p-n junctions
A p-n junction is a metallurgical and electrical junction between p and n materials. When the
materials are the same the result is a HOMOJUNCTION and if they are dissimilar then it is
termed a HETEROJUNCTION.
Junction Formation:
p
NA acceptors -
+
+
+
+
n
ND donors
⊕ ⊕ ⊕
Jdiff
1. Majority carriers diffuse [holes from p to n and electrons from n → p]
2. Bare ionized dopants are exposed on either side of the junction. Positively charged
donors on the n-side and negatively charged acceptors on the p-side.
3. The dopant ions are contained in a region of reduced carrier concentration (as the mobile
majority charges have diffused as stated in (1)). This region is therefore called the
depletion region.
Edepl
p
-xn
⊕ ⊕ ⊕ ⊕ ⊕
- - + +
- - + +
- - + +
W(0)
n
xp
Jdiffusion
Jdrift
4. The process of diffusion continues until the depletion region expands to a width, W(0),
such that the electric field in the depletion region Edepl is large enough to repel the
diffusing carriers. More precisely,
Jdiffusion = Jdrift
(once equilibrium has been established)
FOR EACH CARRIER SEPARATELY
28
5. The driving force for carrier motion is the ELECTRO-CHEMICAL POTENTIAL
DIFFERENCE that exists between the two semiconductors in the bulk prior to junction
formation
In band diagram terms here are the before and after pictures:
EVACUUM
qx
qx
q∅ n
Ec
q∅ p
Eg
q (∅ p − ∅ n )
= E Fn − EFp
Eg
E Fp
Ev
Ec
E Fn
Ev
BEFORE: THE TWO MATERIALS ARE SEPARATE
Definitions:
a) qx = Electron affinity in units of energy. (use eV or Joules)
b) EFp = Fermi Level in the p-type material or electro-chemical potential of the p-type
material.
c) EFn = Fermi Level in the n-type material or electro-chemical potential of the n-type
material.
NOTE: THIS IS THE ELECTRO-CHEMICAL POTENTIAL OF ELECTRONS IN BOTH CASES
d) q∅ p and q∅ n are the work functions of the two materials (p and n) respectively.
(
)
e) Note that the work function difference between the two materials q ∅ p − ∅ n is the
difference between the electro-chemical potentials of the bulk materials EFn and EFp.
29
Ec
Eg
E Fp
Ev
E Fn = EFp = E F
Eg
-xp
W
x=0
Eg
+xn
AFTER: The materials are brought together to form a junction. The fermi levels EFn and EFp now
equalize or EFn = EFp = EF (IN EQUILIBRIUM)
a) Assume the p- material is kept at a constant potential (say ground).
b) The p- material has to increase its electro-chemical potential of electrons (upward motion
of the bands) until the fermi levels line up as shown in the diagram below where the
effect is simulated using two beakers of water in equilibrium with different amounts of
water in each beaker.
p
n
p
n
POTENTIAL
DIFFERENCE
BEFORE: TAP CLOSED
Energy increase
of the p
material
AFTER: TAP OPEN
c) The lowering of the electron energy of the n-type semiconductor is accompanied by the
creation of the depletion region.
30
d) The depletion region has net charge and hence the bands have curvature following Gauss’
Law:
OR
∂E ρ
=
∂x ∈
E = Electric Field
∂ 2V
ρ
=−
2
∂x
∈
E=−
∂V
∂x
where V = Potential energy(of unit positive charge)
OR
∂ 2 Ec ρ
=
∂x 2 ∈
Ec = − qV = Electron energy (Joules)
or –V = Electron energy (eV)
NOTE
NET CHARGE ⇔ CURVATURE OF THE BANDS
NEUTRAL REGIONS ⇒ NO NET CHARGE ⇔ BANDS HAVE NO CURVATURE
DO NOT CONFUSE SLOPE WITH CURVATURE
Neutral regions can have constant slope or equivalently no curvature
31
CALCULATING THE RELEVANT PARAMETERS OF A p- n JUNCTION
- - + +
- - + +
- - + +
+ +
-xp
xn
x=0
ρ (Ccm −3)
ρ =0
+++
+++
qN D
ρ =0
2. CHARGE PROFILE
- - -
−qN A
-xp
E =0
1. SCHEMATIC OF THE JUNCTION
E
xn
E =0
3. ELECTRIC FIELD
∂ E qN D
=
∂x
∈
∂ E − qN A
=
∂x
∈
E = Emax @ the junction, x = 0
qVbi = q (∅ p − ∅ n )
q∅ p
Ec
ELp
EFp
qVbi
Eg
q∅ Fp
qVbi
q∅n
4. BAND DIAGRAM
Ec
EFn
q∅ Fn
Ein
Ev
Ev
NOTE: qVbi = q∅ Fp + q∅ Fn
32
In the analysis depicted in the foru diagrams above, we assume
(i)
that the doping density is constant in the p-region at NA and in the n-region at ND and
that the change is abrupt at x = 0, the junction.
(ii)
the depletion region has only ionized charges –qNA (C cm-3 ) in the p-region and +qND
(Ccm-3 ) in the n-region. [Mobile charges in the depletion region are neglected, i.e. n,
p << NA- and ND+]
(iii)
The transition from the depletion region to the neutral region is abrupt at (- xp ) in the
p-region and (+xn ) in the n-region.
Calculation of the built- in voltage
From the band diagram it is clear that the total band bending is caused by the work function
difference of the two materials. If you follow the vacuum level (which is always reflects the
electrostatic potential energy variation and hence follows the conduction band in our
homojunction), you see that the band bending is the difference of the work function of the p-type
material q∅ p , and the n-type material q∅ n .
qVbi : q∅p − q∅ n
The built- in potential is therefore the internal potential energy required to cancel the diffusive
flow of carriers across the junction and should be exactly equal to the original electro-chemical
potential which caused the diffusion in the first place. THIS IS REASSURING.
To calculate Vbi from parameters such as doping let us follow the intrinsic level from the p-side,
Eip , to the n-side, Ein . Again the total band bending of the intrinsic level is the built in potential
or Eip – Ein = qVbi
or (Eip – EF) – (Ein – EF) = qVbi
Defining
And
Eip – EFp as q∅ Fp
EFn – Ein as q∅ Fn
[Note in equilibrium
EFn = EFp = EF]
We can rewrite qVbi as q∅ Fn + q∅ Fp = qVbi
From Fermi- Dirac Statistics:
q∅
E − EFp
E − EF
p Po = ni e i
or
ni e ip
= ni e kT
RT
kT
Ei − EF
similarly nno = ni e
RT
Assume full ionization
33
p Po = N A and nno = N D
∴ N A = ni e
or
q∅ Fp
kT
q∅ Fp = kT ln
&
NA
ni
N D = ni e
&
q∅ Fn
kT
q∅ Fn = kT ln
ND
ni
 N
N 
∴ q ∅Fp + q ∅Fn = kT  ln A + ln D 
ni 
 ni
N N
or
qVbi = kT ln A 2 D
ni
Vbi =
kT N A N D
ln
q
ni 2
Calculating Depletion Region Widths
From the electric field diagram, Figure 3
−qN A
Emax =
xp
∈
qN A
Emax =
⋅ xp
∈
The magnitude of the area under the electric field versus distance curve (shaded area in figure 3)
is by definition ∫ xn
− xp E ⋅ dx = Voltage difference between –xp and +xn = Vbi
1
⋅W ⋅ Emax
2
1

 2 ⋅ Base ⋅ Height 


1
qN A
Vbi = ⋅ ( xn + x p ) ⋅
xp
2
∈
or Vbi =
We now invoke charge neutrality. Since the original semiconductors were charge neutral the
combined system has to also be charge neutral (since we have not created charges). Now since
the regions beyond the depletion region taken as a whole has to be charge neutral. OR all the
positive charges in the depletion region have to balance all the negative charges.
If the area of the junction is A cm2 then the number of positive charges within the depletion
region from x = 0 to x = xn is
qND
xn
⋅
A = Coulombs
⋅
Coulombs cm-3
cm-3
cm2
Similarly, all the negative charges contained in the region between x = -xp and x = o is
qN A ⋅ x p ⋅ A = Coulombs
34
charge neutrality therefore requires
qN A x p ⋅ A = qND xn ⋅ A or N A x p = ND xn IMPORTANT
To calculate w, xn and xp we use the above relation in the equation for Vbi below
1
Vbi = ⋅ ( xn + xp ) ⋅ Emax
2
p-side
n-side
1
qN
Vbi = ( xn + x p ) A x p
2
∈
Substituting for xn
 qN
1  N + ND
or Vbi =  A
+ xp  A xp
2  ND
 ∈
 N + ND
2∈
Vbi =  A
qN A
 ND
1
qN
Vbi = ( xn + x p ) D xn
2
∈
 qN
1  N + ND
Vbi =  D
+ xn  D xn
2  NA
 ∈
 N + ND  2
2∈
Vbi =  A
 xn
qND
 NA 
qN A
qN D
NOTE: in this analysis Emax was calculated as
x p from the p-side and
xn fom the n-side
∈
∈
2 ∈ ND
1
2 ∈ NA
1
xp =
⋅
⋅
⋅ Vbi
xn =
⋅
⋅
⋅ Vbi
q N A (N A + ND )
q ND (N A + ND )
or
 2
 xp

W = xn + x p


ND
NA
⋅
+

N D ⋅ N A + N D 
 N A ⋅ N A + N D
=
2∈
⋅Vbi
q
=


2∈
NA + ND
⋅ Vbi ⋅ 

q
 N A N D ⋅ N A + N D 
W=
2 ∈ NA + ND
⋅
⋅Vbi
q
NAND
IN EQUILIBRIUM
W (0) =
2∈  1
1 
⋅
+
 ⋅ Vbi
q  NA N D 
IMPORTANT
In general
W (v ) =
2∈  1
1 
⋅
+
 (Vbi ± V
q  N A ND 
V is the applied bias
)
+ V in reverse bias (w expands)
35
− V in forward bias (w shrinks)
FORWARD BIAS
REVERSE BIAS
ZERO
BIAS
ZERO
BIAS
− xPo
− xPo
x=0
xno
x=0
xno
q (Vbi − V )
EFp
qVF
E Fn
qV R
x=0
− xP
qN A
− xP
qN D
- - -
qN A
− xP
E Fn
− xPo
− xn
++
++
Vbi + V
EFp
x=0
+++
+++
- - - - -
xn 0
xn
qN D
− xn
Emax
qNAx p
=
∈
− xP
FEATURES:
1. Total Band bending is now Vbi-V
2. The shaded area ∫ E ⋅ dx = Vbi − V
3. The edges of the depletion region move towards
the junction or W decreases.
4. E Fn − E Fp = VF , the electrochemical potentials
separate by an amount equal to the potential
difference applied, VF.
− xn
Emax =
qNAx p
∈
FEATURES:
1. Total Band bending is now Vbi + V
2. The shaded area ∫ E ⋅ dx = Vbi + V
3. Depletion region expands.
4. E Fp − EFn = VR = amount of potential
36
difference, (VR).
Current flow in p-n junctions
VERY IMPORTANT
A. Forward Bias
JDIFFUSION
JDRIFT
JDIFFUSION
JDRIFT
Vbi − V
Vb
EFn
EFp
EFn
EFp
qV
Zero Bias
Forward Bias
The depletion width is
reduced which increases the
diffusion current. The drift
current remains the same.
The net current is due to the
imbalance of drift and
diffusion
Under forward bias electrons from the n-region and holes from the p-region cross the junction
and diffuse as minority carriers in the p and n-regions respectively. To understand how excess
minority carriers are injected and diffuse one has to understand the LAW OF JUNCTION which
now follows.
p
p
J DIFFUSION + J DRIFT = 0
n
n
J DIFFUSION + J DRIFT = 0
37
Consider the two situations shown below one at zero bias and the other under forward bias.
Ec
qψ ( x′)
ELP
EFp
Ev
qψ ( x′)
qVbi
x′ = 0
Ec ( x′ )
Ec
Ein
Ev
x′ = w
P ( x′ )
Pp 0
nn 0
n ( x′ )
np0
x=0
x′ = 0 is @ x = − x p
Note that Ei is a function of x′ and is Eip in the bulk
p and Ein in the bulk n.
Note that p ( x′) is always given by
E ( x′) − EFp ( x′)
p ( x′ ) = ni e i
kT
Where Ei (x) and EFp (x) are the intrinsic fermi- level
and the fermi level for holes at any place x.
Since we are at zero bias and at equilibrium EFp is
not a function of x′ .
E − EF − Eip − EF
Ei ( x′) − EF
= ni e ip
kT
kT
E − EF −  Eip − Ei (x′) 
= ni e ip
⋅e 

kT
kT


∴ p ( x′ ) = ni e
Pp 0
x ′ = W is @ xn
qψ ( x′)
kT
Eip − Ei ( x′)
or p ( x′ ) = ( pP 0 ) ⋅ e −
where qψ ( x′) ≡
LAW OF
THE
JUNCTION
kT
or p ( x′) decreases exponentially with the local band
bending
38
Note that the edge of the depletion region on the n-side x = xr or x ′ = W the hole concentration is
given by
qψ (x ′ = W )
p ( x′ = W ) = pP 0 ⋅ e−
kT
The total band bending at ( x ′ = W ) is Vbi
qV
∴ p ( x′ = W ) = p p0 e− bi
kT
We know that p ( x′ = W ) is the hole concentration in the n-type semiconductor or np0
qV
So, n p 0 = p p 0e− bi
kT
Let us verify that what we derived does not contradict what we learned in the past.
2
ni
qV
np 0 =
= p pe − bi
nn 0
kT
qVbi
ni 2
∴e
=
kT
p p0 nn0
−
kT nn0 p p0
ln
or with full ionization
2
q
ni
kT N A N D
Vbi =
ln
SAME AS BEFORE
q
ni 2
∴Vbi =
39
Ec
Vbi − VF
Eip
EFp
EFn
VF
Ein
p p0
nn 0
n p (− x p )
On application of a forward bias the electron and
hole concentrations continue to follow the relation
qψ ( x′)
that pn ( x′) = p p 0e−
the law of the junction.
kT
The difference from the zero bias case is that at the
edge of the junction of x ′ = W
ψ (W ) = V bi −V F , not Vbi as in the zero bias case.
∴ pn (W ) = p p0e −
q
np0
Change the
coordinate system
so that xn in
x ′′ coordinate is 0.
kT
pn ( xn )
or pn (W ) = p p0e −
∆ np (− x p )
(Vbi −VF )
∆ np ( − xn )
Consider
only the
n-region.
pn 0
pn (0) = Pn0e
qVb i
kT
qV
⋅ e kT
qV F
kT
THE MINORITY CARRIER CONCENTRATION
IS RAISED FROM ITS ZERO BIAS qv VALUE
BY THE FACTOR e
qV F
kT
.
IMPORTANT
THEN IN THE n-region
∆pn ( x ′′) = ∆pn (0)e
∆pn
− x ′′ / L p
(0)
x = xn
x′ = w
x′′ = 0
40
Similarly the minority carrier concentrations at the edge of the depletion region on the p-side
qV F
n p ( − x p ) = n p0e kT
What happens to these excess carriers? They diffuse away from the edge of the depletion region
to the bulk. The profile that governs the diffusion is set by the recombination rate of the minority
carriers in the bulk. The situation is analyzed by the continuity equation.
∆pn
(0)
x ′′
At any point x ′′ the continuity equation states
1
(x)
diff
∇ ⋅ J p( x ) = G − R Using, J p = J p ( x )
q
′′
′′
d 2 pn( x )
∆pn( x )
by neglecting drift currents (which will be explained later) we get − Dp
=−
τp
dx2
(
)
d2
d2
d2
− pn 0
′′
′′
p
(
x
)
=
p
(
x
)
since
pn 0 = 0
n
n
2
2
2
dx
dx
dx
Note that
d2
d2
′′
p
(
x
)
=
∆pn ( x′′)
n
2
2
dx
dx
∴
d 2 ∆p n ( x′′) ∆pn ( x ′′)
−
=0
2
τp
dx
We know that far away from the junction the excess hole concentration has to be zero since
excess holes have to eventually recombine.
Dp
∴ ∆pn ( x′′) = c1e
+
x′′
Lp
+ c2e
−
x′′
Lp
where Lp = Dpτ p = Diffusion length of the holes
which can be proven to be the average distance a hole diffuses before it recombines with an
electron.
Also, c1 ≡ 0 as ∆pn → 0 as x′′ → ∞
−
∴ ∆pn ( x′′) = c2 e
x′′
Lp
At x ′′ = 0 ∆pn (0) = pn (0) − pn 0 = pn0e
Or ∆pn
(0)

= pn 0  e

qVF
kT

− 1

∆pn ( x ′′) = ∆pn (0)e
−
x′′
Lp
qV F
kT
− pn
IMPORTANT
-(2)
41
This exponential relationship applies to the minority electrons as well where
∆n p ( x ′′′) = ∆n p (0)e
n p ( x′′′)
x′′′
∆n p (0)
x′′′ = 0
−
x′′′
Ln
 qVF

∆n p (0) = np 0  e kT − 1


(New Coordinate System)
42
DERIVATION OF THE DIODE EQUATION UNDER FORWARD BIAS
We now have the minority carrier charge profiles [WHICH IS ALWAYS OBTAINED FROM
THE SOLUTION OF THE CONTINUITY EQUATION]. From this we can calculate the current
across the diode.
Note that the current measured anywhere in the diode has to be the same and equal to the current
in the external circuit.
p
J
J
- +
- ++
- ++
- +
x=0
n
J
p p0
nn 0
np0
pn 0
J T = J p + Jn
JT = J n + J p
Jp
JT ; J p
J T ; Jn
drift
Jn
diffusion→0
Jn
x ′′′ → ∞
Jn
Jp
drift
diffusion→0
Jp
x′′ → ∞
43
J = JT = J n + J p
Jp
drift
Jn
IMPORTANT
FIGURE
Jp
Jn
drift
J p ( x′′)
Observatio n: Far from the junction diffusion current → 0 as J p ( x′′) = −qD p
d
= −qDp
dx
Jn
d
∆pn ( x′′)
dx
x′′
x′′

−

−

Dp 
Lp
Lp
 ∆pn (0)e  = +q
⋅  ∆pn (0) e 
Lp 



Dp
J pdiff (x ′′) = q
LP
∆pn ( x ′′)
-(3)
Since ∆pn ( x′′) → 0 as x′′ → ∞
Jp
diff
( x′′) → 0 as x′′ → ∞
Since J T is always constant and J T = J n + J p in the region far from the junction
JT = Jn
drift
+ Jn
+ Jp
diff
drift
No slope in
electron
profile
nno ≠ f ( x)
So what about J p
drift
+Jp
diff
No slope in
hole
profile
pno ≠ f ( x )
or MINORITY CARRIER DRIFT CURRENTS?
44
We assumed the regions beyond the depletion regions were neutral. However, the application of
a voltage across the diode must result in a field in these regions as shown below.
p
- +
- +
- +
- +
Ε
n
Ε
V
This can be readily understood if you break the diode into three regions, the bulk p, and the bulk
n and depletion region. Since the bulk regions are highly doped they are conductive and hence
have a small resistance. The depletion region, which is devoid of carriers, may be considered a
large resistance. SCHEMATICALLY, we can consider the diode to be as shown below.
Vbulk , p
Vbulk , p
Rbulk , p
Rb u l k, n
I
I
R junction
V
It can be readily seen that the voltage drop in the bulk regions are much smaller than the drop across the junction
∴ JT = Jn
drift
+ Jp
drift
= qµ n ⋅ nn 0 ⋅E + q µ p ⋅ pn0 ⋅ E (in the n-region)
= Jn
drift
J T ≅ Jn
 µ p pn0 
1 + µ ⋅ n 

n
n0 
drift
far from the junction as
Jp
drift
Since we are considering low level injection where
<< J n
drift
by the ratio
µ p pn0
⋅
µn nn0
pn 0 ∆
≡<< 1
nn 0
MINORITY CARRIER DRIFT CURRENTS CAN ALWAYS BE NEGLECTED
Far from the junction the current is carried by majority carrier drift OR
J T = Jn
drift
drift
+ Jn
drift
in the p bulk and
in the n bulk. As is apparent from the figure as you get closer to the junction, because minority carrier
diffusion is significant and
Jn
Jp = Jp
diffusion
J T = J n + J p is always true, J n is always J T − J p
diff
in the n region. (
J n is always
but it is not necessary at the moment to evaluate each component separately).
45
HERE COMES AN IMPORTANT ASSUMPTION:
We assume no recombination or generation in the depletion region. This is valid because the depletion region width,
W, is commonly << Ln and also it will be proven later that recombination only occurs at the junction (x = 0)
because of the carrier concentration profiles and hence is limited in extent. The latter is the correct reason so you
have to accept it.
If this is true then the continuity equation dictates that
1
1
∇⋅ J p = G − R = 0 = ∇ ⋅ J n OR BOTH J n and
q
q
J p are constant across the junction (as shown).
So IF WE KNEW
J n and J p in the depletion region then we could add them to get the total current
J T = J n + J p (everywhere) = J n + J p (depletion region for convenience)
But we know
J p at the edge of the depletion region is only J p
diff
(on the n-side because the minority drift is
negligible).
∴ J p (depletion region) Assumption of no G-R in the depletion region J p
OR
Jp =q
Dp
Lp
diffusion
( x′′ = 0)
∆pn (0) ← from (2), (3)
 qVF

pn 0  e kT − 1 - (4a)
Lp


 qVkTF 
Dn
Similarly J n = q
n  e −1 - (4b)
Ln p0 

Jp = q
Dp
JT = J n + J p

n p0   qVkTF

p
J T = q  D p = n 0 + Dn
  e − 1
Lp
Ln  


IMPORTANT
Note: The assumption of no G-R in the depletion region allowed us to sum the minority diffusion currents at the
edges of the junction to get the total current. This does not mean that only diffusion currents matter. Current is
always carried by carrier drift and diffusion in the device. The assumption allowed us to get the correct expression
without having to calculate the electric field in the structure (a very hard problem).
So
 qV

J T = J s  e kT − 1



n p0 
p
J s = q  D p n 0 + Dn

Lp
Ln 

I T = JT ⋅ A (A = Diode Area)
- (b)
46
REVERSE BIAS CHARACTERISTICS
np0 e
e
The case for reverse bias is very different.
Here the application of bias increases
barriers. The only carriers that can flow are
minority carriers that are aided by the
electric field in the depletion region.
HERE the minority carriers are electrons
injected from the p-region to the n-region
(OPPOSITE TO THE FORWARD BIAS
CASE).
VR
e
⊕ pn 0
e
np0
e n p ( x′′′)
−x p
+ xn
e
pn 0
pn ( x′′) e
First Observation:
Since we are only dealing with minority carrier currents we know that minority carrier drift can
be neglected. Hence only minority carrier diffusion is relevant. To calculate diffusion currents
we need to know the charge profile. Charge profiles are obtained by solving the continuity
equation (in this case equivalently the diffusion equation as drift is negligible).
We assume that the large electric field in the reverse-biased p-n junction sweeps minority
carriers away from the edge of the junction.
Assume n p ( − x p ) = 0 SCHOCKLEY
BOUNDARY - (7)
And
pn ( + xn ) = 0 CONDITIONS
We also know that the minority carrier concentration in the bulk is n p 0 (p-type) and pn 0 (n-type)
respectively. Therefore, the shape of the curve will be qualitatively as shown, reducing from the
bulk value to zero at the depletion region edge.
Consider the flow of minority holes. The charge distribution is obtained by solving
d 2 pn
Dp
+ Gth − R = 0 -(8)
dx ′′2
assuming that the only energy source is thermal.
47
Then D p
d 2 pn pn0 − pn
+
= 0 -(9)
dx′′2
τp
Note that this term is a generation term because pn < pn 0 for all x ′′ . This is natural because both
generation and recombination are mechanisms by which the system returns to its equilibrium
value. When the minority carrier concentration is above the equilibrium minority carrier value
then recombination dominates and when the minority carrier concentration is less than that at
equilibrium then generation dominates. The net generation rate is (analogous to the
recombination rate).
p − pn
Gth − R = n0
(ASSUMING τ p the generation time constant = recombination G-R = 0
τp
which is????)
Note: when pn = pn0 true in equilibrium.
Again using ∆pn ( x′′) = p n0 − pn ( x′′) we get D p
+
x ′′
Lp
−
d 2 ∆p n ( x ′′) ∆pn ( x ′′)
+
=0
dx′′2
τp
x′′
Lp
Again ∆pn ( x ′′) = C1e + C2e
C1 = 0 (for physical reasons)
At x ′′ = ∞ ∆pn → 0
At x ′′ = 0 ∆pn = pn0 − pn (0) = pn0
∴∆pn ( x′′ ) = pn 0e
− x ′′
Lp
pn 0 − pn ( x ′′) = pn 0e
− x ′′
Lp
− x′′


L
OR pn ( x ′′) = pn0  1 − e p 




-(10)
∴
The flux of holes entering the depletion region is J p ( x′′ = 0) = qDp
Jp =q
Dp
⋅ pn0 Similarly J n = q
dpn
( x ′′ = 0 )
dx′′
Dp
np 0
Lp
Lp
Assuming no generation in the depletion region the net current flowing is

n p0 
p
J s = q  D p n 0 + Dn
 -(11)
Lp
Ln 

This is remarkable because we get the same answer if we took the forward bias equation (valid
only in forward bias) and arbitrarily allowed V to be large and negative (for reverse bias)
 qV

i.e. J = J s  e kT − 1 If V is large and negative J R = − J s which is the answer we derived in 11.


48
In summary,
p
⊕
e
Ln
e
⊕
Lp
x=0
n
The equation (11) can be understood as follows. Concentration on the p-region. Any minority
carrier electrons generated within a diffusion length of the n, depletion edge can diffuse to the
edge of the junction and be swept away. Minority electrons generated well beyond a length Ln
will recombine with holes resulting in the equilibrium concentration, n p 0 . Similarly holes
generated within, Lp , a diffusion length, of the depletion region edge will be swept into the
depletion region.
IMPORTANT OBSERVATION:
Look at the first term in equation 11. The slope of the minority carrier profile at the depletion
p
( Difference from Bulk Value ) This is always true when recombination and
edge = n0 =
Lp
Lp
generation dominate.
slope
pn = 0
pn 0
↑ Reverse bias
Recall that even in forward bias (shown below) the slope of the carrier profile is again
Difference From Bulk Value ∆pn (0)
=
Lp
Lp
slope
 qV

∆pn = p n0 e kT − 1


49
In the event that there is light shinning on the p- n junction as shown below
p
n
x=0
then the charge profile is perturbed in the following manner
∆n p GLτ n
∆pn GL τ p
e
np0
e
−x p
x = xn
x=0 e
pn 0
e
where far in the bulk region an excess minority carrier concentration is generated where
∆n p = GLτ n and ∆pn = GLτ p
The new equation to be solved for reverse saturation current differs from equation 8 in that a
light generation term is added.
d2p
d 2 p pn 0 − pn
Dp 2 + Gth − R + G L = 0
or
Dp
+
+ GL = 0
dx′′2
τp
dx
only difference
net thermal generation
with boundary conditions similar to before pn ( ∞) = pn0 + τ p G L and pn (0) = 0
−x


L
we get pn 0 ( x′′) = ( pn0 + τ p GL )  1 − e p 




p
bulk
The slope of the charge profile at the edge of the depletion region is
∂pn (0) pn 0 + τ p GL
=
∂x′′
Lp
 pn 0 + τ p G L 
D
Similarly, J n ( x′′′ = 0) = q n ( GLτ n + n p0 )
∴ J p = qDp ⋅ 



Lp
Ln



n pbulk Dpnbulk 
J Re verse = q  Dn ⋅
+

Ln
L p 

50
By changing the slope of the minority profile at the edge of the junction I can control the reverse
current across a diode.
J
V
Js
J R can be charged
by changing the
slope of minority
carrier profile.
Controlling and monitoring the current flowing across a reverse bias diode forms the bases of a
large numb er of devices.
METHOD OF CONTROLLING THE MINORITY
DEVICE
CARRIER SLOPE
1. TEMPERATURE: Recall that the slope is (for the pn
material) given by p0
Ln
n
J s, n = qDn p 0
Ln
But n p 0
ni2
=
p p0
Assuming full ionization p p 0 ≠ f
2
n
NNe
∴ n p0 = i = c c
NA
NA
−
(T )
and is always ≈ N A
THERMOMETER
Read J s and extract T
Eg
kT
Eg
−
Dn (T )
⋅ Nc (T )N v (T )e kT
Ln (T )
EXPONENTIAL DEPENDENCE ON T
(the other terms have weaker dependence).
∴ J s ,n ( T ) = q ⋅
51
2. Change n p 0 to n p = np0 + GLτ n as described. The
generation rate is dependent on the input photon flux.
D
∴ J s ,n = q n (n p 0 + G Lτ n )
Ln
Measure J s, n
PHOTODETECTOR
(used in optical communications)
Determine GL and input photon flux
3. Change the minority carrier concentration by an
electrical minority carrier injector i.e. a p-n junction.
TRANSISTOR
A TRANSISTOR, short for TRANSfer ResISTOR, is the basic amplifying element in
electronics. The basis of its amplification and its structure is shown below.
FORWARD BIAS REVERSE BIAS
p
E
n
B
pnp
c
C
A transistor consists of a forward bias junction in close proximity to a reverse bias p- n junction
so that carriers
npn
E
n
FORWARD BIAS
C
n
B
p
REVERSE BIAS
injected from forward bias junction (from the emitter labeled E) can travel through the
intermediate layer (called the BASE and labeled B) and across the reverse biased junction into
the COLLECTOR, labeled C.
Schematically for the case of the emitter being a p material and the base being n-type, (a pnp
transistor) the DOMINANT CURRENT FLOW (to be modified later is shown below).
p
E
INJECTED
HOLES
n
p
COLLECTED
HOLES
B
C
52
The holes injected from the p-type emitter contribute to the EMITTER CURRENT, IE, and the
holes collected contribute to the collector current, Ic. These currents in the diagram above are
equal (AN APPROXIMATION TO BE MODIFIED LATER). This situation is equivalent to
Ic
IE
Rin
VIN
applied
Rout
VOUT
measured
The forward bias junction across which the input voltage is applied has a low resistance.
I
Rsmall
V
Rl arg e
The forward bias resistance of a diode can be readily calculated from the I-V characteristics
qV
 qV

I = I s  e kT − 1 ; I s e kT


−1
−1
qV
 q

∂V  ∂I 
kT
 q 
kT
∴R =
=
=
⋅ I s e  OR R =  ⋅ I  OR R =
IMPORTANT

∂I  ∂V 
qI
 kT 
 kT

As I increases R decreases.
25mV
Note that @ room temperature at a current of 25 mA R =
= 1Ω FOR ANY DIODE
25mA
Therefore in a transistor a SMALL INPUT VOLTAGE can generate a large current, IE, because
of the small Rin . This same current flows across a large resistance (as the reverse bias resistance
of the collector junction is large) as Ic.
∴The measured voltage across the output resistance, Rout , is I c ⋅ Rout . The input voltage is
I E ⋅ Rin . ∴ The voltage gain of device, Av , is
V
I R
R
IMPORTANT Av = out = c out ; out
Vin
I E Rin
Rin
In general, if you can control a current source with a small voltage then you can achieve voltage
amplification by passing the current through a large load resistor.
RL = LOAD RESISTOR
I controlled
by small Vin
53
Recall that we could modulate the reverse bias current by changing the minority carrier
concentration flux injected into the junction. The diagram is reproduced below.
I
I s  f ( n p 0 , pn0 ) 
V
Increasing
n p and pn
A transistor achieves modulating the minority concentration by injecting minority carriers using
a forward biased p-n junction. Recall that a p- n junction injects holes from a
⊕
p
en
p-region into the n-region (raising) the minority carrier concentration from pn 0 to a new value
pn ( x )
p
n
pn ( x )
n p ( x)
np0
pn 0
The same applies for the n-region. Let us now consider a p-n-p transistor. It consists of a p- n
junction that injects minority carriers into another but now reverse-biased p-n junction.
p
EMITTER
n
BASE
This junction
injects minority
carriers
p
COLLECTOR
This junction
collects carriers
54