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Unformatted text preview: 6.012  Electronic Devices and Circuits Lecture 5  pn Junction Injection and Flow  Outline • Review
Depletion approximation for an abrupt pn junction
Depletion charge storage and depletion capacitance (Rec. Fri.) qDP(vAB) = – AqNApxp = – A[2εq(φbvAB){NApNDn/(NAp+NDn)}]1/2
Cdp(VAB) ≡ ∂ qDP/∂ vABVAB = A[εq{NApNDn/(NAp+NDn)}/2(φbVAB)]1/2 • Biased pn Diodes
Depletion regions change
Currents flow: two components
– flow issues in quasineutral regions
– boundary conditions on p' and n' at xp and xn (Lecture 4) (Today)
(Lecture 6) • Minority carrier flow in quasineutral regions
The importance of minority carrier diffusion
Boundary conditions
Minority carrier profiles and currents in QNRs
– Short base situations
– Long base situations
– Intermediate situations
Clif Fonstad, 9/24/09 Lecture 5  Slide 1 The Depletion Approximation: an informed first estimate of ρ(x)
Assume full depletion for xp < x < xn, where xp and xn are
two unknowns yet to be determined. This leads to: $0
&
&#qN Ap
"( x ) = %
& qN Dn
&0
' for
x < #x p
for # x p < x < 0
for 0 < x < x n
for
xn < x ρ(x)
qNDn
xp x
xn
qNAp Integrating the charge once gives the electric field $
0
!
& qN
&" Ap ( x + x p )
&#
Si
E ( x) = %
& qN Dn ( x " x )
n
& #Si
&
0
'
Clif Fonstad, 9/24/09 ! for x < "xp for " xp < x < 0 xp Ε(x)
xn for 0 < x < xn for xn < x x E(0) = qNApxp/εSi
= qNDnxn/εSi Lecture 5  Slide 2 The Depletion Approximation, cont.: Integrating again gives the electrostatic potential: %
"p
'
qN Ap
2
' "p +
(x + x p )
'
2$Si
"( x) = &
' " # qN Dn ( x # x ) 2
n
' n 2$Si
'
"n
( for x < #x p for  xp < x < 0 φ(x)
φn
xp
x for
for xn 0 < x < xn φp
φ(0) = φp + qNApxp 2/2εSi
= φn − qNDnxn2/2εSi xn < x Insisting E(x) is continuous at x = 0 yields our first x
equation relating our unknowns, xn and xp: Ε(x) p N Ap x p = N Dn x n xn 1 Requiring that the potential be continuous at x = 0 gives
us our second relationship between xn and xp: !
Clif Fonstad, 9/24/09 qN Ap 2
qN Dn 2
"p +
x p = "n $
xn
2#Si
2#Si x E(0) = qNApxp/εSi
= qNDnxn/εSi 2
Lecture 5  Slide 3 Comparing the depletion approximation
with a full solution:
Example: An unbiased abrupt pn junction with NAp= 1017 cm3, NDn= 5 x 1016 cm3 Charge Potential po(x), no(x) nie±qφ(x)/kT Efield Clif Fonstad, 9/24/09 Lecture 5  Slide 4 Courtesy of Prof. Peter Hagelstein. Used with permission. Depletion approximation: Applied bias
Forward bias, vAB > 0: φ vAB
w p (φbvAB) x
xp 0 xn Reverse bias, vAB < 0: vAB wn φ w p x (φbvAB) xp 0 xn
No drop
in wire
No drop
at contact wn No drop
in wire No drop
in QNR
No drop
in QNR No drop
at contact In a well built diode, all the applied
voltage appears as a change in the
the voltage step crossing the SCL Note: With applied bias we are no longer in thermal equilibrium so
it is no longer true that n(x) = ni eqφ(x)/kT and p(x) = ni eqφ(x)/kT.
Clif Fonstad, 9/24/09 Lecture 5  Slide 5 The Depletion Approximation: Applied bias, cont. Adding vAB to our earlier sketches: assume reverse bias, vAB < 0 2"Si (# b $ v AB ) ( N Ap + N Dn )
w=
q
N Ap N Dn ρ(x)
qNDn w xp xp xn xn x qNAp N Ap w
N Dn w
xp =
, xn =
(N Ap + N Dn )
(N Ap + N Dn ) Ε(x) ! xp xn Epk x E pk ! 2q (" b # v AB ) N Ap N Dn
=
$Si
(N Ap + N Dn ) φ(x) xp (φbvAB) Clif Fonstad, 9/24/09 ! x
xn "# = # b $ v AB
and kT N Dn N Ap
#b =
ln
q
n i2
Lecture 5  Slide 6 The Depletion Approximation: comparison cont.
Example: Same sample, reverse biased 2.4 V Charge Efield Potential p (x)
n (x) Clif Fonstad, 9/24/09 nie±qφ(x)/kT Lecture 5  Slide 7 Courtesy of Prof. Peter Hagelstein. Used with permission. The Depletion Approximation: comparison cont.
Example: Same sample, forward biased 0.6 V Charge Efield Clif Fonstad, 9/24/09 Potential p (x)
n (x) Courtesy of Prof. Peter Hagelstein. Used with permission. nie±qφ(x)/kT Lecture 5  Slide 8 The value of the depletion approximation
The plots look good, but equally important is that
1. It gives an excellent model for making hand calculations
and gives us good values for quantities we care about:
• Depletion region width
• Peak electric field
• Potential step 2. It gives us the proper dependences of these quantities on
the doping levels (relative and absolute) and the bias voltage. Apply bias; what happens?
Two things happen
1. The depletion width changes
• (φb  vAB) replaces φb in the Depletion Approximation Eqs.
2. Currents flow
• This is the main topic of today’s lecture
Clif Fonstad, 9/24/09 Lecture 5  Slide 9 Depletion capacitance: Comparing depletion charge stores with a parallel plate capacitor
ρ(x)
ρ(x) qNDn qA qB
( = qA) xp d/2 x x xn qA d/2 qNAp
qB( = qA) Depletion region charge store Parallel plate capacitor qA ,PP $qA ,PP
C pp (VAB ) #
$v AB
A"
=
d ! qA ,DP (v AB ) = " AqN Ap x p (v AB ) "
= A v AB
d = " A 2q#Si [$ b " v AB ]
v AB = VAB Cdp (VAB ) % Many similarities; important differences. Clif Fonstad, 9/24/09 =A &qA ,DP
&v AB N Ap N Dn [N Ap + N Dn ] v AB = VAB N Ap N Dn
q#Si
2[$ b " VAB ] [ N Ap + N Dn ] = A #Si
w (VAB ) Lecture 5  Slide 10 ! Bias applied, cont.: With vAB ≠ 0, it is not true that n(x) = ni eqφ(x)/kT
and p(x) = ni eqφ(x)/kT because we are no longer in TE. However,
outside of the depletion region things are in quasiequilibrium, and
we can define local electrostatic potentials for which the equilibrium
relationships hold for the majority carriers, assuming LLI. φQNRp Forward bias, vAB > 0:
vAB
vAB w p
(φbvAB) xp vAB w p vAB 0 xn
φQNRp Reverse bias, vAB < 0:
vAB φQNRn wn
φQNRn (φbvAB)
vAB xp 0 xn x vAB x
wn vAB In this region p(x) ≈ ni eqφQNRp/kT
Clif Fonstad, 9/24/09 In this region n(x) ≈ ni eqφQNRn/kT
Lecture 5  Slide 11 Current Flow qφ Unbiased
junction
Population in
equilibrium with
barrier x
qφ Forward bias on junction Barrier lowered so carriers to left can cross over it. x
qφ Reverse bias on junction Barrier raised so the few carriers on top spill back down it. x
Clif Fonstad, 9/24/09 Lecture 5  Slide 12 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
+ Uniform ptype p iD Uniform ntype n  vAB
wp  x p 0 xn Quasineutral
region I
Minority carrier flow
here determines the
electron current
 Today's Lecture  Clif Fonstad, 9/24/09 Ohmic
contact Space charge
region The values of n' at
xp and p' at xn are
established here.
 Lecture 6  wn B x Quasineutral
region II
Minority carrier flow
here determines the
hole current
 Today's Lecture  Lecture 5  Slide 13 Solving the five equations: special cases we can handle
1. Uniform doping, thermal equilibrium (nopo product, no, po): "
"
= 0,
= 0, gL ( x, t ) = 0, J e = J h = 0
"x
"t Lecture 1 2. Uniform doping and Efield (drift conduction, Ohms law): "
"
= 0,
= 0, gL ( x, t ) = 0, E x constant
"x
"t ! Lecture 1 3. Uniform doping and uniform low level optical injection (τmin): "
= 0, gL ( t ), n ' << po
"x ! Lecture 2 3'. Uniform doping, optical injection, and Efield (photoconductivity): ! "
= 0, E x constant, gL ( t )
"x Lecture 2 4. Nonuniform doping in thermal equilibrium (junctions, interfaces) ! "
= 0, gL ( x, t ) = 0, J e = J h = 0
"t 5. Uniform doping, nonuniform LL injection (QNR diffusion) "N d "N a
"n ' "
=
= 0, n ' # p', n' << p o , J e # qDe
,
#0
"x
"x "t
! "x Clif Fonstad, 9/24/09 Lectures 3,4 TODAY
Lecture 5
Lecture 5  Slide 14 QNR Flow: Uniform doping, nonuniform LL injection
What we have: Five things we care about (i.e. want to know):
p( x, t ) and n ( x, t )
Hole and electron concentrations:
Hole and electron currents:
J hx ( x, t ) and J ex ( x, t )
Electric field:
E x ( x, t )
And, five equations relating them: "p( x, t ) 1 "J h ( x, t )
+
= G # R $ Gext ( x, t ) # [ n ( x, t ) p( x, t ) # n i2 ] r( t )
"t !q "x
Electron continuity: "n( x, t ) # 1 "J e ( x, t ) = G # R $ Gext ( x, t ) # [ n ( x, t ) p( x, t ) # n i2 ] r( t )
"t
q "x
"p( x, t )
Hole current density: J h ( x, t ) = qµ h p( x, t ) E ( x, t ) # qDh
"x
"n ( x , t )
Electron current density: J e ( x, t ) = qµ e n ( x, t ) E ( x, t ) + qDe
"x
" [&( x ) E x ( x, t )]
Charge conservation: %( x, t ) =
$ q[ p( x, t ) # n ( x, t ) + N d ( x ) # N a ( x )]
"x
Hole continuity: We can get approximate analytical solutions if 5 conditions are met!
Clif Fonstad, 9/24/09 Lecture 5  Slide 15 ! QNR Flow, cont.: Uniform doping, nonuniform LL injection
Five unknowns, five equations, five flow problem conditions:
1. Uniform doping dn o dpo
# n # n ' # p # p'
=
=0 "
=
,
=
dx
dx
#x #x #x #x po " n o + N d " N a = 0 # $ = q( p " n + N d " N a ) = q( p' "n ') n'
n ' << po " ( np # n ) r $ n ' po r =
(in ptype, for example)
%e
3. Quasineutrality holds n ' " p' , #n' " #p'
# x #x
#n' ( x, t )
!
4. Minority carrier drift is negligible
J e ( x, t ) " qDe
#x
(continuing to assume ptype)
!
Note: It is also always true that "n "n '
"p "p'
=
,
=
"t "t
! "t "t
2. Low level injection
! ! Clif Fonstad, 9/24/09 2
i Lecture 5  Slide 16 QNR Flow, cont.: Uniform doping, nonuniform LL injection
With these first four conditions our five equations become:
(assuming for purposes of discussion that we have a ptype sample) 1, 2 :
3:
4:
5: ! "p' ( x, t ) 1 "J h ( x, t )
"n ' ( x, t ) 1 "J e ( x, t )
n ' ( x, t )
+
=
#
= gL ( x, t ) #
"t
q "x
"t
q "x
$e
"n ' ( x , t )
J e ( x, t ) % +qDe
"x
"p' ( x, t )
J h ( x, t ) = qµ h p( x, t ) E ( x, t ) + qDh
"x
"E ( x, t ) q
= [ p' ( x, t ) # n ' ( x, t )]
"x
& In preparation for continuing to our fifth condition, we note
that combining Equations 1 and 3 yields one equation in
n'(x,t):
2 "n' ( x, t )
" n ' ( x, t )
n ' ( x, t )
# De
= gL ( x, t ) #
"t
"x 2
$e The time dependent diffusion equation
Clif Fonstad, 9/24/09 ! Lecture 5  Slide 17 QNR Flow, cont.: Uniform doping, nonuniform LL injection
The time dependent diffusion equation, which is repeated
below, is in general still very difficult to solve
"n' ( x, t )
" 2 n ' ( x, t )
n ' ( x, t )
# De
= gL ( x, t ) #
"t
"x 2
$e
but things get much easier if we impose a fifth constraint:
5. Quasistatic excitation gL ( x, t ) such that all "
#0
"t !
With this constraint the above equation becomes a second order linear differential equation:
d 2n' ( x )
n' ( x )
!
"De
= gL ( x ) "
2
dx
#e
which in turn becomes, after rearranging the terms : ! d 2 n' ( x )
n' ( x )
1
"
="
gL ( x )
2
dx
De # e
De The steady state diffusion equation Clif Fonstad, 9/24/09 Lecture 5  Slide 18 ! QNR Flow, cont.: Solving the steady state diffusion equation The steady state diffusion equation in ptype material is: d 2 n' ( x )
n' ( x )
1
"
="
gL ( x )
2
2
dx
Le
De
and for ntype 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! diode, we have gL = 0 which means we only need
n
!
the homogenous solutions, i.e. expressions that satisfy:
nside: d 2 p' ( x ) dx 2
Clif Fonstad, 9/24/09 p' ( x )
"
=0
2
Lh pside: d 2 n' ( x )
n' ( x )
"
=0
2
2
dx
Le
Lecture 5  Slide 19 QNR Flow, cont.: Solving the steady state diffusion equation For convenience, we focus on the nside to start with and
find p'(x) for xn ≤ x ≤ wn. p'(x) satisfies d 2 p' ( x )
p' ( x )
=
dx 2
L2
h
subject to the boundary conditions: p' ( w n ) = 0 and p' ( x n ) = something we' ll find next time !
The general solution to this static diffusion equation is: p' ( x ) = Ae" x L h + Be + x L h ! where
where A and B are constants that satisfy the boundary
conditions. Solving for them and putting them into this
equation yields the final general result: ! p' ( x n )e( wn " x n ) Lh
p' ( x n )e"( wn " x n ) Lh
p' ( x ) = ( wn " x n ) Lh
e "( x " x n ) L h " ( w n " x n ) L h
e +( x " x n ) L h
e
" e"( w n " x n ) L h
e
" e"( w n " x n ) L h
for x n # x # w n
Clif Fonstad, 9/24/09 Lecture 5  Slide 20 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  Longbase diode: wn >> Lh
Case II  Shortbase 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 ex//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, 9/24/09 Lecture 5  Slide 21 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, 9/24/09 ! Lecture 5  Slide 22 QNR Flow, cont.: Uniform doping, nonuniform 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) [cm3]
p'n(xn) Jh(x) [A/cm2]
qDhp'n(xn) Lh ex/Lh 0 xn xn+Lh wn x [cm3] ex/Lh 0 xn xn+Lh wn x [cm3] Case II  Short base: wn << Ln (the situation in most Si diodes and transistors)
p'n(x) [cm3]
p'n(xn) Jh(x) [A/cm2]
qDhp'n(xn) [wnxn] 0 xn Clif Fonstad, 9/24/09 wn x [cm3] 0 xn wn x [cm3] Lecture 5  Slide 23 QNR Flow, cont.: Uniform doping, nonuniform LL injection
The four other unknowns
⇒ Solving the steady state diffusion equation gives n’.
⇒ Knowing n'..... we can easily get p’, Je, Jh, and Ex:
First find Je:
Then find Jh: !
Next find Ex:
! J e ( x ) " qDe dn ' ( t )
dx J h ( x ) = JTot " J e ( x ) '
1$
Dh
E x ( x) "
J e ( x ))
&J h ( x ) #
qµ h po %
De
( # dE x ( x )
q dx
!
Finally, go back and check that all of the five conditions are
met by the solution.
Then find p’: Clif Fonstad, 9/24/09 p' ( x ) " n ' ( x ) + Once we solve the diffusion equation to get
!
the minority excess, n', we know everything. Lecture 5  Slide 24 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
+ Uniform ptype p iD Uniform ntype n  vAB
wp  x p 0 xn Quasineutral
region I
Minority carrier flow
here determines the
electron current
Clif Fonstad, 9/24/09 Ohmic
contact Space charge
region The values of n' at
xp and p' at xn are
established here.
 Lecture 6 next Tuesday  wn B x Quasineutral
region II
Minority carrier flow
here determines the
hole current
Lecture 5  Slide 25 The pn 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
w p xp 0
n'(xp) = ? x
0 xn n' p wn p' n
p'(xn) = ?
p' ( w n ) = 0 n'(wp) = 0
x w p 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, 9/24/09 Lecture 5  Slide 26 ….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' ( w n ) = 0 n'(wp) = 0
xp 0 x w p
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, 9/24/09 We will do this in Lecture 6. Lecture 5  Slide 27 6.012  Electronic Devices and Circuits Lecture 5  pn Junction Injection and Flow  Summary • Biased pn Diodes
Depletion regions change
Currents flow: two components (Lecture 4) – flow issues in quasineutral regions
– boundary conditions on p' and n' at xp and xn • Minority carrier flow in quasineutral regions
The importance of minority carrier diffusion
– minority carrier drift is negligible Boundary conditions Minority carrier profiles and currents in QNRs – Short base situations
– Long base situations • Carrier populations across the depletion region (Lecture 6) Potential barriers and carrier populations
Relating minority populations at xp and xn to vAB
Excess minority carriers at xp and xn
Clif Fonstad, 9/24/09 Lecture 5  Slide 28 MIT OpenCourseWare
http://ocw.mit.edu 6.012 Microelectronic Devices and Circuits
Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. ...
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 Fall '09
 CharlesG.Sodini

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