a p-n junction:
p n
p n
p n
V = 0 “zero-bias”
V1 > 0 “Forward-Bias”
p n
V2 < 0 “Reverse-Bias”
The potential simply changes according to =o|V|
- sign: “Forward bias”
+ sign: “Reverse bias”
We are changing the size of the depletion region by applying an external voltage.
o
opn
da
odapn
Vd
eNN
VNNd
12 ,
1
,
The formula for the depletion width can easily be modified to include the effects of a voltage bias:
Zero-bias depletion length
Consider rectification by a p-n junction. The symbol J is used for # of carriers /(area·time). The lower case symbol j is more commonly used for current density and has units j = coulombs/ (area·time). Note j e= -e J e and j
h= e J h.
p-side n-side
(EF)
EC
EV
Consider electron and hole generation currents. These carriers are generated by thermal excitation as we saw before according to
Tk
xeETPxp
Tk
xeETNxn
B
VV
B
CC
)(exp)()(
)(exp)()(
Note: The generation currents involve minority carriers.
e-
h+
In a band diagram, electrons fall down hill and holes “float” uphill (both towards lowest energy) .
In the process of the generation currents, holes (from n-side) p-side, electrons (from p-side) n-side.
Secondly, consider another kind of current, Recombination current. With this kind of current holes (from p-side) n-side. Electrons from the n-side p-side. This current involves majority carriers, and it is made possible by thermal excitation over the barrier, as shown below:
p-side n-side
(EF)
EC
EV
o- |V1| e-
h+
Consider the hole recombination current. The current is approximated well by assuming thermionic emission over the barrier:
TkVJJ
JJ
TkVeJ
Bgenh
rech
genhV
rech
Borech
/exp
|
/exp
0
The middle equation is a statement that no net current can flow during equilibrium when V=0.
The total hole current from the p to the n-side is given by the recombination current minus the generation current:
1exp
1exp
Tk
eVJJej
Tk
eVJJJJ
B
gene
genh
B
genh
genh
rechh
The last equation includes the currents of both holes and electrons, since the same analysis will also apply to electrons. It is obvious to understand the rectifying properties of a p-n junction (diode) from this equation.
p n
V
V
j
gene
genh JJe
Rectifying behavior of a p-n junction (diode).
Saturation
Reverse bias (V<0) Forward bias (V>0)
We still need to calculate this term in terms of fundamental parameters describing transport.
Note: In our notation Jh and Jp will mean the same thing.
In a more general treatment, it is not necessary to separate into generation and recombination currents. Note that to solve for the following five fundamental quantities (Je(x), Jh(x), n(x), p(x), and (x)), we need five equations. In the equilibrium case with V=0, Je(x)= Jh(x)=0, we need three equations to solve the three unknowns, including Poisson’s equation for the potential.
In the presence of a field E and n and p, we can write the electron and hole current densities as:
dx
dpDpEJ
dx
dnDnEJ
pph
nne
Where n and p are the electron and hole mobilities in units of cm2/V·s, and Dn and Dp are the electron and hole diffusion constants in units of cm2/s
When dn/dx=0 ,
m
ne
nEeeJEjcol
ne
2
From the Dude theory (mv=Ft) recall EE
m
nenevj
meEvcol
col
2
/
Therefore ,
p
colp
p
n
coln
n
m
e
m
e
Which expresses in terms of the collision time and masses, based on the Drude model.
Now, recall that
At equilibrium,
Tk
eD
Tk
eD
TkneEDnE
dx
dnDnEJ
Tk
Ene
dx
d
Tk
e
Tk
xeETN
dx
dn
Tk
xeETNxn
B
pp
B
nn
Bnnnne
BBB
CC
B
CC
;
01
)(
)()(exp)(
)(exp)()(
These last two relations are known as the Einstein relations.
Consider continuity equations for transport of charge:
x
J
t
p
x
J
t
n he
; If V0 and carriers are conserved.
We have to include two other processes which act as a source and drain for carriers:
(i )Generation by thermal excitation
(ii )Recombination )electron hole(
EC
EV
(i)
(ii)
As a result the continuity equations need to include these additional terms: x
J
dt
dp
t
p
x
J
dt
dn
t
n h
rg
e
rg
;
The g-r terms act to restore the system to equilibrium when the system deviates from equilibrium. These terms can be further described by electron and hole lifetimes (n and p):
22;
;
io
io
p
o
rgn
o
rg
nnpnpn
pp
dt
dpnn
dt
dn
where no and po are equilibrium concentrations as determined by the law of mass action.
Note that n >>ncol and p >>p
col ; typically n ~10-9 s and ncol ~10-13 s
Very often we deal with a steady state condition in which
This gives
0
t
p
t
n
0;0
p
oh
n
oe pp
dx
dJnn
dx
dJ
p n
V
For V0, we are not in equilibrium but we have a steady state for V = const.
Note that Je = Jh = 0 for V = 0 n = no and p = po.
Now suppose that E0, i.e., the electric field is negligible. Then
dx
dnD
dx
dnDnEJ nnne Fick’s Law for Diffusion
0;0
p
oh
n
oe pp
dx
dJnn
dx
dJ
Therefore, further give the following 2nd order
ordinary differential equations which are referred to as diffusion equations:
p
o
pn
o
n
pp
dx
pdD
nn
dx
ndD
2
2
2
2
;Note no and po are equilibrium values.
The solutions are easily written as:
nnnppp
n
o
n
oo
p
o
p
oo
DLDL
L
xxB
L
xxAnn
L
xxB
L
xxApp
;
)(exp
)(exp
;)(
exp)(
exp
The diffusion lengths Lp and Ln are given by
Note that when E0, the majority carrier density is constant and the carriers in the diffusion equation are for the minority carriers. For example if we have a p-type material pn = ni
2 and p >> n so that electrons (np) are minority carriers. The notation is slightly changed to be more precise:
p
onnn
pn
oppp
n
pp
dx
pdD
nn
dx
ndD
2
2
2
2
;p-type n-type
Examine the meaning of the diffusion lengths, Ln and Lp.
Since
n
ncolnB
n
n
coln
nnB
nnnn
m
TkL
m
e
e
TkDDL
;;
From the equipartion theorem and Drude approximation: ½ mvth2 = 3/2 kBT
Further, the mean free path ln is given by ln=vthcol where vth is the thermal velocity of a carrier.
coln
nn
ncol
coln
nnn
ncol
Bncol
nnB l
m
lmL
m
TkL
lmTk
33
1;
3
12/122
The interpretation here is as follows: When a carrier undergoes N collisions before recombination, the net displacement will be
coln
nn NlN
3;
The factor of 3 comes from the number of degrees of freedom (3D). This is basically a “random walk” problem.
1 2
3
4 Note that n >>ncol and p >>p
col ; typically n ~10-9 s and n
col ~10-13 s N3000 as the number of collisions before recombination in GaAs.
We need to improve our understanding of transport across the p-n junction. Consider an equivalent description of Boltzmann statistics at thermal equilibrium:
Tk
EEnn
B
iFi expRemember that
where ni is the intrinsic carrier concentration EF = is the Fermi-level (or chemical potential), and Ei is the position of the chemical potential for the intrinsic case.
Let us define potentials = -Ei/e and = -EF/e. At thermal equilibrium pn=ni2 (V=0 across
the junction). For the case of V0, we can write the carrier concentrations as
Tk
epp
Tk
enn
B
pi
B
ni
)(exp;
)(exp
Where n and p are called quasi-Fermi levels or imrefs (imaginary references).
Tk
enpn
n
p
e
Tk
n
n
e
Tk
B
npi
i
Bp
i
Bn
)(exp
lnln
2
The advantage of this formalism is that the potential difference V across the junction is just the difference in the quasi-Fermi levels, i.e.
xnxpxnxpV |||| 00
E (eV)EC
EV
EF ()-ep
-en
p-side n-side
-dp 0 dn
Potentialp
n
-dp 0 dn
V
log(n,p)
-dp 0 dn
ppo
pnonpo
nno
pnnp
ni
Forward-Bias Conditions
-ep
-en
-dp 0 dn
p-side n-sideReverse-Bias Conditions
EC
EV
EF ()
p
n
-dp 0 dn
-V
log(n,p)
-dp 0 dn
ppo
pnonpo
nno
pnnp
ni
x x
x
x
x x
Potential
E (eV)
Note that (dp+dn) for reverse bias is greater than (dp+dn) for forward bias.
Tk
eVn
Tk
enpn
Bi
B
npi exp
)(exp 22
For a forward bias V = p-n > 0 and pn > ni2 in the junction. The opposite is the case for a
reverse bias V = p-n < 0 and pn < ni2.
Now consider the current density and write in terms of and :
xne
xxTk
en
e
Tke
xne
x
n
e
TknEeJ n
nn
B
Bnn
Bne
Note that x
E
exE i
1
Our previous notation was .)( constx
E
Similarly ,x
peJ ppp
In the depletion layer –dp < x <dp
00
pe
J
xne
J
x p
pp
n
en
since n and p are sharply decreasing functions in the depletion layer. Also, we assume that passage of carriers across the junction is very fast so that generation and recombination currents in the depletion region are negligible. This leads to assumption that Je and Jp are constant in the depletion region –dp < x <dp
For p and n constant, Je and Jp are 0.
Consider n at x = -dp (on the p-side):
Tk
eVn
Tk
eV
p
nn
Bpo
Bp
idxp p
expexp|2
Note that npo is the electron density on the p-side at x = -.
Similarly, at x = +dn
Tk
eVpp
Bnodxn n
exp| And pno is the hole density on the n-side at x = +
These equations serve as boundary conditions for the I-V equation of the ideal p-n junction .
Recall that just outside of the depletion region (i.e. |x| > dn, dp the region is neutral, E0, and this is called the diffusion region. In these regions, the following diffusion equations apply for the respective minority carrier densities:
p
nonnp
n
poppn
pp
dx
pdD
nn
dx
ndD
2
2
2
2
;
(1)
(2)
Using boundary conditions (1) and (2), the solutions are:
pn
p
Bpopp
np
n
Bnonon
dxL
dx
Tk
eVnnn
dxL
dx
Tk
eVppp
exp1exp
exp1exp
0
Note that the expected limits are observed: np npo at x = - while pn pno at x = +
We can now calculate the electron and hole currents entering the diffusion region where E0.
At x = dn
1exp|
1exp|
Tk
eV
L
neD
x
neDJ
Tk
eV
L
peD
x
peDJ
Bn
pondx
pne
Bp
nopdx
npp
p
nfor the n-side.
for the p-side.
The total current is given by J = Je + Jp
n
pon
p
nopS
BS
L
neD
L
peDJ
Tk
eVJ
1exp
where
This result is the well known Shockley equation. Previously we wrote the current in terms of electron and hole generation currents:
1exp
1exp
2
22
Tk
eV
NL
D
NL
DenJ
NL
nD
L
pDJ
NL
nD
L
nDJ
Tk
eVJJej
BDp
p
An
ni
Dp
ip
p
nopgenh
An
in
n
pongene
B
gene
genh
Dno
Apo
Nn
Np
Note that
(on n-side)(on p-side )and
Contains fundamental parameters describing the rectifying behavior.
Examine the minority carrier densities and current densities for forward and reverse bias:
n,p
-dp 0 dn
pnonpo
pnnp
x
J
-dp 0 dn x
J=Jn+Jp
Jn
Jp
p-side n-side
n,p
-dp 0 dn
pnonpo
pnnp
x
p-side n-side
J
-dp 0 dn x
J=Jn+Jp
Jn
Jp
Note that (dp+dn) for reverse bias is greater than (dp+dn) for forward bias.
Forward Bias (V >0) Reverse Bias (V<0)
Possible to examine details of diffusion, using the Haynes-Shockley Experiment
Consider changes in minority carrier density with time:
p
nonnp
np
n
npnph
p
nonhn
pp
x
pD
x
pE
t
p
x
pDEpJ
pp
x
J
t
p
2
2
;
For E = 0, the solution follows the solution of the diffusion equation:
noppp
n pt
tD
x
tD
Ntxp
4exp
4),(
2
Where N= number of holes/Area generated (laser light pulse).
If E 0, x x - pEt
h
0L
Samplet
V
Oscilloscope
V t1 (E = 0)
t2
t3
t1 t2 (E > 0)
Et1
Et2