6.012 - Microelectronic Devices and Circuits
Lecture 6 - p-n Junctions: I-V Relationship - Outline
• Announcements
First Hour Exam - Oct. 7, 7:30-9:30 pm; thru 10/2/09, PS #4
• Review
Minority carrier flow in QNRs: 1. Lmin << w, 2. Lmin >> w
• I-V relationship for an abrupt p-n junction
Assume: 1. Low level injection
2. All applied voltage appears across junction:
3. Majority carriers in quasi-equilibrium with barrier
4. Negligible SCL generation and recombination
Relate minority populations at QNR edges, -xp and xn, to vAB
Use n'(-xp), p'(xn) to find hole and electron currents in QNRs
Connect currents across SCL to get total junction current, iD
• Features and limitations of the model
Engineering the minority carrier injection across a junction
Deviations at low and high current levels
Deviations at large reverse bias
Clif Fonstad, 9/29/09
Lecture 6 - Slide 1
QNR Flow: Uniform doping, non-uniform LL injection
We use the 5 QNR flow conditions* to simplify our 5 equations...
(assuming a p-type sample)
"p(x,t) 1 "J h (x,t)
"n(x,t) 1 "J e (x,t)
+
=
#
$ gL (x,t) # [ n(x,t) p(x,t) # n i2 ] r(t)
"t
q "x
"t
q "x
LLI
Quasi-static
Quasi-static
1 dJ h (x,t)
1 dJ e (x,t)
n'(x,t)
"
= #
$ gL (x,t) #
q dx
q dx
%e
LLI
#p(x,t)
dn'(x,t)
J h (x,t) = qµ h p(x,t)E(x,t) " qDh
" qµ h po (x)E(x,t) # qDh
#x
dx
Quasi-neutrality
!
Negligible minority drift
!
!
!
"n(x,t)
dn'(x,t)
J e (x,t) = qµ e n(x,t)E(x,t) + qDe
" J e (x,t) # qDe
!
"x
dx
!
q
dE x (x,t)
q Uniform doping
+
#
p(x,t) # n(x,t) + N d (x) # N a (x)] =
" [ p'(x,t) $ n'(x,t)]
[
"(x)
dx
#
!
..and end up with one equation in n': the static diffusion equation!
d 2 n'(x,t)
n'(x,t)
1
"
=
"
gL (x,t)
2
!
dx
De # e
De
Clif Fonstad, 9/29/09
Lecture 6 - Slide 2
* Five assumptions that define flow problems AND should be validated at the end.
QNR Flow, cont.: Solving the steady state diffusion equation
The steady state diffusion equation in p-type material is:
d 2 n'(x)
n'(x)
1
"
=
"
gL (x)
2
2
dx
Le
De
and for n-type material it is:
d 2 p'(x)
p'(x)
1
"
=
"
gL (x)
2
2
dx
Lh
Dh
!
In writing these expressions we have introduced Le and Lh,
the minority carrier diffusion lengths for holes and
electrons, as:
Lh " Dh # h
Le " De # e
!
We'll see that the minority carrier diffusion length tells us how
far the average minority carrier diffuses before it recombines.
In a basic p-n
! diode, we have gL!= 0 which means we only need
the homogenous solutions, i.e. expressions that satisfy:
n-side: d 2 p'(x)
dx 2
Clif Fonstad, 9/29/09
p'(x)
"
= 0
2
Lh
p-side:
d 2 n'(x)
n'(x)
"
= 0
2
2
dx
Le
Lecture 6 - Slide 3
QNR Flow, cont.: Solving the steady state diffusion equation
We seldom care about this general result. Instead, we find
that most diodes fall into one of two cases:
Case I - Long-base diode: wn >> Lh
Case II - Short-base diode: Lh >> wn
Case I: When wn >> Lh, which is the situation in an LED, for
example, the solution is
p'(x) " p'(x n )e#( x#x n ) L h for
x n $ x $ wn
This profile decays from p'(xn) to 0 exponentially as e-x//Lh.
The
! corresponding hole current for xn ≤ x ≤ wn in Case I is
dp'(x) qDh
J h (x) " #qDh
=
p'(x n )e#( x#x n ) Lh
dx
Lh
for x n $ x $ w n
The current decays to zero also, indicating that all of the excess
minority carriers have recombined before getting to the contact.
!
Clif Fonstad, 9/29/09
Lecture 6 - Slide 4
QNR Flow, cont.: Solving the steady state diffusion equation
Case II: When Lh >> wn, which is the situation in integrated
Si diodes, for example, the differential equation simplifies
to:
d 2 p'(x)
p'(x)
dx
2
=
2
h
L
" 0
We see immediately that p'(x) is linear:
p'(x) = A x + B
Fitting the boundary conditions we find:
!
* $ x # x 'n
p'(x) " p'(x n ),1# &
)/! for x n 0 x 0 w n
+ % w n # x n (.
This profile is a straight line, decreasing from p'(xn) at xn to 0 at wn.
!
In Case II the current is constant for xn ≤ x ≤ wn:
dp'(x)
qDh
J h (x) " #qDh
=
p'(x n )
dx
wn # x n
for
x n $ x $ wn
The constant current indicates that no carriers recombine
before reaching the contact.
Clif Fonstad, 9/29/09
!
Lecture 6 - Slide 5
QNR Flow, cont.: Uniform doping, non-uniform LL injection
Sketching and comparing the limiting cases: wn>>Lh, wn<<Lh
Case I - Long base: wn >> Ln (the situation in LEDs)
p'n(x) [cm-3]
p'n(xn)
Jh(x) [A/cm2]
qDhp'n(xn) Lh
e-x/Lh
0
xn
xn+Lh
wn
x [cm]
e-x/Lh
0
xn
xn+Lh
wn
x [cm]
Case II - Short base: wn << Ln (the situation in most Si diodes and transistors)
p'n(x) [cm-3]
p'n(xn)
Jh(x) [A/cm2]
qDhp'n(xn) [wn-xn]
0
xn
Clif Fonstad, 9/29/09
wn
x [cm]
0
xn
wn
x [cm]
Lecture 6 - Slide 6
QNR Flow, cont.: Uniform doping, non-uniform LL injection
The four other unknowns
⇒ In n-type the steady state diffusion equation gives p’.
⇒ Knowing p', we can easily get n’, Je, Jh, and Ex:
dp'(x)
dx
First find Jh:
J h (x) " # qDh
Then find Je:
J e (x) = JTot " J h (x)
!
Next find Ex:
!
Then find n’:
Note: In Lec 5 we saw this
for a p-type sample.
&
1 #
De
E x (x) "
J h (x)(
%J e (x) +
qµ e n o $
Dh
'
n'(x) " p'(x) #
$ dE x (x)
q dx
!
Finally, go back and check that all of the five conditions are
met by the solution.
 Once
! we solve the diffusion equation and get
Clif Fonstad, 9/29/09
the minority carrier excess we know everything.
Lecture 6 - Slide 7
Current flow: finding the relationship between iD and vAB
There are two pieces to the problem:
• Minority carrier flow in the QNRs is what limits the current.
• Carrier equilibrium across the SCR determines n'(-xp) and p'(xn), the
boundary conditions of the QNR minority carrier flow problems.
Ohmic
contact
A
+
iD
Uniform p-type
Uniform n-type
p
n
-
vAB
-wp
-xp 0 xn
Quasineutral
region I
Minority carrier flow
here determines the
electron current
Clif Fonstad, 9/29/09
Ohmic
contact
Space charge
region
The values of n' at
-xp and p' at xn are
established here.
- Today's lecture topic -
wn
B
x
Quasineutral
region II
Minority carrier flow
here determines the
hole current
Lecture 6 - Slide 8
The p-n Junction Diode: the game plan for getting iD(vAB)
We have two QNR's and a flow problem in each:
Quasineutral
region I
Ohmic
contact
A
+
iD
Quasineutral
region II
p
Ohmic
contact
B
n
-
vAB
x
-wp
-xp 0
n'(-xp) = ?
x
0 xn
n'p
wn
p'n
p'(xn) = ?
p'(wn) = 0
n'(-wp) = 0
x
-wp
-xp 0
x
0 xn
wn
If we knew n'(-xp) and p'(xn), we could solve the flow problems
and we could get n'(x) for -wp<x<-xp, and p'(x) for xn<x<wn …
Clif Fonstad, 9/29/09
Lecture 6 - Slide 9
….and knowing n'(x) for -wp<x<-xp, and p'(x) for xn<x<wn, we can
find Je(x) for -wp<x<-xp, and Jh(x) for xn<x<wn.
n'(-xp,vAB) = ?
n'p
p'n
p'(xn,vAB) = ?
p'(wn) = 0
n'(-wp) = 0
-xp 0 x
-wp
Je(-wp<x<-xp)=qDe(dn'/dx)
-wp
Je
0 xn
x
wn
Jh
Jh(xn<x<wn)=-qDh(dp'/dx)
-xp 0
x
0 xn
x
wn
Having Je(x) for -wp<x<-xp, and Jh(x) for xn<x<wn, we can get iD
because we will argue that iD(vAB) = A[Je(-xp,vAB)+Jh(xn,vAB)]…
…but first we need to know n'(-xp,vAB) and p'(xn,vAB).
Clif Fonstad, 9/29/09
We will do this now.
Lecture 6 - Slide 10
The impact of the barrier height change on the carrier populations
and fluxes:
qφ
Unbiased
junction
Population in
equilibrium with
barrier
Forward bias qφ
on junction
Barrier lowered so
carriers to left can
cross over it.
The flux is limited
by how fast they
diffuse in the QNR.
qφ
x
x
Reverse bias
on junction
Barrier raised so
the few carriers on
top spill back down.
x
Clif Fonstad, 9/29/09
Lecture 6 - Slide 11
Hole potential energy, qφ
Majority carriers against the junction barrier
Zero applied bias, vAB = 0; thermal equilibrium barrier
po ( x " x n ) = n ie#q$ n
kT
= n i2 N Dn
q" B
"q# (x ) kT
p
(x
<
x
<
x
)
=
n
e
n
i
! o p
!
po ( x " #x p ) = n ie
!
Notice that:
!
#q$ p kT
-xp
po ( x " x n ) = n ie#q$ n
= N Ap
xn
kT
#q $ #$
= N Ap e ( n p )
kT
= N Ap e#q$ B
kT
"The holes are in equilibrium with the barrier."
Clif Fonstad, 9/29/09
!
Lecture 6 - Slide 12
Boundary condtions at the edges of the space charge layer:
What are n’(-xp) and p’(xn)?
Begin by looking at the situation in thermal equilibrium,
where we have:
po ("x p ) = N Ap and po (x n ) = n i2 N Dn
qφ
!
If the population of holes at the top
of the potential “hill” is related to
the population at the bottom by a
Boltzman factor, then we should
also find that:
po (x n ) = po ("x p )e"q# b / kT
Do we?
!
kT N Ap N Dn
"b =
ln
q
n i2
#
qφb
-xp 0
xn
x
n i2
= N Ap e$q" b / kT
N Dn
n i2
Thus : po (x n ) =
= N Ap e$q" b / kT = po ($x p )e$q" b / kT
N Dn
Clif Fonstad, 9/29/09
!
YES, we do, and the Boltzman relationship holds.
Lecture 6 - Slide 13
Hole potential energy, qφ
Majority carriers against the junction barrier
Forward bias, vAB > 0; barrier lowered, carriers spill over
!
p( x n ) > n i2 N Dn
q(" B # v AB )
po ( x >> x n ) =
n i2 N Dn
po ( x " #x p ) $ N Ap !
!
-xp
xn
!We say the holes are still in equilibrium with the barrier at xn:
p( x n ) = N Ap e"q (# B "v AB ) kT
Clif Fonstad, 9/29/09
n i2 +qv AB
=
e
N Dn
kT
Lecture 6 - Slide 14
Hole potential energy, qφ
Majority carriers against the junction barrier
Reverse bias, vAB < 0; barrier raised, carriers spill back
!
p( x n ) < n i2 N Dn
p( x >> x n ) =
n i2 N Dn
q(" B # v AB )
!
!
p( x < "x p ) # N Ap
-xp
xn
Again the holes maintain equilibrium with the barrier until xn:
!
p( x n ) = N Ap e"q (# B "v AB ) kT
Clif Fonstad, 9/29/09
n i2 +qv AB
=
e
N Dn
kT
And we have the same expression for p(xn).
Lecture 6 - Slide 15
What are n’(-xp) and p’(xn) with vAB applied?
We propose that the majority carrier populations on either side
are still related by the Boltzman factor,* which is now:
exp[-q(φb-vAB)/kT]
Thus:
p(x n ) = p("x p )e"q [ # b "v AB ] / kT
Under low level injection conditions, the majority carrier
population is unchanged, so p(-xp) remains NAp, so:
2
n
"q
#
"v
/
kT
[
]
!
i
p(x n ) = N Ap e b AB
=
e qv AB / kT
N Dn
And the excess population we seek is:
n i2
p'(x n ) = p(x n ) " pon =
e qv AB / kT "1)
(
N Dn
!
Similarly at -xp:
n i2
n'("x p ) =
e qv AB / kT "1)
(
N Ap
!
Clif Fonstad, 9/29/09
* We are assuming that the majority carriers can get across
the SCL much faster than they can diffuse away as minority
carriers, i.e., that diffusion is the bottleneck!
!
Lecture 6 - Slide 16
Biased p-n junctions: current flow, cont.
What is the current, iD?
Knowing p’(xn) and n’(-xp), we know:
J h (x) for
x n < x < wn
and
J e (x) for - w p < x < "x p
But we still don’t know the total current because we don’t
know both currents at the same position, x:
!
i = A J = A [ J (x) + J (x)]
Have to be at same “x”
D
TOT
h
e
To proceed we make the assumption that there is negligible
recombination of holes and electrons in the depletion
!
region, so what goes in comes out and:
J h (x n ) = J h ("x p )
and
J e (x n ) = J e ("x p )
With this assumption, we can write:
!
Clif Fonstad, 9/29/09
iD = A JTOT = A [ J h (x n ) + J e ("x p )]
Values at edges of SCL
Lecture 6 - Slide 17
Biased p-n junctions: current flow, cont.
What is the current, iD, cont,?
Both Jh(xn) and Je(-xp), are proportional to p’(xn) and n’(-xp),
respectively, which in turn are both proportional to (eqv/kT -1):
J h (x n ) " p'(x n ) " [e qv AB / kT -1]
and
J e (-x p ) " n'(x p ) " [e qv AB / kT -1]
Thus the diode current is also proportional to (eqv/kT -1):
!
iD = A [ J h (x n ) + J e ("x p )] # [e qv AB / kT "1] $ iD = Is [e qv AB / kT "1]
(IS is called the reverse saturation current of the diode.)
!
** Notice: The non-linearity, i.e., the exponential dependence of the
diode current on voltage, arises because of the exponential
dependence of the minority carrier populations the edges of the
space charge layer (depletion region). The flow problems
themselves are linear.
Clif Fonstad, 9/29/09
Lecture 6 - Slide 18
Biased p-n junctions: current flow, cont.
The saturation current of three diode types:
IS's dependence on the relative sizes of w and Lmin
p’(x), n’(x)
p’(xn)
Short-base diode, wn << Lh, wp << Le:
n’(-xp)
x
#
-wp
-xp xn
wn
n 2i
Dh
qv AB / kT
J h (x n ) = q
e
-1
[
] %%
'
*
N Dn ( w n " x n )
Dh
De
2
, [e qv AB / kT -1]
+
$ iD = Aqn i )
2
n
De
)( N Dn ( w n " x n ) N Ap ( w p " x p ) ,+
J e (-x p ) = q i
e qv AB / kT -1] %
[
N Ap ( w p " x p )
%&
p’(x), n’(x)
p’(xn)
Long-base diode, wn >> Lh, wp >> Le:
n’(-xp)
"
-wp
-xp xn
wn
n 2i Dh qv AB / kT
J h (x n ) = q
e
-1] $
[
& D
$
N Dn Lh
De ) qv AB / kT
2
h
+
-1]
# iD = Aqn i (
+ [e
n 2i De qv AB / kT
N
L
N
L
' Dn h
Ap e *
J e (-x p ) = q
e
-1] $
[
$%
N Ap Le
!
General diode:
x
" D
De % qv AB / kT
h
iD = Aqn $
+
-1]
' [e
# N Dn w n,eff N Ap w p,eff &
2
i
!
Hole injection into n-side
Clif Fonstad, 9/29/09
!
Electron injection into p-side
Note : w n,eff " Lh tanh( w n - x n ), w p,eff " Le tanh( w p - x p )
Lecture 6 - Slide 19
Biased p-n junctions: current flow, cont.
The ideal exponential diode
108
• General expression:
iD = IS(eqVAB/kT - 1)
• Forward bias, |vAB| > kT/q:
iD ≈ ISeqVAB/kT
Current increases 10x
for every 60 mV increase in vAB.
• Reverse bias, |vAB| > kT/q:
Current saturates at IS.
iD = - IS
107
106
105
104
|J/Js|
60 mV/decade
103
Ideal forward
102
101
Ideal reverse
100
Js
10-1
0
5
10
15
20
q|v|/kT
25
30
Figure by MIT OpenCourseWare.
Ref: Adapted from Figure 18 in S. M. Sze, “Physics
of Semiconductor Devices” 1st. Ed (Wiley, 1969)
Clif Fonstad, 9/29/09
Lecture 6 - Slide 20
Biased p-n junctions: current flow, cont.
Limitations of the model
NOTE: This figure is a bit exagerated,
but it makes the point.
108
• Large forward bias:
Sub-exponential increase
- High level injection (c)
- Series voltage drop (d)
• Large reverse bias:
Abrupt, rapid increase
- Non-destructive breakdown
• Very low bias levels:
Excess current seen
- SCL generation and
recombination (a, e)
(d)
107
106
(c)
105
Forward
(b)
104
|J/Js|
Junction
breakdown
Reverse
103
(a)
(e)
102
Ideal forward
101
Ideal reverse
100
10-1
0
5
10
15
20
q|v|/kT
25
30
Figure by MIT OpenCourseWare.
Ref: Figure 18 in S. M. Sze, “Physics of Semiconductor Devices” 1st. Ed (Wiley, 1969)
Clif Fonstad, 9/29/09
Lecture 6 - Slide 21
Asymmetrically doped junctions: an important special case
Depletion region impacts/issues
A p+-n junction (NAp >> NDn):
x n >> x p , w " x n "
N Ap N Dn
2#Si ($ b % v AB )
,
qN Dn
(N
Ap
E pk "
!
An n+-p junction (NDn >> NAp):
!
x p >> x n , w " x p "
2#Si ($ b % v AB )
,
qN Ap
!
!
N Ap N Dn
(N
E pk "
+ N Dn ) " N Dn
2q ($ b % v AB ) N Dn
#Si
Ap
+ N Dn ) " N Ap
2q ($ b % v AB ) N Ap
#Si
Note that in both cases the depletion region is predominately on the
lightly doped side, and it is the doping level of the more lightly doped
junction that matters (i.e., dominates).
Note also that as the doping level increases the depletion width
decreases and the peak E-field increases. [This is also true in
symmetrical diodes.]
Clif Fonstad, 9/29/09
Two very important and useful observations!!
Lecture 6 - Slide 22
Asymmetrically doped junctions: an important special case
Current flow impact/issues
A p+-n junction (NAp >> NDn):
" D
Dh
De % qv AB / kT
2
2
h
e qv AB / kT -1]
iD = Aqn i $
+
-1] " Aqn i
' [e
[
N Dn w n,eff
# N Dn w n,eff N Ap w p,eff &
Hole injection into n-side
!
An n+-p junction (NDn >> NAp):
" D
De % !qv AB / kT
De
2
h
iD = Aqn i $
+
-1] ( Aqn 2i
e qv AB / kT -1]
' [e
[
N Ap w p,eff
# N Dn w n,eff N Ap w p,eff &
Electron injection into p-side
!
Note that in both cases the minority carrier injection is predominately into
the lightly doped side.
Note also that it is the doping level of the more lightly doped junction that
determines the magnitude of the current, and as the doping level on the
lightly doped side decreases, the magnitude of the current increases.
Two very important and useful observations!!
Clif Fonstad, 9/29/09
Lecture 6 - Slide 23
6.012 - Microelectronic Devices and Circuits
Lecture 6 - p-n Junctions: I-V Relationship - Summary
• I-V relationship for an abrupt p-n junction
Focus is on minority carrier diffusion on either side of SCL
Voltage across SCL sets excess populations -xp and xn:
n'(-xp) = nnoe-q[fb – vAB]/kT-npo = npo(eqvAB/kT-1) = (ni2/NAp)(eqvAB/kT -1)
p'(xn) = ppoe-q[fb – vAB]/kT-pno = pno(eqvAB/kT-1) = (ni2/NDn)(eqvAB/kT -1)
Flow problems in QNR regions give minority currents:
Je(-wp<x<-xp) = q(De/Le)[cosh(wp-x)/sinh(wp-xp)](ni2/NAp)(eqvAB/kT -1)
Jh(xn<x<wn) = q(Dh/Lh)[cosh(wn-x)/sinh(wn-xn)](ni2/NDn)(eqvAB/kT -1)
Total current is found from continuity across SCL:
iD(vAB) = A [Je(-xp) + Jh(xn)] = IS (eqvAB/kT -1), with
IS ≡A q ni2 [(Dh/NDn wn*) + (De/NAp wp*)]
(hole component)
Note: wp
* and
(electron component)
wn*
are the effective widths of the p- and n-sides
If Le >> wp, then wp* ≈ (wp - xp), and if Le << wp, then wp* ≈ Le
If Lh >> wn, then wn* ≈ (wn - xn), and if Lh << wn, then wn* ≈ Lh
• Features and limitations of the model
Exponential dependence enters via boundary conditions
Injection is predominantly into more lightly doped side
Saturation current, IS, goes down as doping levels go up
Limits: 1. SCL g-r may dominate at low current levels
2. Series resistance may reduce junction voltage at high currents
3. Junction may breakdown (conduct) at large reverse bias
Clif Fonstad, 9/29/09
Lecture 6 - Slide 24
MIT OpenCourseWare
http://ocw.mit.edu
6.012 Microelectronic Devices and Circuits
Fall 2009
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